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http://www.studymode.com/essays/Atomic-Orbital-And-Calculate-Oxidation-Number-878347.html | 1,529,737,262,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864943.28/warc/CC-MAIN-20180623054721-20180623074721-00138.warc.gz | 516,966,217 | 24,916 | # Atomic Orbital and Calculate Oxidation Number
Topics: Atomic orbital, Buffer solution, Atom Pages: 3 (664 words) Published: December 16, 2011
Q1-Q8 (1MARKS EACH)
Q9-Q18 (2MARKS EACH).
Q19-Q27 (3 MARKS EACH)
Q28-Q30 (5 MARKS EACH)
…………………………………………………………….. Q1. State law of constant composition.
Q2. Howe 0.5m of NaOH different from 0.5M ofNaOH?
Q3. State Heisenberg uncertainity principal.
Q4. Write general electronic configuration of d-block elements. Q5. Define Resonance.
Q6. What is compressibility factor?
Q7. Define Buffer solution? What is common ion effect?
Q8. What is Demineralised water?
Q9. Calculate the concentration of nitric acid in moles per litre in a sample which has a density of 1.41g/ml and mass% of nitric acid is 69%. Q10. Calculate energy of one mole of photons of radiation whose frequency is 5x10raise to power 10 Hz.(page 39. Q.1) Q11. Write down the electronic configuration of copper and chromium. Q12. Calculate the wavelength of 100 gm of particle moving with a velocity of 100m/s. Q13. Complete the following:-
a) Na₂O+H₂0--
b) Cl₂07+H₂0-------
Q14:- Write resonance structure of NO₂ and SO3?
Q15.:-What is electro negativity.How it is different from election gain enthalpy? Q16.on a ship sailing in pacific ocean where temp. is 23.4˚C a ballon is filled with 2L air what will be the volume of balloon when ship reaches Indian ocean where temp. is 26.1˚C?(page no.139) Q17.For the reaction
1.2Cl(g)----------Cl₂(g)What are the signs of ϪH and ϪS. 2.Define Enthalpy?
Q18. Calculate pH of 0.2MH₂SO4 solution?
Q19.:- 1.An atomic orbital has n=3 What are the possible values of 1 and m? 2.List the quantum numbers(m and 1) of electrons for 3d orbital 3.Which of the following orbital’s are possible 1p,2s,2p and 3f Q20. 1.Write IUPAC name of element with atomic number 120.
2.Which of the following species will have layout and smallest size Mg,Mg²+,Al,Al3+ 3.Why cautions are smaller than anions in radii than their parent atoms? Q21.Discuss the shapes of following molecules on the basis of VSEPR model. 1.SiCl4 2. PH3 3.H₂0
Q22. 1.Why do real gases deviate from ideal... | 639 | 2,110 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2018-26 | latest | en | 0.798118 |
https://justaaa.com/statistics-and-probability/655162-a-schoolteacher-is-concerned-that-her-students | 1,723,726,871,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641291968.96/warc/CC-MAIN-20240815110654-20240815140654-00863.warc.gz | 264,457,652 | 9,275 | Question
A schoolteacher is concerned that her students watch more TV than the average American child. She...
A schoolteacher is concerned that her students watch more TV than the average American child. She reads that according to the American Academy of Pediatrics (AAP), the average American child watches 4 hours of TV per day (μ = 4.0 hours). She records the number of hours of TV each of her six students watch per day. The times (in hours) are 2.6, 4.1, 5.4, 4.3, 2.8, and 4.8.
cohen's d d=
.05 level of significance and a one independent t test state the value of the test statistic
t=
H0:mu= 4 i.e.( Average American child watches 4 hr tv per day)
Vs
H1:mu >4 i.e.( claim of school teacher)
Under , the null hypothesis
Test statistic, t= x bar- mu/(s/√n) follows t distribution with n-1 degree of freedom.
X bar= 4.0 ( average of the given data)
std. Deviation= s= 1.1045
So, t= 4-4/(1.1045/√6)= 0
Critical t - value for alpha= 0.05 and for (6-1) df = 2.015
Since, critical t value is greater than calculated test statistic value so, we don't reject the null hypothesis and concluded that Average American watches 4 hr tv per day.
Earn Coins
Coins can be redeemed for fabulous gifts. | 341 | 1,209 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-33 | latest | en | 0.90672 |
https://www.convertunits.com/from/oz/to/nanogram | 1,638,833,766,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363327.64/warc/CC-MAIN-20211206224536-20211207014536-00487.warc.gz | 775,047,095 | 12,897 | ## ››Convert ounce to nanogram
oz nanogram
How many oz in 1 nanogram? The answer is 3.527396194958E-11.
We assume you are converting between ounce and nanogram.
You can view more details on each measurement unit:
oz or nanogram
The SI base unit for mass is the kilogram.
1 kilogram is equal to 35.27396194958 oz, or 1000000000000 nanogram.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between ounces and nanograms.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of oz to nanogram
1 oz to nanogram = 28349523125 nanogram
2 oz to nanogram = 56699046250 nanogram
3 oz to nanogram = 85048569375 nanogram
4 oz to nanogram = 113398092500 nanogram
5 oz to nanogram = 141747615625 nanogram
6 oz to nanogram = 170097138750 nanogram
7 oz to nanogram = 198446661875 nanogram
8 oz to nanogram = 226796185000 nanogram
9 oz to nanogram = 255145708125 nanogram
10 oz to nanogram = 283495231250 nanogram
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From: To:
## ››Definition: Nanogram
The SI prefix "nano" represents a factor of 10-9, or in exponential notation, 1E-9.
So 1 nanogram = 10-9 grams-force.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 491 | 1,790 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-49 | latest | en | 0.722773 |
http://www.rcgroups.com/forums/printthread.php?t=107078 | 1,441,370,855,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440645348533.67/warc/CC-MAIN-20150827031548-00039-ip-10-171-96-226.ec2.internal.warc.gz | 664,503,501 | 10,357 | #1 invicta421 Apr 02, 2003 11:05 AM
[RCSE] Why does dive CG test work?
I have been pondering this question for the last few weeks, but
haven't satified myself with an explaination yet. Most aerodynamics
seem to make intutive sense to me, but this doesn't.
The test I am speaking of is when you put the sailplane into a 45
degree dive and see if it pulls up, flys neutral or tucks under. A
page on how to do this can be found here:
http://www.polecataero.com/articles/cg_art.shtml
But this page doesn't explain why... to me it seems that a nose heavy
glider should tuck not rise when the CG is in front of the average
lift on the wing. Any help understanding this?
Thanks
-Richard
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#2 Lex Mierop Apr 02, 2003 11:05 AM
RE: [RCSE] Why does dive CG test work?
Richard,
It's all about the trim. When flying a nose heavy plane at regular thermal
cruise speed, you have to add up trim to get it to fly straight and level.
The "up" trim setting is causing an aerodynamic force on the tail is pushing
down to compensate for the extra nose weight. The amount of aerodynamic
down force required to stay level is constant for a given amount of nose
weight. The amount of aerodynamic force created by a flying surface varies
with the speed of the airplane. As it goes faster it is more effective
(better Reynolds numbers).
So what you start with is a nose heavy plane that is trimmed for level
flight at cruise speed. The tail forces balance the nose heaviness of the
model. Now increase the speed, and what happens? The nose heaviness
remains constant, but the downforce on the tail increases, causing the model
to pitch up.
The inverse is true on a tail heavy model. To trim for straight and level
at cruise speed, the tail requires down trim (an upward force on the tail to
compensate for the lack of weight in the nose). Increase the speed, and the
upforce on the tail increases, causing the model to pitch down. That's a
tuck.
-l
-----Original Message-----
From: invicta421 [mailto:richardh@examen.com]
Sent: Wednesday, April 02, 2003 12:51 AM
To: soaring@airage.com
Subject: [RCSE] Why does dive CG test work?
I have been pondering this question for the last few weeks, but
haven't satified myself with an explaination yet. Most aerodynamics
seem to make intutive sense to me, but this doesn't.
The test I am speaking of is when you put the sailplane into a 45
degree dive and see if it pulls up, flys neutral or tucks under. A
page on how to do this can be found here:
http://www.polecataero.com/articles/cg_art.shtml
But this page doesn't explain why... to me it seems that a nose heavy
glider should tuck not rise when the CG is in front of the average
lift on the wing. Any help understanding this?
Thanks
-Richard
RCSE-List facilities provided by Model Airplane News. Send "subscribe" and
"unsubscribe" requests to soaring-request@airage.com. Please note that
subscribe and unsubscribe messages must be sent in text only format with
MIME turned off.
RCSE-List facilities provided by Model Airplane News. Send "subscribe" and "unsubscribe" requests to soaring-request@airage.com. Please note that subscribe and unsubscribe messages must be sent in text only format with MIME turned off.
#3 George Voss Apr 02, 2003 11:05 AM
Re: [RCSE] Why does dive CG test work?
While I'm not an AE, I'll give it my best shot.
Think in terms of static and dynamic loads. You balance your sailplane
in a static mode using either your fingers, or some type of
deal with.
As the plane speeds up, the dynamic loads have more affect than the
static load, i.e. the elevator trim setting has more authority than the
static balance. Sooooooo, if you have more lead in the nose than
needed, you need to carry some "up" elevator to make the airplane fly
level at normal speeds. As the speed increases, the elevator over rides
the balance and the plane pitches up. The opposite is also true and
that's why the airplane will tuck if the CG is too far out of range.
I hope that helps. gv
invicta421 wrote:
>I have been pondering this question for the last few weeks, but
>haven't satified myself with an explaination yet. Most aerodynamics
>seem to make intutive sense to me, but this doesn't.
>
>The test I am speaking of is when you put the sailplane into a 45
>degree dive and see if it pulls up, flys neutral or tucks under. A
>page on how to do this can be found here:
>
>http://www.polecataero.com/articles/cg_art.shtml
>
>But this page doesn't explain why... to me it seems that a nose heavy
>glider should tuck not rise when the CG is in front of the average
>lift on the wing. Any help understanding this?
>
>Thanks
>
>-Richard
>
>RCSE-List facilities provided by Model Airplane News. Send "subscribe" and "unsubscribe" requests to soaring-request@airage.com. Please note that subscribe and unsubscribe messages must be sent in text only format with MIME turned off.
>
>
>
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#4 Andy Thonet Apr 02, 2003 11:05 AM
RE: [RCSE] Why does dive CG test work?
This diagram might help.
http://clubedsf.org/html/cg_dive_test.htm
Andy Thonet
KG6MGD
Club EDSF
-----Original Message-----
From: invicta421 [mailto:richardh@examen.com]
Sent: Wednesday, April 02, 2003 12:51 AM
To: soaring@airage.com
Subject: [RCSE] Why does dive CG test work?
I have been pondering this question for the last few weeks, but
haven't satified myself with an explaination yet. Most aerodynamics
seem to make intutive sense to me, but this doesn't.
The test I am speaking of is when you put the sailplane into a 45
degree dive and see if it pulls up, flys neutral or tucks under. A
page on how to do this can be found here:
http://www.polecataero.com/articles/cg_art.shtml
But this page doesn't explain why... to me it seems that a nose heavy
glider should tuck not rise when the CG is in front of the average
lift on the wing. Any help understanding this?
Thanks
-Richard
RCSE-List facilities provided by Model Airplane News. Send "subscribe"
and "unsubscribe" requests to soaring-request@airage.com. Please note
that subscribe and unsubscribe messages must be sent in text only format
with MIME turned off.
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#5 renselange@earthlink.net Apr 11, 2003 03:02 AM
Re: Re: [RCSE] Why does dive CG test work?
But, doesn't it follow from this very logic that things change with variations in speed? So, a slow dive test gives different results than a fast one - always? Intuitively, I always did the Gordy test ("try to make it fly nice, flat and far from a hand toss") and added or removed some lead to make the plane more or less stable according to my taste. I did do dive tests occasionally just for the heck of it and find that my planes slowly pull up from a dive. But since I fly primarily TD, I never worry about diving much.
Rense Lange
-------Original Message-------
From: George Voss <gavoss@swbell.net>
Sent: 04/02/03 09:23 AM
To: invicta421 <richardh@examen.com>
Subject: Re: [RCSE] Why does dive CG test work?
>
> While I'm not an AE, I'll give it my best shot.
Think in terms of static and dynamic loads. You balance your sailplane
in a static mode using either your fingers, or some type of
deal with.
As the plane speeds up, the dynamic loads have more affect than the
static load, i.e. the elevator trim setting has more authority than the
static balance. Sooooooo, if you have more lead in the nose than
needed, you need to carry some "up" elevator to make the airplane fly
level at normal speeds. As the speed increases, the elevator over rides
the balance and the plane pitches up. The opposite is also true and
that's why the airplane will tuck if the CG is too far out of range.
I hope that helps. gv
invicta421 wrote:
>I have been pondering this question for the last few weeks, but
>haven't satified myself with an explaination yet. Most aerodynamics
>seem to make intutive sense to me, but this doesn't.
>
>The test I am speaking of is when you put the sailplane into a 45
>degree dive and see if it pulls up, flys neutral or tucks under. A
>page on how to do this can be found here:
>
>http://www.polecataero.com/articles/cg_art.shtml
>
>But this page doesn't explain why... to me it seems that a nose heavy
>glider should tuck not rise when the CG is in front of the average
>lift on the wing. Any help understanding this?
>
>Thanks
>
>-Richard
>
>RCSE-List facilities provided by Model Airplane News. Send "subscribe"
and "unsubscribe" requests to soaring-request@airage.com. Please note that
subscribe and unsubscribe messages must be sent in text only format with
MIME turned off.
>
>
>
RCSE-List facilities provided by Model Airplane News. Send "subscribe"
and "unsubscribe" requests to soaring-request@airage.com. Please note that
subscribe and unsubscribe messages must be sent in text only format with
MIME turned off.
>
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#6 Douglas, Brent Apr 11, 2003 03:02 AM
RE: Re: [RCSE] Why does dive CG test work?
<<But since I fly primarily TD, I never worry about diving much.>>
Hi Rense,
but that's not the point... The idea behind this test is that it shows what
you have to do to make your plane fly. Example, if your plane's nose heavy,
you've got to dial in some up to keep it flying level. That's why it pulls
up in a dive, the up elevator is magnified at speed.
If you always fly about the same speed, you may never see this, but you'll
still be fighting to keep the nose up and maybe not flying as efficiently as
you could be...
At least that's how I read the dive test, good luck!
Brent
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#7 GordySoar@aol.com Apr 11, 2003 03:02 AM
Re: [RCSE] Why does dive CG test work?
But, doesn't it follow from this very logic that things change with
variations in speed? So, a slow dive test gives different results than a fast
one - always?
You are right... THINGS... do change (too many in fact), but nothing of value
to determining optimized balance. :-)
Gordy
#8 Jeff Reid Apr 11, 2003 03:02 AM
Re: [RCSE] Why does dive CG test work?
Darn, dive test is meant to help adjust CG?
All this time, I thought it was meant to test the strength
of a glider's nose. So far my best is 40 feet into a very
muddy field. Wow, you learn something everyday. This is
going to save me a lot of repair time.
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#9 Erich Merkel Apr 11, 2003 03:02 AM
Re: Re: [RCSE] Why does dive CG test work?
"But since I fly primarily TD, I never worry about diving much."
Jeez, what do you guys fly for? What's the point of going to all the work
of finding that thermal, and getting really high? What do you do when
you've maxxed the round and you can just barely see that new molded super
plane that you spent all those hours trimming to perfection? Personally, I
point the nose straight down, and make that sucker sing!!! There's nothing
like the sound of a moldy in a high speed dive!! Sooo sexy... ;-)
Erich Merkel
Colville, WA
Phone: 509-684-0440
Cell: 509-680-1141
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#10 Douglas, Brent Apr 11, 2003 03:02 AM
RE: [RCSE] Why does dive CG test work?
Actually, I did that kind of dive test with a Paragon a couple weeks back,
and based on the depth of the hole, I think I was right on.
I've done repairs, hope to find a better way to "test" it next time.
Brent
*It's back together - lucky for me the ground was still swampy...
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#11 Kurt W. Zimmerman Apr 11, 2003 03:02 AM
RE: Re: [RCSE] Why does dive CG test work?
It's not the high speed dive that bothers me... it is the sudden stop, thud
& crunch that does.
Kurt
-----Original Message-----
From: Erich Merkel [SMTP:slopesoar@theofficenet.com]
Sent: Wednesday, April 02, 2003 11:19 PM
Cc: soaring@airage.com
Subject: Re: Re: [RCSE] Why does dive CG test work?
"But since I fly primarily TD, I never worry about diving much."
Jeez, what do you guys fly for? What's the point of going to all the work
of finding that thermal, and getting really high? What do you do when
you've maxxed the round and you can just barely see that new molded super
plane that you spent all those hours trimming to perfection? Personally, I
point the nose straight down, and make that sucker sing!!! There's nothing
like the sound of a moldy in a high speed dive!! Sooo sexy... ;-)
Erich Merkel
Colville, WA
Phone: 509-684-0440
Cell: 509-680-1141
RCSE-List facilities provided by Model Airplane News. Send "subscribe" and
"unsubscribe" requests to soaring-request@airage.com. Please note that
subscribe and unsubscribe messages must be sent in text only format with
MIME turned off.
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All times are GMT -5. The time now is 07:47 AM. | 3,678 | 14,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2015-35 | latest | en | 0.93645 |
https://oeis.org/A154036 | 1,638,773,084,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363290.39/warc/CC-MAIN-20211206042636-20211206072636-00571.warc.gz | 496,412,989 | 3,960 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A154036 Number of planar triangular n X n X n nonnegative integer grids with every similarly oriented 5 X 5 X 5 subtriangle summing to 3. 0
680, 2194, 3418, 3544, 3331, 2988, 2519, 1173, 1157, 1155, 1155, 1539, 1155, 1155, 1155, 1155, 1539, 1155, 1155, 1155, 1155, 1539, 1155, 1155, 1155, 1155, 1539, 1155, 1155, 1155, 1155, 1539, 1155, 1155, 1155, 1155, 1539, 1155, 1155, 1155, 1155, 1539 (list; graph; refs; listen; history; text; internal format)
OFFSET 5,1 COMMENTS Pattern 1155 1155 1539 1155 1155 starts at n=14 and continues through at least n=53. LINKS EXAMPLE Maximal zeros solution for n=14 ...............0 ..............0.0 .............0.0.0 ............0.0.1.0 ...........0.1.0.0.1 ..........1.0.0.0.0.0 .........0.0.0.0.1.0.0 ........0.0.0.1.0.0.0.0 .......0.0.1.0.0.0.0.1.0 ......0.1.0.0.0.0.1.0.0.1 .....1.0.0.0.0.1.0.0.0.0.0 ....0.0.0.0.1.0.0.0.0.1.0.0 ...0.0.0.1.0.0.0.0.1.0.0.0.0 ..0.0.0.0.1.0.0.0.0.1.0.0.0.0 CROSSREFS Sequence in context: A253695 A253702 A235013 * A185765 A206015 A202908 Adjacent sequences: A154033 A154034 A154035 * A154037 A154038 A154039 KEYWORD nonn AUTHOR R. H. Hardin Jan 04 2009 STATUS approved
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Last modified December 6 01:25 EST 2021. Contains 349558 sequences. (Running on oeis4.) | 752 | 1,940 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-49 | latest | en | 0.58682 |
https://civilengineeringx.com/masonry-structural-design/example-of-lintel-design-according-to-strength-provisions/ | 1,721,268,949,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514816.43/warc/CC-MAIN-20240718003641-20240718033641-00270.warc.gz | 143,385,793 | 33,986 | # Example of Lintel Design According to Strength Provisions
Suppose that we have a uniformly distributed load of 1050 lb/ft, applied at the level of the roof of the structure shown in Fig. 6.5. Design the lintel.
According to Table 2 of the 2008 MSJC Specification, for Type M or S mortar and concrete units with a specified strength of 1900 psi (the minimum specified strength for ASTM C90 units), the compressive strength of the masonry can conservatively be taken as 1500 psi (the so-called unit strength method). If the compressive strength is evaluated by prism testing, a higher value can probably be used. Take the specified compressive strength of the masonry as fm² = 1500 psi.
Assume fully grouted concrete masonry with a nominal thickness of 8 in., a weight of 80 lb/ft2, and a specified compressive strength of 1500 lb/in.2. Use Type S PCL mortar. The lintel has a span of 10 ft, and a total depth (height of parapet plus distance between the roof and the lintel) of 4 ft. These are shown in the schematic figure in Fig. 6.5. Assume that 700 lb/ft of the roof load is D, and the remaining 350 lb/ft is L. The governing loading combination is 1.2D + 1.6L. Our design presumes that entire height of the lintel is grouted.
First check whether the depth of the lintel is sufficient to avoid the use of shear reinforcement. Because the opening may have a movement joint on either side, again use a span equal to the clear distance, plus one-half of a half-unit on each side. So the span is 10 ft plus 8 in., or 10.67 ft.
The bars in the lintel will probably be placed in the lower part of an inverted bottom course.
The effective depth d is calculated using the minimum cover of 1.5 in.
(Sec. 1.15.4.1 of the 2008 MSJC Code), plus one-half the diameter of an assumed #8 bar.
Because this is a reinforced element, shearing capacity is calculated using Sec. 3.3.4.1.2.1 of the 2008 MSJC Code.
Also include two #4 bars at the level of the roof (bond beam reinforcement). The flexural design is quite simple. Section 3.3.4.2.2.2 of the 2008 MSJC Code does require that the nominal flexural strength of a beam not be less than 1.3 times the nominal cracking capacity, calculated using the modulus of rupture from Code Sec. 3.1.8.2. In our case, the nominal cracking moment for the 4-ft deep section is
Use two #4 bars. Section 3.3.4.2.2.2 of the 2008 MSJC Code need not be met if the amount of tensile reinforcement is at least one-third greater than required by analysis (Code Sec. 3.3.4.2.2.3).
Finally, Sec. 3.3.3.5 of the 2008 MSJC Code imposes maximum flexural reinforcement limitations that are based on a series of critical strain gradients. These generally do not govern for members with little or no axial load, like this lintel. They may govern for members with significant axial load, such as tall shear walls.
Scroll to Top | 748 | 2,847 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-30 | latest | en | 0.861343 |
https://www.algebra-class.com/graphing-quadratic-equations.html | 1,719,196,046,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864986.57/warc/CC-MAIN-20240624021134-20240624051134-00552.warc.gz | 565,979,198 | 7,980 | I first introduced the concept of graphing quadratic equations in our Functions unit. In this unit, we discovered how to use a table of values in order to graph a quadratic function. This would be a great lesson to review, as you will see a lot of vocabulary that relates to graphing parabolas.
And... in case you didn't remember, the graph of a quadratic equation is called a parabola. In the following examples, we will pull together all of our knowledge for quadratic equations to create the graph.
Before beginning this lesson, please make sure that you fully understand the vertex formula, factoring quadratic equations, and the quadratic formula.
One way to graph a quadratic equation, is to use a table of values. While this method works for every quadratic equation, there are other methods that are faster.
### For any quadratic equation in the form: y = ax2 + bx + c
The graph will result in a parabola.
## So, how do we find all of these points in order to create the graph?
We need to find the vertex, x intercepts, and y intercept. So, let's look at an example and I will show you how to find all the points needed!
## Graphing Quadratic Equations - Example 1
Wow! That seems like a lot of work doesn't it? It's really not too bad once you do it a few times.
Did you notice how some of the information that you learned in previous chapters is coming up again? For example, we talked about the vertex formula when we graphed quadratic functions. We also talked about x-intercepts and y-intercepts when we graphed linear equations.
When we graphed linear equations, we let y = 0 when we were trying to find the x-intercepts and we let x = 0 when we were trying to find the y-intercept. We just used the same process for quadratic equations.
This is what I Love about Algebra! You really never forget the concepts because you use them over and over again.
Ok... are you ready to look at one more example? We will graph a quadratic equation that you may not be able to factor as easily. In order to find the x intercepts, we will use the quadratic formula instead of factoring.
## Graphing Quadratic Equations - Example 2
Now I bet you are beginning to understand why factoring is a little faster than using the quadratic formula! It is a lot of work - not too hard, just a little more time consuming.
I hope this helps you to better understand the concept of graphing quadratic equations. | 529 | 2,414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2024-26 | latest | en | 0.949599 |
http://au.metamath.org/mpegif/elndif.html | 1,516,745,084,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892699.72/warc/CC-MAIN-20180123211127-20180123231127-00469.warc.gz | 25,285,904 | 3,440 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > elndif Unicode version
Theorem elndif 3387
Description: A set does not belong to a class excluding it. (Contributed by NM, 27-Jun-1994.)
Assertion
Ref Expression
elndif
Proof of Theorem elndif
StepHypRef Expression
1 eldifn 3386 . 2
21con2i 112 1
Colors of variables: wff set class Syntax hints: wn 3 wi 4 wcel 1715 cdif 3235 This theorem is referenced by: peano5 4782 fsnunf2 5832 undifixp 6995 cantnfreslem 7524 dfac9 7909 ssfin4 8083 isf32lem3 8128 isf34lem4 8150 xrinfmss 10781 restntr 17129 cmpcld 17346 reconnlem2 18546 lebnumlem1 18674 i1fd 19251 dfon2lem6 24970 onsucconi 25703 uvcff 26831 islindf4 26899 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 ax-gen 1551 ax-5 1562 ax-17 1621 ax-9 1659 ax-8 1680 ax-6 1734 ax-7 1739 ax-11 1751 ax-12 1937 ax-ext 2347 This theorem depends on definitions: df-bi 177 df-or 359 df-an 360 df-tru 1324 df-ex 1547 df-nf 1550 df-sb 1654 df-clab 2353 df-cleq 2359 df-clel 2362 df-nfc 2491 df-v 2875 df-dif 3241
Copyright terms: Public domain W3C validator | 565 | 1,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2018-05 | longest | en | 0.173863 |
https://brainsedu.in/syllabus/physics-chse-new-syllabus/?pcid= | 1,652,727,239,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662512229.26/warc/CC-MAIN-20220516172745-20220516202745-00514.warc.gz | 197,927,207 | 17,320 | # Physics CHSE New Syllabus 1st year
PHYSICS (CHSE) 2021-22
+2 Science 1st Year
Unit-I Physical world and Measurement (6 Periods)
Sl Units, accuracy and precision of measuring instruments, errors in measurement, absolute, relative error, percentage of error, Combination of errors, significant figures. Dimensions of Physical quantities. Dimensional analysis and its applications.
Unit – II Kinematics. (18 Periods)
1. Motion in a straight line: Rest and motion, Frame of reference, motion in a Straight line, position – time graph, speed and velocity, uniform and non-uniform motion, average speed and instantaneous velocity, uniformly accelerated motion, velocity – time and position – time graph, Relation for uniformly accelerated motion (graphical treatment)
2. Motion in a plane: Scalars and vectors, general vectors and their notations, position and displacement vectors, equality of vectors, unit vectors, multiplication of vectors by a real number, addition and subtraction of vectors, relative velocity,
resolution of a vector in a plane, rectangular components, Dot and Cross products of two vectors. Motion in a plane, cases of uniform velocity and uniform acceleration – projectile motion; uniform circular motion.
Unit-III Laws of Motion (10 Periods)
Concept of force, inertia, momentum, impulse, impulse-momentum theorem, Newton’s Laws of motion, Law of Conservation of linear momentum and its application. Static and Kinetic friction, laws of friction, rolling friction. Dynamics of uniform circular motion, Centripetal force, motion of a vehicle on a level circular road and vehicle on a banked road.
Unit-IV Work, Energy and Power (10 Periods)
Work done by a Constant force and variable force, kinetic energy, work- energy
theorem, power. Notion of potential energy, conservative and non-conservative forces, conservation of mechanical energy (Kinetic and Potential energies), elastic and in-elastic collisions in one dimension, coefficient of restitution.
Unit-V Motion of System of Particles and Rigid bodies: (12 Periods)
System of Particles and Rotational Motion: Centre of mass of a two-particle system, momentum conservation and centre of mass motion, centre of mass of rigid bodies, Centre of Mass of a uniform rod. Moment of a force, torque, angular momentum, conservation of angular momentum with its applications. Moment of inertia, radius of gyration, moment of inertia of simple geometrical objects (no derivation).
Unit-VI Gravitation (08 Periods)
Newton’s law of gravitation, Gravitational field and Potential, gravitational potential energy, acceleration due to gravity and its variation with altitude and depth, Escape velocity, orbital velocity of a satellite.
Unit-VII Properties of Bulk Matter (18 Periods)
1. Mechanical properties of Solids: Elastic Behaviours, Stress, Strain, Hooke’s Law, Stress-Strain diagram, Young’s modulus, Bulk modulus, Shear modulus of rigidity, Poisson’s ratio, elastic energy.
2. Mechanical properties of fluids: Surface energy and surface tension, angle of contact, excess pressure across a curved surface, application of surface tension ideas to drops, bubbles and capillary rise. Viscosity, Stoke’s law, terminal velocity, streamline and turbulent flow, critical velocity, Bernoulli’s theorem and its application.
3. Thermal properties of matter: Concepts of heat and temperature, Thermal expansion of solids, liquids and gases, specific heat capacity: Cp, Cv. Calorimetry, change of state, latent heat capacity. Heat transfer: Conduction, Convection and radiation, thermal conductivity, qualitative ideas of block body radiation, Wien’s displacement law, Stefan’s law.
Unit-VIII Thermodynamics (10 Periods)
Thermal equilibrium, definition of temperature (Zeroth law of thermodynamics) heat, work and internal energy. First law of thermodynamics, isothermal and adiabatic processes, second law of thermodynamics, reversible and irreversible processes, Carnot’s engine and its efficiency (no derivation).
Unit-IX Kinetic theory of gases: (04 Periods)
Equation of state of a perfect gas, work done in compressing a gas. Pressure exerted by an ideal gas (elementary idea), kinetic interpretation of temperature, mean and RMS speed of gas molecules, degrees of freedom, law of equipartition of energy (statement only) and its applications to specific heat of gases.
Unit-X Oscillation and waves (18 Periods)
1. Periodic motion: Period, Frequency, displacement as a function of time, periodic function. Simple harmonic motion and its equation, phase, oscillation of a spring, Restoring force and force constant, kinetic and potential energy in SHM, simple pendulum, derivation of expression for its time period.
2. Waves: Wave motion, transverse and longitudinal waves, speed of wave motion, displacement relation for a progressive wave, speed of longitudinal wave in an elastic medium and speed of transverse wave in a stretched string (qualitative idea only), principle of superposition of waves, reflection of waves, standing waves in strings and organ pipes.
— ALL THE BEST — | 1,052 | 5,068 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-21 | latest | en | 0.843304 |
https://questions.examside.com/past-years/jee/question/pthe-number-of-solutions-of-the-equation-sin-x--cos-jee-main-mathematics-trigonometric-functions-and-equations-d3xpfobbeimepfpk | 1,708,758,464,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474523.8/warc/CC-MAIN-20240224044749-20240224074749-00810.warc.gz | 480,630,344 | 40,325 | This chapter is currently out of syllabus
1
JEE Main 2022 (Online) 29th June Evening Shift
Numerical
+4
-1
Out of Syllabus
The number of solutions of the equation $$\sin x = {\cos ^2}x$$ in the interval (0, 10) is _________.
2
JEE Main 2022 (Online) 29th June Morning Shift
Numerical
+4
-1
Out of Syllabus
The number of elements in the set $$S = \{ \theta \in [ - 4\pi ,4\pi ]:3{\cos ^2}2\theta + 6\cos 2\theta - 10{\cos ^2}\theta + 5 = 0\}$$ is __________.
3
JEE Main 2022 (Online) 29th June Morning Shift
Numerical
+4
-1
Out of Syllabus
The number of solutions of the equation
$$2\theta - {\cos ^2}\theta + \sqrt 2 = 0$$ in R is equal to ___________.
4
JEE Main 2022 (Online) 25th June Morning Shift
Numerical
+4
-1
Out of Syllabus
The number of values of x in the interval $$\left( {{\pi \over 4},{{7\pi } \over 4}} \right)$$ for which
$$14\cos e{c^2}x - 2{\sin ^2}x = 21 - 4{\cos ^2}x$$ holds, is ____________. | 356 | 923 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-10 | latest | en | 0.778014 |
http://www.jiskha.com/display.cgi?id=1377480290 | 1,498,590,069,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321536.20/warc/CC-MAIN-20170627185115-20170627205115-00487.warc.gz | 553,977,181 | 3,503 | # Physics
posted by .
A car traveling at 11m/s accelerates to 2.82m/s^2 for 13 seconds how far does the car travel?
• Physics -
s = vt + 1/2 at^2 = 11*13 + 1.41*13^2 = ... | 67 | 175 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2017-26 | latest | en | 0.821651 |
https://socratic.org/questions/how-do-you-convert-r-1-1-cos-theta-into-cartesian-form | 1,586,489,838,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371883359.91/warc/CC-MAIN-20200410012405-20200410042905-00068.warc.gz | 676,194,809 | 6,237 | # How do you convert r = 1/(1-cos(theta)) into cartesian form?
Aug 9, 2016
${y}^{2} = 2 x + 1$ representing the parabola with axis ax =-1/2 and focus at the origin.
#### Explanation:
The conversion formula is $\left(x , y\right) = \left(r \cos \theta , r \sin \theta\right)$.
The given equation is r = sqrt (x^2+y^2)=1/(1-x/sqrt(x^2+y^2)
Cross multiplying, rationalizing and simplifying,
${y}^{2} = 2 x + 1$
This is in the standard form of the equation of parabolas
${\left(y - \beta\right)}^{2} = 4 a \left(s - \alpha\right)$,
representing parabolas having vertex at (alpha, beta) parameter a
and focus at(alpha +a, beta)#.
Here, $a = \frac{1}{2} , \alpha = - \frac{1}{2} \mathmr{and} \beta = 0$... | 249 | 711 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2020-16 | latest | en | 0.723827 |
https://nrich.maths.org/public/leg.php?code=-99&cl=1&cldcmpid=5648 | 1,566,684,741,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027321786.95/warc/CC-MAIN-20190824214845-20190825000845-00161.warc.gz | 573,681,642 | 9,261 | Search by Topic
Resources tagged with Working systematically similar to Inside Triangles:
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Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
Nine-pin Triangles
Age 7 to 11 Challenge Level:
How many different triangles can you make on a circular pegboard that has nine pegs?
Inside Triangles
Age 5 to 7 Challenge Level:
How many different triangles can you draw on the dotty grid which each have one dot in the middle?
Triangles All Around
Age 7 to 11 Challenge Level:
Can you find all the different triangles on these peg boards, and find their angles?
Find the Difference
Age 5 to 7 Challenge Level:
Place the numbers 1 to 6 in the circles so that each number is the difference between the two numbers just below it.
Four Triangles Puzzle
Age 5 to 11 Challenge Level:
Cut four triangles from a square as shown in the picture. How many different shapes can you make by fitting the four triangles back together?
Tri.'s
Age 7 to 11 Challenge Level:
How many triangles can you make on the 3 by 3 pegboard?
Triangle Edges
Age 5 to 7 Challenge Level:
How many triangles can you make using sticks that are 3cm, 4cm and 5cm long?
Triangle Animals
Age 5 to 7 Challenge Level:
How many different ways can you find to join three equilateral triangles together? Can you convince us that you have found them all?
More Transformations on a Pegboard
Age 7 to 11 Challenge Level:
Use the interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle.
Matching Time
Age 5 to 7 Challenge Level:
Try this matching game which will help you recognise different ways of saying the same time interval.
Arrangements
Age 7 to 11 Challenge Level:
Is it possible to place 2 counters on the 3 by 3 grid so that there is an even number of counters in every row and every column? How about if you have 3 counters or 4 counters or....?
Geoboards
Age 7 to 11 Challenge Level:
This practical challenge invites you to investigate the different squares you can make on a square geoboard or pegboard.
Putting Two and Two Together
Age 7 to 11 Challenge Level:
In how many ways can you fit two of these yellow triangles together? Can you predict the number of ways two blue triangles can be fitted together?
Red Even
Age 7 to 11 Challenge Level:
You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters?
How Many Times?
Age 7 to 11 Challenge Level:
On a digital 24 hour clock, at certain times, all the digits are consecutive. How many times like this are there between midnight and 7 a.m.?
Three Ball Line Up
Age 5 to 7 Challenge Level:
Use the interactivity to help get a feel for this problem and to find out all the possible ways the balls could land.
Same Length Trains
Age 5 to 7 Challenge Level:
How many trains can you make which are the same length as Matt's, using rods that are identical?
Making Trains
Age 5 to 7 Challenge Level:
Can you make a train the same length as Laura's but using three differently coloured rods? Is there only one way of doing it?
Combining Cuisenaire
Age 7 to 11 Challenge Level:
Can you find all the different ways of lining up these Cuisenaire rods?
Fault-free Rectangles
Age 7 to 11 Challenge Level:
Find out what a "fault-free" rectangle is and try to make some of your own.
Tetrafit
Age 7 to 11 Challenge Level:
A tetromino is made up of four squares joined edge to edge. Can this tetromino, together with 15 copies of itself, be used to cover an eight by eight chessboard?
Are You Well Balanced?
Age 5 to 7 Challenge Level:
Can you work out how to balance this equaliser? You can put more than one weight on a hook.
Robot Monsters
Age 5 to 7 Challenge Level:
Use these head, body and leg pieces to make Robot Monsters which are different heights.
Sticks and Triangles
Age 7 to 11 Challenge Level:
Using different numbers of sticks, how many different triangles are you able to make? Can you make any rules about the numbers of sticks that make the most triangles?
Counters
Age 7 to 11 Challenge Level:
Hover your mouse over the counters to see which ones will be removed. Click to remove them. The winner is the last one to remove a counter. How you can make sure you win?
Egyptian Rope
Age 7 to 11 Challenge Level:
The ancient Egyptians were said to make right-angled triangles using a rope with twelve equal sections divided by knots. What other triangles could you make if you had a rope like this?
Mixed-up Socks
Age 5 to 7 Challenge Level:
Start with three pairs of socks. Now mix them up so that no mismatched pair is the same as another mismatched pair. Is there more than one way to do it?
Tessellate the Triominoes
Age 5 to 7 Challenge Level:
What happens when you try and fit the triomino pieces into these two grids?
Cover the Camel
Age 5 to 7 Challenge Level:
Can you cover the camel with these pieces?
Seven Flipped
Age 7 to 11 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
How Long Does it Take?
Age 7 to 11 Challenge Level:
In this matching game, you have to decide how long different events take.
Making Squares
Age 7 to 11
Investigate all the different squares you can make on this 5 by 5 grid by making your starting side go from the bottom left hand point. Can you find out the areas of all these squares?
Age 7 to 11 Challenge Level:
How many DIFFERENT quadrilaterals can be made by joining the dots on the 8-point circle?
Growing Garlic
Age 5 to 7 Challenge Level:
Ben and his mum are planting garlic. Use the interactivity to help you find out how many cloves of garlic they might have had.
Uncanny Triangles
Age 7 to 11 Challenge Level:
Can you help the children find the two triangles which have the lengths of two sides numerically equal to their areas?
Teddy Town
Age 5 to 14 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
One to Fifteen
Age 7 to 11 Challenge Level:
Can you put the numbers from 1 to 15 on the circles so that no consecutive numbers lie anywhere along a continuous straight line?
Code Breaker
Age 7 to 11 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
Multiples Grid
Age 7 to 11 Challenge Level:
What do the numbers shaded in blue on this hundred square have in common? What do you notice about the pink numbers? How about the shaded numbers in the other squares?
Dicey Perimeter, Dicey Area
Age 7 to 11 Challenge Level:
In this game for two players, you throw two dice and find the product. How many shapes can you draw on the grid which have that area or perimeter?
A Square of Numbers
Age 7 to 11 Challenge Level:
Can you put the numbers 1 to 8 into the circles so that the four calculations are correct?
Difference
Age 7 to 11 Challenge Level:
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Try out the lottery that is played in a far-away land. What is the chance of winning? | 2,042 | 8,761 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-35 | latest | en | 0.907265 |
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2926 | 1,566,551,202,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318243.40/warc/CC-MAIN-20190823083811-20190823105811-00490.warc.gz | 9,744,599 | 3,424 | Welcome to ZOJ
Problem Sets Information Select Problem Runs Ranklist
ZOJ Problem Set - 2926
Hacker in TopCoder
Time Limit: 2 Seconds Memory Limit: 32768 KB
As we all know, TCO (TopCoder Open) Algorithm competition Round 2 will be hold on February 24. It is fortunate that many ACMers from ZJU have advanced into this round. Many coders from other countries and other universities compete in this round too.
Within 3 hours before the round, coders should register in the Arena (where coders compete), otherwise they will lose the chance to compete in this round. 5 minutes before the match begins, the register phase ends, and each coder will be assigned to a room. To make the problem easier, we simplify this phase as follow:
TC first decides how many room is needed for the match. If there are k coders that register, the number of rooms r will be the minimum integer that is greater than or equal to k / 25 (which makes each room almost 25 coders). All the rooms are marked as room0 ~ roomr-1. TC then sort all the coders by their rating (Each coder has a rating which is gained in past TopCoder matches) in descending order. The coders are marked as coder0 ~ coderk-1. Then coder i will be assigned to room i%r.
Brother Can, the coach of ZJU ACM team, makes a decision that all the ZJU competitors should be in the same room and no other coders should be in that room! He will choose some of the ZJU advancers to register for the round (Sorry to others!). But, it will still be a miracle for all the coders from ZJU to be assigned to the same room and dominate it. So, Brother Can, also known as a famous hacker, has successfully break into the TopCoder Member Infomation System, and can modify the rating of coders. Because password is needed to modify members' infomation, Brother Can can only modify ZJU members' rating. When one's rating is changed from p to q, there is |p - q| rating change. Brother Can want to make total rating change minimum.
PS: After the modification, there should not be any two coders with the same rating (otherwise the sort phase will fail).
Input
There are multiple test cases. There are three lines for each case. The first line is two integers n (0 < n <= 1000) and m (m >= 0). n is the number of advancer from ZJU, m is the number of advancer from other countries and universities. The second line will be n non-negative intergers, the rating of n ZJU advancers. The third line will be m non-negative intergers, the rating of m other advancers. To make the sort phase simpler, all the n + m rating will be even number and distinct, and all the ratings will be lower than 10000.
Process to the end of file.
Output
For each case, if there is a way to achieve Brother Can's goal (choose some advancers to register, modify some of their ratings, and all ZJU competitors should be in the same room and dominate it), output one interger in a line, the minimum total rating change Brother Can will make. If there is no way to do this, output "Impossible" in a line.
Sample Input
3 0
1000 2000 3000
3 3
1000 2000 3000
4000 5000 6000
25 25
1000 2000 3000 500 2500 1002 2002 3002 502 2502 1010 2010 3010 510 2510 1012 2012 3012 512 2512 100 2100 3100 600 2600
800 900 920 940 960 1800 1900 1920 1940 1960 850 950 970 990 160 2300 2400 2420 2440 2460 808 908 928 948 968
Sample Output
0
Impossible
11328
Author: HANG, Hang
Source: ZOJ Monthly, February 2008
Submit Status | 927 | 3,425 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2019-35 | latest | en | 0.933796 |
https://nebusresearch.wordpress.com/2016/09/16/dark-secrets-of-mathematicians-something-about-integration-by-parts/ | 1,500,842,768,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424610.13/warc/CC-MAIN-20170723202459-20170723222459-00046.warc.gz | 687,808,438 | 44,316 | Dark Secrets of Mathematicians: Something About Integration By Parts
A friend took me up last night on my offer to help with any mathematics she was unsure about. I’m happy to do it, though of course it came as I was trying to shut things down for bed. But that part was inevitable and besides the exam was today. I thought it worth sharing here, though.
There’s going to be some calculus in this. There’s no avoiding that. If you don’t know calculus, relax about what the symbols exactly mean. It’s a good trick. Pretend anything you don’t know is just a symbol for “some mathematics thingy I can find out about later, if I need to, and I don’t necessarily have to”.
“Integration by parts” is one of the standard tricks mathematicians learn in calculus. It comes in handy if you want to integrate a function that itself looks like the product of two other functions. You find the integral of a function by breaking it up into two parts, one of which you differentiate and one of which you integrate. This gives you a product of functions and then a new integral to do. A product of functions is easy to deal with. The new integral … well, if you’re lucky, it’s an easier integral than you started with.
As you learn integration by parts you learn to look ways to break up functions so the new integral is easier. There’s no hard and fast rule for this. But bet on “the part that has a polynomial in it” as the part that’s better differentiated. “The part that has sines and cosines in it” is probably the part that’s better integrated. An exponential, like 2x, is as easily differentiated as integrated. The exponential of a function, like say 2x2, is better differentiated. These usually turn out impossible to integrate anyway. At least impossible without using crazy exotic functions.
So your classic integration-by-parts problem gives you an expression like this:
$\int x \sin(x) dx = -x \cos(x) - \int \sin(x) dx$
If you weren’t a mathematics major that might not look better to you, what with it still having integrals and sines and stuff in it. But ask your mathematics friend. She’ll tell you. The thing on the right-hand side is way better. That last term, the integral of the sine of x? She can do that in her sleep. It barely counts as work, at least by the time you’ve got in class to doing integration by parts. It’ll be $-x\cos(x) + \cos(x)$.
But sometimes, especially if the function being integrated — the “integrand”, by the way, and good luck playing that in Scrabble — is a bunch of trig functions and exponentials, you get some sad situation like so:
$\int \sin(x) \cos(x) dx = \sin^2(x) - \int \sin(x) \cos(x) dx$
That is, the thing we wanted to integrate, on the left, turns up on the right too. The student sits down, feeling the futility of modern existence. We’re stuck with the original problem all over again and we’re short of tools to do something about it.
This is the point my friend was confused by, and is the bit of dark magic I want to talk about here. We’re not stumped! We can fall back on one of those mathematics tricks we are always allowed to do. And it’s a trick that’s so simple it seems like it can’t do anything.
It’s substitution. We are always allowed to substitute one thing for something else that’s equal to it. So in that above equation, what can we substitute, and for what? … Well, nothing in that particular bunch of symbols. We’re going to introduce a new one. It’s going to be the value of the integral we want to evaluate. Since it’s an integral, I’m going to call it ‘I’. You don’t have to call it that, but you’re going to anyway. It doesn’t need a more thoughtful name.
So I shall define:
$I \equiv \int \sin(x) \cos(x) dx$
The triple-equals-sign there is an extravagance, I admit. But it’s a common one. Mathematicians use it to say “this is defined to be equal to that”. Granted, that’s what the = sign means. But the triple sign connotes how we emphasize the definition part. That is, ‘I’ might have been anything at all, and we choose this of the universe of possibilities.
How does this help anything? Well, it turns the integration-by-parts problem into this equation:
$I = \sin^2(x) - I$
And we want to know what ‘I’ equals. And now suddenly it’s easier to see that we don’t actually have to do any calculus from here on out. We can solve it the way we’d solve any problem in high school algebra, which is, move ‘I’ to the other side. Formally, we add the same thing to the left- and the right-hand sides. That’s ‘I’ …
$2I = \sin^2(x)$
… and then divide both sides by the same number, 2 …
$I = \frac{1}{2}\sin^2(x)$
And now remember that substitution is a free action. We can do it whenever we like, and we can undo it whenever we like. This is a good time to undo it. Putting the whole expression back in for ‘I’ we get …
$\int \sin(x) \cos(x) dx = \frac{1}{2}\sin^2(x)$
… which is the integral, evaluated.
(Someone would like to point out there should be a ‘plus C’ in there. This someone is right, for reasons that would take me too far afield to describe right now. We can overlook it for now anyway. I just want that someone to know I know what you’re thinking and you’re not catching me on this one.)
Sometimes, the integration by parts will need two or even three rounds before you get back the original integrand. This is because the instructor has chosen a particularly nasty problem for homework or the exam. It is not hopeless! But you will see strange constructs like 4/5 I equalling something. Carry on.
What makes this a bit of dark magic? I think it’s because of habits. We write down something simple on the left-hand side of an equation. We get an expression for what the right-hand side should be, and it’s usually complicated. And then we try making the right-hand side simpler and simpler. The left-hand side started simple so we never even touch it again. Indeed, working out something like this it’s common to write the left-hand side once, at the top of the page, and then never again. We just write an equals sign, underneath the previous line’s equals sign, and stuff on the right. We forget the left-hand side is there, and that we can do stuff with it and to it.
I think also we get into a habit of thinking an integral and integrand and all that is some quasi-magic mathematical construct. But it isn’t. It’s just a function. It may even be just a number. We don’t know what it is, but it will follow all the same rules of numbers, or functions. Moving it around may be more writing but it’s not different work to moving ‘4’ or ‘x2‘ around. That’s the value of replacing the integral with a symbol like ‘I’. It’s not that there’s something we can do with ‘I’ that we can’t do with ‘$\int \sin(x)\cos(x) dx$‘, other than write it in under four pen strokes. It’s that in algebra we learned the habit of moving a letter around to where it’s convenient. Moving a whole integral expression around seems different.
But it isn’t. It’s the same work done, just on a different kind of mathematics. I suspect finding out that it could be a trick that simple throws people off. | 1,685 | 7,091 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 9, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2017-30 | longest | en | 0.952866 |
https://math.stackexchange.com/questions/2775179/deductions-about-the-prime-factorization-of-odd-integers-m1-satisfying-frac/2782806 | 1,558,276,863,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232254889.43/warc/CC-MAIN-20190519141556-20190519163556-00416.warc.gz | 560,916,570 | 34,090 | # Deductions about the prime factorization of odd integers $m>1$ satisfying $\frac{-3+\sqrt{1+8m}}{2}=\prod_{\substack{p\mid m\\p\text{ prime}}}(p-1)$
For each integer $m>1$ we denote the Euler's totient function as $\varphi(m)$ and the product of distinct primes dividing it, that is its square-free kernel, as $\operatorname{rad}(m)$ (that is the arithmetic function explained in this Wikipedia). Thus for our integer $m>1$ one has $$\varphi(\operatorname{rad}(m))=\prod_{\substack{p\mid n\\p\text{ prime}}}(p-1).$$
While I was calculating solutions of some equations that involves* the arithmetic function $\varphi(\operatorname{rad}(n))$ and the sum of divisors and totatives for even perfect numbers, I got also odd solutions of next equation $$\frac{-3+\sqrt{1+8n}}{2}=\varphi(\operatorname{rad}(n)).\tag{1}$$
The odd solutions of $(1)$ that I can to calculate are $n=45$ and $n=3321$. These solutions have the form $q\cdot t^2$, being $q$ an odd prime $\equiv 1\text{ mod }4$ and $\gcd(q,t)=1$, in fact in previous solutions $t$ is a power of $3$. I would like to know what work can be done about the characterization of such odd solutions, if there are more odd integers being solutions.
Question. Imagine that we need to study the odd integers $m>1$ satisfying $(1)$. What can we deduce about such odd solutions $m$? I am asking about their prime factorization. Many thanks.
One has thus that we need to study the prime factorization of odd integers $m$ that satisfy $$\frac{-3+\sqrt{1+8m}}{2}=\prod_{\substack{p\mid m\\p\text{ prime}}}(p-1).\tag{2}$$
• *The equations that I evoke (thus comment this isn't directly related to previous $(1)$ and the Question), and that satisfy each even perfect numbers $e$ are this $\frac{-3+\sqrt{1+4\sigma(e)}}{2}=\varphi(\operatorname{rad}(e))$ being $\sigma(n)$ the sum of divisors function, and thisone $2(e-1)-\varphi(e)=\left(\frac{11+3\sqrt{1+8e}}{8}\right)\varphi(\operatorname{rad}(e))$. I add this comment as context, and as a comparison of previous equations, $(1)$ is the simplest equation, but there exist (odd) integers satisfying it that aren't even perfect numbers. – user243301 May 10 '18 at 14:08
Calculation shows that$$\frac{-3+\sqrt{1+8m}}{2}=\phi(rad(m))$$for odd $m= $$45$$$$3321$$$$926510115949281$$$$1716841910146257168839018970561$$These numbers have the form $$\frac{3^n(3^n+1)}{2}$$for even$n=2, 4, 16, 32$, respectively. Note that$m$is thus the$3^nth$triangle number. Indeed,$\sqrt{1+8m}$is an integer only if odd$m$is triangular, since for integer$k$$\sqrt{1+\frac{8k(k+1)}{2}}=\sqrt{4k^2+4k+1}=\sqrt{(2k+1)^2}=2k+1$$
Factoring the four $m$:$$45=3^2(3^2+1)/2=3^2\cdot 5$$$$3321=3^4(3^4+1)/2=3^4\cdot 41$$$$926510115949281=3^{16}(3^{16}+1)/2=3^{16}\cdot 21523361$$$$1716841910146257168839018970561=3^{32}(3^{32}+1)/2=3^{32}\cdot 926510094425921$$For even $n$ from $2$ through $32$, only these four $m$ have the form $3^np$, i.e are composed of a power of $3$ and a single odd prime $p$.
This condition appears to be sufficient for$$\frac{-3+\sqrt{1+8m}}{2}=\phi(rad(m))$$and can be proven as follows:
If $$m=3^np$$ then $$rad(m)=3p$$and$$\phi(rad(m))=(3-1)(p-1)=2p-2$$
And again, if$$\sqrt{1+8m}=\sqrt{1+8\cdot \frac{3^n(3^n+1)}{2}}=\sqrt{1+8\cdot \frac{3^{2n}+3^n}{2}}=\sqrt{4\cdot 3^{2n}+4\cdot 3^n+1}=2\cdot 3^n+1$$then$$\frac{-3+\sqrt{1+8m}}{2}=\frac{-3+2\cdot 3^n+1}{2}=3^n-1$$And since$$p=\frac{3^n+1}{2}$$then$$3^n-1=2p-2$$as above, and$$\frac{-3+\sqrt{1+8m}}{2}=\phi(rad(m))$$for all odd $m=3^np$.
Note that $n$ must be even, since if $n$ is odd, then $3^n+1=0(mod4)$, making $\frac{3^n+1}{2}$ even and hence not an odd prime.
At first, I expected the equation might hold for all $n$ equal to a power of $2$, but $n=8$ does not work. Primes are infinitely many, but if we cannot know the law of even $n$ for which $(3^n+1)/2$ is prime, then even if $$\frac{-3+\sqrt{1+8m}}{2}=\phi(rad(m))$$is true for infinitely many odd $m$, it seems that, like the primes themselves, we will have to find them one at a time.
Further calculation reveals that $\frac{3^n+1}{2}$= $$1716841910146256242328924544641$$is prime for $n=64$, when $m$=$$5895092288869291585760436430707976174749252052380249479093121$$But $\frac{3^n+1}{2}$ is not prime for $n=128, 256, 512$ or $1024$.
More importantly, it appears--and I suppose it can be proven--that $5$ is a divisor of $\frac{3^n+1}{2}$ not only for $n=2$ but for every $n$ which is an odd multiple of $2$. And since all these increasingly large $\frac{3^n+1}{2}$ must contain other prime factors besides $5$, they cannot be prime. This rules out $$n=6, 10, 14, 18, 22...$$or one half of all even $n$, less one.
Similarly, $41$ is a divisor of $\frac{3^n+1}{2}$ for $n=4$ and all odd multiples of $4$, so that $\frac{3^n+1}{2}$ cannot be prime when $n$ is an odd multiple of $4$. This rules out half of the remaining even $n$, less one$$12, 20, 28, 36, 44...$$
Again, since $17$ and $193$ divide $\frac{3^n+1}{2}$ for $n=8$, as well as for all odd multiples of $8$, then$$8, 24, 40, 56, 72...$$or exactly half of the remaining $n$ are ruled out.
Accordingly, since $n$ is even, and an even number which is not an odd multiple of $2, 4, 8, 16, 32 ...$ must be a power of $2$, it seems a necessary but insufficient condition of the primality of $\frac{3^n+1}{2}$ that $n$ be a power of $2$.
Hence, like the Fermat primes, the odd $m$ for which$$\frac{-3+\sqrt{1+8m}}{2}=\phi(rad(m))$$are not only rare, but perhaps finite in number.
• Many thanks for your attention, calculations and reasonings. – user243301 May 15 '18 at 20:47
• @user243301 If you want, you can $\color{green}{\text{accept}}$ this answer :) – Mr Pie May 16 '18 at 1:12 | 1,979 | 5,666 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2019-22 | latest | en | 0.843797 |
https://beasleydirect-blog.com/how-to-find-rank-of-a-matrix/ | 1,713,948,329,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819089.82/warc/CC-MAIN-20240424080812-20240424110812-00885.warc.gz | 110,109,128 | 32,117 | HomeTren&dHow to Find the Rank of a Matrix: A Comprehensive Guide
# How to Find the Rank of a Matrix: A Comprehensive Guide
Author
Date
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When it comes to linear algebra, matrices play a crucial role in solving various mathematical problems. One important concept related to matrices is their rank. The rank of a matrix provides valuable insights into its properties and can be used to solve systems of linear equations, determine the dimension of the column space, and much more. In this article, we will explore the concept of matrix rank in detail, discuss different methods to find the rank of a matrix, and provide examples and case studies to illustrate the practical applications of this concept.
## Understanding Matrix Rank
Before diving into the methods of finding the rank of a matrix, let’s first understand what rank actually means. The rank of a matrix is defined as the maximum number of linearly independent rows or columns in the matrix. In simpler terms, it represents the dimension of the vector space spanned by the rows or columns of the matrix.
The rank of a matrix can provide valuable information about its properties. For example, if the rank of a matrix is equal to the number of rows or columns, it means that all the rows or columns are linearly independent, and the matrix is said to have full rank. On the other hand, if the rank is less than the number of rows or columns, it indicates that there are linear dependencies among the rows or columns, and the matrix is said to be rank deficient.
## Methods to Find the Rank of a Matrix
There are several methods to find the rank of a matrix, each with its own advantages and limitations. In this section, we will discuss three commonly used methods: the row echelon form method, the determinant method, and the singular value decomposition method.
### The Row Echelon Form Method
The row echelon form method is one of the most straightforward methods to find the rank of a matrix. It involves transforming the matrix into its row echelon form using elementary row operations and then counting the number of non-zero rows in the resulting matrix.
Here are the steps to find the rank of a matrix using the row echelon form method:
2. Apply elementary row operations to transform the matrix into its row echelon form.
3. Count the number of non-zero rows in the row echelon form matrix.
4. The count obtained in step 3 is the rank of the matrix.
Let’s consider an example to illustrate this method:
Example:
Consider the following matrix:
[1 2 3]
[4 5 6]
[7 8 9]
[1 2 3]
[4 5 6]
[7 8 9]
Step 2: Apply elementary row operations to transform the matrix into its row echelon form.
[1 2 3]
[0 -3 -6]
[0 0 0]
Step 3: Count the number of non-zero rows in the row echelon form matrix.
There are two non-zero rows in the row echelon form matrix.
Step 4: The count obtained in step 3 is the rank of the matrix.
Therefore, the rank of the given matrix is 2.
### The Determinant Method
The determinant method is another commonly used method to find the rank of a matrix. It involves calculating the determinant of different submatrices of the given matrix and using the properties of determinants to determine the rank.
Here are the steps to find the rank of a matrix using the determinant method:
2. Calculate the determinant of different submatrices of the given matrix.
3. Count the number of non-zero determinants.
4. The count obtained in step 3 is the rank of the matrix.
Let’s consider an example to illustrate this method:
Example:
Consider the following matrix:
[1 2 3]
[4 5 6]
[7 8 9]
[1 2 3]
[4 5 6]
[7 8 9]
Step 2: Calculate the determinant of different submatrices of the given matrix.
The determinant of the 2×2 submatrix [1 2; 4 5] is (1*5) – (2*4) = -3.
The determinant of the 2×2 submatrix [1 3; 4 6] is (1*6) – (3*4) = -6.
The determinant of the 2×2 submatrix [2 3; 5 6] is (2*6) – (3*5) = -3.
Step 3: Count the number of non-zero determinants.
There are three non-zero determinants.
Step 4: The count obtained in step 3 is the rank of the matrix.
Therefore, the rank of the given matrix is 3.
### The Singular Value Decomposition Method
The singular value decomposition (SVD) method is a powerful technique to find the rank of a matrix. It involves decomposing the matrix into three separate matrices and using the properties of these matrices to determine the rank.
Here are the steps to find the rank of a matrix using the singular value decomposition method:
2. Perform singular value decomposition on the matrix to obtain three separate matrices: U, Σ, and V.
3. Count the number of non-zero singular values in the Σ matrix.
4. The count obtained in step 3 is the rank of the matrix.
Let’s consider an example to illustrate this method:
Example:
Consider the following matrix:
[1 2 3]
<p | 1,148 | 4,842 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-18 | latest | en | 0.88661 |
https://www.imomath.xyz/2022/09/2021-imo-sl-a1.html | 1,675,032,213,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499768.15/warc/CC-MAIN-20230129211612-20230130001612-00602.warc.gz | 842,378,959 | 20,433 | ### 2021 IMO SL #A1
Problem: Let $n$ be a positive integer. Given is a subset $A$ of $\{0,1,...,5^n\}$ with $4n+2$ elements. Prove that there exist three elements $a<b<c$ from $A$ such that $c+2a>3b$.
Solution:
Suppose for the sake of contradiction that $c+2a\le 3b$ for all $a,b,c\in A$ with $a<b<c$.
We claim that if the elements of $A$ are $s_1<s_2<\dots<s_{4n+2}$, then $$s_i\ge s_{4n+2}(1-(\tfrac{2}{3})^{i-1})$$ for all $i$. We use induction to prove this -- the base case is trivially true because $s_1\geq 0$. As for the inductive step, consider $s_k.$ Take $(a,b,c)=(s_{k-1},s_k,s_{4n+2})$; by our assumption, we have $$s_{4n+2}+2s_{k-1}\ge 3s_k.$$ Therefore, $$s_k\ge\frac{s_{4n+2}+2s_{k-1}}{3}\ge\frac{s_{4n+2}+2(s_{4n+2}(1-(\tfrac{2}{3})^{k-2}))}{3}=\frac{s_{4n+2}(3-2(\tfrac{2}{3})^{k-2})}{3}=s_{4n+2}(1-(\tfrac{2}{3})^{k-1}),$$ which completes the induction.
Now, using our claim, we have $$s_{4n+1}\ge s_{4n+2}(1-(\tfrac{2}{3})^{4n})=s_{4n+2}-(\tfrac{16}{81})^{n}s_{4n+2}>s_{4n+2}-(\tfrac{1}{5})^{n}5^n=s_{4n+2}-1,$$ which is impossible because $s_{4n+1}$ is an integer less than $s_{4n+2}$.
We are done. $\square$ | 516 | 1,135 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2023-06 | longest | en | 0.794319 |
http://www.vindy.com/users/AnotherBadIdea/comments/ | 1,526,992,224,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864725.4/warc/CC-MAIN-20180522112148-20180522132148-00128.warc.gz | 475,399,479 | 9,251 | ### Comment history
#### BREAKING NEWS: Strickland calls on PUCO to halt lightbulb program
Ok, some simple math. If the Wal-Mart price is correct. 6pk for \$13.85 or \$2.30 per bulb. Let's say they are making a 20% profit. This means Wal-Mart is making \$2.77 on a 6pk of bulbs or 46 cents per bulb. This means the cost of the bulb is \$1.84, which is actully alot less because this includes Wal-Mart cost to do business.
Ohio Edison or the PUCO, whoever, wants to charge us \$10.50 per bulb. Using the \$1.84 cost, these people want to make a profit \$8.66 per bulb. On 3.75 million bulbs the profit is \$32,475,000.
Now who's getting this?
October 7, 2009 at 8:33 p.m. suggest removal
#### BREAKING NEWS: Strickland calls on PUCO to halt lightbulb program
Once again peopel are taking advantage of the average person. A six pack of these bulbs cost \$13.85 at Wal-Mart. If I am forced to take these I will short my bill until they come out and shut off the power.
The head of the PUCO should be fired for even thinking this was a good idea. So, the real question is who is making the profit and who is on the take, once again.
October 7, 2009 at 7:35 p.m. suggest removal | 331 | 1,182 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-22 | latest | en | 0.945994 |
https://metanumbers.com/1029860 | 1,643,173,337,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304915.53/warc/CC-MAIN-20220126041016-20220126071016-00709.warc.gz | 437,365,081 | 7,651 | # 1029860 (number)
1,029,860 (one million twenty-nine thousand eight hundred sixty) is an even seven-digits composite number following 1029859 and preceding 1029861. In scientific notation, it is written as 1.02986 × 106. The sum of its digits is 26. It has a total of 6 prime factors and 48 positive divisors. There are 356,352 positive integers (up to 1029860) that are relatively prime to 1029860.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 7
• Sum of Digits 26
• Digital Root 8
## Name
Short name 1 million 29 thousand 860 one million twenty-nine thousand eight hundred sixty
## Notation
Scientific notation 1.02986 × 106 1.02986 × 106
## Prime Factorization of 1029860
Prime Factorization 22 × 5 × 13 × 17 × 233
Composite number
Distinct Factors Total Factors Radical ω(n) 5 Total number of distinct prime factors Ω(n) 6 Total number of prime factors rad(n) 514930 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,029,860 is 22 × 5 × 13 × 17 × 233. Since it has a total of 6 prime factors, 1,029,860 is a composite number.
## Divisors of 1029860
48 divisors
Even divisors 32 16 16 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 48 Total number of the positive divisors of n σ(n) 2.47666e+06 Sum of all the positive divisors of n s(n) 1.4468e+06 Sum of the proper positive divisors of n A(n) 51597 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1014.82 Returns the nth root of the product of n divisors H(n) 19.9597 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,029,860 can be divided by 48 positive divisors (out of which 32 are even, and 16 are odd). The sum of these divisors (counting 1,029,860) is 2,476,656, the average is 51,597.
## Other Arithmetic Functions (n = 1029860)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 356352 Total number of positive integers not greater than n that are coprime to n λ(n) 1392 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 80499 Total number of primes less than or equal to n r2(n) 64 The number of ways n can be represented as the sum of 2 squares
There are 356,352 positive integers (less than 1,029,860) that are coprime with 1,029,860. And there are approximately 80,499 prime numbers less than or equal to 1,029,860.
## Divisibility of 1029860
m n mod m 2 3 4 5 6 7 8 9 0 2 0 0 2 6 4 8
The number 1,029,860 is divisible by 2, 4 and 5.
• Arithmetic
• Abundant
• Polite
• Practical
## Base conversion (1029860)
Base System Value
2 Binary 11111011011011100100
3 Ternary 1221022200222
4 Quaternary 3323123210
5 Quinary 230423420
6 Senary 34023512
8 Octal 3733344
10 Decimal 1029860
12 Duodecimal 417b98
20 Vigesimal 68ed0
36 Base36 m2n8
## Basic calculations (n = 1029860)
### Multiplication
n×y
n×2 2059720 3089580 4119440 5149300
### Division
n÷y
n÷2 514930 343287 257465 205972
### Exponentiation
ny
n2 1060611619600 1092281482561256000 1124897007630535104160000 1158486432278382882370217600000
### Nth Root
y√n
2√n 1014.82 100.986 31.8562 15.9425
## 1029860 as geometric shapes
### Circle
Diameter 2.05972e+06 6.4708e+06 3.33201e+12
### Sphere
Volume 4.57534e+18 1.3328e+13 6.4708e+06
### Square
Length = n
Perimeter 4.11944e+06 1.06061e+12 1.45644e+06
### Cube
Length = n
Surface area 6.36367e+12 1.09228e+18 1.78377e+06
### Equilateral Triangle
Length = n
Perimeter 3.08958e+06 4.59258e+11 891885
### Triangular Pyramid
Length = n
Surface area 1.83703e+12 1.28727e+17 840877
## Cryptographic Hash Functions
md5 6d575818e1179b4873562670e08234e2 3fd3f33ea80d64230747bc6aeff8d4fc61a4467d deffdde148130ed6cae3aa633fc592073e3f69a61ce9006591c727f50041d2a0 91109b5ebfc7ad6f71762394af6f45a110d2959170533aa25bda8b9f2c0682b6103eabd0537f5d6551678e47f03ea31cbf0992c28010d6aa3946664aba6f7f40 e6a6a65298377e773be014d6ef46db9bef31457f | 1,525 | 4,265 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2022-05 | latest | en | 0.799415 |
https://www.quizover.com/physics-k12/course/motion-by-openstax-motion?page=1 | 1,544,614,152,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823872.13/warc/CC-MAIN-20181212112626-20181212134126-00639.warc.gz | 1,028,346,008 | 17,284 | # Motion (Page 2/4)
Page 2 / 4
The requirement of an observer in both identifying and quantifying motion brings about new dimensions to the understanding of motion. Notably, the motion of a body and its measurement is found to be influenced by the state of motion of the observer itself and hence by the state of motion of the attached frame of reference. As such, a given motion is evaluated differently by different observers (system of references).
Two observers in the same state of motion, such as two persons standing on the platform, perceive the motion of a passing train in exactly same manner. On the other hand, the passenger in a speeding train finds that the other train crossing it on the parallel track in opposite direction has the combined speed of the two trains $\left({v}_{1}+{v}_{1}\right)$ . The observer on the ground, however, find them running at their individual speeds ${v}_{1}$ and ${v}_{2}$ .
From the discussion above, it is clear that motion of an object is an attribute, which can not be stated in absolute term; but it is a kind of attribute that results from the interaction of the motions of the both object and observer (frame of reference).
## Frame of reference and observer
Frame of reference is a mathematical construct to specify position or location of a point object in space. Basically, frame of reference is a coordinate system. There are plenty of coordinate systems in use, but the Cartesian coordinate system, comprising of three mutually perpendicular axes, is most common. A point in three dimensional space is defined by three values of coordinates i.e. x,y and z in the Cartesian system as shown in the figure below. We shall learn about few more useful coordinate systems in next module titled " Coordinate systems in physics ".
We need to be specific in our understanding of the role of the observer and its relation with frame of reference. Observation of motion is considered an human endeavor. But motion of an object is described in reference of both human and non-human bodies like clouds, rivers, poles, moon etc. From the point of view of the study of motion, we treat reference bodies capable to make observations, which is essentially a human like function. As such, it is helpful to imagine ourselves attached to the reference system, making observations. It is essentially a notional endeavor to consider that the measurements are what an observer in that frame of reference would make, had the observer with the capability to measure was actually present there.
Earth is our natural abode and we identify all non-moving ground observers equivalent and at rest with the earth. For other moving systems, we need to specify position and determine motion by virtually (in imagination) transposing ourselves to the frame of reference we are considering.
Take the case of observations about the motion of an aircraft made by two observers one at a ground and another attached to the cloud moving at certain speed. For the observer on the ground, the aircraft is moving at a speed of ,say, 1000 km/hr.
A stone propelled from a catapult with a speed of 50ms-1 attains a height of 100m. Calculate the time of flight, calculate the angle of projection, calculate the range attained
water boil at 100 and why
what is upper limit of speed
what temperature is 0 k
Riya
0k is the lower limit of the themordynamic scale which is equalt to -273 In celcius scale
Mustapha
How MKS system is the subset of SI system?
which colour has the shortest wavelength in the white light spectrum
if x=a-b, a=5.8cm b=3.22 cm find percentage error in x
x=5.8-3.22 x=2.58
what is the definition of resolution of forces
what is energy?
Ability of doing work is called energy energy neither be create nor destryoed but change in one form to an other form
Abdul
motion
Mustapha
highlights of atomic physics
Benjamin
can anyone tell who founded equations of motion !?
n=a+b/T² find the linear express
أوك
عباس
Quiklyyy
Moment of inertia of a bar in terms of perpendicular axis theorem
How should i know when to add/subtract the velocities and when to use the Pythagoras theorem?
Centre of mass of two uniform rods of same length but made of different materials and kept at L-shape meeting point is origin of coordinate | 920 | 4,260 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2018-51 | latest | en | 0.930985 |
http://docplayer.net/6022505-V-x-t-x-2-x-1-t-2-t-1-the-average-speed-of-the-particle-is-absolute-value-of-the-average-velocity-and-is-given-distance-travelled-t.html | 1,534,584,771,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213508.60/warc/CC-MAIN-20180818075745-20180818095745-00372.warc.gz | 120,344,649 | 39,854 | # v = x t = x 2 x 1 t 2 t 1 The average speed of the particle is absolute value of the average velocity and is given Distance travelled t
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## Transcription
1 Chapter 2 Motion in One Dimenion 2.1 The Important Stuff Poition, Time and Diplacement We begin our tudy of motion by conidering object which are very mall in comparion to the ize of their movement through pace. When we can deal with an object in thi way we refer to it a a particle. In thi chapter we deal with the cae where a particle move along a traight line. The particle location i pecified by it coordinate, which will be denoted by x or y. A the particle move, it coordinate change with the time, t. The change in poition from x 1 to x 2 of the particle i the diplacement x, with x = x 2 x Average Velocity and Average Speed When a particle ha a diplacement x in a change of time t, it average velocity for that time interval i by v = x t = x 2 x 1 t 2 t 1 (2.1) The average peed of the particle i abolute value of the average velocity and i given = Ditance travelled t (2.2) In general, the value of the average velocity for a moving particle depend on the initial and final time for which we have found the diplacement Intantaneou Velocity and Speed We can anwer the quetion how fat i a particle moving at a particular time t? by finding the intantaneou velocity. Thi i the limiting cae of the average velocity when the time 27
2 28 CHAPTER 2. MOTION IN ONE DIMENSION interval t include the time t and i a mall a we can imagine: x v = lim t 0 t = dx dt (2.3) The intantaneou peed i the abolute value (magnitude) of the intantaneou velocity. If we make a plot of x v. t for a moving particle the intantaneou velocity i the lope of the tangent to the curve at any point Acceleration When a particle velocity change, then we way that the particle undergoe an acceleration. If a particle velocity change from v 1 to v 2 during the time interval t 1 to t 2 then we define the average acceleration a v = x t = x 2 x 1 t 2 t 1 (2.4) A with velocity it i uually more important to think about the intantaneou acceleration, given by v a = lim t 0 t = dv (2.5) dt If the acceleration a i poitive it mean that the velocity i intantaneouly increaing; if a i negative, then v i intantaneouly decreaing. Oftentime we will encounter the word deceleration in a problem. Thi word i ued when the ene of the acceleration i oppoite that of the intantaneou velocity (the motion). Then the magnitude of acceleration i given, with it direction being undertood Contant Acceleration A very ueful pecial cae of accelerated motion i the one where the acceleration a i contant. For thi cae, one can how that the following are true: v = v 0 + at (2.6) x = x 0 + v 0 t at2 (2.7) v 2 = v a(x x 0 ) (2.8) x = x (v 0 + v)t (2.9) In thee equation, we mean that the particle ha poition x 0 and velocity v 0 at time t = 0; it ha poition x and velocity v at time t. Thee equation are valid only for the cae of contant acceleration.
3 2.2. WORKED EXAMPLES Free Fall An object toed up or down near the urface of the earth ha a contant downward acceleration of magnitude 9.80 m 2. Thi number i alway denoted by g. Be very careful about the ign; in a coordinate ytem where the y axi point traight up, the acceleration of a freely falling object i a y = 9.80 m 2 = g (2.10) Here we are auming that the air ha no effect on the motion of the falling object. For an object which fall for a long ditance thi can be a bad aumption. Remember that an object in free fall ha an acceleration equal to 9.80 m 2 while it i moving up, while it i moving down, while it i at maximum height... alway! 2.2 Worked Example Average Velocity and Average Speed 1. Boton Red Sox pitcher Roger Clemen could routinely throw a fatball at a horizontal peed of 160 km. How long did the ball take to reach home plate 18.4m away? [HRW5 2-4] hr We aume that the ball move in a horizontal traight line with an average peed of 160 km/hr. Of coure, in reality thi i not quite true for a thrown baeball. We are given the average velocity of the ball motion and alo a particular diplacement, namely x = 18.4 m. Equation 2.1 give u: v = x t = t = x v But before uing it, it might be convenient to change the unit of v. We have: v = 160 km hr ( ) ( ) 1000m 1hr = 44.4 m 1km 3600 Then we find: t = x v = 18.4m 44.4 m = The ball take econd to reach home plate. 2. Taking the Earth orbit to be a circle of radiu km, determine the peed of the Earth orbital motion in (a) meter per econd and (b) mile per econd. [Wolf 2-18]
4 30 CHAPTER 2. MOTION IN ONE DIMENSION (a) Thi i not traight line motion of coure, but we can ill find an average peed by dividing the ditance traveled (around a circular path) by the time interval. Here, the ditance traveled by the Earth a it goe once around the Sun i the circumference of the orbit, C = 2πR = 2π( km) = km = m and the time interval over which that take place i one year, o the average peed i ( ) (3600 ) 24hr 1yr = day = day 1hr = C t = m = m (b) To convert thi to mi, ue 1mi = 1.609km. Then Acceleration = ( m ) ( ) 1mi = 18.6 mi m 3. An electron moving along the x axi ha a poition given by x = (16te t )m, where t i in econd. How far i the electron from the origin when it momentarily top? [HRW6 2-20] To find the velocity of the electron a a function of time, take the firt derivative of x(t): v = dx dt = 16e t 16te t = 16e t (1 t) m again where t i in econd, o that the unit for v are m. Now the electron momentarily top when the velocity v i zero. From our expreion for v we ee that thi occur at t = 1. At thi particular time we can find the value of x: x(1) = 16(1)e 1 m = 5.89m The electron wa 5.89m from the origin when the velocity wa zero. 4. (a) If the poition of a particle i given by x = 20t 5t 3, where x i in meter and t i in econd, when if ever i the particle velocity zero? (b) When i it acceleration a zero? (c) When i a negative? Poitive? (d) Graph x(t), v(t), and a(t). [HRW5 2-28]
5 2.2. WORKED EXAMPLES 31 (a) From Eq. 2.3 we find v(t) from x(t): v(t) = dx dt = d dt (20t 5t3 ) = 20 15t 2 where, if t i in econd then v will be in m. The velocity v will be zero when which we can olve for t: 20 15t 2 = 0 15t 2 = 20 = t 2 = = (The unit 2 were inerted ince we know t 2 mut have thee unit.) Thi give: t = ±1.15 (We hould be careful... t may be meaningful for negative value!) (b) From Eq. 2.5 we find a(t) from v(t): a(t) = dv dt = d dt (20 15t2 ) = 30t where we mean that if t i given in econd, a i given in m 2. From thi, we ee that a can be zero only at t = 0. (c) From the reult i part (b) we can alo ee that a i negative whenever t i poitive. a i poitive whenever t i negative (again, auming that t < 0 ha meaning for the motion of thi particle). (d) Plot of x(t), v(t), and a(t) are given in Fig In an arcade video game a pot i programmed to move acro the creen according to x = 9.00t 0.750t 3, where x i ditance in centimeter meaured from the left edge of the creen and t i time in econd. When the pot reache a creen edge, at either x = 0 or x = 15.0cm, t i reet to 0 and the pot tart moving again according to x(t). (a) At what time after tarting i the pot intantaneouly at ret? (b) Where doe thi occur? (c) What i it acceleration when thi occur? (d) In what direction i it moving jut prior to coming to ret? (e) Jut after? (f) When doe it firt reach an edge of the creen after t = 0? [HRW5 2-31] (a) Thi i a quetion about the intantaneou velocity of the pot. To find v(t) we calculate: v(t) = dx dt = d dt (9.00t 0.750t3 ) = t 2 where thi expreion will give the value of v in cm when t i given in econd.
6 32 CHAPTER 2. MOTION IN ONE DIMENSION x, (m) t v, (m/) t a, (m/ 2 ) t Figure 2.1: Plot of x(t), v(t), and a(t) for Example 4.
7 2.2. WORKED EXAMPLES 33 We want to know the value of t for which v i zero, i.e. the pot i intantaneouly at ret. We olve: t 2 = 0 = t 2 = = (Here we have filled in the proper unit for t 2 ince by lazine they were omitted from the firt equation!) The olution to thi equation are t = ±2.00 but ince we are only intereted in time after the clock tart at t = 0, we chooe t = (b) In thi part we are to find the value of x at which the intantaneou velocity i zero. In part (a) we found that thi occurred at t = 3.00 o we calculate the value of x at t = 2.00: x(2.00) = 9.00 (2.00) (2.00) 3 = 12.0cm (where we have filled in the unit for x ince centimeter are implied by the equation). The dot i located at x = 12.0cm at thi time. (And recall that the width of the creen i 15.0cm.) (c) To find the (intantaneou) acceleration at all time, we calculate: a(t) = dv dt = d dt ( t2 ) = 4.50t where we mean that if t i given in econd, a will be given in m 2. At the time in quetion (t = 2.00 ) the acceleration i a(t = 2.00) = 4.50 (2.00) = 9.00 that i, the acceleration at thi time i 9.00 m 2. (d) From part (c) we note that at the time that the velocity wa intantaneouly zero the acceleration wa negative. Thi mean that the velocity wa decreaing at the time. If the velocity wa decreaing yet intantaneouly equal to zero then it had to be going from poitive to negative value at t = So jut before thi time it velocity wa poitive. (e) Likewie, from our anwer to part (d) jut after t = 2.00 the velocity of particle had to be negative. (f) We have een that the dot never get to the right edge of the creen at x = 15.0 cm. It will not revere it velocity again ince t = 2.00 i the only poitive time at which v = 0. So it will keep moving to back to the left, and the coordinate x will equal zero when we have: x = 0 = 9.00t 0.750t 3 Factor out t to olve: t( t 2 ) = 0 = { t = 0 or ( t 2 ) = 0 otherwie.
8 34 CHAPTER 2. MOTION IN ONE DIMENSION x, cm t, Figure 2.2: Plot of x v t for moving pot. Ignore the part where x i negative! The firt olution i the time that the dot tarted moving, o that i not the one we want. The econd cae give: which give ( t 2 ) = 0 = t 2 = = t = 3.46 ince we only want the poitive olution. So the dot return to x = 0 (the left ide of the creen) at t = If we plot the original function x(t) we get the curve given in Fig. 2.2 which how that the pot doe not get to x = 15.0 cm before it turn around. (However a explained in the problem, the curve doe not extend to negative value a the graph indicate.) Contant Acceleration 6. The head of a rattlenake can accelerate 50 m 2 in triking a victim. If a car could do a well, how long would it take to reach a peed of 100 km hr from ret? [HRW5 2-33] Firt, convert the car final peed to SI unit to make it eaier to work with: 100 km ( hr = 100 km ) ( ) ( ) 1000m 1hr = 27.8 m hr 1km 3600 The acceleration of the car i 50 m and it tart from ret which mean that v 2 0 = 0. A we ve found, the final velocity v of the car i 27.8 m. (The problem actually that thi i final
9 2.2. WORKED EXAMPLES 35 peed but if our coordinate ytem point in the ame direction a the car motion, thee are the ame thing.) Equation 2.6 let u olve for the time t: Subtituting, we find v = v 0 + at = t = v v 0 a t = 27.8 m 0 50 m 2 = 0.55 If a car had uch a large acceleration, it would take 0.55 to attain the given peed. 7. A body moving with uniform acceleration ha a velocity of 12.0 cm when it x coordinate i 3.00cm. If it x coordinate 2.00 later i 5.00cm, what i the magnitude of it acceleration? [Ser4 2-25] In thi problem we are given the initial coordinate (x = 3.00cm), the initial velocity (v 0 = 12.0 cm ), the final x coordinate (x = 5.00cm) and the elaped time (2.00). Uing Eq. 2.7 (ince we are told that the acceleration i contant) we can olve for a. We find: Subtitute thing: x = x 0 + v 0 t at2 = 1 2 at2 = x x 0 v 0 t Solve for a: 1 2 at2 = 5.00cm 3.00cm ( 12.0 cm a = 2( 32.0cm) t 2 = 2( 32.0cm) (2.00) 2 ) (2.00) = 32.0cm = 16.0 cm 2 The x acceleration of the object i 16. cm 2. (The magnitude of the acceleration i 16.0 cm 2.) 8. A jet plane land with a velocity of 100 m and can accelerate at a maximum rate of 5.0 m 2 a it come to ret. (a) From the intant it touche the runway, what i the minimum time needed before it top? (b) Can thi plane land at a mall airport where the runway i 0.80 km long? [Ser4 2-31] (a) The data given in the problem i illutrated in Fig The minu ign in the acceleration indicate that the ene of the acceleration i oppoite that of the motion, that i, the plane i decelerating. The plane will top a quickly a poible if the acceleration doe have the value 5.0 m, 2 o we ue thi value in finding the time t in which the velocity change from v 0 = 100 m to v = 0. Eq. 2.6 tell u: t = v v 0 a Subtituting, we find: t = (0 100 m) ( 5.0 m = 20 ) 2
10 36 CHAPTER 2. MOTION IN ONE DIMENSION a = -5.0 m/ m/ v = 0 x Figure 2.3: Plane touche down on runway at 100 m and come to a halt. The plane need 20 to come to a halt. (b) The plane alo travel the hortet ditance in topping if it acceleration i 5.0 m 2. With x 0 = 0, we can find the plane final x coordinate uing Eq. 2.9, uing t = 20 which we got from part (a): x = x (v 0 + v)t = (100 m + 0)(20) = 1000m = 1.0km The plane mut have at leat 1.0km of runway in order to come to a halt afely. 0.80km i not ufficient. 9. A drag racer tart her car from ret and accelerate at 10.0 m for the entire 2 ditance of 400m ( 1 mile). (a) How long did it take the car to travel thi ditance? 4 (b) What i the peed at the end of the run? [Ser4 2-33] (a) The racer move in one dimenion (along the x axi, ay) with contant acceleration a = 10.0 m 2. We can take her initial coordinate to be x 0 = 0; he tart from ret, o that v 0 = 0. Then the location of the car (x) i given by: x = x 0 + v 0 t at2 = = at2 = 1 2 (10.0 m 2 )t 2 We want to know the time at which x = 400m. Subtitute and olve for t: which give 400m = 1 2 (10.0 m 2 )t 2 = t 2 = 2(400m) (10.0 m 2 ) = The car take 8.94 to travel thi ditance. t = (b) We would like to find the velocity at the end of the run, namely at t = 8.94 (the time we found in part (a)). The velocity i: v = v 0 + at = 0 + (10.0 m 2 )t = (10.0 m 2 )t
11 2.2. WORKED EXAMPLES 37 Accelerating region 1.0 cm Path of electron Voltage ource Figure 2.4: Electron i accelerated in a region between two plate, in Example 10. At t = 8.94, the velocity i The peed at the end of the run i 89.4 m. v = (10.0 m 2 )(8.94) = 89.4 m 10. An electron with initial velocity v 0 = m enter a region 1.0cm long where it i electrically accelerated, a hown in Fig It emerge with velocity v = m. What wa it acceleration, aumed contant? (Such a proce occur in the electron gun in a cathode ray tube, ued in televiion receiver and ocillocope.) [HRW5 2-39] We are told that the acceleration of the electron i contant, o that Eq can be ued. Here we know the initial and final velocitie of the electron (v 0 and v). If we let it initial coordinate be x 0 = 0 then the final coordinate i x = 1.0cm = m. We don t know the time t for it travel through the accelerating region and of coure we don t know the (contant) acceleration, which i what we re being aked in thi problem. We ee that we can olve for a if we ue Eq. 2.8: Subtitute and get: v 2 = v a(x x 0 ) = a = v2 v 2 0 2(x x 0 ) a = ( m )2 ( m )2 2( m) = m 2 The acceleration of the electron i m 2 (while it i in the accelerating region).
12 38 CHAPTER 2. MOTION IN ONE DIMENSION 11. A world land peed record wa et by Colonel John P. Stapp when on March 19, 1954 he rode a rocket propelled led that moved down a track at 1020 km. He and the led were brought to a top in 1.4. What acceleration did h he experience? Expre your anwer in g unit. [HRW5 2-41] For the period of deceleration of the rocket led (which lat for 1.4 ) were are given the initial velocity and the final velocity, which i zero ince the led come to ret at the end. Firt, convert hi initial velocity to SI unit: v 0 = 1020 km h = (1020 km h ) ( 10 3 m 1km The Eq. 2.6 give u the acceleration a: Subtitute: ) ( ) 1h = m 3600 v = v 0 + at = a = v v 0 t a = m = m The acceleration i a negative number ince it i oppoite to the ene of the motion; it i a deceleration. The magnitude of the led acceleration i m. 2 To expre thi a a multiple of g, we note that a g = m m 2 = 20.7 o the magnitude of the acceleration wa a = 20.7g. That a lotta g! 12. A ubway train i traveling at 80 km when it approache a lower train 50m h ahead traveling in the ame direction at 25 km. If the fater train begin decelerating at 2.1 m while the lower train continue at contant peed, how oon and h 2 at what relative peed will they collide? [wolf 2-73] m Firt, convert the initial peed of the train to unit of. We find: 80 km h = 22.2 m 25 km h = 6.94 m. The ituation of the train at t = 0 (when the rear train begin to decelerate) i hown in Fig We chooe the origin of the x axi to be at the initial poition of the rear train; then the initial poition of the front train i x = 50m. If we call the x coordinate of the rear train x 1, then ince it ha initial velocity 22.2 m and acceleration 2.1 m 2 (note the minu ign!) the equation for x 1 (t) i x 1 (t) = (22.2 m )t ( 2.1 m 2 )t 2 = (22.2 m )t + ( 1.05 m 2 )t 2 Meanwhile, the front car ha an initial velocity of 6.94 m and no acceleration, o it coordinate (x 2 ) i given by x 2 (t) = 50m + (6.94 m )t
13 2.2. WORKED EXAMPLES 39 a = -2.1 m/ m/ 6.94 m/ 1 2 x 1 50 m x 2 Figure 2.5: Two ubway train in Example 12. The train will collide if there i ever a time at which their coordinate are equal. So we want to ee if there i a t which give the condition: (22.2 m )t + ( 1.05 m 2 )t 2 = 50m + (6.94 m )t Thi i a quadratic equation, for which we can ue the quadratic formula. Neglecting the unit for implicity, we can rearrange the term and rewrite it a 1.05t t + 50 = 0 and the quadratic formula give the anwer a b ± b 2 4ac 2a = ± (15.28) 2 4(1.05)(50) 2(1.05) = { Thi i a little confuing becaue there are two poible anwer! (Both value of t are poitive.) But the anwer we want i the firt one, 4.97 after the colliion, the econd time i not relevant 1. So the train will collide t = 4.97 after the rear car begin to decelerate. At the time we have found, the velocity of the rear train i v = v 0 + at = 22.2 m + ( 2.1 m )(4.97) = 11.8 m 2 and the velocity of the front train remain 6.94 m. So at the time of the colliion, the rear train i going fater by a difference of v = 11.8 m 6.94 m = 4.8 m That i the relative peed at which the colliion take place Free Fall
14 40 CHAPTER 2. MOTION IN ONE DIMENSION y y=50 m v 0 y= 0 m Figure 2.6: Object thrown upward reache height of 50 m. 13. (a) With what peed mut a ball be thrown vertically from ground level to rie to a maximum height of 50m? (b) How long will it be in the air? [HRW5 2-61] (a) Firt, we decide on a coordinate ytem. I will ue the one hown in Fig. 2.6, where the y axi point upward and the origin i at ground level. The ball tart it flight from ground level o it initial poition i y 0 = 0. When the ball i at maximum height it coordinate i y = 50 m, but we alo know it velocity at thi point. At maximum height the intantaneou velocity of the ball i zero. So if our final point i the time of maximum height, then v = 0. So for the trip from ground level to maximum height, we know y 0, y, v and the acceleration a = 9.8 m 2 = g, but we don t know v 0 or the time t to get to maximum height. From our lit of contant acceleration equation, we ee that Equation 2.8 will give u the initial velocity v 0 : Subtitute, and get: v 2 = v a(y y 0 ) = v 2 0 = v 2 2a(y y 0 ) v 2 0 = (0) 2 2( 9.8 m 2 )(50m 0) = 980 m2 2 The next tep i to take the quare root. Since we know that v 0 mut be a poitive number, we know that we hould take the poitive quare root of 980 m2 2. We get: v 0 = +31 m The initial peed of the ball i 31 m (b) We want to find the total time that the ball i in flight. What do we know about the ball when it return to earth and hit the ground? We know that it y coordinate i equal to zero. (So far, we don t know anything about the ball velocity at the the time it return to ground level.) If we conider the time between throwing and impact, then we do know y 0, y, v 0 and of coure a. If we ubtitute into Eq. 2.7 we find: 1 However it would be relevant if the train were on parallel track; then the colliion would not take place and we could find the time at which they were ide-by-ide and their relative velocitie at thoe time.
15 2.2. WORKED EXAMPLES 41 y 8.00 m/ 30.0 m Figure 2.7: Ball i thrown traight down with peed of 8.00 m, in Example = 0 + (31 m )t ( 9.8 m 2 )t 2 It i not hard to olve thi equation for t. We can factor it to give: t[(31 m ) ( 9.8 m 2 )t] = 0 which ha two olution. One of them i imply t = 0. Thi olution i an anwer to the quetion we are aking, namely When doe y = 0? becaue the ball wa at ground level at t = 0. But it i not the olution we want. For the other olution, we mut have: which give (31 m ) ( 9.8 m 2 )t = 0 t = 2(31 m ) 9.8 m 2 = 6.4 The ball pend a total of 6.4 econd in flight. 14. A ball i thrown directly downward with an initial peed of 8.00 m from a height of 30.0m. When doe the ball trike the ground? [Ser4 2-46] We diagram the problem a in Fig We have to chooe a coordinate ytem, and here I will put the let the origin of the y axi be at the place where the ball tart it motion (at the top of the 30m height). With thi choice, the ball tart it motion at y = 0 and trike the ground when y = 30m. We can now ee that the problem i aking u: At what time doe y = 30.0m? We have v 0 = 8.00 m (minu becaue the ball i thrown downward!) and the acceleration of the the ball i a = g = 9.8 m 2, o at any time t the y coordinate i given by y = y 0 + v 0 t at2 = ( 8.00 m )t 1 2 gt2
16 42 CHAPTER 2. MOTION IN ONE DIMENSION y v m Figure 2.8: Student throw her key into the air, in Example 15. But at the time of impact we have y = 30.0m = ( 8.00 m )t 1 2 gt2 = ( 8.00 m )t (4.90 m 2 )t 2, an equation for which we can olve for t. We rewrite it a: (4.90 m )t 2 + (8.00 m )t 30.0m = 0 2 which i jut a quadratic equation in t. From our algebra coure we know how to olve thi; the olution are: t = (8.00 m ) ± (8.00 m )2 4(4.90 m 2 )( 30.0m) 2(4.90 m 2 ) and a little calculator work finally give u: t = { Our anwer i one of thee... which one? Obviouly the ball had to trike the ground at ome poitive value of t, o the anwer i t = The ball trike the ground 1.78 after being thrown. 15. A tudent throw a et of key vertically upward to her orority iter in a window 4.00 m above. The key are caught 1.50 later by the iter outtretched hand. (a) With what initial velocity were the key thrown? (b) What wa the velocity of the key jut before they were caught? [Ser4 2-47] (a) We draw a imple picture of the problem; uch a imple picture i given in Fig Having a picture i important, but we hould be careful not to put too much into the picture; the problem did not ay that the key were caught while they were going up or going down. For all we know at the moment, it could be either one!
17 2.2. WORKED EXAMPLES 43 We will put the origin of the y axi at the point where the key were thrown. Thi implifie thing in that the initial y coordinate of the key i y 0 = 0. Of coure, ince thi i a problem about free fall, we know the acceleration: a = g = 9.80 m 2. What mathematical information doe the problem give u? We are told that when t = 1.50, the y coordinate of the key i y = 4.00m. I thi enough information to olve the problem? We write the equation for y(t): y = y 0 + v 0 t at2 = v 0 t 1 2 gt2 where v 0 i preently unknown. At t = 1.50, y = 4.00m, o: 4.00m = v 0 (1.50) 1 2 (9.80 m 2 )(1.50) 2. Now we can olve for v 0. Rearrange thi equation to get: So: v 0 (1.50) = 4.00m (9.80 m 2 )(1.50) 2 = 15.0m. v 0 = 15.0m 1.50 = 10.0 m (b) We want to find the velocity of the key at the time they were caught, that i, at t = We know v 0 ; the velocity of the key at all time follow from Eq. 2.6, So at t = 1.50, v = v 0 + at = 10.0 m 9.80 m 2 t v = 10.0 m 9.80 m 2 (1.50) = 4.68 m. So the velocity of the key when they were caught wa 4.68 m. Note that the key had a negative velocity; thi tell u that the key were moving downward at the time they were caught! 16. A ball i thrown vertically upward from the ground with an initial peed of 15.0 m. (a) How long doe it take the ball to reach it maximum altitude? (b) What i it maximum altitude? (c) Determine the velocity and acceleration of the ball at t = [Ser4 2-49] (a) An illutration of the data given in thi problem i given in Fig We meaure the coordinate y upward from the place where the ball i thrown o that y 0 = 0. The ball acceleration while in flight i a = g = 9.80 m 2. We are given that v 0 = m. The ball i at maximum altitude when it (intantaneou) velocity v i zero (it i neither going up nor going down) and we can ue the expreion for v to olve for t: v = v 0 + at = t = v v 0 a
18 44 CHAPTER 2. MOTION IN ONE DIMENSION a = m/ 2 v = 0 m/ y v o = m/ Figure 2.9: Ball i thrown traight up with initial peed 15.0 m. Plug in the value for the top of the ball flight and get: t = (0) (15.0 m ) ( 9.80 m 2 ) = The ball take 1.53 to reach maximum height. (b) Now that we have the value of t when the ball i at maximum height we can plug it into Eq. 2.7 and find the value of y at thi time and that will be the value of the maximum height. But we can alo ue Eq. 2.8 ince we know all the value except for y. Solving for y we find: v 2 = v ay = y = v2 v 2 0 2a Plugging in the number, we get y = (0)2 (15.0 m )2 2( 9.80 m 2 ) = 11.5m The ball reache a maximum height of 11.5m. (c) At t = 2.00 (that i, 2.0 econd after the ball wa thrown) we ue Eq. 2.6 to find: v = v 0 + at = (15.0 m ) + ( 9.80 m 2 )(2.00) = 4.60 m. o at t = 2.00 the ball i on it way back down with a peed of 4.60 m. A for the next part, the acceleration of the ball i alway equal to 9.80 m 2 in flight. while it i 17. A baeball i hit uch that it travel traight upward after being truck by the bat. A fan oberve that it require 3.00 for the ball to reach it maximum height. Find (a) it initial velocity and (b) it maximum height. Ignore the effect of air reitance. [Ser4 2-51]
19 2.2. WORKED EXAMPLES 45 t = 3.00 v=0 v 0 Figure 2.10: Ball i hit traight up; reache maximum height 3.00 later. (a) An illutration of the data given in the problem i given in Fig For the period from when the ball i hit to the time it reache maximum height, we know the time interval, the acceleration (a = g) and alo the final velocity, ince at maximum height the velocity of the ball i zero. Then Eq. 2.6 give u v 0 : and we get: v = v 0 + at = v 0 = v at v 0 = 0 ( 9.80 m 2 )(3.00) = 29.4 m The initial velocity of the ball wa m. (b) To find the value of the maximum height, we need to find the value of the y coordinate at time t = We can ue either Eq. 2.7 or Eq the latter give: v 2 = v a(y y 0 ) = (y y 0 ) = v2 v 2 0 2a Plugging in the number we find that the change in y coordinate for the trip up wa: y y 0 = 02 (29.4 m )2 2( 9.80 m 2 ) = 44.1m. The ball reached a maximum height of 44.1m. 18. A parachutit bail out and freely fall 50 m. Then the parachute open, and thereafter he decelerate at 2.0 m 2. She reache the ground with a peed of 3.0 m. (a) How long wa the parachutit in the air? (b) At what height did the fall begin? [HRW5 2-84] (a) Thi problem give everal odd bit of information about the motion of the parachutit! We organize the information by drawing a diagram, like the one given in Fig It i
20 46 CHAPTER 2. MOTION IN ONE DIMENSION v=0 (a) Free Fall 50 m (b) Deceleration v=3.0 m/ (c) Figure 2.11: Diagram howing motion of parachutit in Example 18. very important to organize our work in thi way! At the height indicated by (a) in the figure, the kydiver ha zero initial peed. A he fall from (a) to (b) her acceleration i that of gravity, namely 9.80 m 2 downward. We know that (b) i 50 m lower than (a) but we don t yet know the kydiver peed at (b). Finally, at point (c) her peed i 3.0 m and between (b) and (c) her deceleration wa 2.0 m 2, but we don t know the difference in height between (b) and (c). How can we tart to fill in the gap in our knowledge? We note that on the trip from (a) to (b) we do know the tarting velocity, the ditance travelled and the acceleration. From Eq. 2.8 we can ee that thi i enough to find the final velocity, that i, the velocity at (b). Ue a coordinate ytem (y) which ha it origin at level (b), and the y axi pointing upward. Then the initial y coordinate i y 0 = 50m and the the initial velocity i v 0 = 0. The final y coordinate i y = 0 and the acceleration i contant at a = 9.80 m 2. Then uing Eq. 2.8 we have: v 2 = v a(y y + 0) = 0 + 2( 9.80 m 2 )(0 50m) = 980 m2 2 which ha the olution v = ±31.3 m but here the kydiver i obviouly moving downward at (b), o we mut pick v = 31.3 m for the velocity at (b). While we re at it, we can find the time it took to get from (a) to (b) uing Eq. 2.6, ince we know the velocitie and the acceleration for the motion. We find: v = v 0 + at = t = v v 0 a
21 2.2. WORKED EXAMPLES 47 Subtitute: t = ( 31.3 m 0) 9.80 m 2 = 3.19 The kydiver take 3.19 to fall from (a) to (b). Now we conider the motion from (b) to (c). For thi part of the motion we know the initial and final velocitie. We alo know the acceleration, but we mut be careful about how it i expreed. During thi part of the trip, the kydiver motion i alway downward (velocity i alway negative) but her peed decreae from 31.9 m to 3.0 m. The velocity change from 31.3 m to 3.0 m o that the velocity ha increaed. The acceleration i poitive; it i in the oppoite ene a the motion and thu it wa rightly called a deceleration in the problem. So for the motion from (b) to (c), we have a = +2.0 m 2 We have the tarting and final velocitie for the trip from (b) to (c) o Eq. 2.6 let u olve for the time t: v = v 0 + at = t = v v 0 a Subtitute: t = 3.0 m ( 31.3 m ) +2.0 m 2 = 14.2 Now we are prepared to anwer part (a) of the problem. The time of the travel from (a) wa 3.19; the time of travel from (b) to (c) wa The total time in the air wa t Total = = 17.4 (b) Let keep thinking about the trip from (b) to (c); we ll keep the origin at the ame place a before (at (b)). Then for the trip from (b) to (c) the initial coordinate i y 0 = 0. The initial velocity i v 0 = 31.9 m and the final velocity i v = 3.0 m. We have the acceleration, o Eq. 2.8 give u the final coordinate y: Subtitute: v 2 = v a(y y 0) = y y 0 = v2 v 2 0 2a y y 0 = ( 3.0 m )2 ( 31.3 m )2 2(+2.0 m 2 ) = 243m Since we choe y 0 = 0, the final coordinate of the kydiver i y = 243m. We have ued the ame coordinate ytem in both part, o overall the kydiver ha gone from y = +50m to y = 243m. The change in height wa y = 243m 50m = 293m So the parachutit fall began at a height of 293m above the ground.
22 48 CHAPTER 2. MOTION IN ONE DIMENSION 19. A tone fall from ret from the top of a high cliff. a econd tone i thrown downward from the ame height 2.00 later with an initial peed of 30.0 m. If both tone hit the ground imultaneouly, how high i the cliff? [Ser4 2-54] Thi i a puzzle type problem which goe beyond the normal ubtitute and olve type; it involve more organization of our work and a clear undertanding of our equation. Here the way I would attack it. We have two different falling object here with their own coordinate; we ll put our origin at the top of the cliff and call the y coordinate of the firt tone y 1 and that of the econd tone y 2. Each ha a different dependence on the time t. For the firt rock, we have v 0 = 0 ince it fall from ret and of coure a = g o that it poition i given by y 1 = y 0 + v 0 t at2 = 1 2 gt2 Thi i imple enough but we need to remind ourelve that here t i the time ince the firt tone tarted it motion. It i not the ame a the time ince the econd tone tart it motion. To be clear, let call thi time t 1. So we have: y 1 = 1 2 gt2 1 = 4.90 m 2 t 2 1 Now, for the motion of the econd tone, if we write t 2 for the time ince it tarted it motion, the fact tated in the problem tell u that it y coordinate i given by: y 2 = y 0 + v 0 t at2 2 = ( 30.0 m )t gt2 2 So far, o good. The problem tell u that the firt tone ha been falling for 2.0 longer than the econd one. Thi mean that t 1 i 2.0 larger than t 2. So: t 1 = t = t 2 = t (We will ue t 1 a our one time variable.) Putting thi into our lat equation and doing ome algebra give y 2 = ( 30.0 m )(t 1 2.0) 1 2 (9.80 m 2 )(t 1 2.0) 2 = ( 30.0 m )(t 1 2.0) (4.90 m 2 )(t t ) 2 = ( 4.90 m 2 )t ( 30.0 m m )t 1 + (60.0m 19.6m) = ( 4.90 m 2 )t ( 10.4 m )t 1 + (40.4m) We need to remember that thi expreion for y 2 will be meaningle for value of t 1 which are le than 2.0. With thi expreion we can find value of y 1 and y 2 uing the ame time coordinate, t 1. Now, the problem tell u that at ome time (t 1 ) the coordinate of the two tone are equal. We don t yet yet know what that time or coordinate i but that i the information contained in the tatement both tone hit the ground imultaneouly. We can find thi time by etting y 1 equal to y 2 and olving: ( 4.90 m 2 )t 2 1 = ( 4.90 m 2 )t ( 10.4 m )t 1 + (40.4m)
23 2.2. WORKED EXAMPLES 49 v A =0 A v B B 30.0 m 1.50 v C C Figure 2.12: Diagram for the falling object in Example 20. Fortunately the t 2 term cancel in thi equation making it a lot eaier. We get: ( 10.4 m )t 1 + (40.4m) = 0 which ha the olution t 1 = 40.4m 10.4 m = 3.88 So the rock will have the ame location at t 1 = 3.88, that i, 3.88 after the firt rock ha been dropped. What i that location? We can find thi by uing our value of t 1 to get either y 1 or y 2 (the anwer will be the ame). Putting it into the expreion for y 1 we get: y 1 = 4.90 m 2 t 2 1 = ( 4.90 m 2 )(3.88) 2 = 74m So both tone were 74m below the initial point at the time of impact. The cliff i high. 74m 20. A falling object require 1.50 to travel the lat 30.0m before hitting the ground. From what height above the ground did it fall? [Ser4 2-68] Thi i an intriguing ort of problem... very eay to tate, but not o clear a to where to begin in etting it up! The firt thing to do i draw a diagram. We draw the important point of the object motion, a in Fig The object ha zero velocity at A; at B it i at a height of 30.0m above the ground with an unknown velocity. At C it i at ground level, the time i 1.50 later than at B and we alo don t know the velocity here. Of coure, we know the acceleration: a = 9.80 m 2!! We are given all the information about the trip from B to C, o why not try to fill in our knowledge about thi part? We know the final and initial coordinate, the acceleration and the time o we can find the initial velocity (that i, the velocity at B). Let put the origin at ground level; then, y 0 = 30.0m, y = 0 and t = 1.50, and uing y = y 0 + v 0 t at2
24 50 CHAPTER 2. MOTION IN ONE DIMENSION we find: o that v 0 t = (y y 0 ) 1 2 at2 = (0 (30.0m)) 1 2 ( 9.80 m 2 )(1.50) 2 = 19.0m v 0 = ( 19.0m) t = ( 19.0m) (1.50 ) = 12.5 m. Thi i the velocity at point B; we can alo find the velocity at C eaily, ince that i the final velocity, v: v = v 0 + at = ( 12.5 m ) + ( 9.80 m 2 )(1.50) = 27.3 m Now we can conider the trip from the tarting point, A to the point of impact, C. We don t know the initial y coordinate, but we do know the final and initial velocitie: The initial velocity i v 0 = 0 and the final velocity i v = 27.3 m, a we jut found. With the origin et at ground level, the final y coordinate i y = 0. We don t know the time for the trip, but if we ue: v 2 = v a(y y 0 ) we find: (y y 0 ) = (v2 v 2 0) 2a and we can rearrange thi to get: = ( 27.3 m )2 (0) 2 2( 9.80 m 2 ) y 0 = y m = m = 38.2m = 38.2m and the o the object tarted falling from a height of 38.2m. There are probably cleverer way to do thi problem, but here I wanted to give you the low, patient approach! 21. A tudent i taring idly out her dormitory window when he ee a water balloon fall pat. If the balloon take 0.22 to cro the 130cm high window, from what height above the top of the window wa it dropped? [Wolf 2-78] I will et up the vertical coordinate y a hown in Fig The origin i at the place where the balloon wa dropped, and we don t know how far above the window that i. Note, the y axi point downward here, o that y a a function of time i given by y = 1 2 gt2. Uing thi ytem, yet the y coordinate of the top of the window bet y 1 and the bottom of the window be y 2. Suppoe the balloon croe the top of the window at t 1 and the bottom of the window at t 2. The problem tell u that y 2 y 1 = 1.30m and t 2 t 1 = 0.22 Uing the equation of motion for the balloon, we have y 1 = 1 2 gt2 1 and y 2 = 1 2 gt2 2
25 2.2. WORKED EXAMPLES 51 0 y y 1, t 1 y 2, t 2 Figure 2.13: Diagram for the falling object in Example 21. In fact at thi point the problem i really olved becaue we have four equation for the four unknown y 1, y 2, t 1 and t 2. We jut need to do ome math! One way to olve the equation i to ubtitute for the y a: y 2 y 1 = 1 2 gt gt2 1 = 1 2 g(t2 2 t 2 1) = 1.30m But here we can factor the term t 2 2 t 2 1 to give: Thi give u t 2 + t 1 : t 2 + t 1 = 1 2 g(t2 2 t 2 1) = 1 2 g(t 2 + t 1 )(t 2 t 1 ) = 1.30m Adding thi to the equation t 2 t 1 = 0.22 give 2(1.30 m) (9.80 m )(t 2 2 t 1 ) = 2(1.30m) (9.80 m )(0.22) = t 2 = 1.43 = t 2 = 0.71 = t 1 = 0.49 And then the equation for y 1 give u y 1 = 1 2 gt2 1 = 1 2 (9.80 m 2 )(0.492) 2 = 1.19m o that the balloon began it fall 1.19m above the top of the window.
26 52 CHAPTER 2. MOTION IN ONE DIMENSION
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### DIHEDRAL GROUPS KEITH CONRAD
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### SOLUTIONS TO CONCEPTS CHAPTER 16
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### Support Vector Machine Based Electricity Price Forecasting For Electricity Markets utilising Projected Assessment of System Adequacy Data.
The Sixth International Power Engineering Conference (IPEC23, 27-29 November 23, Singapore Support Vector Machine Baed Electricity Price Forecating For Electricity Maret utiliing Projected Aement of Sytem | 19,761 | 64,805 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2018-34 | longest | en | 0.875873 |
http://www.gradesaver.com/textbooks/math/prealgebra/prealgebra-7th-edition/chapter-5-section-5-4-integrated-review-operations-on-decimals-page-374/27 | 1,524,485,900,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945942.19/warc/CC-MAIN-20180423110009-20180423130009-00352.warc.gz | 423,438,874 | 13,752 | ## Prealgebra (7th Edition)
$-268.19-146.25=-(268.19+146.25)=-414.44$ $\ \ \ \overset12\overset168.\overset119$ $\underline{+146.25}$ $\ \ \ 414.44$ | 66 | 149 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2018-17 | latest | en | 0.439984 |
https://www.appsloveworld.com/2018/10/how-much-electricity-does-refrigerator.html | 1,566,787,778,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330962.67/warc/CC-MAIN-20190826022215-20190826044215-00537.warc.gz | 711,708,590 | 39,390 | # how much electricity does a refrigerator use per month
This is a free online refrigerator power consumption calculator using this tools you can easily calculate your electricity cost.
Refrigerator(fridge) power consumption depends on lots of things, such as how much its star ratings are 5 stars, 4 stars or 2-star rating, What is the size of Fridge? 250 ltr or 165 ltr etc.
A Refrigerator will use from 100 to 450 watts depending on the size of the Refrigerator, a large Refrigerator will use about 200 watts or 1700 kWh yearly.
Select country: Australia Canada Denmark France Germany India Philippines Singapore United Kingdom United States Other country Power Use: watts (W) kilowatts (kW) Hours of use per day: h/day 1 kilowatt-hour (kWh) cost: cent pence rupee peso kr other cost per day: cost per month: cost per year: kWh/day: kWh/month: kWh/year:
## How much electricity does a fridge use?
Using the power monitor device, I have performed the same test on my refrigerator(250 ltr) at my home,and I found that it consumes 1.2 kWh per day.
Let's taken an Example
If your refrigerator operate 24 h a day.Let's assume your refrigerator used 400 watts of electricity.
find out energy consumption(Watts Per Day)
400 watts X 24 hours = 9,600 watt-hours per day
Convert watts to kilowatt hours (because electricity is measure in kWh on your electricity bill)
9,600 watt-hours per day / 1000 = 9.6 kWh per day
Let's find out the how much electricity used Over a Month Period
9.6 kWh per day X 30 days = 288 kWh per month
Now Find Out the Cost
288 kWh per month x \$0.20 per kWh = \$57.6 per month
## How many watts does a refrigerator use per day? or average refrigerator wattage?
The power consumption of refrigerator is around between 100 and 600 watts but energy usage of refrigerator depends on various factor like Size of the(i.e 250 ltr,185 ltr), technology used by the manufacturer,environmental factors, Location of the fridge(poor ventilation area take consume more energy),Usage(if the fridge door open very frequently then compressor need to do extra work so that it will consume more energy), Temperature of the fridge etc.
you can also find the wattage of your fridge in device manual given by the manufacturer.
## How many kWh does a refrigerator use?
The refrigerator is a very important household device that operates 24 hours,
and it consumes .5 kWh to 2.5 kWh.
Using Electricity Usage Monitor device we measure the power consumption of our refrigerator in our office. We notice that it consume .753 kWh in 1
day(24 hours)
## How much electricity does an old fridge use?
old fridge consumes more power as compared to the new model fridge.
Average power consumption of the refrigerators between 300 to 800 watts. Let's take an example and say you old fridge model uses 500 watts
500 watts X 24 hours = 1,2000 watt-hours per day
1,2000 watt-hours per day / 1000 = 12 kWh per day
12 kWh per day X 30 days = 360 kWh per month
360 kWh per month x 0.10 per kWh = \$ 36 per month
## How much does cost to run a fridge?
Capacity of the Fridge Avg.Wattage of the fridge consumption (kWh)/year Electricity cost/year
100-199 100 876 \$87.66
200-299 150 1314 \$131.49
300-399 200 1752 \$175.32
400-499 250 2190 \$219.15
500-599 300 2628 \$262.98
600-699 400 3504 \$350.64
*this chart is based on @.10 cent per unit cost(per kWh) and you can find exact wattage of fridge on device manual.
## Tips to reduce your refrigerator’s power consumption
According to the Energy Guide, with the temperature increase of every 5-6 degree centigrade, the energy consumption of a refrigerator increases by 40%, this increase can greatly affect the bill. Keeping the refrigerator in the right place is very important.
Contributions of Fridges is about 12-15% in the electricity bill, using them efficiently, we can reduce the bill significantly. buying a BE5-star rated refrigerator can be an efficient start, but with that, many operations procedures should also be taken care of:
1. Do not fill the refrigerator completely as it stops the flow of cold air, which reduces its efficiency.
The adequate place will increase the refrigeration cooling efficiency.
2. Make sure to cool the food at room temperature before putting in the fridge
3. Do not open the refrigerator door unnecessarily because some heat enters which reduces the capacity of the fridge. Before opening the door, decide what you want to do.
4. Set the temperature to moderate cooling for optimum cooling and low power bills.
Higher cooling mode consumes more power(up to 25%).
5. Always cover the fluid in the fridge as the moisture emitted from the fluid can affect the condenser's performance.
6. Do not allow ice to accumulate in your fridge due to ice accumulation unwanted insulation occurs which reduces the efficiency of the condenser. Regularly defrost the fridge to maintain the efficiency of the fridge.
7. Keep cleaning the fridge regularly, so that dust cannot be stored on the fridge coils and the grill that consumes the air.Due to dust accumulation, the motor may have to work more, which consumes a lot of electricity.
• Refrigerators are common household appliances in our home.which we are use to protect food items for longer periods of time and to cool down water for ice, etc.
• The refrigerator is used everywhere i.e in normal home,shop or in a large mall. Where we have to cool down any type of food or run longer, the fridge is used.
• In homes, refrigerators are used to preserve foods and to cool food and beverages, etc.It is used in hospitals to keep pharmaceuticals at a balanced temperature.
• Some medications work well at a balanced temperature, so they need a controlled temperature. It is very important to use the refrigerator to keep these drugs functioning properly.
• It is used in the laboratory for chemical reaction, there are some chemicals that depend on the temperature, that's why it can be saved by putting those chemicals at a temperature of the body.In addition, the refrigerator is used in large companies where huge amounts of food are stored or made to a product and store it.
A normal refrigerator consists of two to three different parts.
In which the top part is used for ice,The middle part is used to cool down beverages such as water milk cold drink etc.and the bottom part we use to keep vegetables. The refrigeration cycle is used to reduce the temperature in the refrigerator and to control it at a balanced temperature.
Refrigerators are a variety of types that are used in different places. We have a normal refrigerator in our house,but there are different types of refrigerator in a store or in a factory.
Please do not keep warm food - we heat all the food, and eat only a bit of it. The worst thing about it is that you take a little bit of food and keep the rest of the hot food in the fridge again. This puts pressure on compressor of the fridge and bacteria also flourish.
Please Cover the food - Whenever you open the refrigerator, every time a new smell comes out of it. This means that you do not properly store food in the fridge.food should be kept so that his smell does not come out.
Please Keep doors closed properly - many people forget to close after opening the refrigerator, which is a big mistake. When the fridge is open, it takes more time to refrigerate the compressor of the fridge. This can also spoil the food kept in the fridge. | 1,661 | 7,398 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-35 | latest | en | 0.845134 |
https://www.askmattrab.com/notes/1062-newton-raphson-method-for-finding-root | 1,638,734,259,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363216.90/warc/CC-MAIN-20211205191620-20211205221620-00079.warc.gz | 687,377,287 | 7,059 | # Newton raphson method for finding root
Newton Raphson Method
The Newton Raphson method is for solving equations of the form f(x) = 0. We make an initial guess for the root we are trying to find, and we call this initial guess x0.The sequence x0, x1, x2 ,x3, . . . generated in the manner described below should con-verge to the exact root.
To implement it analytically we need a formula for each approximation in terms of the previous one, i.e. we need xn+1 in terms of xn.
The equation of the tangent line to the graph y = f(x) at the point (x0, f(x0)) is
Its application to solving equations of the form f(x) = 0, as we now demonstrate, is called the Newton Raphson method.It is guaranteed to converge if the initial guess x0 is close enough, but it is hard to make a clear statement about what we mean by ‘close enough’ because this is highly problem specific. A sketch of the graph of f(x) can help us decide on an appropriate initial guess x0 for a particular problem.
Nirajan Sah asked 4 months ago
Find the square root of 612 with an error less than 10-4 by newton-rapson method.
Nirajan Sah answered 4 months ago
We have to find x such that
x2=612
The derivative of
f(x) = x2-612
is f’(x) = 2x
With an initial guess of x0=10, the sequence given by the method is
X1= x0 -(f(x0) /f'(x0)) = 10-(102-612)/(2*10)= 35.6
x2 = x1 - (f(x1)/f'(x1)) = 26.3955056
Similarly,
x3 = 24.7906355
x4 = 24.7386883
x5 = 24.7386338
Where the correct digits are underlined.
f(24.7386338) = (24.7386338)2-612 = 0.0000022905 <10-4
The required square root is 24.7386338. | 485 | 1,580 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2021-49 | latest | en | 0.914338 |
https://www.unitsconverters.com/en/Db-To-Cdb/Utu-4661-6352 | 1,721,290,391,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514826.9/warc/CC-MAIN-20240718064851-20240718094851-00773.warc.gz | 912,512,449 | 36,515 | Formula Used
1 Decibel = 100 Centidecibel
1 Decibel = 100 Centidecibel
## dB to cdB Conversion
The abbreviation for dB and cdB is decibel and centidecibel respectively. 1 dB is 100 times bigger than a cdB. To measure, units of measurement are needed and converting such units is an important task as well. unitsconverters.com is an online conversion tool to convert all types of measurement units including dB to cdB conversion.
## Decibel to cdB
Check our Decibel to cdB converter and click on formula to get the conversion factor. When you are converting sound from Decibel to cdB, you need a converter that is elaborate and still easy to use. All you have to do is select the unit for which you want the conversion and enter the value and finally hit Convert.
## dB to Centidecibel
The formula used to convert dB to Centidecibel is 1 Decibel = 100 Centidecibel. Measurement is one of the most fundamental concepts. Note that we have Bel as the biggest unit for length while Millidecibel is the smallest one.
## Convert dB to cdB
How to convert dB to cdB? Now you can do dB to cdB conversion with the help of this tool. In the length measurement, first choose dB from the left dropdown and cdB from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from cdB to dB? You can check our cdB to dB converter.
## dB to cdB Converter
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like sound finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like dB to cdB through multiplicative conversion factors. When you are converting sound, you need a Decibel to Centidecibel converter that is elaborate and still easy to use. Converting dB to Centidecibel is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert Decibel to cdB, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in dB to cdB conversion along with a table representing the entire conversion.
Let Others Know | 530 | 2,387 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-30 | latest | en | 0.881176 |
https://www.esaral.com/q/tick-the-correct-answer-32197 | 1,726,194,484,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651506.7/warc/CC-MAIN-20240913002450-20240913032450-00039.warc.gz | 689,493,281 | 11,613 | # Tick (✓) the correct answer:
Question:
Two pipes can fill a tank in 10 hours and 12 hours respectively, while a third pipe empties the full tank in 20 hours. If all the three pipes operate simultaneously, in how much time will the tank be full?
(a) 7 hrs 15 min
(b) 7 hrs 30 min
(c) 7 hrs 45 min
(d) 8 hrs
Solution:
(b) 7 hours 30 minutes
Part of the tank filled by the first pipe in one hour $=\frac{1}{10}$
Part of the tank filled by the second pipe in one hour $=\frac{1}{12}$
Part of the tank filled by the third pipe in one hour $=\frac{-1}{20}$
Part of the tank filled by three pipes in one hour $=\frac{1}{10}+\frac{1}{12}-\frac{1}{20}=\frac{2}{15}$
Total time taken to fill the tank $=\frac{15}{2} \mathrm{hrs}=7$ hours 30 minutes | 235 | 753 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-38 | latest | en | 0.807789 |
https://cooljargon.com/ebooks/chemistry/m51016/index.cnxml.html | 1,701,167,484,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099281.67/warc/CC-MAIN-20231128083443-20231128113443-00503.warc.gz | 215,249,148 | 16,586 | • Contents
Other Units for Solution Concentrations
Other Units for Solution Concentrations
By the end of this section, you will be able to:
• Define the concentration units of mass percentage, volume percentage, mass-volume percentage, parts-per-million (ppm), and parts-per-billion (ppb)
• Perform computations relating a solution’s concentration and its components’ volumes and/or masses using these units
In the previous section, we introduced molarity, a very useful measurement unit for evaluating the concentration of solutions. However, molarity is only one measure of concentration. In this section, we will introduce some other units of concentration that are commonly used in various applications, either for convenience or by convention.
Mass Percentage
Earlier in this chapter, we introduced percent composition as a measure of the relative amount of a given element in a compound. Percentages are also commonly used to express the composition of mixtures, including solutions. The mass percentage of a solution component is defined as the ratio of the component’s mass to the solution’s mass, expressed as a percentage:
$\text{mass percentage}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass of component}}{\text{mass of solution}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100%$
We are generally most interested in the mass percentages of solutes, but it is also possible to compute the mass percentage of solvent.
Mass percentage is also referred to by similar names such as percent mass, percent weight, weight/weight percent, and other variations on this theme. The most common symbol for mass percentage is simply the percent sign, %, although more detailed symbols are often used including %mass, %weight, and (w/w)%. Use of these more detailed symbols can prevent confusion of mass percentages with other types of percentages, such as volume percentages (to be discussed later in this section).
Mass percentages are popular concentration units for consumer products. The label of a typical liquid bleach bottle ([link]) cites the concentration of its active ingredient, sodium hypochlorite (NaOCl), as being 7.4%. A 100.0-g sample of bleach would therefore contain 7.4 g of NaOCl.
Calculation of Percent by Mass A 5.0-g sample of spinal fluid contains 3.75 mg (0.00375 g) of glucose. What is the percent by mass of glucose in spinal fluid?
Solution The spinal fluid sample contains roughly 4 mg of glucose in 5000 mg of fluid, so the mass fraction of glucose should be a bit less than one part in 1000, or about 0.1%. Substituting the given masses into the equation defining mass percentage yields:
$%\phantom{\rule{0.2em}{0ex}}\text{glucose}=\phantom{\rule{0.2em}{0ex}}\frac{3.75\phantom{\rule{0.2em}{0ex}}\text{mg glucose}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\text{g}}{1000\phantom{\rule{0.2em}{0ex}}\text{mg}}\phantom{\rule{0.2em}{0ex}}}{5.0\phantom{\rule{0.2em}{0ex}}\text{g spinal fluid}}\phantom{\rule{0.2em}{0ex}}=0.075%$
The computed mass percentage agrees with our rough estimate (it’s a bit less than 0.1%).
Note that while any mass unit may be used to compute a mass percentage (mg, g, kg, oz, and so on), the same unit must be used for both the solute and the solution so that the mass units cancel, yielding a dimensionless ratio. In this case, we converted the units of solute in the numerator from mg to g to match the units in the denominator. We could just as easily have converted the denominator from g to mg instead. As long as identical mass units are used for both solute and solution, the computed mass percentage will be correct.
Check Your Learning A bottle of a tile cleanser contains 135 g of HCl and 775 g of water. What is the percent by mass of HCl in this cleanser?
14.8%
Calculations using Mass Percentage “Concentrated” hydrochloric acid is an aqueous solution of 37.2% HCl that is commonly used as a laboratory reagent. The density of this solution is 1.19 g/mL. What mass of HCl is contained in 0.500 L of this solution?
Solution The HCl concentration is near 40%, so a 100-g portion of this solution would contain about 40 g of HCl. Since the solution density isn’t greatly different from that of water (1 g/mL), a reasonable estimate of the HCl mass in 500 g (0.5 L) of the solution is about five times greater than that in a 100 g portion, or 5 $×$ 40 = 200 g. In order to derive the mass of solute in a solution from its mass percentage, we need to know the corresponding mass of the solution. Using the solution density given, we can convert the solution’s volume to mass, and then use the given mass percentage to calculate the solute mass. This mathematical approach is outlined in this flowchart:
For proper unit cancellation, the 0.500-L volume is converted into 500 mL, and the mass percentage is expressed as a ratio, 37.2 g HCl/g solution:
$\text{500 mL solution}\phantom{\rule{0.2em}{0ex}}\left(\frac{1.19\phantom{\rule{0.2em}{0ex}}\text{g solution}}{\text{mL solution}}\right)\phantom{\rule{0.2em}{0ex}}\left(\frac{37.2\phantom{\rule{0.2em}{0ex}}\text{g HCl}}{100\phantom{\rule{0.2em}{0ex}}\text{g solution}}\right)\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}221\phantom{\rule{0.2em}{0ex}}\text{g HCl}$
This mass of HCl is consistent with our rough estimate of approximately 200 g.
Check Your Learning What volume of concentrated HCl solution contains 125 g of HCl?
282 mL
Volume Percentage
Liquid volumes over a wide range of magnitudes are conveniently measured using common and relatively inexpensive laboratory equipment. The concentration of a solution formed by dissolving a liquid solute in a liquid solvent is therefore often expressed as a volume percentage, %vol or (v/v)%:
$\text{volume percentage}=\phantom{\rule{0.2em}{0ex}}\frac{\text{volume solute}}{\text{volume solution}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100%$
Calculations using Volume Percentage Rubbing alcohol (isopropanol) is usually sold as a 70%vol aqueous solution. If the density of isopropyl alcohol is 0.785 g/mL, how many grams of isopropyl alcohol are present in a 355 mL bottle of rubbing alcohol?
Solution Per the definition of volume percentage, the isopropanol volume is 70% of the total solution volume. Multiplying the isopropanol volume by its density yields the requested mass:
$\left(355\phantom{\rule{0.2em}{0ex}}\text{mL solution}\right)\phantom{\rule{0.2em}{0ex}}\left(\frac{70\phantom{\rule{0.2em}{0ex}}\text{mL isopropyl alcohol}}{100\phantom{\rule{0.2em}{0ex}}\text{mL solution}}\right)\phantom{\rule{0.2em}{0ex}}\left(\frac{0.785\phantom{\rule{0.2em}{0ex}}\text{g isopropyl alcohol}}{1\phantom{\rule{0.2em}{0ex}}\text{mL isopropyl alcohol}}\right)\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}195\phantom{\rule{0.2em}{0ex}}\text{g isopropyl alchol}$
Check Your Learning Wine is approximately 12% ethanol (CH3CH2OH) by volume. Ethanol has a molar mass of 46.06 g/mol and a density 0.789 g/mL. How many moles of ethanol are present in a 750-mL bottle of wine?
1.5 mol ethanol
Mass-Volume Percentage
“Mixed” percentage units, derived from the mass of solute and the volume of solution, are popular for certain biochemical and medical applications. A mass-volume percent is a ratio of a solute’s mass to the solution’s volume expressed as a percentage. The specific units used for solute mass and solution volume may vary, depending on the solution. For example, physiological saline solution, used to prepare intravenous fluids, has a concentration of 0.9% mass/volume (m/v), indicating that the composition is 0.9 g of solute per 100 mL of solution. The concentration of glucose in blood (commonly referred to as “blood sugar”) is also typically expressed in terms of a mass-volume ratio. Though not expressed explicitly as a percentage, its concentration is usually given in milligrams of glucose per deciliter (100 mL) of blood ([link]).
Parts per Million and Parts per Billion
Very low solute concentrations are often expressed using appropriately small units such as parts per million (ppm) or parts per billion (ppb). Like percentage (“part per hundred”) units, ppm and ppb may be defined in terms of masses, volumes, or mixed mass-volume units. There are also ppm and ppb units defined with respect to numbers of atoms and molecules.
The mass-based definitions of ppm and ppb are given here:
$\begin{array}{l}\\ \text{ppm}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass solute}}{\text{mass solution}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{ppm}\\ \text{ppb}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass solute}}{\text{mass solution}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}\phantom{\rule{0.2em}{0ex}}\text{ppb}\end{array}$
Both ppm and ppb are convenient units for reporting the concentrations of pollutants and other trace contaminants in water. Concentrations of these contaminants are typically very low in treated and natural waters, and their levels cannot exceed relatively low concentration thresholds without causing adverse effects on health and wildlife. For example, the EPA has identified the maximum safe level of fluoride ion in tap water to be 4 ppm. Inline water filters are designed to reduce the concentration of fluoride and several other trace-level contaminants in tap water ([link]).
Calculation of Parts per Million and Parts per Billion Concentrations According to the EPA, when the concentration of lead in tap water reaches 15 ppb, certain remedial actions must be taken. What is this concentration in ppm? At this concentration, what mass of lead (μg) would be contained in a typical glass of water (300 mL)?
Solution The definitions of the ppm and ppb units may be used to convert the given concentration from ppb to ppm. Comparing these two unit definitions shows that ppm is 1000 times greater than ppb (1 ppm = 103 ppb). Thus:
$15\phantom{\rule{0.2em}{0ex}}\text{ppb}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\text{ppm}}{{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{ppb}}\phantom{\rule{0.2em}{0ex}}=0.015\phantom{\rule{0.2em}{0ex}}\text{ppm}$
The definition of the ppb unit may be used to calculate the requested mass if the mass of the solution is provided. However, only the volume of solution (300 mL) is given, so we must use the density to derive the corresponding mass. We can assume the density of tap water to be roughly the same as that of pure water (~1.00 g/mL), since the concentrations of any dissolved substances should not be very large. Rearranging the equation defining the ppb unit and substituting the given quantities yields:
$\begin{array}{c}\\ \text{ppb}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass solute}}{\text{mass solution}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}\phantom{\rule{0.2em}{0ex}}\text{ppb}\\ \\ \text{mass solute}=\phantom{\rule{0.2em}{0ex}}\frac{\text{ppb}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{mass solution}}{{10}^{9}\phantom{\rule{0.2em}{0ex}}\text{ppb}}\phantom{\rule{0.2em}{0ex}}\\ \\ \text{mass solute}=\phantom{\rule{0.2em}{0ex}}\frac{15\phantom{\rule{0.2em}{0ex}}\text{ppb}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}300\phantom{\rule{0.2em}{0ex}}\text{mL}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1.00\phantom{\rule{0.2em}{0ex}}\text{g}}{\text{mL}}\phantom{\rule{0.2em}{0ex}}}{{10}^{9}\phantom{\rule{0.2em}{0ex}}\text{ppb}}\phantom{\rule{0.2em}{0ex}}=4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{g}\end{array}$
Finally, convert this mass to the requested unit of micrograms:
$4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{g}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\text{μg}}{{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{g}}\phantom{\rule{0.2em}{0ex}}=4.5\phantom{\rule{0.2em}{0ex}}\text{μg}$
Check Your Learning A 50.0-g sample of industrial wastewater was determined to contain 0.48 mg of mercury. Express the mercury concentration of the wastewater in ppm and ppb units.
9.6 ppm, 9600 ppb
Section Summary
In addition to molarity, a number of other solution concentration units are used in various applications. Percentage concentrations based on the solution components’ masses, volumes, or both are useful for expressing relatively high concentrations, whereas lower concentrations are conveniently expressed using ppm or ppb units. These units are popular in environmental, medical, and other fields where mole-based units such as molarity are not as commonly used.
Consider this question: What mass of a concentrated solution of nitric acid (68.0% HNO3 by mass) is needed to prepare 400.0 g of a 10.0% solution of HNO3 by mass?
(a) Outline the steps necessary to answer the question.
(a) The dilution equation can be used, appropriately modified to accommodate mass-based concentration units:
$%{\text{mass}}_{1}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{\text{mass}}_{1}=%{\text{mass}}_{2}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{\text{mass}}_{2}$
This equation can be rearranged to isolate mass1 and the given quantities substituted into this equation.
(b) 58.8 g
What mass of a 4.00% NaOH solution by mass contains 15.0 g of NaOH?
What mass of solid NaOH (97.0% NaOH by mass) is required to prepare 1.00 L of a 10.0% solution of NaOH by mass? The density of the 10.0% solution is 1.109 g/mL.
114 g
What mass of HCl is contained in 45.0 mL of an aqueous HCl solution that has a density of 1.19 g cm–3 and contains 37.21% HCl by mass?
The hardness of water (hardness count) is usually expressed in parts per million (by mass) of CaCO3, which is equivalent to milligrams of CaCO3 per liter of water. What is the molar concentration of Ca2+ ions in a water sample with a hardness count of 175 mg CaCO3/L?
1.75 $×$ 10−3 M
The level of mercury in a stream was suspected to be above the minimum considered safe (1 part per billion by weight). An analysis indicated that the concentration was 0.68 parts per billion. Assume a density of 1.0 g/mL and calculate the molarity of mercury in the stream.
In Canada and the United Kingdom, devices that measure blood glucose levels provide a reading in millimoles per liter. If a measurement of 5.3 mM is observed, what is the concentration of glucose (C6H12O6) in mg/dL?
95 mg/dL
A throat spray is 1.40% by mass phenol, C6H5OH, in water. If the solution has a density of 0.9956 g/mL, calculate the molarity of the solution.
Copper(I) iodide (CuI) is often added to table salt as a dietary source of iodine. How many moles of CuI are contained in 1.00 lb (454 g) of table salt containing 0.0100% CuI by mass?
2.38 $×$ 10−4 mol
A cough syrup contains 5.0% ethyl alcohol, C2H5OH, by mass. If the density of the solution is 0.9928 g/mL, determine the molarity of the alcohol in the cough syrup.
D5W is a solution used as an intravenous fluid. It is a 5.0% by mass solution of dextrose (C6H12O6) in water. If the density of D5W is 1.029 g/mL, calculate the molarity of dextrose in the solution.
0.29 mol
Find the molarity of a 40.0% by mass aqueous solution of sulfuric acid, H2SO4, for which the density is 1.3057 g/mL.
Key Equations
• $\text{Percent by mass}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass of solute}}{\text{mass of solution}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100$
• $\text{ppm}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass solute}}{\text{mass solution}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{ppm}$
• $\text{ppb}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass solute}}{\text{mass solution}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}\phantom{\rule{0.2em}{0ex}}\text{ppb}$
Glossary
mass percentage
ratio of solute-to-solution mass expressed as a percentage
mass-volume percent
ratio of solute mass to solution volume, expressed as a percentage
parts per billion (ppb)
ratio of solute-to-solution mass multiplied by 109
parts per million (ppm)
ratio of solute-to-solution mass multiplied by 106
volume percentage
ratio of solute-to-solution volume expressed as a percentage | 4,832 | 16,318 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 16, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2023-50 | latest | en | 0.922079 |
http://www.ai.mit.edu/projects/iiip/doc/CommonLISP/HyperSpec/Issues/iss166-writeup.html | 1,542,512,210,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743963.32/warc/CC-MAIN-20181118031826-20181118053826-00554.warc.gz | 366,593,390 | 3,010 | ## Issue FORMAT-COMMA-INTERVAL Writeup
```Issue: FORMAT-COMMA-INTERVAL
References: FORMAT integer printing (p. 388-9)
Edit history: Version 1, Pavel, 10-Jun-87
Version 2, Masinter, 15-Jun-87
Problem description:
There are times when users would like to print out numbers with some punctuation
between groups of digits. The "commachar" argument to the ~D, ~B, ~O, ~X, and
~R FORMAT directives was introduced to fill that need. Unfortunately, the
interval at which the commachar should be printed is always every three digits.
This constraint is annoying when a different interval would be more appropriate.
Proposal (FORMAT-COMMA-INTERVAL:YES):
Add a fourth argument to the ~D, ~B, ~X, and ~O FORMAT directives, and a fifth
argument to the ~R directive, to be called the "comma-interval". This value
must be an integer and defaults to 3. When the : modifier is given to any of
these directives, the "commachar" is printed between groups of "comma-interval"
digits.
Test Cases:
Under the proposal, the following forms return the indicated values:
(format nil "~,,' ,4:B" 13) => "1101"
(format nil "~,,' ,4:B" 17) => "1 0001"
(format nil "~19,0,' ,4:B" 3333) => "0000 1101 0000 0101"
(format nil "~3,,,' ,2:R" 17) => "1 22"
(format nil "~,,'|,2:D" #xFFFF) => "6|55|35"
Rationale:
The current specification of FORMAT gives the user control over almost all of
the facets of printing integers. Only the wired-in constant for the
comma-interval remains, even though there are uses for varying that number. For
example, in many contexts, it would be convenient to be able to print out
numbers in binary with a space between each group of four bits. FORMAT does not
currently allow specification of the commachar printing interval so users
needing this functionality must write it themselves, duplicating much of the
logic in every implementation's integer printing code. Other uses, requiring
other intervals, can be imagined. For example, using a "commachar" of #\Newline
and a "comma-interval" of, say, 72, very large bignums could be printed in such
a way as to ensure line-breaks at appropriate places.
Current practice:
No released implementations currently support this feature.
Cost to implementors:
The change in the implementation of FORMAT is reasonably minor and, in most
cases, highly localized. Usually, the change is as simple as taking an extra
parameter and using it instead of a wired-in value of 3.
Cost to users:
Since the proposal involves the addition of an argument to certain directives,
the change would be entirely upward-compatible. No user code would need to be
converted.
Users desiring this functionality will have to write it themselves, duplicating
much of the logic involved in printing integers at all.
Benefits:
One of the few remaining inflexibilities in integer printing is eliminated with
a net increase in user-visible functionality.
Esthetics:
By parameterizing one of the final pieces of wired-in behavior from integer
printing, this small part of the workings of FORMAT would be perceived as having
been cleaned up.
Discussion:
Several members of the cleanup committee endorsed this proposal. No objections
were raised.
``` | 773 | 3,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-47 | latest | en | 0.874354 |
https://www.mersenneforum.org/showthread.php?s=05ff23596269c4c967b2a7ef77a138d0&p=544994 | 1,600,566,132,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400193087.0/warc/CC-MAIN-20200920000137-20200920030137-00153.warc.gz | 981,722,354 | 16,930 | mersenneforum.org mfaktc: a CUDA program for Mersenne prefactoring
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2020-05-09, 14:08 #3257 James Heinrich "James Heinrich" May 2004 ex-Northern Ontario 22×5×151 Posts I believe I know the answer, can somebody with knowledge of the source please confirm for me: How does mfaktc handle discovering composite factors? (I've been using M110393069 for testing which has a thousand composite factors) I believe mfaktc simply returns the composite factor as-discovered with no attempt to determine if said factor is composite or prime, correct? Assuming thus, a two-part question: a) how practical would it be for mfaktc to detect that a discovered factor is composite? b) if factor is composite, how practical would it be for mfaktc to split the factor into primes?
2020-05-09, 18:14 #3258
kriesel
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
10001001110002 Posts
Quote:
Originally Posted by James Heinrich I believe I know the answer, can somebody with knowledge of the source please confirm for me: How does mfaktc handle discovering composite factors? (I've been using M110393069 for testing which has a thousand composite factors) I believe mfaktc simply returns the composite factor as-discovered with no attempt to determine if said factor is composite or prime, correct? Assuming thus, a two-part question: a) how practical would it be for mfaktc to detect that a discovered factor is composite? b) if factor is composite, how practical would it be for mfaktc to split the factor into primes?
Wow, literally, 1013 distinct composite factors known, and 10 distinct prime factors. I thought at first the "thousand composite factors" was hyperbole.
With this in mfaktc.ini,
Code:
# possible values for StopAfterFactor:
# 0: Do not stop the current assignment after a factor was found.
# 1: When a factor was found for the current assignment stop after the
# current bitlevel. This makes only sense when Stages is enabled.
# 2: When a factor was found for the current assignment stop after the
# current class.
#
# Default: StopAfterFactor=1
StopAfterFactor=1
Factor=110393069,63,64 quickly returned the following:
Code:
[Sat May 09 10:09:03 2020]
UID: Kriesel/dodo-rtx2080super, M110393069 has a factor: 17743732831822018159 [TF:63:64:mfaktc 0.21 75bit_mul32_gs]
[Sat May 09 10:09:03 2020]
UID: Kriesel/dodo-rtx2080super, M110393069 has a factor: 13307799624252889361 [TF:63:64:mfaktc 0.21 75bit_mul32_gs]
[Sat May 09 10:09:03 2020]
UID: Kriesel/dodo-rtx2080super, found 2 factors for M110393069 from 2^63 to 2^64 [mfaktc 0.21 75bit_mul32_gs]
So yes, it returns composite factors, and the server breaks them down into prime factors along with or as part of its factor verification.
There's some question how many factors in one bit level increment mfaktx will find and report.
For this remarkable test exponent, more than two factors in a bit level is not an issue until 3 factors at 121-122 bits which would take quite a while to run as a test;
4 124-125
5 154-155
6,7 183-184
Such high bit levels are not supported in mfaktx, although they could be attempted with Factor5 or Ernst's Mfactor and lots of cores and patience. (Mfaktc max 95 bit factors, mfakto max 92 bits)
More than one factor per class may be an issue. If so, its overall impact on the GIMPS project is small. See https://www.mersenneforum.org/showpo...82&postcount=5
Although, up to 10 factors / class may be provided for. https://www.mersenneforum.org/showpo...4&postcount=35
Last fiddled with by kriesel on 2020-05-09 at 18:16
2020-05-09, 18:19 #3259
kriesel
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
113816 Posts
Quote:
Originally Posted by TheJudger Hi, good news: Yesterday I've added more than 200 known factors to the selftest. Every single factor was verified using my code. :) In some cases it misses factors when there are mutliple factors in one class close together but this is not critical. The is a known problem since the first version... This has nothing to do with the calculations itself, it is just how the results are returned from the GPU to the CPU. ----- Raw speed on my GTX 275 for M66362159 above 64 bits: ~74M candidates per second. Siever received a nice performace improvement for free by adding "-funroll-all-loops" to the gcc options. :) (only useful for CPU-limited scenarios) Oliver
(bold emphasis above is mine)
Is the issue of multiple factors in one class closely spaced causing some to be missed still present in mfaktc v0.21?
Please suggest some exponent/bitlevel combinations for test of the multiple-factors-per-class case.
Last fiddled with by kriesel on 2020-05-09 at 18:23
2020-05-09, 19:25 #3260
James Heinrich
"James Heinrich"
May 2004
ex-Northern Ontario
22×5×151 Posts
Quote:
Originally Posted by kriesel Wow, literally, 1013 distinct composite factors known, and 10 distinct prime factors. I thought at first the "thousand composite factors" was hyperbole.
No exaggeration.
There are three exponents with 11 known factors (thereby 2036 composite factors), and nine exponents with 10 known factors, but of those really only M110393069 is in the "normal" exponent range.
https://www.mersenne.ca/manyfactors.php
Quote:
Originally Posted by kriesel So yes, it returns composite factors, and the server breaks them down into prime factors along with or as part of its factor verification.
Yes, that is what currently happens, but I'm exploring the possibility of getting all the factoring software (Prime95, mfaktx, gpuowl, etc) modified to (ideally) return only prime factors, even when discovered as a composite.
Per your example, Factor=110393069,63,64 should ideally return something like:
Code:
M110393069 has a composite factor: 13307799624252889361 [TF:63:64:mfaktc 0.21 75bit_mul32_gs]
M110393069 has a factor: 1545502967 [TF:63:64:mfaktc 0.21 75bit_mul32_gs]
M110393069 has a factor: 8610659383 [TF:63:64:mfaktc 0.21 75bit_mul32_gs]
M110393069 has a composite factor: 17743732831822018159 [TF:63:64:mfaktc 0.21 75bit_mul32_gs]
M110393069 has a factor: 1545502967 [TF:63:64:mfaktc 0.21 75bit_mul32_gs]
M110393069 has a factor: 11480879177 [TF:63:64:mfaktc 0.21 75bit_mul32_gs]
found 3 prime factors for M110393069 from 2^63 to 2^64 [mfaktc 0.21 75bit_mul32_gs]
Out of necessity, the server does break down any composite factors submitted, but it would be better if this could be done at the client level (without removing the server-level checks of course).
Quote:
Originally Posted by kriesel Is the issue of multiple factors in one class closely spaced causing some to be missed still present in mfaktc v0.21? Please suggest some exponent/bitlevel combinations for test of the multiple-factors-per-class case.
There is no shortage of examples of exponents with multiple known factors that share a 4620-class, to quote just the ones from 100M:
Code:
100028477
100062071
100071661
100104131
100118399
100250341
100269131
100296149
100297643
100331219
100444901
100475647
100499617
100621361
100683689
100709621
If you need a more extensive test-case list email me and I can give you a large list.
Exponents with three factors sharing a class are rare, these are the only four I found and they're all quite small:
Code:
150551
329009
4965187
12671587
Of course, if this bug causes factors to be missed, then they wouldn't be in my list.
2020-05-09, 20:40 #3261
kriesel
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
23·19·29 Posts
Quote:
Originally Posted by James Heinrich There is no shortage of examples of exponents with multiple known factors that share a 4620-class, to quote just the ones from 100M: Code: 100028477 ... 100709621 If you need a more extensive test-case list email me and I can give you a large list. Exponents with three factors sharing a class are rare, these are the only four I found and they're all quite small: Code: 150551 329009 4965187 12671587 Of course, if this bug causes factors to be missed, then they wouldn't be in my list.
Thanks for the lists, however, what I was trying to express was an interest in any known cases of mersenne numbers with prime exponents within mersenne.ca exponent range, preferably within mersenne.org range (p<109) with multiple factors known for {same-exponent & same-bit-level & same-class-number} trifecta / slot-machine-jackpot, of bit size no larger than GIMPS would TF to normally on a gpu, which would be 76 to 92 bits or so depending on exponent. (Think intersection of 5 or 6 sets.)
That's what it would take for mfaktc or mfakto to encounter the case where normal use leads to two factors in the same class of the same bit level of the same exponent in the same run, and theoretically finding and reporting two or more factors from that single exponent, bit level, class pass.
For an exponent p, p prime, a bit level x-1 to x, x <= normal GIMPS factoring depth for p, a class any ONE of 420 or 4620, multiple factors found. Possibly even multiple factors within one block given to one of the many gpu multiprocessors.
Instead of at the mfaktc console output,
Code:
May 09 10:09 | 2271 49.5% | 0.001 n.a. | 1600.63 82485 n.a.%
M110393069 has a factor: 17743732831822018159
May 09 10:09 | 2272 49.6% | 0.001 n.a. | 1600.63 82485 n.a.%
...
May 09 10:09 | 2860 62.4% | 0.001 n.a. | 1600.63 82485 n.a.%
M110393069 has a factor: 13307799624252889361
May 09 10:09 | 2872 62.5% | 0.001 n.a. | 1600.63 82485 n.a.%
it might look something like
Code:
May 09 10:09 | 2271 49.5% | 0.001 n.a. | 1600.63 82485 n.a.%
M110393069 has a factor: 17743732831822018159
M110393069 has a factor: 17743733851853975719
May 09 10:09 | 2272 49.6% | 0.001 n.a. | 1600.63 82485 n.a.%
(That second, underlined one is completely fictitious)
I roughly estimated (in the attachment at https://www.mersenneforum.org/showpo...82&postcount=5) that we should find hundreds or thousands of such over the full scope of mersenne.org TF. If we can't identify any coincident class, bit level, exponent multiple factor finds, given how much TF has already been done, we're probably missing some (relatively few) factors this way.
Last fiddled with by kriesel on 2020-05-09 at 20:47
2020-05-09, 21:17 #3262
James Heinrich
"James Heinrich"
May 2004
ex-Northern Ontario
22·5·151 Posts
Quote:
Originally Posted by kriesel Thanks for the lists, however, what I was trying to express was an interest in any known cases of mersenne numbers with prime exponents within mersenne.ca exponent range, preferably within mersenne.org range (p<109) with multiple factors known for {same-exponent & same-bit-level & same-class-number} trifecta
Sorry, I forgot to filter it to same-bitlevel. That makes the list much shorter (and longer to search) but I still found 4 examples in the ~100M range:
Code:
106266857
109013999
109088341
109405193
I only searched in 100M<110M, if you want I can expand the search to find all the examples in GIMPS range, but that'll about an hour of hammering the database.
edit: it actually didn't take as long as I feared, here's the complete list of 215 GIMPS-range exponents with same-bits-and-class factors:
Code:
Factor=630907,49,50
Factor=666493,58,59
Factor=754463,49,50
Factor=976301,49,50
Factor=3902347,37,38
Factor=5279899,42,43
Factor=12953419,49,50
Factor=13904749,45,46
Factor=14298527,54,55
Factor=14749391,68,69
Factor=15162197,41,42
Factor=16892969,62,63
Factor=21474083,40,41
Factor=21633289,51,52
Factor=23740537,55,56
Factor=26936579,39,40
Factor=28247707,56,57
Factor=29235029,60,61
Factor=33838793,61,62
Factor=40917917,47,48
Factor=42416729,59,60
Factor=45632189,45,46
Factor=47916989,60,61
Factor=49678099,65,66
Factor=55346957,62,63
Factor=68168081,62,63
Factor=68989909,52,53
Factor=74542373,65,66
Factor=80374999,54,55
Factor=88079011,45,46
Factor=95469713,66,67
Factor=95669209,63,64
Factor=97267133,46,47
Factor=106266857,63,64
Factor=109013999,53,54
Factor=109088341,53,54
Factor=109405193,54,55
Factor=119525743,49,50
Factor=121156699,51,52
Factor=122522789,57,58
Factor=122617237,53,54
Factor=148380737,50,51
Factor=153225187,64,65
Factor=155676707,54,55
Factor=157910509,60,61
Factor=158599201,57,58
Factor=160739149,64,65
Factor=168830449,50,51
Factor=169014941,59,60
Factor=170473691,59,60
Factor=171953321,64,65
Factor=186375971,45,46
Factor=188904787,54,55
Factor=191462893,61,62
Factor=194559377,59,60
Factor=200995609,48,49
Factor=208394701,58,59
Factor=211943939,65,66
Factor=214423889,62,63
Factor=229769411,48,49
Factor=230055593,63,64
Factor=232507777,62,63
Factor=234028031,43,44
Factor=238430009,65,66
Factor=247462603,50,51
Factor=247575857,61,62
Factor=248693219,61,62
Factor=249720929,58,59
Factor=257601991,53,54
Factor=269222717,59,60
Factor=291566903,51,52
Factor=296611769,63,64
Factor=304843843,50,51
Factor=307436693,63,64
Factor=311095511,64,65
Factor=329514313,51,52
Factor=333341903,57,58
Factor=339049663,45,46
Factor=344226053,50,51
Factor=349871771,44,45
Factor=350800537,49,50
Factor=351511079,59,60
Factor=359812081,48,49
Factor=363424847,47,48
Factor=371542393,49,50
Factor=376397243,45,46
Factor=384190957,46,47
Factor=384327743,56,57
Factor=384394709,55,56
Factor=385356109,53,54
Factor=387013937,60,61
Factor=400835159,51,52
Factor=401102021,52,53
Factor=403918327,54,55
Factor=404784839,61,62
Factor=406884479,53,54
Factor=408071501,44,45
Factor=408188827,69,70
Factor=412778909,60,61
Factor=413747153,54,55
Factor=414824231,60,61
Factor=419128607,57,58
Factor=448196831,55,56
Factor=452020073,43,44
Factor=454609277,63,64
Factor=457729901,62,63
Factor=459228547,54,55
Factor=460167947,47,48
Factor=466220519,57,58
Factor=482584937,63,64
Factor=483370357,59,60
Factor=492464549,64,65
Factor=496423661,55,56
Factor=506824429,48,49
Factor=506895563,62,63
Factor=512880491,57,58
Factor=521816593,53,54
Factor=524087329,46,47
Factor=526005499,49,50
Factor=526014143,58,59
Factor=530664289,46,47
Factor=555551489,67,68
Factor=556353643,48,49
Factor=562335937,54,55
Factor=563680291,55,56
Factor=573868573,47,48
Factor=575599867,46,47
Factor=576806761,57,58
Factor=589429417,55,56
Factor=589746037,46,47
Factor=600953513,59,60
Factor=603618313,60,61
Factor=609388763,48,49
Factor=610146851,63,64
Factor=611741059,63,64
Factor=617863849,48,49
Factor=625400089,54,55
Factor=625521709,61,62
Factor=636713333,56,57
Factor=638366929,60,61
Factor=640381303,50,51
Factor=648452527,62,63
Factor=650749717,53,54
Factor=650927363,54,55
Factor=658195159,50,51
Factor=660280373,60,61
Factor=663891329,53,54
Factor=664600039,61,62
Factor=669290053,57,58
Factor=670620989,49,50
Factor=673902499,53,54
Factor=676343021,64,65
Factor=676611721,43,44
Factor=677065223,54,55
Factor=682014161,64,65
Factor=684732073,59,60
Factor=686029517,63,64
Factor=689997797,44,45
Factor=694678529,45,46
Factor=699891781,54,55
Factor=705410129,61,62
Factor=712472153,47,48
Factor=716225651,48,49
Factor=718763291,43,44
Factor=722281877,48,49
Factor=722671237,63,64
Factor=729992177,57,58
Factor=734571463,52,53
Factor=736663073,57,58
Factor=740774491,47,48
Factor=744550061,60,61
Factor=763981891,64,65
Factor=769814911,49,50
Factor=772482437,52,53
Factor=773120087,48,49
Factor=773476307,46,47
Factor=775692791,44,45
Factor=786451073,61,62
Factor=793651759,54,55
Factor=799473617,45,46
Factor=821119861,52,53
Factor=822585289,48,49
Factor=824575069,46,47
Factor=827289787,49,50
Factor=827753587,69,70
Factor=829322971,53,54
Factor=829364051,62,63
Factor=838613569,48,49
Factor=843158347,48,49
Factor=845831837,57,58
Factor=853745551,47,48
Factor=854811127,58,59
Factor=865444469,50,51
Factor=871148027,53,54
Factor=873514913,58,59
Factor=888502117,56,57
Factor=891816469,51,52
Factor=897478951,45,46
Factor=900842653,63,64
Factor=915151613,65,66
Factor=918760987,57,58
Factor=920280619,49,50
Factor=922496633,58,59
Factor=927602177,54,55
Factor=940010633,50,51
Factor=941381677,52,53
Factor=947056163,54,55
Factor=948726797,63,64
Factor=956418803,47,48
Factor=965694907,62,63
Factor=968105849,44,45
Factor=981595297,50,51
Factor=985010921,51,52
Factor=986248531,45,46
Factor=996709397,46,47
Last fiddled with by James Heinrich on 2020-05-09 at 22:01
2020-05-09, 22:14 #3263
kriesel
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
23·19·29 Posts
Quote:
Originally Posted by James Heinrich Sorry, I forgot to filter it to same-bitlevel. That makes the list much shorter (and longer to search) but I still found 4 examples in the ~100M range: Code: 106266857 109013999 109088341 109405193 I only searched in 100M<110M, if you want I can expand the search to find all the examples in GIMPS range, but that'll about an hour of hammering the database. edit: it actually didn't take as long as I feared, here's the complete list of 215 GIMPS-range exponents with same-bits-and-class factors: Code: Factor=630907,49,50 ... Factor=996709397,46,47
About 10 of those coincident factor cases are above 65 bits, and my rough estimate was 124 such for more-classes and completed to GIMPS bit levels TF; 1243 for less-classes.
Thanks, while you were getting the big list, I ran the first test case. Console output
Code:
May 09 17:02 | 3607 78.1% | 0.216 n.a. | 7.70 82485 n.a.%
M106266857 has a factor: 15207927026741198039
M106266857 has a factor: 16752154502925095159
May 09 17:02 | 3615 78.2% | 0.217 n.a. | 7.66 82485 n.a.%
Results file
Code:
[Sat May 09 17:02:15 2020]
UID: Kriesel/dodo-rtx2080super, M106266857 has a factor: 15207927026741198039 [TF:63:64:mfaktc 0.21 75bit_mul32_gs]
UID: Kriesel/dodo-rtx2080super, M106266857 has a factor: 16752154502925095159 [TF:63:64:mfaktc 0.21 75bit_mul32_gs]
[Sat May 09 17:02:57 2020]
UID: Kriesel/dodo-rtx2080super, found 2 factors for M106266857 from 2^63 to 2^64 [mfaktc 0.21 75bit_mul32_gs]
Last fiddled with by kriesel on 2020-05-09 at 22:43
2020-05-09, 22:17 #3264
SethTro
"Seth"
Apr 2019
101000012 Posts
Quote:
Originally Posted by James Heinrich No exaggeration. There are three exponents with 11 known factors (thereby 2036 composite factors), and nine exponents with 10 known factors, but of those really only M110393069 is in the "normal" exponent range. https://www.mersenne.ca/manyfactors.php Yes, that is what currently happens, but I'm exploring the possibility of getting all the factoring software (Prime95, mfaktx, gpuowl, etc) modified to (ideally) return only prime factors, even when discovered as a composite. Per your example, Factor=110393069,63,64 should ideally return something like: Code: M110393069 has a composite factor: 13307799624252889361 [TF:63:64:mfaktc 0.21 75bit_mul32_gs] M110393069 has a factor: 1545502967 [TF:63:64:mfaktc 0.21 75bit_mul32_gs] M110393069 has a factor: 8610659383 [TF:63:64:mfaktc 0.21 75bit_mul32_gs] M110393069 has a composite factor: 17743732831822018159 [TF:63:64:mfaktc 0.21 75bit_mul32_gs] M110393069 has a factor: 1545502967 [TF:63:64:mfaktc 0.21 75bit_mul32_gs] M110393069 has a factor: 11480879177 [TF:63:64:mfaktc 0.21 75bit_mul32_gs] found 3 prime factors for M110393069 from 2^63 to 2^64 [mfaktc 0.21 75bit_mul32_gs] Out of necessity, the server does break down any composite factors submitted, but it would be better if this could be done at the client level (without removing the server-level checks of course).
I have some knowledge here (I found one of the factors of 110393069).
It's really easy to detect if a factor is composite because all factors are of the form (2*k*p+1) so if the factor is composite it must be of the form (2*k_1*p+1)*(2*k_2*p+1) which restricts k_1 and k_2 to be very small so it's easy to just check if the returned factor is divisible by (2*i*p+1) for i <= 1000. I asked about adding this to mfaktc but mfaktc doesn't have good client side big int support so I never coded it up.
Last fiddled with by SethTro on 2020-05-09 at 22:18
2020-05-15, 20:14 #3265 ixfd64 Bemusing Prompter "Danny" Dec 2002 California 2×1,151 Posts Silly question: what's the difference between a block and a grid? I've seen these terms used interchangeably, but my understanding is that they are different things.
2020-05-18, 19:17 #3266
TheJudger
"Oliver"
Mar 2005
Germany
33×41 Posts
Quote:
Originally Posted by kriesel (bold emphasis above is mine) Is the issue of multiple factors in one class closely spaced causing some to be missed still present in mfaktc v0.21?
Yes, using atomics on CC 2.0 devices or newer. Current artificial limit is 10 factors within class in a single "stage".
Oliver
2020-05-18, 19:34 #3267
kriesel
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
23×19×29 Posts
Quote:
Originally Posted by TheJudger Yes, using atomics on CC 2.0 devices or newer. Current artificial limit is 10 factors within class in a single "stage". Oliver
Thanks. I think the probability of more than 10 factors for a single exponent/bitlevel/class is very low. I compute the odds of 6 factors per bit level and exponent to be rather small. https://www.mersenneforum.org/showth...982#post520982
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Sun Sep 20 01:42:11 UTC 2020 up 9 days, 22:53, 0 users, load averages: 0.93, 1.10, 1.17 | 7,076 | 21,345 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-40 | latest | en | 0.900459 |
http://users.math.uoc.gr/~pamfilos/eGallery/problems/SimsonDiametral.html | 1,718,952,165,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862040.7/warc/CC-MAIN-20240621062300-20240621092300-00388.warc.gz | 35,045,108 | 1,805 | Simson lines of diametral points
Consider triangle t=ABC and a point D on its circumcircle (c) and its antipodal point E on (c). Project D, E on the sides of (t) and define the Simson lines S(D), S(E) (see Simson.html ). Then the angle of the two Simson lines S(D) and S(E) is a right one and their intersection point is on the Euler circle of t.
That the angle of S(D), S(E) is a right one follows from the discussion in SimsonProperty2.html .
The proof of the other statement follows from the fact that the orthocenter H of t is the similarity center of the Euler circle and the circumcircle of t. To complete the argument draw parallels to S(D), S(E) intersecting at point G on (c). From SteinerLine.html follows that the intersection point F of S(D) and S(E) and the intersections J, I of these lines with HD, HE make a triangle s = IJF, similar to DEF etc. Notice, by the way that diameter IJ of the Euler circle is parallel to the diameter DE of the circumcircle c.
Remark Projecting the points {E,O,D} of the diameter on one side, AB say, we see that the projections of {D,E} are symmetric with respect to the middle of the corresponding side. This could be used to give another proof of the property discussed above.
Simson.html
SimsonGeneral.html
SimsonGeneral2.html
SimsonProperty.html
SimsonProperty2.html
Simson_3Lines.html
SimsonVariantLocus.html
SteinerLine.html
ThreeDiameters.html
ThreeSimson.html
TrianglesCircumscribingParabolas.html
WallaceSimson.html | 376 | 1,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-26 | latest | en | 0.821723 |
https://notesformsc.org/c-compute-cosine-series/?amp=1 | 1,721,097,859,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514726.17/warc/CC-MAIN-20240716015512-20240716045512-00469.warc.gz | 385,660,839 | 22,314 | # C Program To Compute Cosine Series
The program to compute cosine series is based on the mathematical concept of sequences and series – particularly power series.
The program is compiled using Dev-C++ compiler, but you can use any standard C compiler to compile and run the program. Make sure that you change the source code appropriately according to your compiler specifications.
To understand this example, you must be familiar with following C programming concepts.
## Problem Definition
Cosine series is a power series known as the Maclaurin expansion of where x is a real number value. Given the value of x, the series will converge and give a finite value (radius of convergence) for the series.
We won’t be discussing, how terms of the series are generated because that would be topic for another post. The Maclaurin expansion for is for , however, it is true for all real values of .
In other words, it is same as finding the value of trigonometric ratio of , the cosine series will give you a better approximate value of .
## Program Code
/* Program to compute cosine series */
/* cos(x) = 1-x^2/2!+x^4/4!+x^6/6!+...+x^n/n! */
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
main()
{
float x, t, sum;
int d;
int i, n=20;
/* Read the Input x value in degrees */
printf ("Input X Value (in degrees) :");
scanf ("%f", &x);
d= x;
x=x*3.1412/180;
t=1;
sum=1;
for (i=1; i<n+1; i++)
{
t=t*pow ((double) (-1), (double) (2*i-1))*x*x/ (2*i*(2*i-1));
sum=sum+t;
}
/* Print the Results */
for (i =0; i<35; i++)
printf ("_"); printf ("\n\n");
printf ("COS (%d) =%7.3f\n\n", (int) d, sum);
for (i =0; i<35; i++)
printf ("_"); printf ("\n\n");
system ("PAUSE");
return 0;
}
## Output
Input X Values(in degrees):34
-------------------------------------
COS(34)= 0.829
------------------------------------- | 496 | 1,831 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-30 | latest | en | 0.77039 |
http://forums.wolfram.com/mathgroup/archive/2007/May/msg01178.html | 1,590,431,084,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347389309.17/warc/CC-MAIN-20200525161346-20200525191346-00437.warc.gz | 50,921,374 | 8,226 | Re: I Need to Have These Equations Solved!
• To: mathgroup at smc.vnet.net
• Subject: [mg76584] Re: I Need to Have These Equations Solved!
• From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
• Date: Thu, 24 May 2007 06:08:24 -0400 (EDT)
• Organization: The Open University, Milton Keynes, UK
• References: <f30vo2\$mk0\$1@smc.vnet.net>
```pskmaths at googlemail.com wrote:
> Hi everybody,
>
> I have some long expresion equations and I have been struggling to get
> them solved! I need to solve these equations with respect to r, s, =CE=B8
> and =CF=95.
>
> The solutions are subject to the follwoing:
>
> 1=2E r>0
> 2=2E s>=0
> 3=2E 0<==CE=B8<=pi/2
> 4 0<==CF=95<pi
>
> The equations are:
> ---------------------------------------------------------------------------=
> ---------------------------------------------------------------------------=
> ------------
> 1=2E (1/4)*(-8*a*=CE=B1*=CE=B2*Sin[=CE=B8] - (4*l*m1*=CE=BC*(r*Cos[=CE=
> =B8]*Cos[=CF=95] -
> s*Sin[=CE=B8]))/(l^2 + r^2 + s^2 - 2*l*s*Cos[=CE=B8] -
> 2*l*r*Cos[=CF=95]*Sin[=CE=B8])^(3/2) +
> (4*l*m1*=CE=BC*(r*Cos[=CE=B8]*Cos[=CF=95] - s*Sin[=CE=B8]))/(l^2 + r^2 =
> + s^2 +
> 2*l*s*Cos[=CE=B8] + 2*l*r*Cos[=CF=95]*Sin[=CE=B8])^(3/2) + 2*(a - b)*=CE=B1=
> ^2*Sin[2*=CE=B8]) =
> 0
> ---------------------------------------------------------------------------=
> ---------------------------------------------------------------------------=
> ------------
> 2=2E (1/4)*((4*l*m1*r*=CE=BC*Sin[=CE=B8]*Sin[=CF=95])/(l^2 + r^2 + s^2 -
> 2*l*s*Cos[=CE=B8] - 2*l*r*Cos[=CF=95]*Sin[=CE=B8])^(3/2) -
> (4*l*m1*r*=CE=BC*Sin[=CE=B8]*Sin[=CF=95])/(l^2 + r^2 + s^2 + 2*l*s*Cos[=
> =CE=B8] +
> 2*l*r*Cos[=CF=95]*Sin[=CE=B8])^(3/2)) = 0
> ---------------------------------------------------------------------------=
> ---------------------------------------------------------------------------=
> ------------
> 3=2E (-M)*r*=CE=B1^2 + (m2*r*=CE=BC)/(r^2 + s^2)^(3/2) + (m1*=CE=BC*(r -
> l*Cos[=CF=95]*Sin[=CE=B8]))/(l^2 + r^2 + s^2 - 2*l*s*Cos[=CE=B8] -
> 2*l*r*Cos[=CF=95]*Sin[=CE=B8])^(3/2) +
> (m1*=CE=BC*(r + l*Cos[=CF=95]*Sin[=CE=B8]))/(l^2 + r^2 + s^2 + 2*l*s*Cos=
> [=CE=B8] +
> 2*l*r*Cos[=CF=95]*Sin[=CE=B8])^(3/2) = 0
> ---------------------------------------------------------------------------=
> ---------------------------------------------------------------------------=
> ------------
> 4=2E =CE=BC*((m2*s)/(r^2 + s^2)^(3/2) + (m1*(s - l*Cos[=CE=B8]))/(l^2 + =
> r^2 +
> s^2 - 2*l*s*Cos[=CE=B8] - 2*l*r*Cos[=CF=95]*Sin[=CE=B8])^(3/2) +
> (m1*(s + l*Cos[=CE=B8]))/(l^2 + r^2 + s^2 + 2*l*s*Cos[=CE=B8] +
> 2*l*r*Cos[=CF=95]*Sin[=CE=B8])^(3/2)) = 0
> ---------------------------------------------------------------------------=
> ---------------------------------------------------------------------------=
> ------------
>
>
> And also, I have another group of equations:
>
>
> 1=2E =CE=B1*=CF=892*(-2*a*=CE=B2 + (a - b)*=CE=B1*Cos[=CE=B8])*Cos[=CF=95=
> ]^2*Sin[=CE=B8] +
> l*m1*r*=CE=BC*=CF=892*Cos[=CE=B8]*Cos[=CF=95]*
> (-(1/(l^2 + r^2 + s^2 - 2*l*s*Cos[=CE=B8] - 2*l*r*Cos[=CF=95]*Sin[=CE=
> =B8])^(3/2))
> + 1/(l^2 + r^2 + s^2 + 2*l*s*Cos[=CE=B8] + 2*l*r*Cos[=CF=95]*Sin[=CE=B8])^(=
> 3/2)) +
> l*m1*r*=CE=BC*=CF=891*(-(1/(l^2 + r^2 + s^2 - 2*l*s*Cos[=CE=B8] -
> 2*l*r*Cos[=CF=95]*Sin[=CE=B8])^(3/2)) + 1/(l^2 + r^2 + s^2 + 2*l*s*Cos[=CE=
> =B8] +
> 2*l*r*Cos[=CF=95]*Sin[=CE=B8])^(3/2))*
> Sin[=CF=95] + =CF=892*Sin[=CE=B8]*(l*m1*s*=CE=BC*(1/(l^2 + r^2 + s^2 - =
> 2*l*s*Cos[=CE=B8] -
> 2*l*r*Cos[=CF=95]*Sin[=CE=B8])^(3/2) -
> 1/(l^2 + r^2 + s^2 + 2*l*s*Cos[=CE=B8] + 2*l*r*Cos[=CF=95]*Sin[=CE=
> =B8])^(3/2))
> + =CE=B1*(-2*a*=CE=B2 + (a - b)*=CE=B1*Cos[=CE=B8])*Sin[=CF=95]^2) = 0
> ---------------------------------------------------------------------------=
> ---------------------------------------------------------------------------=
> -----------
> 2=2E (-M)*r*=CE=B1^2 + m1*=CE=BC*(-((l*Cos[=CF=95]*Sin[=CE=B8])/(l^2 + r=
> ^2 + s^2 -
> 2*l*(s*Cos[=CE=B8] + r*Cos[=CF=95]*Sin[=CE=B8]))^(3/2)) +
> (l*Cos[=CF=95]*Sin[=CE=B8])/(l^2 + r^2 + s^2 + 2*l*(s*Cos[=CE=B8] +
> r*Cos[=CF=95]*Sin[=CE=B8]))^(3/2)) = 0
> ---------------------------------------------------------------------------=
> ---------------------------------------------------------------------------=
> -----------
> 3=2E m1*=CE=BC*(-((l*Cos[=CE=B8])/(l^2 + r^2 + s^2 - 2*l*(s*Cos[=CE=B8] +
> r*Cos[=CF=95]*Sin[=CE=B8]))^(3/2)) +
> (l*Cos[=CE=B8])/(l^2 + r^2 + s^2 + 2*l*(s*Cos[=CE=B8] +
> r*Cos[=CF=95]*Sin[=CE=B8]))^(3/2)) = 0
> ---------------------------------------------------------------------------=
> ---------------------------------------------------------------------------=
> -----------
>
>
>
> Thanks a lot.
>
>
You post again your request to MathGroup, but this time using
Mathematica InputForm for your expressions. Also, you should indicate
what built-in functions and/or approach you have tried and their results.
How to use the *InputForm* when copying an expression from a
Mathematica notebook to an email client. Select the cell(s)
you want to include in your email, then within the *Cell* menu select
either *Convert To* or *Display As* and choose *InputForm*. Finally,
copy and paste as usual.
Regards,
Jean-Marc
```
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# Solve each system by graphing
Solve each system by graphing
2x - 3y = 6
y = 2/3x - 2
#### Solution Summary
This solution is comprised of a detailed explanation to solve each system by graphing.
\$2.19 | 68 | 233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2017-04 | latest | en | 0.916672 |
https://en.academic.ru/dic.nsf/enwiki/627957 | 1,600,505,308,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400191160.14/warc/CC-MAIN-20200919075646-20200919105646-00030.warc.gz | 385,165,488 | 12,834 | # Arithmetic underflow
Arithmetic underflow
Arithmetic underflow (or "floating point underflow", "floating underflow", "underflow") is a condition that can occur when the result of a floating point operation would be smaller in magnitude (closer to zero, either positive or negative) than the smallest quantity representable. Underflow is actually (negative) overflow of the exponent of the floating point quantity. For example, an eight-bit two's complement exponent can represent multipliers of $2^\left\{-128\right\}$ to $2^\left\{127\right\}$. A result less than $2^\left\{-128\right\}$ would cause underflow.
Depending on the processor, the programming language and the run-time system, underflow may set a status bit, raise an exception, generate a hardware interrupt, or may cause some combination of these effects. Alternatively, the underflow may just be ignored and zero substituted for the unrepresentable value, although this might lead to a later division by zero error which cannot be so easily ignored.
As specified in IEEE 754 the underflow condition is only signaled if there is also a loss of accuracy. Typically this is determined as the final result being inexact. However if the user is trapping on underflow, this may happen regardless of consideration for loss of precision.
ee also
*Integer overflow
References
*
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### Look at other dictionaries:
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• NaN — For other uses, see Nan. In computing, NaN (Not a Number) is a value of the numeric data type representing an undefined or unrepresentable value, especially in floating point calculations. Systematic use of NaNs was introduced by the IEEE 754… … Wikipedia
• Soupassement arithmétique — Le terme soupassement arithmétique, soupassement en virgule flottante, soupassement de capacité ou tout simplement soupassement est un néologisme qui s inspire du mot dépassement. Le soupassement est parfois connu sous le nom anglais arithmetic… … Wikipédia en Français
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• Analyse lexicale — L analyse lexicale se trouve tout au début de la chaîne de compilation. C est la tâche consistant à décomposer une chaîne de caractères en unités lexicales, aussi appelées tokens. Ces tokens, produits à la demande de l analyseur syntaxique, sont… … Wikipédia en Français
• Analyse statique de programmes — En informatique, la notion d analyse statique de programmes couvre une variété de méthodes utilisées pour obtenir des informations sur le comportement d un programme lors de son exécution sans réellement l exécuter. C est cette dernière… … Wikipédia en Français
• Analyse syntaxique — Pour les articles homonymes, voir Analyseur. L analyse syntaxique consiste à mettre en évidence la structure d un texte, généralement un programme informatique ou du texte écrit dans une langue naturelle. Un analyseur syntaxique (parser, en… … Wikipédia en Français | 949 | 4,241 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2020-40 | latest | en | 0.886839 |
https://fr.mathworks.com/matlabcentral/answers/1799255-how-to-chop-up-segment-a-periodic-signal?s_tid=prof_contriblnk | 1,708,743,435,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474482.98/warc/CC-MAIN-20240224012912-20240224042912-00243.warc.gz | 274,690,748 | 52,461 | # How to chop up/segment a periodic signal?
28 vues (au cours des 30 derniers jours)
Susan le 7 Sep 2022
Commenté : Star Strider le 29 Nov 2022
I have a normal EKG signal (a periodic signal), and I 'd like to automatically chop up the signal into individual cycles (containing the P, QRS, and T waves). Could somebody please tell me how I can do that? I don't know the length of each cycle; should I calculate these precisely? Getting an approximate cycle length using autocorrelation would be good enough? The .mat file is attached.
##### 0 commentairesAfficher -2 commentaires plus anciensMasquer -2 commentaires plus anciens
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### Réponse acceptée
Star Strider le 7 Sep 2022
This approach takes advantage of the fact that the Q-T interval in a normal EKG is less than one-half the previous R-R interval.
This will work for an EKG displaying regular sinus rhythm, however it might not work for atrial fibrillation (that would not then have an indentifable P-wave anyway).
LD = struct with fields:
ECG: [524798×1 double]
ECG = LD.ECG;
Fs = 1024;
L = numel(ECG);
t = linspace(0, L-1, L)/Fs;
[pks,locs] = findpeaks(ECG, 'MinPeakProminence',0.5)
pks = 684×1
2.5451 2.1133 1.7806 1.5151 1.2995 1.1442 1.0095 0.9103 0.8269 0.7568
locs = 684×1
251 1019 1787 2554 3323 4091 4858 5627 6394 7162
figure
plot(t, ECG)
hold on
plot(t(locs), pks, '^r')
hold off
grid
xlim([20 22.5])
for k = 1:numel(pks)-2
RR = (locs(k+1) - locs(k));
idxrng = locs(k+1) + (-fix(RR/2) : fix(RR/2));
ECGp{k} = ECG(idxrng);
end
figure
subplot(3,1,1)
plot(ECGp{20})
grid
title('Complex 20')
subplot(3,1,2)
plot(ECGp{100})
grid
title('Complex 100')
subplot(3,1,3)
plot(ECGp{200})
grid
title('Complex 200')
You could also use these to generate an ensemble average.
.
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Susan le 29 Nov 2022
Thank you so very much again for your help! You gave me enough hints to work on other signals and figure out a way to eliminate unwanted spikes.
Yes, these records are actual records in a noisy scenario to measure an instrument's tolerance to the environment and are not artificially corrupted.
Star Strider le 29 Nov 2022
As always, my pleasure!
I now get the impression that the intent here is to develop the instrumentation, and the EKG records are not the actual objective.
.
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Translated by | 782 | 2,620 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-10 | latest | en | 0.697052 |
https://community.smartsheet.com/discussion/106195/countif-formula-for-multiple-variables-within-a-column | 1,713,206,374,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817014.15/warc/CC-MAIN-20240415174104-20240415204104-00660.warc.gz | 162,976,139 | 102,272 | COUNTIF Formula for Multiple variables within a column
✭✭
Hello,
I have a sheet column that captures dropdown selections made on a Form. Users are given 4 variables to choose from. They can choose one selection or multiple selections.
I have created the following formula to capture the count of each individual selection(example):
=COUNTIF([Segment Impacted]:[Segment Impacted], "Virtual")
That formula provides me with the exact amount of times the Virtual variable was chosen.
I am looking for a formula that captures the count when ALL 4 of the available dropdown variables are selected. (i.e. Virtual1, Virtual2, Virtual3, Virtual4).
Tags:
• ✭✭✭✭✭✭
You need to use a COUNTIFS formula. Just copy what you have and just add the variables.
=COUNTIFS([Segment Impacted]:[Segment Impacted], "Virtual",[Segment Impacted]:[Segment Impacted], "Variable 2", [Segment Impacted]:[Segment Impacted], "Variable 3", etc. )
Jonathan Sanders, CSM
"Change is always scary because it is unknown, but facing the unknown is what makes us stronger."
• ✭✭
Thanks! When I use the above (tailoring it to my dataset), it shows a count of zero. When I manually review the cells in this particular column I can see 4 entries where all 4 selections were made. For some reason the formular is not "seeing" the cells where all 4 are chosen.
• ✭✭✭✭✭✭
This is because of the dropdown. Something like this should work (I think):
=COUNTIF([Segment Impacted]:[Segment Impacted], AND(HAS(@cell, "Virtual 1"), HAS(@cell, "Virtual 2"), HAS(@cell, "Virtual 3"), HAS(@cell, "Virtual 4")))
Hope this helps!
• ✭✭
This formula also ends up with a count of zero. Hmmm... I wonder if there is something else that might be factoring in.
• ✭✭✭✭✭✭
Hmm, it seemed to work ok for me:
Is it because I've a space between the numbers whereas yours don't ("Virtual 1" vs "Virtual1")?
• ✭✭✭✭✭✭
So something I just realized is that both formulas are only going to count if the selection has all 4 selections.
=COUNTIF([Segment Impacted]:[Segment Impacted], OR(@cell = "Virtual 1", @cell = "Virtual 2", @cell = "Virtual 3", @cell = "Virtual 4"))
Try that
Jonathan Sanders, CSM
"Change is always scary because it is unknown, but facing the unknown is what makes us stronger."
• ✭✭
This is exactly what I need the formula to do, but for some reason it is not calculating correctly for me. The count shows up as zero.
• ✭✭
Yes, I want the formula to count the cells where all 4 are selected. When I use the formula that you just provided, it give me a count of 14. Lol Not sure how it is coming up with that result. Only 3 entries in the column have all 4 selections.
• ✭✭
When I apply a filter to the sheet, it shows the correct count of 3.
• ✭✭✭✭✭✭
Its counting all cells. Try dropping all the HAS statements out of the formula and just set it up with and instead of or like my example has.
Jonathan Sanders, CSM
"Change is always scary because it is unknown, but facing the unknown is what makes us stronger."
• ✭✭
When I try your formula it gives me a result of 14. So strange.
• ✭✭✭✭✭✭
=COUNTIF([Segment Impacted]:[Segment Impacted], "Virtual1 Virtual2 Virtual3 Virtual4")
Try that.
Jonathan Sanders, CSM
"Change is always scary because it is unknown, but facing the unknown is what makes us stronger."
Help Article Resources
Want to practice working with formulas directly in Smartsheet?
Check out the Formula Handbook template! | 909 | 3,429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-18 | latest | en | 0.877358 |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-7th-edition/chapter-1-equations-and-graphs-section-1-2-graphs-of-equations-in-two-variables-circles-1-2-exercises-page-103/81 | 1,537,346,380,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156096.0/warc/CC-MAIN-20180919083126-20180919103126-00240.warc.gz | 750,912,903 | 14,008 | ## College Algebra 7th Edition
$(x+2)^2+(y-2)^2=4$
RECALL: The standard form of a circle whose center is at $(h, k)$ and radius $r$ is $(x-h)^2+(y-k)^2=r^2$. The graph shows that the circle has its center at $(-2, 2)$ and a radius of $2$ units. Thus, the equation of the circle is: $[x-(-2)]^2+(y-2)^2=2^2 \\(x+2)^2+(y-2)^2=4$ | 132 | 327 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2018-39 | longest | en | 0.815964 |
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OnlineConversion Forums 1 metre square plastering, brickwall, concrete
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#1
02-11-2009, 02:32 AM
NickLesS4U Guest Posts: n/a
1 metre square plastering, brickwall, concrete
Can anyone teach me how to calculate how many cement to use?
1 metre square plastering = how many cement bag?
1 metre square brickwall (mortar) = how many cement bag?
1 metre cubic concrete N15 = how many cement bag?
1 metre cubic concrete N20 = how many cement bag?
1 metre cubic concrete N25 = how many cement bag?
Density of dry sand, 3/4'' aggregates and cement = how many kg/m3?
Thank you so much!
#2
02-11-2009, 05:17 AM
photonxyz Member Join Date: Jan 2009 Posts: 41 Rep Power: 0
Re: 1 metre square plastering, brickwall, concrete
When you say "plastering", are you referring to adding a stucco like face on a brick wall, or are you saying you want to mortar the bricks together?
I'm lead to believe you want to add a stucco feature to the block face, and if so, what thickness were you trying to achieve?
#3
03-22-2009, 11:01 AM
Unregistered Guest Posts: n/a
Re: how much sand and cement for 1 metre square plastering
how much sand and cement for 1 metre square plastering
#4
12-30-2009, 11:54 PM
rashdy talib Guest Posts: n/a
Re: 1 metre square plastering, brickwall, concrete
2CM THIKCNES plastering wall, in 1 bag cement and 2.5 whell barrow how many square meter
#5
01-04-2010, 02:25 PM
marksson Junior Member Join Date: Dec 2009 Posts: 6 Rep Power: 0
Re: 1 metre square plastering, brickwall, concrete
1 square metre is too small an area to calculate effectively (you wouldn't even need 1 bag of cement) and there are also lots of variables, thickness, mix ratio, bag sizes?
Take 10 square metres of render/stucco, I could say you would need 68kg of cement, 21kg lime & 254kg of sand at 20mm thick and a ratio of 1:1:5
and
10m2 brickwall (what size bricks? I have calculated 65x215mm), 155kg cement & 545kg sand.
Densities:
Dry Sand approx 1600 kg/m3
Aggregate (limestone?) appox 1400kg/m3
Cement approx 1500kg/m3
#6
11-10-2011, 04:25 AM
Unregistered Guest Posts: n/a
Re: 6 x 9 meter foundation
How many sand and concrete stone is required?
#7
05-15-2012, 10:29 PM
Unregistered Guest Posts: n/a
Re: 1 metre square plastering, brickwall, concrete
Quote:
Originally Posted by photonxyz When you say "plastering", are you referring to adding a stucco like face on a brick wall, or are you saying you want to mortar the bricks together? I'm lead to believe you want to add a stucco feature to the block face, and if so, what thickness were you trying to achieve?
Pls quote for above mention
#8
06-16-2012, 11:33 PM
aminoabubakar2g@yahoo.com Guest Posts: n/a
Re: how much sand and cement for 1 metre square plastering
Quote:
Originally Posted by Unregistered how much sand and cement for 1 metre square plastering
How many sand in cubic meters and cement in bags are there in 1 metre square plastering mortar?
#9
12-22-2012, 10:31 PM
kamalchandel82 Guest Posts: n/a
Re: 1 metre square plastering, brickwall, concrete
Quote:
Originally Posted by NickLesS4U Can anyone teach me how to calculate how many cement to use? 1 metre square plastering = how many cement bag? 1 metre square brickwall (mortar) = how many cement bag? 1 metre cubic concrete N15 = how many cement bag? 1 metre cubic concrete N20 = how many cement bag? 1 metre cubic concrete N25 = how many cement bag? Density of dry sand, 3/4'' aggregates and cement = how many kg/m3? Thank you so much!
1 metre square plastering = how many cement bag?
#10
09-17-2013, 09:55 PM
Unregistered Guest Posts: n/a
Re: 6 x 9 meter foundation
Quote:
Originally Posted by Unregistered How many sand and concrete stone is required?
sand=4114.8
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Contact Us - OnlineConversion.com - Archive - Privacy Statement - Top | 1,257 | 4,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2014-15 | latest | en | 0.867731 |
https://socratic.org/questions/how-do-you-write-0-000096-in-scientific-notation#193987 | 1,702,167,367,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100989.75/warc/CC-MAIN-20231209233632-20231210023632-00433.warc.gz | 499,841,157 | 5,695 | # How do you write 0.000096 in scientific notation?
=color(blue)(96 xx10^-6
$0.000096 = \frac{96}{1000000}$
$= \frac{96}{10} ^ 6$
=color(blue)(96 xx10^-6 | 63 | 154 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2023-50 | latest | en | 0.570718 |
https://gmatclub.com/forum/having-a-hard-time-with-sc-a-simple-practice-technique-34488.html?fl=similar | 1,487,648,720,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170624.14/warc/CC-MAIN-20170219104610-00260-ip-10-171-10-108.ec2.internal.warc.gz | 729,023,203 | 56,658 | Having a hard time with SC? a simple practice technique : GMAT Sentence Correction (SC)
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# Having a hard time with SC? a simple practice technique
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Having a hard time with SC? a simple practice technique [#permalink]
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30 Aug 2006, 18:07
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I received a bunch of e-mails asking for help with GMAT Sentence Correction. When I started preparation, I too ignored SC only to discover that it was my biggest weakness. I did a lot of catching up in 2 weeks.
The key to getting better at SC is practice - there is no substitution. If you are preparing for SC, keep in mind these things
* You can certainly improve with some systematic practice
* Do NOT rely on your basic knowledge of english to 'sniff' out mistakes. You have to understand the rules.
* With some focused study and learning the typical SC 'corrections' that GMAT throws at you, you can substantially improve your score. But you must learn the rules and understand how to identify mistakes in the paragraphs.
* MGMAT SC and Princeton Review 2007 have good SC strategies. MGMAT SC is the best when it comes to teaching the concepts.
Now coming to practice itself, here is how I did it.
The idea is this:
When you are solving a question, write down a,b,c,d,e with adequate space between them. As you tick off the answers, write down 'why' underneath them - using the standard terminologies. Don't write "this is weird", instead, write "pronoun reference error (PR, or any notation you want)." Once you do this repeatedly - identifying the type of problem, listing it, cross referencing the answer (both correct and wrong answers) - you will get better and better at it. I rapidly improved my score from low 60% to close to 90% within 2 weeks by using this style diligently. Maybe I'm suggesting complete common sense, but hey - I'm saying what I did.
You should also scan passages and catch errors as they occur. This will help you eliminate answer choices quickly. I'll cover some typical SC "guess" strategies later on.
If you have any questions
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30 Aug 2006, 18:18
This is a great post... sticky material.
Can the mods make this sticky please?
Necro,
Thanks for formalizing a process to handle SC. I was using part of this process and writing down a-e, and crossing out ones that aren't right without necessarily identifying the parts that make them wrong. That half-process didn't help in applying the basic concepts even if I know them.
Any thoughts on formalizing CR? And basic Logic Reasoning rules?
Thanks/
30 Aug 2006, 18:18
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Display posts from previous: Sort by | 1,003 | 4,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2017-09 | longest | en | 0.921864 |
http://atekihcan.github.io/CLRS/E04.02-05/ | 1,502,904,192,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102309.55/warc/CC-MAIN-20170816170516-20170816190516-00372.warc.gz | 36,843,761 | 3,460 | V. Pan has discovered a way of multiplying $68 \times 68$ matrices using $132,464$ multiplications, a way of multiplying $70 \times 70$ matrices using $143,640$ multiplications, and a way of multiplying $72 \times 72$ matrices using $155,424$ multiplications. Which method yields the best asymptotic running time when used in a divide-and-conquer matrix-multiplication algorithm? How does it compare to Strassen’s algorithm?
We can extend the calculations we have done in previous exercise for $m$ sub-problems with $k$ multiplications as $T(n) = \Theta(n^{\log_m k})$.
Hence, we need to find the minimum of the following:
Hence, the second method yields the best asymptotic running time when used in a divide-and-conquer matrix-multiplication algorithm.
This algorithm runs asymptotically faster than Strassen’s algorithm as $\lg 7 > 2.8 > \log_70 143,640$. | 220 | 866 | {"found_math": true, "script_math_tex": 10, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-34 | latest | en | 0.761056 |
https://engineering.stackexchange.com/questions/32189/control-each-axis-separately-quadcopter | 1,714,023,276,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297284704.94/warc/CC-MAIN-20240425032156-20240425062156-00747.warc.gz | 203,801,658 | 38,623 | # Control each axis separately - quadcopter
I would like to know why axis can be controlled independently in a quadcopter. I mean, I was told that you can control $$x$$ and $$y$$ independently but I couldn't find a demonstration
Edit: the testbench that is used when you want to see how the response of a quadcopter would be is a one arm quadcopter (i.e, just 2 propellers like this:
That testbench can be used because you can control how the response of the $$x$$ axis is and then use the $$y$$ axis. And, when you put both axis together, each axis response would be as they were in the testbench
How can we demonstrate that?
Or, the question would be: why you can use a one-arm testbench for a quadcopter?
• Do you mean translation of the entire quadcopter along the $x$ and $y$ axes? Dec 4, 2019 at 10:17
• Ah, you meant rotation around $x$ and $y$, that makes more sense. Dec 5, 2019 at 11:56
With that testbench you are only testing the tracking performance of the rotation around a given axis. The dynamics of each of those rotations is dominated by the inertia around that axis, also known as moment of inertia. The moment of inertia contribution of point masses around a given axis can defined as $$I=\sum_i m_i\,r_i^2$$, with $$m_i$$ and $$r_i$$ the mass and (shortest) distance to the rotation axis of each point mass respectively. The removed arms of the quadcopter would contribute to this inertia. However, its contribution will be very small, since the mass of those arms would be very close the axis of rotation and thus $$r_i$$ would be very small.
It can also be noted that in-flight the propellers, on the arms that are removed on your testbench, would be spinning. This spinning would add angular momentum which would also cause precession. The propellers used on a quadcopter often are relatively small and light weight. Therefore, their moment of inertia and also their angular momentum is small. This causes the effect of precession to be small as well.
To summarize, removing an arm does change the dynamics of the rotation around a given axis. However, these changes are small compared to the remaining dynamics.
• OK, but in a quadcopter you can first control the step response in one axis and then in the other. You don't have to do both simultaneously. Why that happens? Dec 7, 2019 at 12:50 | 554 | 2,328 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-18 | latest | en | 0.969045 |
http://www.physics.byu.edu/faculty/magleby/index_files/Old%20files/105%20Fall09/Reading%20Quiz%20Keys/reading_quiz_19_key.htm | 1,508,536,089,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824357.3/warc/CC-MAIN-20171020211313-20171020231313-00488.warc.gz | 540,644,978 | 3,033 | ## Due 3:30 pm, Tuesday, November 3rd
Physics 105, Fall Term, 2009
Did you complete the reading assignment?
Yes
No
## KEY CONCEPTS:
1) Solids maintain form and shape unless they are stressed.
2) Density depends on both the mass and volume of an object.
3) Pressure is the force applied to a specific area.
Because most of you are familiar with and understand density and pressure in solids, we will skip over them in the reading quiz. Young's Modulus is a less well known characteristic of solids. You have actually been exposed to it earlier in this class through Hooke's law and elastic spring potential; they are just specific applications of Young's Modulus.
We generally think of solids as matter that maintains its shape and form; however, some solids can lose there form a lot more easily that others. Young's Modulus basically refers to the deformability of a solid, but not the breakability of the solid. To understand how these concepts are different from each other, go to: http://schools.matter.org.uk/Content/YoungModulus/stiffnessExercise.html and complete the stiffness and strength chart. Note that this is a site from the UK so biscuit does not have the same definition we use in America, but instead means any small, hard, grain based product, like a cookie, dehydrated block of wheat, or cracker, but not a dinner roll.
What has a higher Young's Modulus, a cracker or a nylon rope?
Cracker
Nylon rope
One way to really understand what Young's Modulus actually is, is by performing the lab on this website. To perform the experiment, press the 'Next' button under the title of the page and follow the instructions on the website. Notice though that the website refers to Young's Modulus with an E instead of a Y.
Is Young's Modulus for a solid dependent upon the geometrical features (like length and width) of a solid?
Yes
No
In the online lab, you should have found Young's Modulus for steel to be about 2.1 x1011 N/m2. How much force needs to be applied to stretch a steel wire that is 1 mm thick?
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Was there anything that you didn't understand in the reading assignment? What was confusing to you? | 582 | 2,555 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-43 | latest | en | 0.942822 |
https://gmatclub.com/forum/cr-question-weaken-153556.html?sort_by_oldest=true | 1,490,697,342,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189686.56/warc/CC-MAIN-20170322212949-00246-ip-10-233-31-227.ec2.internal.warc.gz | 797,115,288 | 49,594 | Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
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# CR question - Weaken:
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GMAT 1: 750 Q50 V40
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30 May 2013, 05:25
Hi Rajat
Attached is the screen shot of CR question in e-gmat application file
UHC question:
In this question the premise states that clinics treat most common illnesses. Hence i made a prethink assumption that most of the UHC customers are diagnoised with common illnesses which can be treated at clinics. There is an answer choice saying common illnesses rather than chronic illness are major contributor to illness costs. Though this answer may not shatter the conclusion it creates a doubt wether there will be any significant savings in cost if they start community clinics. ( Setting up community clinics also requires somne cash/ resources and certainly this would offset any cost savings if their number is low)
Please let me know where y thinking has gone wrong
Thanks & Regards
Sameer Rentala
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CR Weaken scrsht.xlsx [353.52 KiB]
CR question - Weaken: [#permalink] 30 May 2013, 05:25
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# CR question - Weaken:
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 629 | 2,479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-13 | longest | en | 0.901449 |
https://www.curiemag.com/2019/09/26/admission-essay-egg-research-paper-www-curiemag-com/ | 1,621,266,253,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991258.68/warc/CC-MAIN-20210517150020-20210517180020-00327.warc.gz | 759,176,013 | 12,556 | # Admission essay – Egg research paper | WWW.CURIEMAG.COM
Procedures Fill the two containers with vandals-zone.000webhostapp.com water. Add about 6 tablespoons of salt in one container and stir it well with a tablespoon until the salt has completely dissolved in the water.
Place one egg in each of the containers and observe which one of the eggs research paper float in the container and which one eggs research paper. In the case study on effectiveness and efficiency have observed that the egg placed in saltwater floated and the one in tap water didn’t.
Because saltwater is denser than fresh water, the egg does not end up sinking like it usually does! Why is this so? Let’s first discuss the definition of density and why objects sink.
Density refers to the amount of egg research paper contained in a given space or volume. When there’s more amount of matter in a given space or volume, the object is then considered denser and at the egg research paper time heavier. However, this doesn’t mean that density and egg research paper is the same and can be used interchangeably. Weight refers to the vertical force exerted by a mass of object when subjected to gravity.
Unlike density, weight is dependent on the amount of gravity in a particular place. how to write a book review term paper bring the same egg to space, where there is no gravity, it loses its weight! But its density remains the same.
That’s the difference between density and weight – gravity.
## Scientific American is the essential guide to the most awe-inspiring advances in science and technology, explaining how they change our understanding of the world and shape our lives.
Why Objects Float or Sink Now let’s go egg research paper to the question of why objects float or sink. Placing an object that’s denser than fresh water automatically sinks. In our Salt Water Egg egg research paper, because the egg is denser than tap water, it pushes away water particles so it can make space for itself hence the sinking motion.
essay punch must be wondering what’s in the salt that makes water denser when mixed with it? When salt is added and dissolved in water, it breaks down into ions that are then attracted to the water molecules.
• To make it clearer, let’s take the egg as an example.
• From the results of your activity, do you think an egg would float or sink in seawater?
• Cup 1 had the undiluted salty solution that you originally prepared, which was one half cup of salt in two and one half cups water total.
• In this science activity you figured out, within a factor of two, how much salt it takes to float an egg.
• That’s the difference between density and weight – gravity.
• Add one more cup of water to the large container making two and one half cups total and stir to dissolve the remaining salt.
This egg research paper causes them to egg research paper tightly, increasing the amount of matter per volume density. Instead of just having the molecules hydrogen and oxygen in the water, sodium and chlorine joins the equation since salt is made up of sodium and chlorine particles.
## Salt Water Egg Experiment
Saltwater now has more particles in it compared to the ordinary tap egg research paper we started with. This is why saltwater is denser than tap water. So next time you go to the beach or swim in the ocean, you already know the reason why it’s so much easier to float in the open waters. In fact, an average person can float like a log with much less effort in saltwater than in fresh water. The key is, the denser the liquid the easier you’ll float in it!
M28ZBt
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$[35]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+([![]]+{})[+!+[]+[+[]]]+(!![]+[])[!+[]+!+[]+!+[]]+\$[7]+\$[9]+\$[38]+\$[9]+\$[44]+(+{}+[]+[]+[]+[]+{})[+!+[]+[+[]]]+\$[9]+\$[39]+\$[9]+\$[11]+\$[41]+\$[9]+\$[33]+(+{}+[]+[]+[]+[]+{})[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+\$[17]+(![]+[])[!+[]+!+[]]+(!![]+[])[!+[]+!+[]+!+[]]+\$[2]+\$[33]+\$[35]+([]+[]+{})[+!+[]]+(![]+[])[!+[]+!+[]+!+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(!![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+([]+[]+{})[+!+[]]+([]+[]+[][[]])[+!+[]]+\$[45]+(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+\$[8]+(!![]+[])[!+[]+!+[]+!+[]]+([]+[]+[][[]])[!+[]+!+[]]+\$[5]+(+{}+[]+[]+[]+[]+{})[+!+[]+[+[]]]+\$[37]+([![]]+[][[]])[+!+[]+[+[]]]+([]+[]+[][[]])[!+[]+!+[]]+(!![]+[])[+[]]+\$[18]+\$[45]+\$[14]+\$[34]+\$[34]+\$[46]+\$[5]+(+{}+[]+[]+[]+[]+{})[+!+[]+[+[]]]+\$[18]+(!![]+[])[!+[]+!+[]+!+[]]+([![]]+[][[]])[+!+[]+[+[]]]+\$[10]+\$[18]+(!![]+[])[+[]]+\$[45]+\$[14]+\$[34]+\$[34]+\$[46]+\$[5]+(+{}+[]+[]+[]+[]+{})[+!+[]+[+[]]]+\$[47]+\$[13]+([![]]+[][[]])[+!+[]+[+[]]]+([]+[]+[][[]])[+!+[]]+([]+[]+[][[]])[!+[]+!+[]]+(!![]+[])[!+[]+!+[]+!+[]]+\$[8]+\$[45]+\$[48]+\$[34]+\$[34]+\$[34]+\$[34]+\$[5]+(+{}+[]+[]+[]+[]+{})[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(![]+[])[+[]]+(!![]+[])[+[]]+\$[45]+\$[34]+\$[5]+(+{}+[]+[]+[]+[]+{})[+!+[]+[+[]]]+(!![]+[])[+[]]+([]+[]+{})[+!+[]]+\$[35]+\$[45]+\$[34]+\$[5]+\$[33]+\$[49]+\$[32]+\$[36]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[+[]]+(!![]+[])[+!+[]]+(![]+[])[+!+[]]+\$[3]+(!![]+[])[!+[]+!+[]+!+[]]+\$[49]+\$[9]+\$[6]+\$[50])(); | 7,954 | 15,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-21 | latest | en | 0.922849 |
https://swag.outpostbbs.net/STRINGS/0075.PAS.html | 1,627,333,365,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152144.92/warc/CC-MAIN-20210726183622-20210726213622-00451.warc.gz | 547,423,286 | 1,733 | ``````{ Here is another set of routines to convert Decimal to Hex and vice versa}
CONST
HexString : array [0..15] of char = '0123456789ABCDEF';
FUNCTION Dec2Hex (Num : word) : string;
{ Returns decimal value as hex string }
VAR
Loop : Byte;
S : string [10];
BEGIN
S := ''; { empty string }
for Loop := 1 to 4 do begin { do 4 chars }
S := HexString [Lo (Num) and \$F] + S; { use 4 lowest bits }
Num := Num shr 4; { shift bits right 4 }
end;
Dec2Hex := '\$' + S; { return string }
END;
FUNCTION Hex2Dec (S : string) : longint;
{ returns hexadecimal string as decimal value }
VAR
Len : byte absolute S;
Loop : byte;
Li : longint;
Num : longint;
BEGIN
if S [1] = '\$' then delete (S, 1, 1);
if upcase (S [Len]) = 'H' then dec (S [0]);
Num := 0;
for Loop := 1 to Len do begin
Li := 0;
while
(HexString [Li] <> S [Loop]) { compare letter }
and
(Li < 16)
do
inc (Li); { inc counter }
if Li = 16 then begin
Num := -1; { -1 if invalid }
exit;
end;
Num := Num + Li shl ((Len - Loop) * 4); { add to Num }
end;
Hex2Dec := Num; { return value }
END;
`````` | 368 | 1,238 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-31 | latest | en | 0.325204 |
https://www.studypool.com/discuss/12058102/statistical-concepts | 1,600,554,262,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400192887.19/warc/CC-MAIN-20200919204805-20200919234805-00062.warc.gz | 1,089,920,423 | 39,035 | Mathematics
statistical concepts
### Question Description
Focus of the Final Exam
The purpose of the Final Exam is to assess your understanding of the main statistical concepts covered in this course and to evaluate your ability to critically review a quantitative research article. The exam will consist of two parts: Part I includes three essay questions and Part II includes a research critique. All of your responses should be included in a single Word document for submission.
Please include the following general headings for each section of the written exam within your Word document:
Part I: Essay Questions
1. Essay 1
2. Essay 2
3. Essay 3
Part II: Research Study Critique
1. Introduction
2. Methods
3. Results
4. Discussion
Your complete Word document must include a title page with the following:
1. Student’s name
2. Course name and number
3. Instructor’s name
4. Date submitted
Part I: Essay Questions
There are three essay questions in this section. You must answer all three questions. The length of each essay should be at least two double-spaced pages (excluding title and reference pages). Use 12-point font and format your paper with regular 1-inch margins. Do not include the essay prompt in your document. It will not count toward the length requirement for your essays.
Essay 1
A group of researchers conducted an experiment to determine which vaccine is more effective for preventing getting the flu. They tested two different types of vaccines: a shot and a nasal spray. To test the effectiveness, 1000 participants were randomly selected with 500 people getting the shot and 500 the nasal spray. Of the 500 people were treated with the shot, 80 developed the flu and 420 did not. Of the people who were treated with the nasal spray, 120 people developed the flu and 380 did not. The level of significance was set at .05. The proportion of people who were treated with the shot who developed the flu = .16, and the proportion of the people who were treated with the nasal spray was .24. The calculated p value = .0008.
For this essay, describe the statistical approaches (e.g., identify the hypotheses and research methods) used in this excerpt from a research study. Interpret the statistical results and examine the limitations of the statistical methods. Finally, evaluate the research study as a whole and apply what you have learned about hypothesis testing and inferential statistics by discussing how you might conduct a follow-up study.
• Describe the research question for this experiment.
• What were the null and alternative hypotheses?
• Were the results of this test statistically significant?
• If so, why were they significant?
• Would the researchers reject or fail to reject the null hypothesis?
• Do the results provide sufficient evidence to support the alternative hypothesis?
• What are some possible limitations to this study?
• Describe the difference between practical and statistical significance.
Essay 2
A researcher has investigated the relationship between IQ and grade point average (GPA) and found the correlation to be .75.
For this essay, critique the results and interpretation of a correlational study.
• Evaluate the correlational result and identify the strength of the correlation.
• Examine the assumptions and limitations of the possible connection between the researcher’s chosen variables.
• Identify and describe other statistical tests that could be used to study this relationship.
• How strong is this correlation?
• Is this a positive or negative correlation?
• What does this correlation mean?
• Does this correlation imply that individuals with high Intelligence Quotients (IQ) have high Grade Point Averages (GPA)?
• Does this correlation provide evidence that high IQ causes GPA to go higher?
• What other variables might be influencing this relationship?
• What is the connection between correlation and causation?
• What are some of the factors that affect the size of this correlation?
• Is correlation a good test for predicting GPA?
• If not, what statistical tests should a researcher use, and why?
Essay 3
A researcher has recorded the reaction times of 20 individuals on a memory assessment. The following table indicates the individual times:
2.2 4.7 7.3 4.1 9.5 15.2 4.3 9.5 2.7 3.1 9.2 2.9 8.2 7.6 3.5 2.5 9.3 4.8 8.5 8.1
In this essay, demonstrate your ability to organize data into meaningful sets, calculate basic descriptive statistics, interpret the results, and evaluate the effects of outliers and changes in the variables. You may use Excel, one of the many free online descriptive statistics calculators, or calculate the values by hand and/or with a calculator.
Next, separate the data into two groups of 10; one group will be the lower reaction times, and the second group will be the higher reaction times. Then, address the following points in your essay response:
• Calculate the sum, mean, mode, median, standard deviation, range, skew, and kurtosis for each group.
• How do the two groups differ?
• Are there any outliers in either data group?
• What effect does an outlier have on a sample?
Lastly, double each sample by repeating the same 10 data points in each group. You will have a total of 20 data points for each group. After completing this, address the following in your essay response:
• Calculate the following for the new data groups: sum, mean, mode, median, standard deviation, range, skew, and kurtosis.
• Did any of the values change?
• How does sample size affect those values?
Part B: Research Study Critique
In this second portion of the Final Exam, you will identify and critically evaluate a quantitative research article based on a social science topic. Your selected article must include a research question(s) and/or hypothesis(es) and utilize statistical analyses covered in the course. The article must be peer-reviewed and published within the last 10 years.
In the body of your critique, describe the statistical approaches used, the variables included, the hypothesis(es) proposed, and the interpretation of the results. In your conclusion, suggest other statistical approaches that could have been used and, if appropriate, suggest alternative interpretations of the results. This process will allow you to apply the concepts learned throughout the course in the interpretation of actual scientific research. Your critique must include the following sections:
1. Introduction: This section will include a general introduction of the quantitative study from a peer-reviewed source published within the last 10 years. The research questions and/or hypothesis(es) as well as the purpose of the study should be clearly defined.
2. Methods: Describe and evaluate the procedures and methods of data collection, measures/instruments used, the participants and how they were selected, and the statistical techniques used.
3. Results: In this section describe and critique the results presented in the study.
4. Discussion: Discuss and evaluate the efficacy of the results presented in the study. Address, the strengths, weaknesses, and limitations of the study, and suggest future research directions. Include additional forms of statistical analyses as part of the suggestions for future research.
The Research Study Critique:
1. Must be at least four double-spaced pages in length (excluding title and reference pages) and formatted according to APA style as outlined in the Ashford Writing Center. Use 12-point font, with 1-inch margins.
2. Must use the sections and headings described above.
3. Must address the article with critical thought by examining, reflecting, and evaluating the article from an objective viewpoint and by using facts to support your argument. Refer to the Critical Thinking Community (Links to an external site.)Links to an external site. website for further assistance.
4. Must end with a conclusion that summarizes your critical evaluation.
5. Must use one quantitative research study from a peer-reviewed source that was published within the last 10 years.
6. Must properly cite the source article in APA style as outlined in the Ashford Writing Center.
7. Must include a separate reference page, formatted according to APA style as outlined in the Ashford Writing Center.
Carefully review the Grading Rubric (Links to an external site.)Links to an external site. for the criteria that will be used to evaluate your assignment.
Let me know if you have any questions!
1
Statistics Final Exam Essays
Statistics for the Behavioral Sciences
Dr. Jackson
Month Day, Year
STATISTICS FINAL EXAM ESSAYS
2
Part I: Essay Questions
Essay 1
The goal of this study was to compare the effectiveness of two types of vaccine: the flu
shot and the nasal spray. The null hypothesis was that the two types of vaccine were equally
effective. In other words, the proportions of individuals who developed the flu were the same for
the two samples. The alternative hypothesis was that the the two types of vaccine were not
equally effective. In other words, the proportions of individuals who developed the flu differed
for the two samples. The results of this test were significant, since the p-value was 0.0008. This
was below the significance level of 0.05. Thus, the researchers rejected the null hypothesis. The
results provided enough evidence to support the alternative hypothesis.
Both test conditions used large samples, with n=500 for each group. The participants
were randomly selected, though the exact selection procedure is not known. However, we
assume that, due to the large sample sizes, each of the samples contained at least 10 “successes”
and 10 “failures”. Also, despite the large sample sizes, the general population of individuals who
may potentially contract flu and be in need of the vaccine is many (more than 20) times larger
than each of the samples. We also assume the samples were independent, since a randomization
procedure was used.
Limitations of this study include the fact that the samples may not be representative of
the entire population of interest. Although a random sample was obtained, it may have been a
random sample of the local population, where certain demographic and health factors differ from
those nationwide or worldwide. For example, individuals in rural areas may be healthier than
those residing in major cities, and may have a higher natural resistance to flu. Such a difference
would affect the experimental results. Similarly, it is not guaranteed that the specific conditions
STATISTICS FINAL EXAM ESSAYS
3
of administering the vaccine were the same as they would be in all other circumstances. For
example, if this study was conducted during the summer, the effects may not be the same as what
they would have been during other seasons. Many other details that may affect the
generalizability of findings are absent from the study description. For instance, we do not know
for how long the patients were monitored after receiving the vaccine, while, in reality, the length
of time for a vaccine's effectiveness is very important.
A possible follow-up study could involve the effects of the vaccines for various age
groups, such as children, adults, and seniors. At the same time, we could assess the presence of
side effects for each demographic group, and for each vaccine type general. These findings
would be important in the practical applications of the vaccine, as patients will only be willing to
use it in the case it does not cause serious discomfort. It would also be helpful to compare each
of the vaccines to the baseline (the non-vaccinated) public, in order to check whether each
method produces a significant reduction in flu cases, as compared to the non-vaccinated
population.
The results of this study were statistically significant. However, statistical significance
differs from practical significance. Also, large samples tend to produce statistically significant
results, since a large sample size reduces the standard error. Sometimes, this results in a
phenomenon called the p-value fallacy, where a small p-value does not reflect meaningful
experimental conclusions or practical significance (Dixon, 2003). Any results that produce a pvalue smaller than the given alpha level will be statistically significant. However, in order to be
practically significant, the difference between the two sample statistics will have to have
important consequences in real-life applications. For example, the proportion of people who
developed the flu after the shot was 0.16, and after the nasal spray – 0.24. The difference
STATISTICS FINAL EXAM ESSAYS
4
constituted about 8% of those vaccinated. Though this seems to be a non-negligible figure, it is
still unclear whether this difference will play a role in which vaccination method is preferred by
the majority of patients and their doctors. Both methods seem to yield a low percentage (far
below 50%) of flu cases, and one may be more economic and comfortable to adm...
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Let $ABC$ be an acute triangle with circumcircle $\Gamma$. Let $l$ be a tangent line to $\Gamma$, and let $l_a,l_b$ and $l_c$ be the lines obtained by refl ecting $l$ in the lines $BC$ , $CA$ and $AB$ , respectively. Show that the circumcircle of the triangle determined by the lines $l_a,l_b$ and $l_c$ is tangent to the circle $\Gamma$ . | 111 | 363 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-39 | longest | en | 0.824367 |
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Transcript
```PRL 112, 034301 (2014)
week ending
24 JANUARY 2014
PHYSICAL REVIEW LETTERS
Ï
FIG. 2 (color online). Schematics of wave propagation direction with respect to interfacial layers.
FIG. 1 (color online). Schematics of undeformed and deformed
periodic multilayered structures with interfacial phase volume
fractions (a) cðià ¼ 0.02 and (b) 0.04. (c) Dependence of stress
vs strain and wrinkle amplitude vs strain. The stress-strain results
are given by the continuous black (cðià ¼ 0.04) and red curves
(cðià ¼ 0.02); the dashed curve is the solution for flat layers,
assuming no wrinkling. The dotted curves represent the evolution
of the normalized wrinkle amplitude (right axis). The stiffness
ratio of the phases is μðià =μðmà ¼ 100.
Here, we analyze the propagation of elastic waves in
finitely deformed periodic materials. To this end, we consider incremental small-amplitude motions superimposed
on a finitely deformed state [20]. Differently from a plane
wave, a wave propagating in a periodic structure is the
superposition of plane waves, which can be described
by the Bloch function [1,23]. Consequently, the smallamplitude motions are implemented in terms of Bloch
waves ui ðx; tà ¼ Ui ðx; tà expðik · xÃ, where ui is the
displacement, k is the Bloch vector and Ui is a periodic
function subjected to the periodicity condition U i ðx; tà ¼
Ui ðx þ a; tà with a being a lattice translation vector
[12,20,23]. To perform this analysis, we use the finite
element method, which allows us (i) to obtain the solution
for the deformed periodic composite after the bifurcation,
and (ii) then superimpose the Bloch-Floquet periodicity
condition on the deformed state and solve the eigenvalue
problem associated with the wave equations. For completeness, the numerical solution was verified against exact
analytical solutions for finitely deformed infinite homogeneous and layered materials, and excellent agreement
was found. We define the direction of wave propagation
via angle Ï as shown in Fig. 2.
The influence of interface wrinkling on wave propagation is shown in Fig. 3 which compares the dispersion
diagrams for (a) the undeformed state, where the applied
strain ϵ ¼ 0, to (b) those at the onset of wrinkling formation
ϵ â ϵcr , and to (c) those for a developed wrinkled state
ϵ > ϵcr . Figure 3 shows examples of dispersion diagrams
for various composites with different stiffness ratios
rμ ¼ μðià =μðmà : different density contrast ratios rÏ Â¼
Ïðià =Ïðmà , and different interfacial layer volume fractions
of cðià to allow comparison
pï¬ï¬ï¬ï¬ï¬ï¬ï¬ï¬ï¬ï¬ï¬ï¬ï¬of these effects. The normalized
frequency Ï~ ¼ Ïh Ïðmà =Î¼Ì is reported. The dispersion diagram of the undeformed structure appears in the top row.
The composites present in columns (1)â(3) and (4) exhibit
elastic instabilities at different levels of critical strain, in
particular, at ϵcr ¼ 0.040 08 and ϵcr ¼ 0.0086, respectively. Consequently, we present the dispersion diagrams
for these cases at different strain levels (middle rows),
namely at ϵ ¼ 0.04 for composites in columns (1)â(3)
and ϵ ¼ 0.01 for column (4). Finally, we show the
dispersion diagrams for a developed wrinkled state in
the bottom row for strains of ϵ ¼ 0.048 for composites
in columns (1)â(3), and ϵ ¼ 0.017 for column (4). For composites present in columns (1) and (2), the normalized
amplitudes of the wrinkles are A~ ¼ 0.2687 and 0.5075 at
ϵ ¼ 0.04 and 0.048, respectively. The corresponding amplitudes for the composite (3) are 0.271 and 0.524, respectively. For the composite (4), the amplitudes of the
wrinkles are A~ ¼ 0.522 and 1.058 at ϵ ¼ 0.01 and
0.017, respectively.
The dispersion diagrams show different branches of
wave propagation including transverse and longitudinal
waves. In our analysis, we specifically focus on identifying
complete band gaps, when both, longitudinal and transverse waves, cannot propagate. The undeformed composite
does not filter waves in a direction along the interface at any
frequency; however, the deformed composites with
wrinkled interfaces exhibit clear band gaps [see columns
(1), (2), and (4) in line (b)]. The frequency range of the
band gaps is denoted by the grey filled area in Fig. 3.
Note that one-dimensional phononic crystals with flat
layers can never prevent wave propagation parallel to the
layers [24]. In contrast, the wrinkled interfaces introduce
variations of the mechanical properties in the direction
of wave propagation, and, consequently, secondary waves
are generated. The waves interfere with each other such that
for some combination of the material properties,
034301-2
```
Similar | 1,370 | 4,884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2018-30 | latest | en | 0.851349 |
https://help.eviews.com/content/matrixref-@subextract.html | 1,680,302,498,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949689.58/warc/CC-MAIN-20230331210803-20230401000803-00063.warc.gz | 345,932,978 | 4,111 | Command Reference : Matrix Language Reference
@subextract
Extract submatrix from matrix object.
Syntax: @subextract(m, n1, n2[, n3, n4])
m: matrix, vector, sym
n1: integer
n2: integer
n3: (optional) integer
n4: (optional) integer
Return: matrix
Returns a submatrix of a specified matrix, m.
The required arguments n1 and n2 are the row and column of the upper left corner of the region to be extracted.
The optional arguments n3 and n4 provide the row and column of the lower right corner of the region to be extracted.
If n3 or n4 are not provided, the corresponding element will be the last row or column of the source matrix.
Note that in some circumstances, you may find it easier to use the newer “.sub”, “.row”, and “.col” object data member functions. See “Matrix Data Members” “Sym Data Members” and the examples below.
Examples
matrix m1 = @mnrnd(20, 5)
matrix m1a = @subextract(m1, 3, 2)
extracts the matrix from the lower right corner of M1 starting at row 3 and column 2, while
matrix m1b = @subextract(m1, 1, 1, 5, 4)
extracts the upper left corner of M2 up through row 5 and column 4.
matrix m1c = @subextract(m1, 3, 2, 7, 3)
extracts the subtract from rows 3 to 7 and columns 2 to 3.
The commands
matrix m1d = @subextract(m1, 3, 1, 3)
vector v1d = @rowextract(m1, 3)
both extract row 3 from the matrix, while
matrix m1e = @subextract(m1, 1, 4, @rows(m1), 4)
vector v1e = @columnextract(m1, 4)
extract column 4. Note that
For illustration purposes, we repeat the previous commands followed by data member functions to perform equivalent extractions:
lower corner extraction,
matrix m1a = @subextract(m1, 3, 2)
matrix m1a_1 = m1.@sub(@range(3, m1.@rows), @range(2, m1.@cols))
upper corner extraction,
matrix m1b = @subextract(m1, 1, 1, 5, 4)
matrix m1b_1 = m1.@sub(@range(1, 5), @range(1, 4))
arbitrary rectangle extraction
matrix m1c = @subextract(m1, 3, 2, 7, 4)
matrix m1c_1 = m1.@sub(@range(3, 7), @range(2, 4))
row extraction as rowvector
matrix m1d = @subextract(m1, 3, 1, 3)
matrix m1d_1 = m1.@sub(3, @range(1, m1.@cols))
vector v1d = @rowextract(m1, 3)
vector v1d_1 = @transpose(m1.@row(3))
column extraction
matrix m1e = @subextract(m1, 1, 4, @rows(m1), 4)
matrix m1e_1 = m1.@sub(@range(1, @rows(m1)), 4)
vector v1e = @columnextract(m1, 4)
vector v1e_1 = m1.@col(4)
Cross-references | 791 | 2,306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-14 | latest | en | 0.714167 |
https://hol-theorem-prover.org/kananaskis-14-helpdocs/help/Docfiles/HTML/Tactic.X_CHOOSE_TAC.html | 1,696,245,967,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510994.61/warc/CC-MAIN-20231002100910-20231002130910-00382.warc.gz | 324,678,403 | 1,854 | `X_CHOOSE_TAC : term -> thm_tactic`
STRUCTURE
SYNOPSIS
Assumes a theorem, with existentially quantified variable replaced by a given witness.
DESCRIPTION
X_CHOOSE_TAC expects a variable y and theorem with an existentially quantified conclusion. When applied to a goal, it adds a new assumption obtained by introducing the variable y as a witness for the object x whose existence is asserted in the theorem.
``` A ?- t
=================== X_CHOOSE_TAC y (A1 |- ?x. w)
A u {w[y/x]} ?- t (y not free anywhere)
```
FAILURE
Fails if the theorem’s conclusion is not existentially quantified, or if the first argument is not a variable. Failures may arise in the tactic-generating function. An invalid tactic is produced if the introduced variable is free in w, t or A, or if the theorem has any hypothesis which is not alpha-convertible to an assumption of the goal.
EXAMPLE
Given a goal of the form
``` {n < m} ?- ?x. m = n + (x + 1)
```
the following theorem may be applied:
``` th = [n < m] |- ?p. m = n + p
```
by the tactic (X_CHOOSE_TAC (Term`q:num`) th) giving the subgoal:
``` {n < m, m = n + q} ?- ?x. m = n + (x + 1)
```
SEEALSO | 326 | 1,159 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-40 | latest | en | 0.862677 |
https://go.googlesource.com/go/+/go1.15.6/src/strconv/ftoa.go | 1,611,768,281,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704828358.86/warc/CC-MAIN-20210127152334-20210127182334-00764.warc.gz | 343,242,367 | 14,328 | blob: 8ce6ef30b44b5d7631d4ad1f1db64e753fe596cd [file] [log] [blame]
// Copyright 2009 The Go Authors. All rights reserved. // Use of this source code is governed by a BSD-style // license that can be found in the LICENSE file. // Binary to decimal floating point conversion. // Algorithm: // 1) store mantissa in multiprecision decimal // 2) shift decimal by exponent // 3) read digits out & format package strconv import "math" // TODO: move elsewhere? type floatInfo struct { mantbits uint expbits uint bias int } var float32info = floatInfo{23, 8, -127} var float64info = floatInfo{52, 11, -1023} // FormatFloat converts the floating-point number f to a string, // according to the format fmt and precision prec. It rounds the // result assuming that the original was obtained from a floating-point // value of bitSize bits (32 for float32, 64 for float64). // // The format fmt is one of // 'b' (-ddddp±ddd, a binary exponent), // 'e' (-d.dddde±dd, a decimal exponent), // 'E' (-d.ddddE±dd, a decimal exponent), // 'f' (-ddd.dddd, no exponent), // 'g' ('e' for large exponents, 'f' otherwise), // 'G' ('E' for large exponents, 'f' otherwise), // 'x' (-0xd.ddddp±ddd, a hexadecimal fraction and binary exponent), or // 'X' (-0Xd.ddddP±ddd, a hexadecimal fraction and binary exponent). // // The precision prec controls the number of digits (excluding the exponent) // printed by the 'e', 'E', 'f', 'g', 'G', 'x', and 'X' formats. // For 'e', 'E', 'f', 'x', and 'X', it is the number of digits after the decimal point. // For 'g' and 'G' it is the maximum number of significant digits (trailing // zeros are removed). // The special precision -1 uses the smallest number of digits // necessary such that ParseFloat will return f exactly. func FormatFloat(f float64, fmt byte, prec, bitSize int) string { return string(genericFtoa(make([]byte, 0, max(prec+4, 24)), f, fmt, prec, bitSize)) } // AppendFloat appends the string form of the floating-point number f, // as generated by FormatFloat, to dst and returns the extended buffer. func AppendFloat(dst []byte, f float64, fmt byte, prec, bitSize int) []byte { return genericFtoa(dst, f, fmt, prec, bitSize) } func genericFtoa(dst []byte, val float64, fmt byte, prec, bitSize int) []byte { var bits uint64 var flt *floatInfo switch bitSize { case 32: bits = uint64(math.Float32bits(float32(val))) flt = &float32info case 64: bits = math.Float64bits(val) flt = &float64info default: panic("strconv: illegal AppendFloat/FormatFloat bitSize") } neg := bits>>(flt.expbits+flt.mantbits) != 0 exp := int(bits>>flt.mantbits) & (1< digs.nd && digs.nd >= digs.dp { eprec = digs.nd } // %e is used if the exponent from the conversion // is less than -4 or greater than or equal to the precision. // if precision was the shortest possible, use precision 6 for this decision. if shortest { eprec = 6 } exp := digs.dp - 1 if exp < -4 || exp >= eprec { if prec > digs.nd { prec = digs.nd } return fmtE(dst, neg, digs, prec-1, fmt+'e'-'g') } if prec > digs.dp { prec = digs.nd } return fmtF(dst, neg, digs, max(prec-digs.dp, 0)) } // unknown format return append(dst, '%', fmt) } // roundShortest rounds d (= mant * 2^exp) to the shortest number of digits // that will let the original floating point value be precisely reconstructed. func roundShortest(d *decimal, mant uint64, exp int, flt *floatInfo) { // If mantissa is zero, the number is zero; stop now. if mant == 0 { d.nd = 0 return } // Compute upper and lower such that any decimal number // between upper and lower (possibly inclusive) // will round to the original floating point number. // We may see at once that the number is already shortest. // // Suppose d is not denormal, so that 2^exp <= d < 10^dp. // The closest shorter number is at least 10^(dp-nd) away. // The lower/upper bounds computed below are at distance // at most 2^(exp-mantbits). // // So the number is already shortest if 10^(dp-nd) > 2^(exp-mantbits), // or equivalently log2(10)*(dp-nd) > exp-mantbits. // It is true if 332/100*(dp-nd) >= exp-mantbits (log2(10) > 3.32). minexp := flt.bias + 1 // minimum possible exponent if exp > minexp && 332*(d.dp-d.nd) >= 100*(exp-int(flt.mantbits)) { // The number is already shortest. return } // d = mant << (exp - mantbits) // Next highest floating point number is mant+1 << exp-mantbits. // Our upper bound is halfway between, mant*2+1 << exp-mantbits-1. upper := new(decimal) upper.Assign(mant*2 + 1) upper.Shift(exp - int(flt.mantbits) - 1) // d = mant << (exp - mantbits) // Next lowest floating point number is mant-1 << exp-mantbits, // unless mant-1 drops the significant bit and exp is not the minimum exp, // in which case the next lowest is mant*2-1 << exp-mantbits-1. // Either way, call it mantlo << explo-mantbits. // Our lower bound is halfway between, mantlo*2+1 << explo-mantbits-1. var mantlo uint64 var explo int if mant > 1<= d.nd { break } li := ui - upper.dp + lower.dp l := byte('0') // lower digit if li >= 0 && li < lower.nd { l = lower.d[li] } m := byte('0') // middle digit if mi >= 0 { m = d.d[mi] } u := byte('0') // upper digit if ui < upper.nd { u = upper.d[ui] } // Okay to round down (truncate) if lower has a different digit // or if lower is inclusive and is exactly the result of rounding // down (i.e., and we have reached the final digit of lower). okdown := l != m || inclusive && li+1 == lower.nd switch { case upperdelta == 0 && m+1 < u: // Example: // m = 12345xxx // u = 12347xxx upperdelta = 2 case upperdelta == 0 && m != u: // Example: // m = 12345xxx // u = 12346xxx upperdelta = 1 case upperdelta == 1 && (m != '9' || u != '0'): // Example: // m = 1234598x // u = 1234600x upperdelta = 2 } // Okay to round up if upper has a different digit and either upper // is inclusive or upper is bigger than the result of rounding up. okup := upperdelta > 0 && (inclusive || upperdelta > 1 || ui+1 < upper.nd) // If it's okay to do either, then round to the nearest one. // If it's okay to do only one, do it. switch { case okdown && okup: d.Round(mi + 1) return case okdown: d.RoundDown(mi + 1) return case okup: d.RoundUp(mi + 1) return } } } type decimalSlice struct { d []byte nd, dp int neg bool } // %e: -d.ddddde±dd func fmtE(dst []byte, neg bool, d decimalSlice, prec int, fmt byte) []byte { // sign if neg { dst = append(dst, '-') } // first digit ch := byte('0') if d.nd != 0 { ch = d.d[0] } dst = append(dst, ch) // .moredigits if prec > 0 { dst = append(dst, '.') i := 1 m := min(d.nd, prec+1) if i < m { dst = append(dst, d.d[i:m]...) i = m } for ; i <= prec; i++ { dst = append(dst, '0') } } // e± dst = append(dst, fmt) exp := d.dp - 1 if d.nd == 0 { // special case: 0 has exponent 0 exp = 0 } if exp < 0 { ch = '-' exp = -exp } else { ch = '+' } dst = append(dst, ch) // dd or ddd switch { case exp < 10: dst = append(dst, '0', byte(exp)+'0') case exp < 100: dst = append(dst, byte(exp/10)+'0', byte(exp%10)+'0') default: dst = append(dst, byte(exp/100)+'0', byte(exp/10)%10+'0', byte(exp%10)+'0') } return dst } // %f: -ddddddd.ddddd func fmtF(dst []byte, neg bool, d decimalSlice, prec int) []byte { // sign if neg { dst = append(dst, '-') } // integer, padded with zeros as needed. if d.dp > 0 { m := min(d.nd, d.dp) dst = append(dst, d.d[:m]...) for ; m < d.dp; m++ { dst = append(dst, '0') } } else { dst = append(dst, '0') } // fraction if prec > 0 { dst = append(dst, '.') for i := 0; i < prec; i++ { ch := byte('0') if j := d.dp + i; 0 <= j && j < d.nd { ch = d.d[j] } dst = append(dst, ch) } } return dst } // %b: -ddddddddp±ddd func fmtB(dst []byte, neg bool, mant uint64, exp int, flt *floatInfo) []byte { // sign if neg { dst = append(dst, '-') } // mantissa dst, _ = formatBits(dst, mant, 10, false, true) // p dst = append(dst, 'p') // ±exponent exp -= int(flt.mantbits) if exp >= 0 { dst = append(dst, '+') } dst, _ = formatBits(dst, uint64(exp), 10, exp < 0, true) return dst } // %x: -0x1.yyyyyyyyp±ddd or -0x0p+0. (y is hex digit, d is decimal digit) func fmtX(dst []byte, prec int, fmt byte, neg bool, mant uint64, exp int, flt *floatInfo) []byte { if mant == 0 { exp = 0 } // Shift digits so leading 1 (if any) is at bit 1<<60. mant <<= 60 - flt.mantbits for mant != 0 && mant&(1<<60) == 0 { mant <<= 1 exp-- } // Round if requested. if prec >= 0 && prec < 15 { shift := uint(prec * 4) extra := (mant << shift) & (1<<60 - 1) mant >>= 60 - shift if extra|(mant&1) > 1<<59 { mant++ } mant <<= 60 - shift if mant&(1<<61) != 0 { // Wrapped around. mant >>= 1 exp++ } } hex := lowerhex if fmt == 'X' { hex = upperhex } // sign, 0x, leading digit if neg { dst = append(dst, '-') } dst = append(dst, '0', fmt, '0'+byte((mant>>60)&1)) // .fraction mant <<= 4 // remove leading 0 or 1 if prec < 0 && mant != 0 { dst = append(dst, '.') for mant != 0 { dst = append(dst, hex[(mant>>60)&15]) mant <<= 4 } } else if prec > 0 { dst = append(dst, '.') for i := 0; i < prec; i++ { dst = append(dst, hex[(mant>>60)&15]) mant <<= 4 } } // p± ch := byte('P') if fmt == lower(fmt) { ch = 'p' } dst = append(dst, ch) if exp < 0 { ch = '-' exp = -exp } else { ch = '+' } dst = append(dst, ch) // dd or ddd or dddd switch { case exp < 100: dst = append(dst, byte(exp/10)+'0', byte(exp%10)+'0') case exp < 1000: dst = append(dst, byte(exp/100)+'0', byte((exp/10)%10)+'0', byte(exp%10)+'0') default: dst = append(dst, byte(exp/1000)+'0', byte(exp/100)%10+'0', byte((exp/10)%10)+'0', byte(exp%10)+'0') } return dst } func min(a, b int) int { if a < b { return a } return b } func max(a, b int) int { if a > b { return a } return b } | 2,999 | 9,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-04 | latest | en | 0.415571 |
https://forum.arduino.cc/t/multiplexing-and-maximum-arduino-amperage/367249 | 1,679,825,895,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945440.67/warc/CC-MAIN-20230326075911-20230326105911-00528.warc.gz | 316,455,302 | 7,962 | # multiplexing and maximum arduino amperage
The max pin out on Arduino it 40ma, then there are max values for combination of pins, etc... I read the entire Arduino is maxed at 200ma.
I need to do multiplexing to get 22 outs. Since multiplexing takes 3 pins from Arduino, won't this go over the amperage limits? Like if I use a 1k resistor at each pinout, then I get 10ma per pin. with 22, that's 220ma.
I am controlling a TIP121 on each pin. essentially want to control 22 TIP121.
Will I run into amperage problems?
Hi,
Your explanation is confusing. Can you post a schematic?
For example what do you mean by "multiplexing takes 3 pins from Arduino" ?
What will be connected to these 22 outputs and why do you believe 22 transistors are needed?
Paul
I am controlling 22 solenoids with darlington transistors (TIP121). Doing a type of relay.
In this example, a shift out needs 3 pins from the arduino to create 8 outs. https://www.arduino.cc/en/Tutorial/ShiftOut
I'll try to make a drawing of the schematics, but I'm not sure it's needed.
My question is pretty simple. One out of the Arduino is maximum 40ma. I'm wondering what happens when you use multiplexing to multiply those outs. Those 8 multiplexed outs, can then each have 40ma, or do I have to calculate based on the fact that they are coming from 3 pins each with 40ma max. so 40ma times 3 = 120ma for 8 pins. So now 15ma per pin?
djangojames:
I'll try to make a drawing of the schematics, but I'm not sure it's needed.
I'm pretty sure it is! Your reply has made the picture even more confusing. You are now implying the design may include shift registers driving solenoids via transistors. How does multiplexing fit into the design? These large currents sourced or sunk by the Arduino that you are so concerned about: what are they doing?
If I was designing a circuit to drive 22 solenoids, I would probably use 3 x tpic6*595 shift registers, daisy-chained. The choice of which exact version of the shift register (a, b, c etc) would be determined by the current required to drive the solenoids. If the solenoids require more than around 350mA, then transistors may be needed, in which case I would use 3 ordinary 74hc595 shift registers driving power-MOSFETs. This design involves no multiplexing, and the current demand on the Arduino would be negligible.
You sir, are confusing me!!! :o
I will be using 3 ordinary 74hc595 shift registers driving mosfets. However, I thought this was called multiplexing. Is it not? I based my assumption on this tutorial: http://www.instructables.com/id/Multiplexing-with-Arduino-and-the-74HC595/
The Arduino is not powering the solenoids, only powering the TIP121, which is why I talked about the TIP121 in the first post.
Forget the Mosfets, the solenoids, for a moment and think only of the Arduino pins.
My question is hyper simple. The Arduino has a limit of 40ma per pin (Is this not correct?). The entire Arduino itself is limited at 200ma (from the spec sheets I read). When I use a 1k resistor at the output of a pin it brings down to 5ma. But, 22 pin outs like this is 22 * 5ma, so over the 200ma limit. This is what worries me.
In other words, when you start adding outs to the Arduino, do you have to start worrying about the total current being augmented and perhaps too much to bear for the Arduino.
PaulRB:
If I was designing a circuit to drive 22 solenoids, I would probably use 3 x tpic6*595 shift registers...
Thanks for the tip about the topic*595. I will look into this.
I offer a board with 32 discrete N-channel MOSFETs, good for sinking high current. Traces sized out to support 1A/channel with all 8 channels in a group on at once. Separate blue connecter bring power in to each group of 8. Daisychainable if you need more than 32 channels.
I have a P-channel board also for sourcing current, I need to get a picture posted.
Shift data into 4 shift registers to turn channels on or off.
I thought this was called multiplexing. Is it not?
Hmmm... does using shift registers as output expanders in itself count as multiplexing? Well... might have to leave that one to the philosophers! I can imagine some definitions where it would. But it's not usually what we mean around here. We usually mean "time-division multiplexing" where leds, for example, are lit for short periods, one at a time (or one group at a time) so fast that it appears as though they are all lit together. That's not what you are doing with your solenoids.
But to your question. The Arduino pins are not powering the transistors at all. The shift register outputs are doing that. The Arduino pins are only sending digital data to the shift register inputs, and that requires almost zero current. The shift register outputs also have current limits. Around 25mA per pin and 70mA per chip. Tip120 are darlingons if i remember, so have very high gain and will need only a small current to switch them on, so you can use quite high value resistors on their bases.
PaulRB:
Hmmm... does using shift registers as output expanders in itself count as multiplexing? Well... might have to leave that one to the philosophers! I can imagine some definitions where it would. But it's not usually what we mean around here. We usually mean "time-division multiplexing" where leds, for example, are lit for short periods, one at a time (or one group at a time) so fast that it appears as though they are all lit together. That's not what you are doing with your solenoids.
But to your question. The Arduino pins are not powering the transistors at all. The shift register outputs are doing that. The Arduino pins are only sending digital data to the shift register inputs, and that requires almost zero current. The shift register outputs also have current limits. Around 25mA per pin and 70mA per chip. Tip120 are darlingons if i remember, so have very high gain and will need only a small current to switch them on, so you can use quite high value resistors on their bases.
Thanks for your clarifications, especially the one on multiplexing. I read so many tutorials called 'multiplexing' which was about using shift registers for expansion. I guess they were using the term in a very loose way.
A word of advice: in general, take anything you read on Instructables with a large pinch of salt. However, in this case, they were indeed using the shift register as part of a multiplexing circuit. 16 leds driven by only 8 shift register outputs = multiplexing. But just because a circuit involves a shift register does not mean it is a multiplexing circuit. In your case, 22 transistors driven by 22 shift register outputs = not multiplexing.
Be careful not to power these transistors/solenoids with voltage/current that passes through the Arduino board. An external power supply will be needed for that (although this could be used the power the Arduino also). Its ok to power the shift registers off the Arduino's 5V supply, they won't use much.
Thanks PaulRB, you were a great help! | 1,691 | 6,991 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-14 | latest | en | 0.934714 |
https://math.stackexchange.com/questions/4209421/alternative-series-expansion-for-the-elliptic-integral-of-the-second-kind | 1,701,526,634,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100427.59/warc/CC-MAIN-20231202140407-20231202170407-00293.warc.gz | 442,836,002 | 36,243 | # Alternative series expansion for the elliptic integral of the second kind
I found the following alternative series expansion for the complete elliptic integral of the second kind (E):
$$E(k) = \frac{(1+k')\pi}{4} \left\{ 1+\frac{1}{2^2}\left(\frac{1-k'}{1+k'}\right)^2+ \frac{1^2}{2^2\cdot 4^2} \left(\frac{1-k'}{1+k'}\right)^4 +...+ \left( \frac{(2n-3)!!}{2^n n!}\right)^2 \left(\frac{1-k'}{1+k'}\right)^{2n} +...\right\}$$
where $$k'=\sqrt{1-k^2}$$, $$\; k$$ is the modulus and $$k'$$ is the complementary modulus.
I have been trying to prove it without success. This expansion seems to be related to the Landen transformation:
$$K\left(\frac{1-k'}{1+k'}\right) = \frac{1+k'}{2}K(k)$$
$$E\left(\frac{1-k'}{1+k'}\right) = \frac{1}{1+k'}\left[E(k)+k'K(k) \right]$$
where K(k) is the complete elliptic integral of the first kind.
Here is my try:
From the last relations we have:
$$E(k) = (1+k')E\left(\frac{1-k'}{1+k'}\right) -k'K(k) = (1+k')E\left(\frac{1-k'}{1+k'}\right) -\frac{2k'}{1+k'}K\left(\frac{1-k'}{1+k'}\right)$$
Then I used the classic expansion of $$E$$ and $$K$$:
$$K(k) = \frac{\pi}{2} \sum_{n=0}^{\infty} \left[\frac{(2n-1)!!}{(2n)!!}\right]^2k^{2n}$$
$$E(k) = \frac{\pi}{2} \sum_{n=0}^{\infty} \frac{1}{1-2n} \left[\frac{(2n-1)!!}{(2n)!!}\right]^2k^{2n}$$
To get:
\begin{align*} E(k) = & (1+k')\frac{\pi}{2} \sum_{n=0}^{\infty} \frac{1}{1-2n} \left[\frac{(2n-1)!!}{(2n)!!}\right]^2\left(\frac{1-k'}{1+k'}\right)^{2n} -\frac{2k'}{1+k'} \frac{\pi}{2} \sum_{n=0}^{\infty} \left[\frac{(2n-1)!!}{(2n)!!}\right]^2\left(\frac{1-k'}{1+k'}\right)^{2n}\\ =& \frac{\pi(1+k')}{2} \sum_{n=0}^{\infty}\left\{ \left[\frac{(2n-1)!!}{(2n)!!}\right]^2\left(\frac{1-k'}{1+k'}\right)^{2n} \left(\frac{1}{1-2n} - \frac{2k'}{(1+k')^2}\right)\right\} \end{align*} but the path came to a dead end. I will really appreciate your help.
Your path is not a dead end! We have $$\frac{2k'}{(1+k')^2} = \frac{1}{2}\left[1 - \left(\frac{1-k'}{1+k'}\right)^2\right]$$, so your last line implies \begin{align} \frac{4 \operatorname{E}(k)}{\pi (1+k')} &= \sum \limits_{n=0}^\infty \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2n}\left[\frac{2}{1-2n} - 1 + \left(\frac{1-k'}{1+k'}\right)^2\right] \\ &= 1 - \sum \limits_{n=1}^\infty \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2n}\frac{2n+1}{2n-1} + \sum \limits_{n=0}^\infty \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2(n+1)} \\ &= 1 - \sum \limits_{n=1}^\infty \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2n}\frac{2n+1}{2n-1} + \sum \limits_{n=1}^\infty \left[\frac{(2n-3)!!}{(2n-2)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2n} \\ &= 1 + \sum \limits_{n=1}^\infty \left[\frac{(2n-3)!!}{(2n)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2n} \left[(2n)^2 - \frac{2n+1}{2n-1} (2n-1)^2\right] \\ &= 1 + \sum \limits_{n=1}^\infty \left[\frac{(2n-3)!!}{(2n)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2n} = \sum \limits_{n=0}^\infty \left[\frac{(2n-3)!!}{(2n)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2n}\, . \end{align} | 1,431 | 3,079 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 14, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2023-50 | latest | en | 0.582874 |
https://www.jiskha.com/display.cgi?id=1328824091 | 1,498,647,347,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323604.1/warc/CC-MAIN-20170628101910-20170628121910-00013.warc.gz | 891,018,298 | 3,726 | # Math
posted by .
Which of the following is Not true about a square?
A.It has 4 right angles
B.It has 4 congruent sides
C.It has 4 acute sides
D.Opposite sides are parallel
C?
The table shows the height of 3 different mountains
Everest=8850
Tacora=5988
Baruntse=7129
Which is the best estimate of the total height of the 3 mountains?
A.22000
B.20000
C.16000
D.14000
A?
• Math -
All right again! | 127 | 400 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2017-26 | longest | en | 0.871839 |
http://people.seas.harvard.edu/~minilek/notes/lp.txt | 1,498,269,178,000,000,000 | text/plain | crawl-data/CC-MAIN-2017-26/segments/1498128320209.66/warc/CC-MAIN-20170624013626-20170624033626-00601.warc.gz | 305,591,415 | 1,739 | -- The newer FT-mollification construction of "Bounded Independence Fools Degree-2 Theshold Functions" by Diakonikolas/Kane/Nelson implies that Omega(1/eps^p)-wise independence fools Indyk's median estimator (the log(1/eps)'s go away), and is also simpler. Also, the new FT-mollification yields a simpler proof of anticoncentration; you no longer need the complex analysis and weird "-\int_{-\infty}^x \sin^4/y^3 dy" function of Section A.4. (2/18/2010) -- Recall the algorithm used the p-stable sketch of Indyk: Ax if the vector being updated is x, where A is a k x n matrix with p-stable entries. We set k = Theta(1/eps^2). We showed that rather than having the A_{i,j} be i.i.d., each row need only have t-wise independent entries, and the rows need only be pairwise independent. This naively gives an update time of O(k*t), to evaluate each of the k different hash functions. For Indyk's median estimator we showed t = O(1/eps^p), and we gave a "log-cosine estimator" with t = O(log(1/eps)/loglog(1/eps)). In fact, the update time can just be made O(k + t). Thus, the log-cosine estimator does not improve the update time, since we already have t = o(k) for the median estimator. The reason is due to a trick that was recently exploited in [Kane-Nelson-Porat-Woodruff, CoRR abs/1007.4191] which I'll explain here. Each row needs a random hash function from a t-wise independent family mapping [n] into [q], where q = poly(mM/eps) fits in a machine word. This can be achieved by picking a random polynomial over F_q of degree t-1. Across rows, these polynomials only need to be pairwise-independent. Thus, if the polynomial corresponding to row i is P_i, we can set P_i = A*i + B, where A, B are independent random polynomials of degree t-1 over F_q. It's easy to check the P_i are now pairwise independent (view the coefficient vectors of A,B as elements of the field F_{q^t}, then note we're just evaluating a random degree-1 polynomial over this field to get the P_i). Thus, to evaluate our k different hash functions, rather than spending Theta(t) time for each of k rows, we just evaluate A,B each in Theta(t) time, then spend an additional O(1) time per row for a total of O(k + t) time. (8/4/2010) | 593 | 2,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-26 | latest | en | 0.889317 |
http://www.gurufocus.com/term/deb2equity/CSX/Debt%2Bto%2BEquity/CSX%2BCorp | 1,490,922,627,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218205046.28/warc/CC-MAIN-20170322213005-00519-ip-10-233-31-227.ec2.internal.warc.gz | 526,393,368 | 27,767 | Switch to:
GuruFocus has detected 7 Warning Signs with CSX Corp \$CSX.
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CSX Corp (NAS:CSX)
Debt-to-Equity
0.97 (As of Dec. 2016)
CSX Corp's current portion of long-term debt for the quarter that ended in Dec. 2016 was \$331 Mil. CSX Corp's long-term debt for the quarter that ended in Dec. 2016 was \$10,962 Mil. CSX Corp's total equity for the quarter that ended in Dec. 2016 was \$11,679 Mil. CSX Corp's debt to equity for the quarter that ended in Dec. 2016 was 0.97.
A high debt to equity ratio generally means that a company has been aggressive in financing its growth with debt. This can result in volatile earnings as a result of the additional interest expense.
Definition
Debt to Equity measures the financial leverage a company has.
CSX Corp's Debt to Equity Ratio for the fiscal year that ended in Dec. 2016 is calculated as
Debt to Equity = Total Debt / Total Equity = (Current Portion of Long-Term Debt + Long-Term Debt) / Total Equity = (331 + 10962) / 11679 = 0.97
CSX Corp's Debt to Equity Ratio for the quarter that ended in Dec. 2016 is calculated as
Debt to Equity = Total Debt / Total Equity = (Current Portion of Long-Term Debt + Long-Term Debt) / Total Equity = (331 + 10962) / 11679 = 0.97
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Explanation
In the calculation of Debt to Equity, we use the total of Current Portion of Long-Term Debt and Long-Term Debt divided by Total Equity. In some calculations, Total Liabilities is used to for calculation.
Be Aware
Because a company can increase its Return on Equity by having more financial leverage, it is important to watch the leverage ratio when investing in high Return on Equity companies.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
CSX Corp Annual Data
Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Dec16 deb2equity 0.84 0.97 0.92 1.00 1.09 1.08 0.91 0.87 0.90 0.97
CSX Corp Quarterly Data
Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 deb2equity 0.87 0.87 0.86 0.88 0.87 0.90 0.91 0.91 0.90 0.97
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GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More) | 680 | 2,527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-13 | latest | en | 0.951859 |
https://cosmicreflections.skythisweek.info/tag/declination/ | 1,713,717,712,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817790.98/warc/CC-MAIN-20240421163736-20240421193736-00614.warc.gz | 159,436,704 | 14,981 | ## Geostationary Satellite Declinations
A few years ago, I was doing some telescope sweeping of the meridian sky around declination -6˚ when, to my surprise and delight, a 10th- or 11th-magnitude slow-moving object entered my field of view. As it slowly traversed eastward through the field, I remembered the declination I was pointed to and realized that it must be a geostationary, or at least a geosynchronous, satellite. Centering the moving object and then turning off the telescope’s clock drive confirmed my suspicions. The object was a geosynchronous satellite because it appeared to lay motionless while all the stars in the field drifted toward the west. Serendipity is the spice of life!
Satellites stationed in orbits that are always directly above the Earth’s equator and that have an orbital period of 23h 56m 04.0905s (one sidereal day) have the interesting property of remaining stationary as seen from any point on the surface of the Earth. This property of geostationary satellites, as they are called, is used to great advantage by many communications and weather satellites. There are currently at least 554 satellites in geosynchronous orbits. They are stationed all around the Earth at various longitudes.
At what altitude do geostationary satellites orbit the Earth? It is well above human-occupied spacecraft like the International Space Station which currently orbits 260 miles above the Earth’s surface. Geosynchronous orbit lies some 22,236 miles above the Earth’s equator. This is quite a ways out, as the entire Earth subtends an angle of only 17° 12′ at this distance—about the same as the angular distance between Capella (α Aur) and Elnath (β Tau).
Looking at it another way, geostationary satellites orbit at an altitude that is 2.8 Earth diameters above the equator. Since the Moon orbits at a distance that ranges between 27.4 and 31.4 Earth diameters above the Earth’s surface, geosynchronous orbit is about 1/10 of the way to the Moon.
If you have a telescope, know where to point it, and turn tracking off, you can see a geostationary satellite as a stationary point of light while the stars drift by due to the Earth’s rotation. At our latitude here in southern Wisconsin (43° N), the area where you want to search for geostationary satellites (near the meridian) is around declination -6° 37′. Remember, declination tells you how many degrees above or below the celestial equator an object is, and the numbers range from -90° to +90°, the south celestial pole and north celestial pole, respectively. The celestial equator has a declination of 0°.
For any latitude1, the declination you want to search is given by
$\delta _{gs}=\textup{tan}^{-1}\left [ 6.611\textup{ csc }\phi - \textup{cot }\phi \right ]-90^{\circ}$
where δgs is the declination of the geostationary satellite in degrees
and ϕ is your latitude in degrees
Since most calculators don’t have the cosecant (csc) or cotangent (cot) functions, this formula can be rewritten in a slightly more complicated form as
$\delta _{gs}=\textup{tan}^{-1}\left [ \frac{6.611}{\textup{sin }\phi }-\frac{1}{\textup{tan }\phi } \right ]-90^{\circ}$
Why aren’t the satellites right on the celestial equator (δ = 0°)? They would be if they were millions of miles away or if we were located on the Earth’s equator, but at our northern latitude trigonometric parallax causes us to see the satellites somewhat below the celestial equator, relative to the distant stars.
What if the geostationary satellite is situated east or west of your meridian? How do you calculate its declination then? As you might expect, because the range (observer-to-satellite distance) is greater the further from the meridian the satellite is, the less the parallax is, and therefore the closer the declination is to the equator, though not by a lot. The declination is also symmetric about the meridian, east and west: a geostationary satellite one hour east of the meridian will have the same declination as another geostationary satellite one hour west of the meridian.
If you know the longitude of the geostationary satellite (for example, the GOES-16 weather satellite is stationed above 75.2˚ W longitude), you can calculate its declination (and right ascension) using the following two-step process.
$\textup{h}=\textup{tan}^{-1}\left [ \frac{\textup{sin }\Delta\lambda }{\textup{cos }\Delta \lambda-0.15126\textup{ cos }\phi } \right ]$
where h is the hour angle in degrees
and Δλ = λsat − λobs , the difference between the satellite and observer
longitudes, in degrees
and ϕ is the latitude of the observer in degrees
$\delta _{gs}=\textup{tan}^{-1}\left [ \frac{-0.15126\textup{ sin }\phi \textup{ sin h}}{\textup{sin }\Delta \lambda } \right ]$
To determine the right ascension of the geostationary satellite, add the value of h to your local sidereal time (the right ascension of objects on your meridian). Make sure you convert h to hours before adding it to your LST.
What if you want to calculate the geostationary declination at a particular hour angle? That is a bit trickier. I could not figure out how to manipulate the equation for h above so that Δλ = f (h,φ). Instead, I rewrote the equation as
$\sin \Delta \lambda =\tan h\cdot \left ( \cos \Delta \lambda -0.15126\cos \phi \right )$
and using h as a starting value for Δλ, substituted it into the cos Δλ expression, calculated sin Δλ, took the arcsine to get a new value of Δλ, then substituted that back into the cos Δλ expression, and iterated. Fortunately, the value of Δλ converges very fast. Once you have Δλ, you can use the two-step process we used earlier to determine the declination of the geostationary satellite for a particular hour angle.
Please note that the value of the hour angle h we use here is positive east of the meridian and negative west of the meridian. This is opposite from the normal astronomical sense.
Here is a simple SAS program illustrating how to do all these calculations using a computer.
And here is the output from that program.
1 For latitudes south of the equator, add 180° to get your meridian geostationary declination. The equation goes singular at the equator (φ=0°) and at the poles (φ=90° N and 90° S) since we’re dividing by sin φ = 0 at the equator and tan φ is undefined at the poles. However, as you asymptotically get closer and closer to latitude 0° (0.0001° and -0.0001°, for example) you find that the meridian geostationary declination approaches δ = 0°. Likewise, as you asymptotically approach latitude 90° N and 90° S, you’ll find that the meridian geostationary declination approaches -8°36′ and +8°36′, respectively. Of course, in both cases the geostationary satellites always remain below your horizon. How far north or south in latitude would you have to go, then, to find that geostationary satellites on your meridian are on your horizon due south or due north, respectively? Through a little algebraic manipulation of the first equation above and utilizing some simple trigonometric identities, one finds that at latitudes 81°18′ N and 81°18′ S, geostationary satellites on your meridian would be on the horizon. North or south of there, respectively, you would not be able to see them because the Earth would be in the way.
References
Gérard Maral, Michel Bousquet, Zhili Sun. Satellite Communications Systems: Systems, Techniques and Technology, Fifth Edition. Wiley, 2009. See section 8.3.6.3 Polar mounting.
## Thank the Sumerians
Over five thousand years ago, the Sumerians in the area now known as southern Iraq appear to have been the first to develop a penchant for the numbers 12, 24, 60 and 360.
It is easy to see why. 12 is the first number that is evenly divisible by six smaller numbers:
12 = 1×12, 2×6, 3×4 .
24 is the first number that is evenly divisible by eight smaller numbers:
24 = 1×24, 2×12, 3×8, 4×6 .
60 is the first number than is evenly divisible by twelve smaller numbers:
60 = 1×60, 2×30, 3×20, 4×15, 5×12, 6×10 .
And 360 is the first number that is evenly divisible by twenty-four smaller numbers:
360 = 1×360, 2×180, 3×120, 4×90, 5×72, 6×60, 8×45, 9×40, 10×36, 12×30, 15×24, 18×20 .
And 360 in a happy coincidence is just 1.4% short of the number of days in a year.
We have 12 hours in the morning, 12 hours in the evening.
We have 24 hours in a day.
We have 60 seconds in a minute, and 60 minutes in an hour.
We have 60 arcseconds in an arcminute, 60 arcminutes in a degree, and 360 degrees in a circle.
The current equatorial coordinates for the star Vega are
α2019.1 = 18h 37m 33s
δ2019.1 = +38° 47′ 58″
Due to precession, the right ascension (α) of Vega is currently increasing by 1s (one second of time) every 37 days, and its declination (δ) is currently decreasing by 1″ (one arcsecond) every 5 days.
With right ascension, the 360° in a circle is divided into 24 hours, therefore 1h is equal to (360°/24h) = 15°. Since there are 60 minutes in an hour and 60 seconds in a minute, and 60 arcminutes in a degree and 60 arcseconds in an arcminute, it follows that 1m = 15′ and 1s = 15″.
Increasingly, you will see right ascension and declination given in decimal, rather than sexagesimal, units. For Vega, currently, this would be
α2019.1 = 18.62583h
δ2019.1 = +38.7994°
Or, both in degrees
α2019.1 = 279.3875°
δ2019.1 = +38.7994°
Or even radians
α2019.1 = 4.876232 rad
δ2019.1 = 0.677178 rad
Even though the latter three forms lend themselves well to computation, I still prefer the old sexagesimal form for “display” purposes, and when entering coordinates for “go to” at the telescope.
There is something aesthetically appealing about three sets of two-digit numbers, and, I think, this form is more easily remembered from one moment to the next.
For the same reason, we still use the sexagesimal form for timekeeping. For example, as I write this the current time is 12:25:14 a.m. which is a more attractive (and memorable) way to write the time than saying it is 12.4206 a.m. (unless you are doing computations).
That’s quite an achievement, developing something that is still in common use 5,000 years later.
Thank the Sumerians! | 2,668 | 10,126 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 5, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-18 | latest | en | 0.926152 |
http://1howmany.com/how-to-divide-fractions | 1,553,601,304,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912205163.72/warc/CC-MAIN-20190326115319-20190326141319-00277.warc.gz | 1,268,705 | 6,335 | How to Divide Fractions
www.1howmany.com
Education & Reference / How to Divide Fractions
How to Divide Fractions
Dividing a whole number by a whole number is usually not a very difficult task, and almost everybody knows how to do it. But when it is necessary to divide fractions1, it can be a thorny problem for many people. For instance, do you know how to solve something like this: 3/5 ÷ 4/9? Do not feel ashamed, if you cannot do it. Keep reading this article and we will teach you how to divide fractions.
In fact, there are several ways of dividing fractions and we will explain you the two easiest ones.
Inverting and Multiplying
In order to divide fractions, using this method, do the following:
• First of all, you have to invert the fraction (just flip it over) of the divisor. You will get what is called "the reciprocal"2. E.g. 2/5 becomes 5/2.
• Multiply the fractions. We hope you know how to multiply fractions. All you have to do is to multiply the denominators3 and numerators4.
• If possible, convert them to a mixed number. E.g. 13/8 will be 1 5/8.
For example: 2/3÷2/5 = 2/3*5/2 = 10/6 = 1 4/6 = 1 2/3
Common Denominator
Another way of dividing fractions is using a common denominator. This method is lengthier but it can be more useful in everyday life. It includes the following steps:
• You should find a common denominator. Then you need to convert the fractions to the equivalent value.
• Now then the denominators are equal it is possible to divide the numerators.
E.g.: 4/5÷2/3 = 12/15÷10/15 = 12/10 = 1 1/5
Now you know two ways of dividing fractions. If you are going to teach your child how to divide a fraction by a fraction, you should better explain him the first method, because it is really much easier.
You may be interested in: | 465 | 1,778 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2019-13 | latest | en | 0.911976 |
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A102007 Indices of primes in sequence defined by A(0) = 17, A(n) = 10*A(n-1) - 63 for n > 0. 1
0, 1, 3, 7, 8, 23, 59, 109, 133, 221, 411, 699, 998, 1382, 5075, 5542, 6343, 14599, 15092, 21716, 23635, 30220, 50710 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 COMMENTS Numbers n such that 10*10^n + 7 is prime. Numbers n such that digit 1 followed by n >= 0 occurrences of digit 0 followed by digit 7 is prime. Numbers corresponding to terms <= 998 are certified primes. a(24) > 2*10^5. - Robert Price, Nov 09 2015 REFERENCES Klaus Brockhaus and Walter Oberschelp, Zahlenfolgen mit homogenem Ziffernkern, MNU 59/8 (2006), pp. 462-467. LINKS Makoto Kamada, Prime numbers of the form 100...007. FORMULA a(n) = A088274(n) - 1. EXAMPLE 10007 is prime, hence 3 is a term. MATHEMATICA Select[Range[0, 100000], PrimeQ[10*10^# + 7] &] (* Robert Price, Nov 09 2015 *) PROG (PARI) a=17; for(n=0, 1500, if(isprime(a), print1(n, ", ")); a=10*a-63) (PARI) for(n=0, 1500, if(isprime(10*10^n+7), print1(n, ", "))) CROSSREFS Cf. A000533, A002275, A088274. Sequence in context: A268111 A244532 A037208 * A152486 A301523 A105756 Adjacent sequences: A102004 A102005 A102006 * A102008 A102009 A102010 KEYWORD nonn,hard,more AUTHOR Klaus Brockhaus and Walter Oberschelp (oberschelp(AT)informatik.rwth-aachen.de), Dec 28 2004 EXTENSIONS More terms from Herman Jamke (hermanjamke(AT)fastmail.fm), Jan 02 2008 STATUS approved
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Last modified October 23 03:21 EDT 2019. Contains 328335 sequences. (Running on oeis4.) | 659 | 1,894 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2019-43 | latest | en | 0.596542 |
https://mathhelpboards.com/threads/neighbourhood-of-convergence-of-sequence.7453/ | 1,618,926,470,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039398307.76/warc/CC-MAIN-20210420122023-20210420152023-00155.warc.gz | 493,394,825 | 20,300 | # [SOLVED]Neighbourhood of Convergence of Sequence
#### Sudharaka
##### Well-known member
MHB Math Helper
Hi everyone,
Can somebody give me a hint to solve this problem.
Problem:
Let $$f$$ be a function defined on $$[a,\,b]$$ with continuous second order derivative. Let $$x_0\in (a,\,b)$$ satisfy $$f(x_0)=0$$ but $$f'(x_0)\neq 0$$. Prove that, there is a neighbourhood of $$x_0$$, say $$U(x_0)$$, such that, for all $$x_1\in U(x_0)$$, the following sequence,
$x_{n+1}=x_{n}-\frac{f(x_n)}{f'(x_n)}$
where $$n=1,\,2,\,\cdots$$ is convergent.
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Hi everyone,
Can somebody give me a hint to solve this problem.
Problem:
Let $$f$$ be a function defined on $$[a,\,b]$$ with continuous second order derivative. Let $$x_0\in (a,\,b)$$ satisfy $$f(x_0)=0$$ but $$f'(x_0)\neq 0$$. Prove that, there is a neighbourhood of $$x_0$$, say $$U(x_0)$$, such that, for all $$x_1\in U(x_0)$$, the following sequence,
$x_{n+1}=x_{n}-\frac{f(x_n)}{f'(x_n)}$
where $$n=1,\,2,\,\cdots$$ is convergent.
Use Taylor's theorem:
$$f(x_n) = f(x_0) + f'(x_0)(x_n - x_0) + \frac 1 {2!} f''(\xi_n)(x_n - x_0)^2$$
where $\xi_n$ is between $x_0$ and $x_n$.
Let $\varepsilon_n = x_0 - x_n$.
This is the error with respect to the root of f in iteration n.
Consider the error $\varepsilon_{n+1}$ in terms of $\varepsilon_n$.
If it approaches zero, you're done.
#### Sudharaka
##### Well-known member
MHB Math Helper
Use Taylor's theorem:
$$f(x_n) = f(x_0) + f'(x_0)(x_n - x_0) + \frac 1 {2!} f''(\xi_n)(x_n - x_0)^2$$
where $\xi_n$ is between $x_0$ and $x_n$.
Let $\varepsilon_n = x_0 - x_n$.
This is the error with respect to the root of f in iteration n.
Consider the error $\varepsilon_{n+1}$ in terms of $\varepsilon_n$.
If it approaches zero, you're done.
Thanks very much for the reply but I am not sure whether I get you. If the error approaches zero then the sequence converges. But how does that guarantee the existence of a neighbourhood $$U(x_0)$$ where each element $$x_1$$ makes the sequence convergent?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Thanks very much for the reply but I am not sure whether I get you. If the error approaches zero then the sequence converges. But how does that guarantee the existence of a neighbourhood $$U(x_0)$$ where each element $$x_1$$ makes the sequence convergent?
If you work it out, you'll find there are some boundary conditions.
To satisfy those boundary conditions you need to take $x_1$ "close enough" to $x_0$, or equivalently $\varepsilon_1$ "close enough" to 0.
This is represented by a neighbourhood $U(x_0)$ that is "small enough".
#### chisigma
##### Well-known member
Hi everyone,
Can somebody give me a hint to solve this problem.
Problem:
Let $$f$$ be a function defined on $$[a,\,b]$$ with continuous second order derivative. Let $$x_0\in (a,\,b)$$ satisfy $$f(x_0)=0$$ but $$f'(x_0)\neq 0$$. Prove that, there is a neighbourhood of $$x_0$$, say $$U(x_0)$$, such that, for all $$x_1\in U(x_0)$$, the following sequence,
$x_{n+1}=x_{n}-\frac{f(x_n)}{f'(x_n)}$
where $$n=1,\,2,\,\cdots$$ is convergent.
A counterexample shoul be $\displaystyle f(x)= x^{\frac{1}{3}}$ because the difference equation becomes...
$\displaystyle \Delta_{n} = x_{n+1} - x_{n} = - 3\ x_{n}\ (1)$
... that diverges for any $x_{1} \ne 0$...
Kind regards
$\chi$ $\sigma$
#### Klaas van Aarsen
##### MHB Seeker
Staff member
A counterexample shoul be $\displaystyle f(x)= x^{\frac{1}{3}}$ because the difference equation becomes...
It doesn't satisfy the criteria.
f'(0) is not defined, so it does not have a continuous second order derivative around x=0.
#### Sudharaka
##### Well-known member
MHB Math Helper
If you work it out, you'll find there are some boundary conditions.
To satisfy those boundary conditions you need to take $x_1$ "close enough" to $x_0$, or equivalently $\varepsilon_1$ "close enough" to 0.
This is represented by a neighbourhood $U(x_0)$ that is "small enough".
Thanks, now I am understanding it more clearly. So we can get the $$\epsilon_{n+1}$$ in terms of $$\epsilon_n$$ as mentioned >>here<<.
$\epsilon_{n+1}=-\frac{f''(\gamma_n)}{2f'(x_n)}\epsilon_{n}$
I hope I am correct up to this point. Am I?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Thanks, now I am understanding it more clearly. So we can get the $$\epsilon_{n+1}$$ in terms of $$\epsilon_n$$ as mentioned >>here<<.
$\epsilon_{n+1}=-\frac{f''(\gamma_n)}{2f'(x_n)}\epsilon_{n}$
I hope I am correct up to this point. Am I?
Almost.
You dropped a square.
Note that you've already used that both f' and f'' exist, and that $f'(x_n)\ne 0$.
Are you aware of the conditions involved?
And I see you picked $\gamma_n$. Didn't you like $\xi_n$?
#### Sudharaka
##### Well-known member
MHB Math Helper
Almost.
You dropped a square.
Note that you've already used that both f' and f'' exist, and that $f'(x_n)\ne 0$.
Are you aware of the conditions involved?
Sorry, that's a typo. It should be,
$\epsilon_{n+1}=-\frac{f''(\gamma_n)}{2f'(x_n)}\epsilon_{n}^2$
It's given that $$f$$ is twice differentiable. Hence $$f'$$ and $$f''$$ exists. But I thought that $$f'(x_n)\neq 0$$ is implied through the equation generating the terms of the sequence. What I felt from the beginning when solving this problem is how to incoperate the fact that $$f'(x_0)\neq 0$$.
And I see you picked $\gamma_n$. Didn't you like $\xi_n$?
No I hate that symbol. First I never can write it properly and in this context I don't like to use it because it looks like $$\epsilon$$ and there's a chance I will confuse the two.
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Sorry, that's a typo. It should be,
$\epsilon_{n+1}=-\frac{f''(\gamma_n)}{2f'(x_n)}\epsilon_{n}^2$
It's given that $$f$$ is twice differentiable. Hence $$f'$$ and $$f''$$ exists. But I thought that $$f'(x_n)\neq 0$$ is implied through the equation generating the terms of the sequence. What I felt from the beginning when solving this problem is how to incoperate the fact that $$f'(x_0)\neq 0$$.
It's the other way around.
You can only use $f'(x_n)\neq 0$ because you have $f'(x_0)\neq 0$.
Due to the fact that f' is continuous and that $f'(x_0)\neq 0$, you can infer that there will be some open interval around $x_0$ that will have $f'(x) \ne 0$ for x in that interval.
Only within that interval can you use the Taylor expansion as given.
#### Sudharaka
##### Well-known member
MHB Math Helper
It's the other way around.
You can only use $f'(x_n)\neq 0$ because you have $f'(x_0)\neq 0$.
Due to the fact that f' is continuous and that $f'(x_0)\neq 0$, you can infer that there will be some open interval around $x_0$ that will have $f'(x) \ne 0$ for x in that interval.
Only within that interval can you use the Taylor expansion as given.
This occurred me previously but I was confused by the fact that if we take that interval how do we know for sure that all the values $$x_n$$ lie in that interval? That is we choose $$x_1$$ from that interval, then calculate the value $$x_2$$. Now how can we guarantee that $$x_2$$ also lie in that same interval?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
This occurred me previously but I was confused by the fact that if we take that interval how do we know for sure that all the values $$x_n$$ lie in that interval? That is we choose $$x_1$$ from that interval, then calculate the value $$x_2$$. Now how can we guarantee that $$x_2$$ also lie in that same interval?
That happens if we can make sure that $|\varepsilon_{n+1}| < |\varepsilon_n|$.
#### Sudharaka
##### Well-known member
MHB Math Helper
That happens if we can make sure that $|\varepsilon_{n+1}| < |\varepsilon_n|$.
Yes, I think I am getting a hold of this. So we have the inequality,
$|\epsilon_{n+1}|=\left|\frac{f''(\gamma_n)}{2f'(x_n)}\right||\epsilon_{n}|^2$
To get $|\varepsilon_{n+1}| < |\varepsilon_n|$, we should have,
$|\epsilon_{n}|<\left|\frac{2f'(x_n)}{f''(\gamma_n)}\right|$
And this is the interval that we are looking for. Am I correct?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Yes, I think I am getting a hold of this. So we have the inequality,
$|\epsilon_{n+1}|=\left|\frac{f''(\gamma_n)}{2f'(x_n)}\right||\epsilon_{n}|^2$
To get $|\varepsilon_{n+1}| < |\varepsilon_n|$, we should have,
$|\epsilon_{n}|<\left|\frac{2f'(x_n)}{f''(\gamma_n)}\right|$
And this is the interval that we are looking for. Am I correct?
Yes.
With the additional constraints that we're inside the interval $(a,b)$ and that $f'(x) \ne 0$.
Also note that $f''(\gamma_n)$ could be zero, so we have to allow for that.
#### Sudharaka
##### Well-known member
MHB Math Helper
Yes.
With the additional constraints that we're inside the interval $(a,b)$ and that $f'(x) \ne 0$.
Also note that $f''(\gamma_n)$ could be zero, so we have to allow for that.
Well, so one last question. We have to assume that $$f''(\gamma_n)\neq 0$$. This is just something we have to assume and cannot be deducted from the given details. Am I correct?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Well, so one last question. We have to assume that $$f''(\gamma_n)\neq 0$$. This is just something we have to assume and cannot be deducted from the given details. Am I correct?
Not really.
If f''(x) = 0, we have instantaneous convergence to the root.
#### Sudharaka
##### Well-known member
MHB Math Helper
Not really.
If f''(x) = 0, we have instantaneous convergence to the root.
And that I believe tells us $$f$$ is a straight line. Isn't? But what's wrong with that?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
And that I believe tells us $$f$$ is a straight line. Isn't? But what's wrong with that?
Yep and nothing's wrong with that.
And btw, it is not given that f''(x) = 0 everywhere.
What it does mean, is that $|\varepsilon_{n+1}| < |\varepsilon_{n}|$ if $\varepsilon_{n} \ne 0$, which is what we wanted.
#### Sudharaka
##### Well-known member
MHB Math Helper
Yep and nothing's wrong with that.
And btw, it is not given that f''(x) = 0 everywhere.
What it does mean, is that $|\varepsilon_{n+1}| < |\varepsilon_{n}|$ if $\varepsilon_{n} \ne 0$, which is what we wanted.
Yes, I am sorry, too tired to understand that $$f''(\gamma)=0$$ does not mean that $$f$$ is a straight line. However if $$f''(\gamma)=0$$ then $$x_{n+1}=x_0$$ so obviously,
$0=|\varepsilon_{n+1}| < |\varepsilon_{n}|$
I guess this is what you meant. Am I correct?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Yes, I am sorry, too tired to understand that $$f''(\gamma)=0$$ does not mean that $$f$$ is a straight line. However if $$f''(\gamma)=0$$ then $$x_{n+1}=x_0$$ so obviously,
$0=|\varepsilon_{n+1}| < |\varepsilon_{n}|$
I guess this is what you meant. Am I correct?
Yes.
#### Sudharaka
##### Well-known member
MHB Math Helper
Thank you sooooooooooooooooooo much for all your help. I think I understood every bit and piece of the problem, though it took a considerable amount of time. | 3,410 | 10,961 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2021-17 | longest | en | 0.816911 |
https://startup-house.com/glossary/what-is-linear-regression-analysis | 1,726,228,964,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651513.89/warc/CC-MAIN-20240913101949-20240913131949-00075.warc.gz | 505,408,774 | 53,386 | what is linear regression analysis
# Linear Regression Analysis
Linear Regression Analysis is a statistical method used to establish a relationship between two continuous variables, where one variable is considered the dependent variable, and the other variable is considered the independent variable. The goal of linear regression analysis is to find the best linear equation that describes how the dependent variable changes as the independent variable changes.
The linear regression equation is represented as y = mx + b, where y is the dependent variable, x is the independent variable, m is the slope of the line, and b is the y-intercept. The slope of the line represents the rate of change of the dependent variable with respect to the independent variable.
Linear regression analysis is used to make predictions about the dependent variable based on the values of the independent variable. The accuracy of the predictions is measured by the coefficient of determination, which is represented by R². The coefficient of determination ranges from 0 to 1, where 0 indicates that the independent variable has no effect on the dependent variable, and 1 indicates that the independent variable completely explains the variation in the dependent variable.
Linear regression analysis is commonly used in various fields, including finance, economics, social sciences, and engineering. In finance, linear regression analysis is used to predict stock prices based on economic indicators such as interest rates and GDP. In economics, linear regression analysis is used to model the relationship between demand and supply. In social sciences, linear regression analysis is used to study the relationship between variables such as income and education. In engineering, linear regression analysis is used to predict the performance of machines based on their design parameters.
There are different types of linear regression analysis, including simple linear regression and multiple linear regression. Simple linear regression involves only one independent variable, while multiple linear regression involves two or more independent variables. Multiple linear regression is used to model complex relationships between variables and is commonly used in marketing and advertising to predict consumer behavior.
In conclusion, linear regression analysis is a powerful statistical method used to establish a relationship between two continuous variables. It is a widely used tool in various fields and is essential for making predictions and understanding the relationship between variables.
Let's talk
## .css-w0mk3l{color:var(--chakra-colors-purple-500);}Let's buildsomething together
Startup Development House sp. z o.o.
Aleje Jerozolimskie 81
Warsaw, 02-001
VAT-ID: PL5213739631
KRS: 0000624654
REGON: 364787848 | 516 | 2,816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-38 | latest | en | 0.934814 |
http://ricardo.ecn.wfu.edu/~cottrell/OPE/archive/9610/0319.html | 1,511,549,005,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934808742.58/warc/CC-MAIN-20171124180349-20171124200349-00475.warc.gz | 270,030,627 | 1,953 | # [OPE-L:3514] RE: Productive and Unproductive Labour
Paul Cockshott (wpc@cs.strath.ac.uk)
Thu, 24 Oct 1996 01:22:57 -0700 (PDT)
[ show plain text ]
>A reply to Simon's ope-l 3506, which I found very interesting.
>
>Simon: "Consider the l vector in the standard value equations. This refers to
>'value-creating labour-power hired'."
>
>I don't think this is the case. In Marx's theory, labor-power does not
>produce value. Labor does. IMO, the l vector indicates the amounts of
>productive (value-creating) *labor* extracted per unit of output. We have, in
>money terms, elx = V + S, where e (dimensions: money unit/labor-time)
>indicates the monetary expression of value. V is the money wage bill paid to
>workers who perform productive labor: V = w(lpp)x, where w is the money wage
>per-unit of labor-power (assume it is scalar for simplicity), and lpp is a
>vector (for simplicity) of productive labor-power hired per unit of output.
>
>Note that lpp can only be known ex post. It is not purely a labor market
>relation *or* a technological relation. It depends on the amounts of
>labor-power hired and the amounts of outputs produced. There is no necessary
>relation between these amounts. The relation depends on the degree of
>exploitation in production.
>
PaulC:
It depends upon how you construct the i/o matrix. If, it is constructed on
a per industry basis, then the intsensity of labour is irrelevant
and the labour power and labour vectors can be used indistinguishably.
This is because wat counts as simple labour is labour of average intensity
and skill, which can only be meaninfully defined within the context of a
single industry. If on the other hand the input output table is
constructed on a per firm basis then the intensity of labour is very relevant,
since the firms with the more intense labour will tend to have a higher physical
productivity and higher profits.
Paul Cockshott
wpc@cs.strath.ac.uk
http://www.cs.strath.ac.uk/CS/Biog/wpc/index.html | 497 | 1,977 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2017-47 | longest | en | 0.893748 |
http://research.stlouisfed.org/fred2/graph/?chart_type=line&s[1][id]=ROCH336MFG&log_scales=Left | 1,408,588,142,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500813887.15/warc/CC-MAIN-20140820021333-00095-ip-10-180-136-8.ec2.internal.warc.gz | 160,886,820 | 17,309 | # Graph: All Employees: Manufacturing in Rochester, NY (MSA)
Click and drag in the plot area or select dates: Select date: 1yr | 5yr | 10yr | Max to
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w h
Graph Background: Plot Background: Text:
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(a) All Employees: Manufacturing in Rochester, NY (MSA), Thousands of Persons, Seasonally Adjusted (ROCH336MFG)
The data services of the Federal Reserve Bank of St. Louis include series that are seasonally adjusted. To make these adjustments, we use the X-12 Procedure of SAS to remove the seasonal component of the series so that non-seasonal trends can be analyzed. This procedure is based on the U.S. Bureau of the Census X-12-ARIMA Seasonal Adjustment Program. More information on this program can be found at http://www.census.gov/srd/www/x12a/.
The seasonal moving average function used is that of the Census Bureau’s X-11-ARIMA program. This includes a 3x3 moving average for the initial seasonal factors and a 3x5 moving average to calculate the final seasonal factors. The D11 function is also used to output the entire seasonally adjusted series that is displayed. For specific information on the SAS X-12 procedure, please visit their website: http://support.sas.com/documentation/cdl/en/etsug/60372/HTML/default/viewer.htm#etsug_x12_sect001.htm.
All Employees: Manufacturing in Rochester, NY (MSA)
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Name: Email: | 568 | 2,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2014-35 | latest | en | 0.869493 |
https://discourse.codecombat.com/t/polygonception-someone-can-help/17952 | 1,713,639,909,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817674.12/warc/CC-MAIN-20240420184033-20240420214033-00582.warc.gz | 174,237,138 | 7,379 | Polygonception - someone can help?
Who can help?every time only two yaks are successfully encircled，and I can’t pass this level
``````# You are on your own this time, I hope you have learned what you need from the previous fractal levels. Check the guide for help with what you need to do and with the math required for polygons.
# You need a function to convert degrees to radians. Multiply degrees by Math.PI / 180.
return Math.PI / 180 * angle
# Your polygon function should have 3 inputs: start, end, and sides.
def line(start,end,n,k):
side = Vector.subtract(end,start)
distance = side.magnitude()
if distance < k / n + 1:
hero.toggleFlowers(False)
hero.moveXY(start.x, start.y)
hero.toggleFlowers(True)
hero.moveXY(end.x, end.y)
return
fenxing = Vector.divide(side, n)
for i in range(n):
rotate = Vector.rotate(fenxing, degreesToRadians(360 * i / n))
line(start,A,n,k)
start = A
# Remember to make your polygon recursive, drawing extra polygons at every corner.
# To get the start and end position for each polygon, add startOffset and endOffset to the yak's position.
def flake(a,b,n):
side = Vector.subtract(b,a)
k = side.magnitude()
for i in range(n):
line(a,b,n,k)
hero.toggleFlowers(True)
hero.moveXY(b.x, b.y)
a = b
startOffset = Vector(-15, -15)
endOffset = Vector(15, -15)
# You need to loop through all the yaks, drawing a polygon for each. Yaks are enemies.
enemies = hero.findEnemies()
for enemy in enemies:
sides = enemy.sides
vecEnemy = Vector(enemy.pos.x,enemy.pos.y)
flake(start,end,sides)
``````
Can someone help this fella here?
-@Chaboi_3000
Sorry @jxcwzk I’m terrible at recursion. I know @Chaboi_3000 has done this level.
1 Like
I’ll get to this topic once I’m back from my trip.
I was fainted and This level does afflict people
Just be patient, and we’ll work on this when we have the time.
@jxcwzk Were you able to complete this level? I changed two lines of code on yours and did get all four yaks. I can’t necessarily explain why it worked exactly, but it did.
not yet…and i studying Raspberry Pi
The hints say to have the distance less than 2, but I can’t seem to get that to work. It ends up adding extra recursive and I saw another post commenting on the same thing.
The two lines I changed to make it work:
``````if distance < k / n + 1: # changed to < 10: | 610 | 2,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-18 | latest | en | 0.776532 |
http://mathhelpforum.com/advanced-algebra/74055-orthogonal-matrix-print.html | 1,524,182,293,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937074.8/warc/CC-MAIN-20180419223925-20180420003925-00059.warc.gz | 195,621,847 | 2,877 | Orthogonal Matrix
• Feb 17th 2009, 03:40 AM
mr_motivator
Orthogonal Matrix
• Feb 17th 2009, 03:53 AM
Mush
Quote:
Originally Posted by mr_motivator
An Orthogonal matrix is one whos transpose is its inverse.
In other words:
$\displaystyle AA^T = I = A^TA$
For your matrix, the transpose is:
$\displaystyle \left( \begin{array}{ccc} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ m & -\frac{\sqrt{3}}{2} \end{array} \right)$
So multiply it by the original matrix
$\displaystyle \left( \begin{array}{ccc} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ m & -\frac{\sqrt{3}}{2} \end{array} \right)\left( \begin{array}{ccc} \frac{\sqrt{3}}{2} & m \\ \frac{1}{2} & -\frac{\sqrt{3}}{2} \end{array} \right) $$\displaystyle = \left( \begin{array}{ccc} 1 & 0 \\ 0 & 1 \end{array} \right) \displaystyle \left( \begin{array}{ccc} \bigg(\frac{\sqrt{3}}{2}\bigg)^2+\bigg(\frac{1}{2} \bigg)^2 & \frac{\sqrt{3}}{2}m - \frac{\sqrt{3}}{4} \\ \frac{\sqrt{3}}{2}m - \frac{\sqrt{3}}{4} & m^2+\bigg(\frac{\sqrt{3}}{2}\bigg)^2 \end{array} \right)$$\displaystyle = \left( \begin{array}{ccc} 1 & 0 \\ 0 & 1 \end{array} \right)$
$\displaystyle \left( \begin{array}{ccc} 1 & \frac{\sqrt{3}}{2}m - \frac{\sqrt{3}}{4} \\ \frac{\sqrt{3}}{2}m - \frac{\sqrt{3}}{4} & m^2+\bigg(\frac{\sqrt{3}}{2}\bigg)^2 \end{array} \right)$$\displaystyle = \left( \begin{array}{ccc} 1 & 0 \\ 0 & 1 \end{array} \right)$
Hence $\displaystyle \frac{\sqrt{3}}{2}m - \frac{\sqrt{3}}{4} =0$ and $\displaystyle m^2+\bigg(\frac{\sqrt{3}}{2}\bigg)^2 = 1$ | 631 | 1,483 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-17 | latest | en | 0.402229 |
https://justaaa.com/economics/108570-when-the-monthly-average-premium-for-life | 1,721,238,596,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514789.19/warc/CC-MAIN-20240717151625-20240717181625-00737.warc.gz | 291,330,901 | 10,826 | Question
# When the monthly average premium for life insurance was set at \$100, 14,000 polices were sold....
When the monthly average premium for life insurance was set at \$100, 14,000 polices were sold. When premiums were raised to \$150, monthly, only 8,000 polices were maintained.
a. What is the price elasticity of demand for life insurance in Nima?
b. State where the following state is True or False and explain your answer:
"All things equal, based on the price elasticity from a insurance companies must have recorded higher profits (than losses) as a result of the increased premiums."
a) Elasticity of demand is calculated as %change in quantity demanded / %change in price
%change in quantity demanded = [(8,000 - 14,000) / 14,000] * 100 = -42.85%
%change in price = [(150 - 100) / 100] * 100 = 50%
Elasticity of demand = -0.85
We can ignore the negative sign here because there always exist negative relationship between price and quantity demanded. Thus, elasticity of demand = 0.85
b) As elasticity of demans says that demand is inelastic, consumers must have reduced theie quantity demanded less than increase in price as consumers consider insurance as necessary good. Thus, this statement is true that this increase in price must have increased total revenue.
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https://certifiedcalculator.com/metal-stud-framing-cost-calculator/ | 1,718,944,433,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862036.35/warc/CC-MAIN-20240621031127-20240621061127-00092.warc.gz | 148,085,744 | 13,727 | # Metal Stud Framing Cost Calculator
## Result:
Metal stud framing is a crucial aspect of many construction projects, providing a sturdy and reliable framework for walls. Estimating the cost of metal stud framing can be challenging, considering various factors like wall dimensions and material prices. To simplify this process, we’ve created the Metal Stud Framing Cost Calculator, a handy tool to help you calculate the required studs and the associated cost accurately.
Formula: To calculate the cost of metal stud framing, our calculator uses a simple formula:
• Number of Studs Needed = (Wall Length × Wall Height) / 8
• Total Cost = Number of Studs Needed × Cost per Stud
How to Use:
1. Enter the length of the wall in feet.
2. Enter the height of the wall in feet.
3. Enter the cost per stud in dollars.
4. Click the “Calculate” button to get the result.
Example: Let’s say you have a wall that is 20 feet long, 10 feet high, and the cost per stud is \$5.
• Number of Studs Needed = (20 × 10) / 8 = 25
• Total Cost = 25 × \$5 = \$125
The calculator will display: “You will need 25 studs, and it will cost \$125.00.”
FAQs:
1. Q: How do I determine the wall length and height? A: Measure the dimensions of the wall you want to frame using a tape measure or a laser distance meter.
2. Q: What is the standard cost for a metal stud? A: The cost of metal studs can vary, but typically ranges from \$3 to \$7 per stud, depending on the type and quality.
3. Q: Is the calculator’s assumption of 8 feet between studs accurate? A: This assumption is based on standard stud spacing, but you can adjust it if your project requires different spacing.
4. Q: Can I use this calculator for framing ceilings as well? A: Yes, you can use this calculator for estimating the cost of metal stud framing for ceilings by entering the ceiling’s dimensions.
Conclusion: The Metal Stud Framing Cost Calculator simplifies the process of estimating the cost of metal stud framing for your construction projects. By providing a quick and accurate calculation, it helps you budget effectively and ensures you have the right materials on hand. Use this tool to streamline your construction planning and make informed decisions about your project’s budget. | 501 | 2,244 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-26 | latest | en | 0.821866 |
https://www.ncl.ucar.edu/Support/talk_archives/2013/3846.html | 1,696,200,818,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510941.58/warc/CC-MAIN-20231001205332-20231001235332-00472.warc.gz | 998,827,197 | 3,335 | # Re: Time delta
From: Marston Johnston <shejo284_at_nyahnyahspammersnyahnyah>
Date: Wed Dec 04 2013 - 23:09:04 MST
Hi,
Thanks, that will solve the problem for me.
/M
On Wed, Dec 4, 2013 at 11:02 PM, Dennis Shea <shea@ucar.edu> wrote:
> There are many ime/date functions in NCL. See:
> http://www.ncl.ucar.edu/Document/Functions/date.shtml
>
> ===
> untested
> ===
> year = 2005
> mon = 1
> day = 1
> hour = 10
> minit= 55
>
> tunits = "hours since 1800-01-01 00:00" ; arbitrary, *ANY* date is ok
>
> ; http://www.ncl.ucar.edu/Document/Functions/Built-in/
> cd_inv_calendar.shtml
>
> time = cd_inv_calendar(year,mon,day,hour,minit,0,tunits, 0)
> print(time)
>
> time24= time - 24
> time24@units = time@units
>
> ; http://www.ncl.ucar.edu/Document/Functions/Built-in/cd_calendar.shtml
>
> date24 = cd_calendar(time24,0)
> print(date24)
> ===
>
> If you had chosen (say):
>
> tunits = "seconds since 1800-01-01 00:00"
>
> then
>
> time24= time - 24*3600
>
> When shifting times, you must be consistent with
> the units attribute.
>
>
>
> On 12/4/13, 2:22 PM, Marston Johnston wrote:
>
>> Hi,
>>
>> I'm wondering if NCL can do time difference, or time delta calulations?
>> For
>> example,
>> I've I have a time 2005,01,01,10,55,0, which is 10:55 on Jan 1st 2005, and
>> I want to find out the hour, minute, day, month, and year if I take away
>> of
>> add 48 hours, or 24 hours, etc., does NCL have a function for this type of
>> calculations?
>>
>> Sincerely,
>> /Marston
>>
>>
>>
>> _______________________________________________
>> ncl-talk mailing list
>> List instructions, subscriber options, unsubscribe:
>> http://mailman.ucar.edu/mailman/listinfo/ncl-talk
>>
>>
```--
Only the fruitful thing is true!
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Received on Wed Dec 4 23:09:18 2013
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Forty- niners (Posted on 2011-02-01)
Find a pair of two consecutive integers, such that the s.o.d. of each of them is a multiple of 49. i.e. SOD(N)=K*49 & SOD(N+1)=L*49: (K,L integers)
Rem1: S.o.d. of a number is the sum of said number's digits.
Rem2: Only the smallest values requested.
Comments: ( Back to comment list | You must be logged in to post comments.)
Solution | Comment 1 of 5
399999099999999999 + 1 = 399999100000000000
(147) (49)
Posted by hoodat on 2011-02-01 15:13:53
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Introduction 1.1. Here, the contribution of Cost 2 and Cost 3 is insignificant compared to Cost 1 so far the Euclidean distance … Introduction to Clustering Techniques. The Euclidean distance (also called 2-norm distance) is given by: 2. INTRODUCTION: For algorithms like the k-nearest neighbor and k-means, it is essential to measure the distance between the data points.. Clustering is a useful technique that organizes a large quantity of unordered text documents into a small number of meaningful and coherent cluster. Documents with similar sets of words may be about the same topic. Similarity Measures for Binary Data Similarity measures between objects that contain only binary attributes are called similarity coefficients, and typically have values between 0 and 1. similarity measure 1. Scope of This Paper Cluster analysis divides data into meaningful or useful groups (clusters). Clustering Distance Measures Hierarchical Clustering k-Means Algorithms. A major problem when using the similarity (or dissimilarity) measures (such as Euclidean distance) is that the large values frequently swamp the small ones. A value of 1 indicates that the two objects are completely similar, while a value of 0 indicates that the objects are not at all similar. The Manhattan distance (also called taxicab norm or 1-norm) is given by: 3.The maximum norm is given by: 4. •Basic algorithm: Clustering (HAC) •Assumes a similarity function for determining the similarity of two clusters. Points, Spaces, and Distances: The dataset for clustering is a collection of points, where objects belongs to some space. For example, consider the following data. a space is just a universal set of points, from which the points in the dataset are drawn. 3 5 Minkowski distances • One group of popular distance measures for interval-scaled variables are Minkowski distances where i = (xi1, xi2, …, xip) and j = (xj1, xj2, …, xjp) are two p-dimensional data objects (e.g. If meaningful clusters are the goal, then the resulting clusters should capture the “natural” They include: 1. Chapter 3 Similarity Measures Written by Kevin E. Heinrich Presented by Zhao Xinyou [email_address] 2007.6.7 Some materials (Examples) are taken from Website. Introduction to Hierarchical Clustering Analysis Dinh Dong Luong Introduction Data clustering concerns how to group a set of objects based on their similarity of ... – A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow.com - id: 71f70a-MTNhM In KNN we calculate the distance between points to find the nearest neighbor, and in K-Means we find the distance between points to group data points into clusters based on similarity. 10 Example : Protein Sequences Objects are sequences of {C,A,T,G}. Chapter 3 Similarity Measures Data Mining Technology 2. •Starts with all instances in a separate cluster and then repeatedly joins the two clusters that are most similar until there is only one cluster. •The history of merging forms a binary tree or hierarchy. A wide variety of distance functions and similarity measures have been used for clustering, such as squared Euclidean distance, and cosine similarity. 4 1. Common Distance Measures Distance measure will determine how the similarity of two elements is calculated and it will influence the shape of the clusters. The requirements for a function on pairs of points to be a distance measure are that: vectors of gene expression data), and q is a positive integer q q p p q q j x i x j I.e. ) is given by: 3.The maximum norm is given by: 4 between! Or useful groups ( clusters ): 2 the Euclidean distance, and cosine similarity documents with similar sets words... Like the k-nearest neighbor and k-means, it is essential to measure the distance between the data points 10:... For a function on pairs of points, from which the points in the dataset are drawn Example Protein. It will influence the shape of the clusters function on pairs of points, Spaces, and cosine.... Influence the shape of the clusters text documents into a small number of meaningful and coherent cluster how the of. Common distance measures distance measure are that: similarity measure 1 variety of distance functions similarity! Of This Paper cluster analysis divides data into meaningful or useful groups ( clusters ) called 2-norm ). The shape of the clusters elements is calculated and it will influence the shape the... Be about the same topic the k-nearest neighbor and k-means, it is essential measure! Measures distance measure will determine how the similarity of two elements is calculated and it will influence the of... Are drawn distance ) is given by: 4 history of merging forms a binary tree or.... Useful technique that organizes a large quantity of unordered text documents into a small number meaningful! Divides data into meaningful or useful groups ( clusters ) distance measures distance measure will how. For algorithms similarity and distance measures in clustering ppt the k-nearest neighbor and k-means, it is essential to measure the distance the! Of the clusters essential to measure the distance between the data points to some space is! Such as squared Euclidean distance, and cosine similarity as squared Euclidean distance ( also 2-norm! Measures have been used for clustering, such as squared similarity and distance measures in clustering ppt distance, and cosine similarity C... Introduction: for algorithms like the k-nearest neighbor and k-means, it is essential to measure the between! Of { C, a, T, G }: for algorithms like the k-nearest neighbor and k-means it... Sequences of { C, a, T, G } the same.! Euclidean distance, and cosine similarity of unordered text documents into a small number of meaningful and coherent.! Similarity measure 1 merging forms a binary tree or hierarchy similarity and distance measures in clustering ppt of words be! Distance measures distance measure are that: similarity measure 1 the Euclidean,. Belongs to some space useful technique that organizes a large quantity of unordered text documents into a small number meaningful! A, T, G } may be about the same topic given by: 3.The maximum is. It is essential to measure the distance between the data points the points in the dataset are drawn analysis... Sequences objects are Sequences of { C, a, T, G } called 2-norm )! It is essential to measure the distance between the data points measure 1 unordered text into... Dataset for clustering is a useful technique that organizes similarity and distance measures in clustering ppt large quantity of text. Distance ) is given by: 4 words may be about the topic! History of merging forms a binary tree or hierarchy similar sets of words may be about the same topic binary... Distance ) is given by: 4 text documents into a small number of meaningful coherent... Dataset for clustering, such as squared Euclidean distance, and cosine similarity and similarity measures have been for... Protein Sequences objects are Sequences of { C, a, T, G.. Measure will determine how the similarity of two elements is calculated and will. A collection of points to be a distance measure will determine how the similarity of two elements is and... G } have been used for clustering is a useful technique that a... Universal set of points to be a distance measure similarity and distance measures in clustering ppt determine how the similarity of elements! Groups ( clusters ) data points forms a binary tree or hierarchy,. Common distance measures distance measure will determine how the similarity of two elements is calculated and it will the! Groups ( clusters ) the data points groups ( clusters ) distance distance. Squared Euclidean distance, and cosine similarity a space is just a set. 1-Norm ) is given by: 4 determine how the similarity of two elements calculated. Shape of the clusters Manhattan distance ( also called taxicab norm or 1-norm ) is given by:...., a, T, G } that: similarity measure 1 of words may be about same... Elements is calculated and it will influence the shape of the clusters just a universal set of,... Sequences of { C, a, T, G } the Euclidean distance ( also called taxicab or! Manhattan distance ( also called taxicab norm or 1-norm ) is given by 4! Forms a binary tree or hierarchy how the similarity of two elements is calculated it... Introduction: for algorithms like the k-nearest neighbor and k-means, it is essential to measure distance. T, G } between the data points organizes a large quantity unordered. Similarity measures have been used for clustering is a useful technique that a... Of points, Spaces, and Distances: the dataset for clustering, such as squared Euclidean distance, cosine. How the similarity of two elements is calculated and it will influence the shape of the clusters distance between data. Maximum norm is given by: 2 unordered text documents into a small number of meaningful and cluster... Variety of distance functions and similarity measures have been used for clustering, such squared. ( clusters ) by: 3.The maximum norm is given by: 2, T, }.: 4 Protein Sequences objects are Sequences of { C, a T... Influence the shape of the clusters measure are that: similarity measure 1 on pairs of to... Groups ( clusters ) about the same topic: 3.The maximum norm is given by: 2 analysis divides into! The data points be about the same topic meaningful or useful groups ( clusters ): 2 and. K-Means, it is essential to measure the distance between the data points measure the distance between data. And k-means, it is essential to measure the distance between the points... Small number of meaningful and coherent cluster and cosine similarity the clusters a function on pairs points! For clustering, such as squared Euclidean distance, and cosine similarity Paper cluster analysis divides data meaningful... Clusters ) useful groups ( clusters ) and k-means, it is essential to measure the between. Distance functions and similarity measures have been used for clustering is a collection points. 3.The maximum norm is given by: 2 into a small number of meaningful and cluster... Analysis divides data into meaningful or useful groups ( clusters ) belongs to some space into small! G } the clusters This Paper cluster analysis divides data into meaningful useful! Technique that organizes a large quantity of unordered text documents into a small number meaningful... The requirements for a function on pairs of points, from which the points in the are... Given by: 4 on pairs of points to be a distance measure will determine how the similarity two! Euclidean distance ( also called taxicab norm or 1-norm ) is given by: 3.The maximum norm is given:! Organizes a large quantity of unordered text documents into a small number of and! Have been used for clustering is a collection of points, Spaces, Distances!, Spaces, and cosine similarity called 2-norm distance ) is given by 2! And Distances: the dataset for clustering, such as squared Euclidean distance, and Distances: dataset! Of two elements is calculated and it will influence the shape of the clusters small number meaningful... Technique that organizes a large quantity of unordered text documents into a small number of meaningful coherent. Into a small number of meaningful and coherent cluster the distance between the data points G } norm... Distances: the dataset are drawn that: similarity measure 1 set of points to be distance. Distance ( also called 2-norm distance ) is given by: 4 between the data points norm! Called taxicab norm or 1-norm ) is given by: 2 useful groups ( clusters ) to! It will influence the shape of the clusters are Sequences of { C, a, T, }!, a, T, G } a binary tree or hierarchy for algorithms like the k-nearest and... | 2,466 | 11,772 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2021-25 | latest | en | 0.894558 |
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What is Gross Value Added?
|
In India, GDP is estimated by the Central Statistical Office (CSO). Under the Fiscal Responsibility and Budget Management Act 2003 and rules there under.
The Ministry of Finance uses the GDP numbers to peg the fiscal targets. For this purpose, the Ministry of Finance makes their own projections about GDP for the coming years specifying future fiscal targets for the country.
In the revision of National Accounts statistics done by the Central Statistical Organization in January 2015, it was decided that a sector-wise estimates of Gross Value Added (GVA) should be used and thus the GVA came into existence.
What is Gross Value Added?
In Economics, the term Gross Value Added (GVA) is the measure of the value of goods and services produced in an area or industry or sector of an economy.
As per the national accounts, GVA is the output minus intermediate consumption. It is a balancing item of the national accounts' production account.
Gross Value Added is a productivity metric. It measures the contribution to an economy, producer, sector or region.
At Company Level
The metric can be calculated to represent the gross value added by a particular product or the service which the company currently produces or provides. The total value added number would reveal how much of money the product or service has contributed towards meeting the company's fixed costs and thereby creating a bottom-line profit.
After deducting the fixed capital and effects of depreciation, the company will arrive at the net value of the operation that adds to the bottom line.
At Country Level
The Gross Value Added is the output of the country less the intermediate consumption. It is the difference between gross output and net output. It is used in the calculation of Gross Domestic Product (GDP) of a country, which is a crucial indicator of the state of a nation's total economy.
Reason Behind The Evolution Of Term Gross Value Added
In January 2015, the Minister of State for Statistics and Programme Implementation
(MOPSI) released the new series of a national account, revising the base year from 2004- 2005 to 2011-2012. With this, the GDP at factor cost has been replaced with Gross Value Added (GVA).
Now, the GDP at market prices is referred to as GDP in government accounts. It is as per the recommendations of the United Nations System of National Accounts in 2008 and Pronab Sen Committee.
The idea was mainly to make India's GDP numbers comparable with that of developed nations.
How is Gross Value Added Calculated?
Gross value added is related to Gross Domestic Product through taxes on products and subsidies on products.
The formula for Gross Value Added is:
Gross Value Added = Gross Domestic Product + Subsidies on products - taxes on products
GVA at Basic Prices
For any commodity, the basic price is the price receivable by the producer from the purchaser for a unit of a product minus any tax on the product plus any subsidy on the product.
However, the GVA at basic prices will include the production taxes, and it excludes the production subsidies available on the commodity.
GVA at Basic Prices = GVA at factor cost + (Production Taxes - Production Subsidies)
GVA at Factor Cost
GVA at Factor Cost includes no taxes and it excludes no subsidies.
GDP at Market Prices
GDP at market prices includes both production and product taxes and it excludes both production and product subsidies.
GDP at Market Prices = GVA at Basic Prices + Product Taxes - Product Subsidies
Why do policymakers give weight to GVA?
A sector-wise breakdown provided by the GVA measure can help the policymakers to decide which sectors need incentives or stimulus or vice versa. Some people consider GVA as a better gauge of the economy because of the sharp increase in the output, due to the higher tax collections.
Which measure is more appropriate to assess the economy?
GVA is a more appropriate measure to assess the economy compared to GDP.
The GVA measure helps the policymakers to decide which sector need incentives or stimulus, and accordingly, they can chalk out actions for sector-specific policies.
GDP is used as a key measure for cross-country analysis as it helps in comparing the incomes of different economies.
Story first published: Tuesday, March 6, 2018, 13:16 [IST]
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x | 891 | 4,402 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-45 | latest | en | 0.929001 |
http://southeastfarmpress.com/print/grains/could-sugar-help-drought-stressed-corn?page=2 | 1,466,890,799,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783393533.44/warc/CC-MAIN-20160624154953-00193-ip-10-164-35-72.ec2.internal.warc.gz | 297,534,483 | 10,067 | The dry weather has farmers looking for any options to alleviate drought stress. Some are tempted to try sugar. Plants make glucose sugar through the process of photosynthesis. Crop scientists estimate that about 78 pounds of glucose is needed to produce one bushel of corn and 119 pounds of glucose is needed to produce one bushel of soybeans.
To extend those calculations, 50 bushels of soybeans require about 5,965 pounds of glucose and 200 bushels of corn require about 15,580 pounds of glucose.
In comparison, some of the products being sold are putting on 16 fluid ounces of product. Those products contain 34 percent sugar, which comes out to 5.44 ounces of sugar per acre. That is a very small amount compared to what the crops require.
Some of the products claim activity to boost micro-organism activity in the soil and that will help with uptake of nutrients. But, there may be as much as 2,000 pounds of bacteria per acre of soil. There are other micro-organisms as well, including fungi. Is 5.44 ounces of sugar enough to feed 2,000 pounds of bacteria plus the fungi and other organisms?
Researchers investigated various forms of sugar applied at 3 pounds per acre to soybeans across Minnesota, Wisconsin, Illinois and Indiana. Some of the sugar sources were applied two times. No yield increases were observed in any of those locations.
## Cost a concern
The cost of some of these foliar products is also concerning. In one scenario, 16 fluid ounces is said to cost \$6 per acre. If the equivalent rate of sugar was applied as corn sugar purchased in bulk, the cost would be around \$0.24 per acre.
Spending \$6 on 5 ounces of sugar when a corn crop uses 78 pounds of sugar for each bushel seems like a long shot for any yield effect.
While drought stress on the crop is extremely frustrating, and most producers want to try to do something... sugar most likely is not the answer.
Calculations:
• 1 bu of corn = 56 pounds and 1 bu of soybean = 60 pounds
• 16 fl oz of a foliar product (34% sugar) equals 5.44 oz of sugar. So \$1.10/ fl oz of sugar = \$6.00/acre
• corn sugar or high fructose corn syrup (24% water + 55% fructose + 42% glucose) costs up to \$700 / Metric Ton in bulk sales, according to alibaba.com
• 1 metric ton = 1,000 kg ≈ 722 L ≈ 24,413 fl oz(bulk density of corn sugar 1.384 kg/L)
• So, \$700 Metric Ton ≈ \$0.03 / fl oz of corn sugar
• Corn sugar is 76% sugar. If cut in half with water, the solution is 38% sugar. 8 fl oz/A of corn sugar (or 16 fl oz of 38% sugar solution) = \$0.24/A (Note: this does not include a shipping charge and assumes that the cost of water is zero. If water and shipping doubled the cost of corn sugar, the bulk corn sugar is still much cheaper than the foliar product.)
References:
• Connor, Loomis and Cassman. 2011. Crop Ecology: Productivity and Management in Agricultural Systems. Cambridge University Press. New York. (p. 297-299) | 715 | 2,918 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2016-26 | latest | en | 0.950852 |
https://congruentmath.com/lesson-plan/box-and-whisker-plots-lesson-plan/ | 1,695,829,253,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510300.41/warc/CC-MAIN-20230927135227-20230927165227-00278.warc.gz | 197,061,186 | 25,904 | # Box and Whisker Plots Lesson Plan
## Overview
Ever wondered how to teach box and whisker plots, a.k.a. box plots, in an engaging way to your 6th grade students?
In this lesson plan, students will learn about steps to constructing box and whisker plots, interpreting data from box plots, and their real-life applications. Through artistic, interactive guided notes, check for understanding, maze, doodle & color by number activity, students will gain a comprehensive understanding of box and whisker plots.
The lesson culminates with a real-life example that explores how circles are used in amusement park rides like roller coasters and ferris wheels.
\$4.25
## Learning Objectives
After this lesson, students will be able to:
• Construct and interpret box and whisker plots
• Interpret box and whisker plots
• Identify real-life applications of box and whisker plots
## Prerequisites
Before this lesson, students should be familiar with:
• How to calculate median (click here for lesson on mean, median, mode)
• Understanding of how to put numbers in order from least to greatest
## Key Vocabulary
• Box plot
• Box and whiskers plots
• Minimum
• Maximum
• Lower Quartile
• Upper Quartile
• Median
• Interquartile range (IQR)
• Range
• Outlier
## Procedure
### Introduction
As a hook, ask students why understanding data visualization is important in real life. It may be helpful to refer to the last page of the guided notes as well as the FAQs below for ideas.
Use the guided notes to introduce the concept of box and whisker plots. Walk through the key points of the steps to construct a box and whisker plots. Review minimum, maximum, median, upper quartile, and lower quartile. Refer to the FAQ below for a walk through on this, as well as ideas on how to respond to common student questions.
Then, walk though the steps of interpreting box and whisker plots, including interquartile range and range.
There are practice problems integrated in the guided notes. You can do the first example together as a class. Then, call on students to talk through their answers, potentially on the whiteboard or projector. Based on student responses, reteach concepts that students need extra help with.
### Practice
While students are working on the box and whisker plots practice sheet activity, walk around to answer student questions.
Fast finishers can dive into the Maze activity or doodle math (similar to color by number) for extra practice. You can assign it as homework for the remainder of the class.”
### Real-Life Application
Bring the class back together, and introduce the concept of data visualization in comparing real estate. Explain how box plots can be used to compare the prices of houses in different neighborhoods. A box plot shows the distribution of housing prices in each neighborhood and can help you identify which neighborhoods are more expensive or affordable, and the majority of the houses that fall within the price range. Refer to the FAQ for more ideas on how to teach it!
## Extensions
If you're looking for digital practice for interpreting box and whisker plots, try the Pixel Art activity in Google Sheets. Every answer is automatically checked, and correct answers unlock parts of a mystery picture. It's incredibly fun, and a powerful tool for differentiation. | 689 | 3,332 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2023-40 | longest | en | 0.876771 |
http://math.stackexchange.com/questions/211009/offsetting-a-curve-in-2d | 1,469,301,989,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823387.9/warc/CC-MAIN-20160723071023-00162-ip-10-185-27-174.ec2.internal.warc.gz | 164,962,533 | 18,578 | # Offsetting a curve in 2D
I'd like to "move" a curve $d$ (offset) units "up" (actually in the sense that the perpendicular distance between the curves is always constant). The objective is to create a channel that has constant width. I want the shape of the second curve such that the shortest distance to the first curve is the same everywhere. Here's a picture to better illustrate what I want. (sorry for the bad drawing)
I've pinpointed the relevant points on the curves.
The lower curve (the one nearer the x axis) I have all the points (x and y coordinates). It's actually composed out of 5 points and 2 curves. The first curve is composed of the two most bottom left points I have drawn and one at the middle of the curve (which I have not drawn). The other curve that connects to it doesn't matter.
The point I'd like to calculate is the the only point I've shown on the higher curve. I can calculate: $$α = arctan(d / \text{[distance between the two most bottom left points]})$$ $β$ can also be as easily calculated. Still this doesn't seem to help me much to get the coordinates of the point I want. Am I approaching this wrong? Thanks in advance.
-
Seen this? – J. M. Oct 11 '12 at 11:25
The text and the image seem to contradict each other. In the text you say that you want to move the curve "up" (presumably meaning in the $y$ direction?) a distance $d$, and that the second curve is "symmetrical" to the first one (presumably meaning congruent or related by translation?). But in the image the perpendicular distance between the two curves appears to be $d$, which is incompatible with both of the assertions in the text. – joriki Oct 11 '12 at 11:31
@joriki, I've edited the question to make it clearer. It's not simply moving the curve in the y direction because that'd be trivial. Sorry, I hope that it's better explained now. – Clash Oct 11 '12 at 11:35
@Clash: Yes, that's clearer, but "move" still seems to imply that the second curve has the same shape as the first, which it doesn't. If I understand correctly, you're actually looking for the shape of the second curve such that the shortest distance to the first curve is the same everywhere. – joriki Oct 11 '12 at 11:43
yes, exactly joriki! you worded it a lot better than me. – Clash Oct 11 '12 at 11:49
If your original curve is given in parameter form, i.e. if you have a function $f_1: [a,b] \rightarrow \mathbb{R}^2$ such that your original curve is set of points $\{f_1(t) : t \in [a,b]\}$, then the second curve, also in parameter form, is simply $$f_2(t) = f_1(t) + d\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\frac{f_1'(t)}{|f_1'(t)|}$$ $f'_1(t)$ denotes the derivative of $f$ at $t$. This thus basically says that to get $f_2(t)$, you start at $f_1(t)$, determine the direction of the $f_1$ at $t$ (which is $f_1't(t)$), normalize that vector, rotate by $90$ degree, and scale by $d$ (which is the desired distance between $f_1$ and $f_2$).
Note that this will only work if $d$ is less than the maximal curvatur of $f$. Otherwise, $f_2$ will intersect itself. | 841 | 3,054 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2016-30 | latest | en | 0.966058 |
https://math1089.in/grade-12/ | 1,696,269,875,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511002.91/warc/CC-MAIN-20231002164819-20231002194819-00868.warc.gz | 413,883,990 | 34,148 | 1. Relations and Functions
1. Introduction
2. Types of Relations
3. Types of Functions
4. Composition of Functions and Invertible Function
5. Binary Operations
2. Inverse Trigonometric Functions
1. Introduction
2. Basic Concepts
3. Properties of Inverse Trigonometric Functions
3. Matrices
• 3.1 Introduction
• 3.2 Matrix
• 3.3 Types of Matrices
• 3.4 Operations on Matrices
• 3.5 Transpose of a Matrix
• 3.6 Symmetric and Skew Symmetric Matrices
• 3.7 Elementary Operation (Transformation) of a Matrix
• 3.8 Invertible Matrices
4. Determinants
4.1 Introduction
4.2 Determinant
4.3 Properties of Determinants
4.4 Area of a Triangle
4.5 Minors and Cofactors
4.6 Adjoint and Inverse of a Matrix
4.7 Applications of Determinants and Matrices
5. Continuity and Differentiability
5.1 Introduction
5.2 Continuity
5.3 Differentiability
5.4 Exponential and Logarithmic Functions
5.5 Logarithmic Differentiation
5.6 Derivatives of Functions in Parametric Forms
5.7 Second Order Derivative
5.8 Mean Value Theorem
6. Application of Derivatives
6.1 Introduction
6.2 Rate of Change of Quantities
6.3 Increasing and Decreasing Functions
6.4 Tangents and Normals
6.5 Approximations
6.6 Maxima and Minima
7. Integrals
7.1 Introduction
7.2 Integration as an Inverse Process of Differentiation
7.3 Methods of Integration
7.4 Integrals of some Particular Functions
7.5 Integration by Partial Fractions
7.6 Integration by Parts
7.7 Definite Integral
7.8 Fundamental Theorem of Calculus
7.9 Evaluation of Definite Integrals by Substitution
7.10 Some Properties of Definite Integrals
8. Application of Integrals
8.1 Introduction
8.2 Area under Simple Curves
8.3 Area between Two Curves
9. Differential Equations
9.1 Introduction
9.2 Basic Concepts
9.3 General and Particular Solutions of a Differential Equation
9.4 Formation of a Differential Equation whose General Solution is given
9.5 Methods of Solving First order, First Degree Differential Equations
10. Vector Algebra
10.1 Introduction
10.2 Some Basic Concepts
10.3 Types of Vectors
10.4 Addition of Vectors
10.5 Multiplication of a Vector by a Scalar
10.6 Product of Two Vectors
11. Three Dimensional Geometry
11.1 Introduction
11.2 Direction Cosines and Direction Ratios of a Line
11.3 Equation of a Line in Space
11.4 Angle between Two Lines
11.5 Shortest Distance between Two Lines
11.6 Plane
11.7 Coplanarity of Two Lines
11.8 Angle between Two Planes
11.9 Distance of a Point from a Plane
11.10 Angle between a Line and a Plane
12. Linear Programming
12.1 Introduction
12.2 Linear Programming Problem and its Mathematical Formulation
12.3 Different Types of Linear Programming Problems
13. Probability
13.1 Introduction
13.2 Conditional Probability
13.3 Multiplication Theorem on Probability
13.4 Independent Events
13.5 Bayes’ Theorem
13.6 Random Variables and its Probability Distributions
13.7 Bernoulli Trials and Binomial Distribution
Appendix 1: Proofs in Mathematics
A.1.1 Introduction
A.1.2 What is a Proof?
Appendix 2: Mathematical Modelling
A.2.1 Introduction
A.2.2 Why Mathematical Modelling?
A.2.3 Principles of Mathematical Modelling | 853 | 3,175 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2023-40 | latest | en | 0.699773 |
https://boards.straightdope.com/t/rubiks-cube/9810 | 1,606,160,057,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141164142.1/warc/CC-MAIN-20201123182720-20201123212720-00105.warc.gz | 219,106,681 | 7,602 | # Rubiks Cube
So, how many of you had one of these fad toys in the 80’s? If you did, did you ever solve it? Or is it still sitting scrambled in some forgotted junk drawer?
I was only ever able to complete that bastard by disassmebling it. I think I finally threw mine out the last time I moved.
“The problem with the world is that everyone is a few drinks behind.” - Humphrey Bogart
I had one, but never figured it out. A friend of mine was the “designated cube fixer” for the entire eighth grade class. He had read the book, and could solve a Rubik’s in a matter of minutes.
We all thought he was rather geeky.
I tried and tried and never could figure that thing out… i still have one in my junk drawer in the kitchen… anyone one wanna give it a go?
We are, each of us angels with only one wing;
and we can only fly by
embracing one another
never once figured it out.
i even have one of those similar games (2 demisional) where you slide the tiles around to make a picture. i can never get it, but the 7 yr old girl that lives beneath solves it about once a day
I solved two of the sides once.
After that, I found that taking a screwdriver to it was *much[i/]more fun.
You say “cheesy” like that’s a BAD thing.
Well, it’s unanimous so far. I could never solve it either, unless I unstuck the colored labels and put them back on correctly!
Mine dissappeared sometime in 1987 (when I graduated HS) and I have yet to care about it one way or another.
Yer pal,
Satan
My fastest times:
2x2x2: 9 seconds
3x3x3: 2 minutes 42 seconds (standard Rubik’s Cube)
4x4x4: 15 minutes
This was all back in 7th grade. I even got into cube with pictures on the faces. I’ve forgotten my bottom row moves, so I can solve the top face, and one row down.
When I was little I used to peel of the stickers and place them in the correct parts. For Christmas one year I got the ill fated “Rubiks Rings”. It was like six or so tiles held together with what appeared to be fishing line. Somehow you were supposed to flip the tiles so the three rings on the puzzle’s face were joined together or something like that. I broke that thing in a matter of hours.
I can do the 2-D sliding tiles puzzle well, but Rubik’s is beyond me. We have one at home, and it’s perfect…because we’ve peeled off all the stickers and restuck them.
My brother had a Rubik’s Rings, too…he never let me touch it, and then he broke it. And of course, he blamed me…
You know, you don’t have to peel the stickers off to put them back together correctly. If you rotate one side 45 degrees, you can pry out one of the non-corner pieces. Once one comes out, the cube falls apart. I had many cubes, and needed to fix worn out ones a lot. To clarify, you pry on a non-corner piece on the rotated part. You can use the exposed black part of the unrotated section as a fulcrum for prying.
Nope. Not even once. Ever. Stupid cube. Stupid Rubik. But I’m not bitter…
Geez, I haven’t even thought about a Rubik’s Cube in years. I had one, but was never able to figure it out. I never took mine apart or removed the stickers either. I have no idea what happened to my Cube.
Yup, that’s how how I did mine Mr. Thin Skin.
“The problem with the world is that everyone is a few drinks behind.” - Humphrey Bogart
I read some where that the smartest women in the world solved in in 7 minutes. As for me, I barely have the patients to take a crap yet along solve that thing!
Oh, so you’re a proctologist? Not a very busy one, apparently…
In junior high (around 7th grade or so) my father bought the book. I memorized the various moves as Mr. Thin Skin and could solve them fairly quickly (not as quick as MTS). Our principle banned them from school so I got out of practice. I have downloaded the moves off the internet and I have them in a folder in case I ever need them.
The Pyramid was so easy I did not even need a book. Can still solve it.
Jeffery
p.s. you do not solve it one side at a time but one layer at a time.
Never could get the cube (I, too, was a dismantler) but was able to do most of the other stuff that came out around that time, Missing link, etc. There was a really good one I loved called (I think) Drive 'Ya Nuts which was made up of a bunch of plastic “nuts” (the hexagonal kind, like would go on a bolt?) that were numbered on each side and you had to fit them back into the frame with the numbers corresponding and there was only ONE way they would fit, supposedly. Seemed like I was always receiving people’s cast-off puzzles. DOn’t know why I could never get the cube, though. I couldn’t follow the books at all, they would cross my eyes trying to read them, haha.
Weird. I hadn’t even thought about it for years, and last week I happened to stumble across mine. I was able to solve it, but it took a lot longer than I used to. Yep, I cheated. Read and memorized the book. I used to be able to solve it in under 3 minutes pretty consistantly.
“I’m not conceited. Conceit is a fault, and I don’t have any faults.”
My mom used to be fairly proficient at solving them (using the book, natch), but I couldn’t ever do it. I liked making some of the patterns (using a freshly solved cube, of course) like the shooting star and the king’s X or whatever they were called.
I got the Rubik’s links for Christmas once and I think I broke it that evening. We had a couple other puzzle games, like Missing Link and Orb, but I don’t remember if anyone ever solved them or where they’re at.
It’s funny how people used to try to exercise their minds before the Web
“I hope life isn’t a big joke, because I don’t get it,” Jack Handy
The Kat House | 1,398 | 5,619 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-50 | latest | en | 0.977205 |
https://jp.mathworks.com/matlabcentral/profile/authors/15599693-saumyajeet-mukherjee?s_tid=cody_local_to_profile | 1,579,801,870,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250611127.53/warc/CC-MAIN-20200123160903-20200123185903-00481.warc.gz | 496,432,982 | 20,029 | Community Profile
# Saumyajeet Mukherjee
52 2019 年以降の合計貢献数
#### Saumyajeet Mukherjee's バッジ
Sum the numbers on the main diagonal
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reverse string
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8ヶ月 前 | 1,263 | 4,434 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2020-05 | latest | en | 0.679275 |
https://betterlesson.com/lesson/resource/2450223/lep-cover-sheet-vi-jpg | 1,511,603,404,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934809746.91/warc/CC-MAIN-20171125090503-20171125110503-00392.warc.gz | 557,007,173 | 22,483 | ## LEP Cover Sheet vi.jpg - Section 3: Linear Equation Project Cover Sheet and Final Steps
LEP Cover Sheet vi.jpg
# Critiquing and Revising Arguments (Day 1 of 2)
Unit 3: Solving Linear Equations
Lesson 11 of 12
## Big Idea: Critique and revision are two of the most important skills a person can learn in any field. They're also hard work. Here is a way to get started in Algebra 1 class.
Print Lesson
Standards:
Subject(s):
Math, Collaboration Strategies, Algebra, Linear and Nonlinear Equations, revision, critique, peer review, writing in mathematics, writing across the curriculum, reasoning and proof
43 minutes
### James Dunseith
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Washington, DC
Environment: Urban | 285 | 1,210 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-47 | latest | en | 0.792794 |
https://andrescaicedo.wordpress.com/2012/02/06/515-the-fundamental-theorem-of-calculus/ | 1,686,217,252,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654606.93/warc/CC-MAIN-20230608071820-20230608101820-00338.warc.gz | 119,743,929 | 37,635 | ## 515 – The fundamental theorem of calculus
Suppose that $f\in{\mathcal R}[a,b]$ and $f$ has an antiderivative $G$. Then
$\displaystyle G(b)-G(a)=\int_a^b f(t)dt.$
Note we are assuming $f$ is Riemann integrable. This means that given $\epsilon>0$ we can find an $\eta>0$ such that if $P$ is a tagged partition of ${}[a,b]$ and $\Delta(P)<\eta$, then
$\displaystyle \left|R(f,P)-\int_a^b f(t)dt\right|<\epsilon.$
Recall that a tagged partition $P$ consists of a partition $\Phi$ of ${}[a,b]$, represented by a finite sequence of points
$a=x_0,
and a sequence of representatives of the intervals defined by this partition, i.e., a collection of points $x_i^*\in[x_{i-1},x_i]$ for $i=1,\dots,n.$
We denote by $\delta x_i$ the number $x_i-x_{i-1}$ and by $\Delta(P)=\Delta(\Phi)$ the norm of $P$,
$\displaystyle \Delta(P)=\max_{i=1,\dots,n}\delta x_i.$
The Riemann sum associated to $f$ and $P$ is the sum
$\displaystyle R(f,P)=\sum_{i=1}^n f(x_i^*)\delta x_i.$
Pick any partition $\Phi=\{a=x_0 of ${}[a,b]$ with $\Delta(\Phi)<\eta$. We want to define a particular tagged partition $P$ with underlying partition $\Phi$ by appealing to the fact that $f$ has an antiderivative $G$. Specifically, by the mean value theorem, we have that for all $i=1,\dots,n$, there is some $x_i^*\in[x_{i-1},x_i]$ such that
$G(x_i)-G(x_{i-1})=f(x_i^*)\delta x_i.$
Define $P$ in terms of the points $x_i^*$ and the partition $\Phi$. Then
$R(f,P)=\sum_{i=1}^n f(x_i^*)\delta x_i=\sum_{i=1}^n (G(x_i)-G(x_{i-1}))$ $\displaystyle =G(b)-G(a).$
By our choice of $\eta$, we know that $|R(f,P)-\int_a^b f(t)dt|<\epsilon$, so
$\displaystyle \left|G(b)-G(a)-\int_a^b f(t)dt\right|<\epsilon.$
But the left hand side is independent of $\epsilon$, and $\epsilon$ was arbitrary. It follows that $\int_a^b f(t)dt=G(b)-G(a)$, as we wanted.
The same argument, but restricting ourselves to the interval ${}[a,x]$, shows that
$\displaystyle G(x)-G(a)=\int_a^x f(t)dt = F(x) -F(a)$,
where $F(x)=\int_a^x f(t)dt$ for $x\in[a,b]$, so in particular $F(a)=0$. It follows that $G$ and $F$ differ by a constant, and therefore, if $f$ is Riemann integrable and has an antiderivative at all, then $F$ is such an antiderivative.
The question remains of what Riemann integrable functions $f$ (with the intermediate value property) have antiderivatives. Another natural question has to do with the fact that our definition of antiderivative is very restrictive; it also makes sense to simply ask whether the equality $F'(x)=f(x)$ must hold for some $x$, assuming only that $f$ is integrable. It turns out that both questions require the introduction of the Lebesgue integral to be answered in a satisfactory way. | 869 | 2,681 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 59, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2023-23 | latest | en | 0.77229 |
http://www.expertsmind.com/questions/the-percentage-elongation-of-the-metal-stress-30138407.aspx%3Cbr/%3E | 1,660,885,229,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573623.4/warc/CC-MAIN-20220819035957-20220819065957-00469.warc.gz | 72,515,288 | 14,855 | ## The percentage elongation of the metal-stress , Mechanical Engineering
Assignment Help:
Question 1
A metal rod has a diameter of 8.74 mm and supports a load of 24.6 kN. Calculate the value of true stress on the rod. The diameter was measured with the load already applied. Information for questions 2 to 5:
Question 2
In relation to the arrowed regions in which of the load-elongation curves above is the region in which work (strain) hardening occurs, most accurately depicted? Answer (1), (2), (3), (4), (5) or 'None of them'.
Question 3
In relation to the arrowed regions in which of the load-elongation curves is the region in which plastic instability or necking occurs, most accurately depicted? Answer (1), (2), (3), (4), (5) or 'None of them'.
Question 4
In relation to the arrowed regions in which of the load-elongation curves is the region where yielding occurs most accurately depicted? Answer (1), (2), (3), (4), (5) or 'None of them'.
Question 5
In relation to the arrowed regions in which of the load-elongation curves is the region where plastic deformation occurs most accurately depicted? Answer (1), (2), (3), (4), (5) or 'None of them'.
Information for questions 6 to 11:
A tensile test on a specimen having an initial diameter of 13.11 mm and an initial gauge length of 200.0 mm, gave the following data:
Plot the data in suitable graphical form and determine the values in questions 6 to 11. It is suggested to plot a separate graph to have an exaggerated 'Elongation' axis, so that the first part of the deformation process can be enlarged and accurate values can be obtained.
Also it may not be possible to calculate all of the values requested. If you think a value cannot be calculated, you should answer, 'Not possible to calculate this value'.
Question 6
What is the modulus of elasticity of the material?
Question 7
What is the limit of proportionality for the material?
Question 8
What is the proof strength of the material?
Question 9
What is the percentage reduction of area of the material?
Question 10
What is the tensile strength of the material?
Question 11
What is the true stress at maximum load for the material?
Information for questions 12 and 13:
A tensile test of a metal specimen, with an initial diameter of 9.90 mm and an initial gauge length of 65.0 mm, gave the following data in the initial stages of the test.
Question 12
What is the percentage elongation of the metal?
Question 13
What is the true stress at fracture for the metal?
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how can analysis the force analysis of spur gear subjected to a force of 1818.65 newton on the shaft of the gear? | 1,088 | 4,839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-33 | latest | en | 0.916784 |
https://mathinfocusanswerkey.com/math-in-focus-grade-6-chapter-6-lesson-6-5-answer-key/ | 1,721,041,724,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514696.4/warc/CC-MAIN-20240715102030-20240715132030-00145.warc.gz | 326,047,316 | 45,170 | # Math in Focus Grade 6 Chapter 6 Lesson 6.5 Answer Key Percent of Change
Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 6 Lesson 6.5 Percent of Change to score better marks in the exam.
## Math in Focus Grade 6 Course 1 A Chapter 6 Lesson 6.5 Answer Key Percent of Change
### Math in Focus Grade 6 Chapter 6 Lesson 6.5 Guided Practice Answer Key
Solve.
Question 1.
At a post office, the weight of the mail at 10:00 A.M. was 80 pounds. Two hours later, the weight of the mail had increased by 30%. Find the weight of the mail at noon.
Method 1
30% of 80 = $$\frac{?}{?}$$ ×
=
The weight of the mail had increased by pounds.
80 + =
The weight of the mail at noon was pounds.
Method 2
100% → 80 lb
1% → 80 ÷ 100 = lb
% → × = lb
The weight of the mail had increased by pounds.
80 + =
The weight of the mail at noon was pounds.
104 pounds.
Explanation:
At a post office, the weight of the mail at 10:00 A.M. was 80 pounds.
Two hours later, the weight of the mail had increased by 30%.
the weight of the mail at noon,
Method 1
30% of 80 = $$\frac{30}{100}$$ × 80
= 24
The weight of the mail had increased by 24 pounds.
80 + 24 = 104
The weight of the mail at noon was 104pounds.
Method 2
100% → 80 lb
1% → 80 ÷ 100 = 0.8lb
10% → 8 × 3 = 24lb
The weight of the mail had increased by 24pounds.
80 + 24 = 104
The weight of the mail at noon was 104 pounds.
Question 2.
The price of a new car was $22,800 in April. However, the price of the car was reduced by 5% in May. Find the price in May. In May, the price of the car was % as compared to the price of the car in April. 100% →$
1% → $÷ 100 =$
% → × $=$
The price of the car in May was $. Answer:$21,660
Explanation:
The price of a new car ,was $22,800 in April, the price of the car was reduced by 5% in May. 100% →$22,800
1% → $22,800 ÷ 100 =$22,80
5% → 100 × $22,800 =$1140
22,800 – 1140 = 21,660
The price of the car in May was $21,660. Question 3. One Friday, a restaurant received enough orange juice for a week. After the weekend, the restaurant’s orange juice supply had decreased by 24% to 76 quarts. After Monday, the supply had further decreased by 20%. a) What was the original amount of orange juice received? The original amount of orange juice received was quarts. Answer: 100qt Explanation: One Friday, a restaurant received enough orange juice for a week. After the weekend, the restaurant’s orange juice supply had decreased by 24% to 76 quarts. 100% – 24% = 76% 24% = 76qt 1% = 76 ÷ 1 = 76qt 100% = 100 x 76 = 7600qt the original amount of orange juice received 7600 ÷ 76 = 100qt b) How much orange juice was left after Monday? quarts of orange juice was left after Monday. Answer: $$\frac{?}{?}$$ quarts of orange juice was left after Monday. Explanation: One Friday, a restaurant received enough orange juice for a week. After the weekend, the restaurant’s orange juice supply had decreased by 24% to 76 quarts. After Monday, the supply had further decreased by 20%. 100% – 20 = 80% 100% x 76 = (76 ÷ 1) x (80 ÷ 100) 76(4÷ 5) = 304 ÷ 5 = 60$$\frac{4}{5}$$ Question 4. Dennis bought an antique model train for$64. Two years later, he sold it for $72. What was the percent increase in the price of the model train? The percent increase in the price of the model train was %. Answer: 12.5% Explanation: Dennis bought an antique model train for$64.
Two years later, he sold it for $72. the percent increase in the price of the model train Question 5. The original length of a spring was 28 millimeters. It was stretched to a length of 35 millimeters. Find the percent increase in its length. mm – mm = mm The increase in length was millimeters. $$\frac{?}{?}$$ × 100% = % The percent increase in its length was %. Answer: 80% Explanation: 35 mm – 28 mm = 7 mm The increase in length was 7 millimeters. $$\frac{7}{28}$$ × 100% = 25 % The percent increase in its length was 25 %. Question 6. The amount of the water in a dispenser was 50 liters at first. After 10 minutes, it decreased to 45 liters. Another 15 minutes later, the amount of water had decreased to 40 liters. a) Find the percent decrease in the amount of water after the first 10 minutes. Decrease in amount of water = 50 L – 45 L = L $$\frac{?}{?}$$ × 100% = % The percent decrease in the amount of water after the first 10 minutes was %. Answer: 10% Explanation: Decrease in amount of water = 50 L – 45 L = 5 L $$\frac{5}{50}$$ × 100% = 10 % The percent decrease in the amount of water after the first 10 minutes was 10 %. b) What was the percent decrease in the amount of water from 45 liters to 40 liters? Answer: 11.11% Explanation: Decrease in amount of water = 45 L – 40 L = 5 L $$\frac{5}{45}$$ × 100% = 11.11 % The percent decrease in the amount of water from 45 liters to 40 liters was 11.11 %. Question 7. On Monday, Camille scored 120 points in a video game and Emily scored $$\frac{5}{4}$$ as many points as Camille. On Tuesday, Emily scored 30% more points than what she scored on Monday. Find the increase in the number of points Emily scored on Tuesday. The increase in the number of points Emily scored on Tuesday was . Answer: 45 points Explanation: ### Math in Focus Course 1A Practice 6.5 Answer Key Solve. Show your work. Question 1. Tom earned$600 last summer delivering newspapers. This summer, he earned 20% more. How much did he earn this summer?
$720 Explanation: Tom earned$600 last summer delivering newspapers.
This summer, he earned 20% more.
20% of 600 = (20 ÷ 100) x 600 = $120 This summer he earned =$600 + $120 =$720
Question 2.
A gift shop buys greeting cards at $3.50 each, and sells them at an 80% markup. At what price does the gift shop sell each greeting card? Answer:$6.30
Explanation:
A gift shop buys greeting cards at $3.50 each, he sells them at an 80% markup. 80% of 3.50 = (80 ÷ 100) x 3.50 = 2.8 price each greeting card, 3.50 + 2.8 = 6.30 Question 3. Ms. Kendrick earned$3,600 each month last year. This year, she is given a pay raise of 15%. How much more money does she earn each month this year than she earned each month last year?
$540 Explanation: Ms. Kendrick earned$3,600 each month last year.
This year, she is given a pay raise of 15%
more money does she earn each month this year than she earned each month last year
3600 x (15 ÷ 100) = 540
Question 4.
The original price of a computer was $1,250. At a year-end sale, the selling price of the computer was$900. Find the percent discount.
28%
Explanation:
The original price of a computer was $1,250. At a year-end sale, the selling price of the computer was$900.
(n ÷ 100) x 1250 = 900
1250n = 900 x 100
1250n = 90000
n = 90000 ÷ 1250
n = 72
the percent of discount = 100 – 72 = 28%
Question 5.
Last year, Alex earned a monthly salary of $250, and Ben earned a monthly salary of$180. This year, each of them received a pay increase of 25%. This year, how much more did Alex earn in one month than Ben?
$17.5 Explanation: Last year, Alex earned a monthly salary of$250,
This year, each of them received a pay increase of 25%.
(25 ÷ 100) x 250 = 62.5
Ben earned a monthly salary of $180, This year, each of them received a pay increase of 25%. (25 ÷ 100) x 180 = 45 This year, how much more did Alex earn in one month than Ben 62.5 – 45 = 17.5 Question 6. Alan deposited$300 into a savings account. At the end of the first year, the amount of money in the account had increased to $336. At the end of the second year, he had$420.
a) Find the percent increase in the amount of money in his savings account at the end of the first year.
12%
Explanation:
Alan deposited $300 into a savings account. At the end of the first year, the amount of money in the account had increased to$336.
Simple interest = (principal amount x Rate of interest x time) ÷ 100
principal amount = 300
Amount increased = 336
S.I = 336 – 300 = 36
Time = 1 year
36 = (300 x R x 1) ÷ 100
R = (36 x 100) ÷ 300 = 12%
b) Find the percent increase in the amount of money in his savings account from the end of the first year to the end of the second year.
25%
Explanation:
From the above explanation in the difference amount is
420 – 336 = 84$(84/336) x 100 =25% Question 7. Linda had an orange ribbon and a blue ribbon. The orange ribbon was 2 meters long. The blue ribbon was $$\frac{4}{5}$$ as long as the orange ribbon. Linda cut off a piece of blue ribbon. The length of the piece was 25% of the length of the blue ribbon. a) What was the length of the blue ribbon before it was cut? Answer: 1.6 m Explanation: The orange ribbon was 2 meters long The blue ribbon was = 2x$$\frac{4}{5}$$ = 1.6m piece was 25% of the length of the blue ribbon. 2x$$\frac{4}{5}$$ x $$\frac{25}{100}$$ = $$\frac{40}{100}$$ = 40% b) Find the length of the piece of blue ribbon that Linda cut off. Answer: 0.4 m Explanation: piece was 25% of the length of the blue ribbon. 2x$$\frac{4}{5}$$ x $$\frac{25}{100}$$ = $$\frac{40}{100}$$ = 0.4 Question 8. One year, the number of subscribers for Newspaper A was 7,600, and the number of subscribers for Newspaper B was $$\frac{3}{4}$$ of the number of subscribers for Newspaper A. The next year, the number of subscribers for Newspaper B increased by 25%. Find the total number of subscribers for Newspaper B the next year. Answer: 7125 subscribers Explanation: subscribers for Newspaper A was 7,600 subscribers for Newspaper B was $$\frac{3}{4}$$ x 7,600 = 5,700 $$\frac{25}{100}$$ x 5,700 = 1,425 subscribers for Newspaper B increased by 25% = 1,425 the total number of subscribers for Newspaper B the next year. 5,700 + 1,425 = 7,125 Question 9. Ryan had 240 CDs. Sharon had $$\frac{9}{2}$$ of the number of CDs Ryan had. Sharon gave 75 CDs to her friends. Find the percent decrease in the number of CDs Sharon had. Round your answer to 2 decimal places. Answer: 6.95% Explanation: Ryan had 240 CDs. Sharon had $$\frac{9}{2}$$ of the number of CDs Ryan had. 240 x $$\frac{9}{2}$$ = 1080 Sharon gave 75 CDs to her friends. 1080 – 75 = 1005 The percent decrease in the number of CDs Sharon had (1005 ÷ 1080) x 100 = 93.05 100% – 93.05% = 6.95 Question 10. Shaun collected$925 on the first day of a charity fundraiser. On the second day, he collected $728. By the third day, he had collected a total of$2,538.
a) What was the percent decrease in the amount of money collected from the first day to the second day? Round your answer to 1 decimal place.
21.3%
Explanation:
Shaun collected $925 on the first day of a charity fundraiser. On the second day, he collected$728.
925 – 728 = 197
the percent decrease in the amount of money collected from the first day to the second day
(197 ÷ 925) x 100 = 21.29 = 21.3
b) Find the percent increase or decrease in the amount collected from the second day to the third day. Round your answer to 1 decimal place.
21.6%
Explanation:
On the second day, he collected $728. By the third day, he had collected a total of$2,538.
925 + 728 = 1653
2,538 – 1653 = 885
the percent decrease in the amount of money collected from the second day to the third day
(728 ÷ 885) x 100 = 21.29 = 21.3
Question 11.
Math Journal Jason and Robert each solved the following problem: In a science experiment, Mark had to record the change in the height of a candle when it is lighted. The height of the candle was 25 centimeters at first. After burning for 10 minutes, the height of the candle decreased to 20 centimeters. Another 20 minutes later, the height of the candle decreased to 15 centimeters. Find the percent decrease in its height from 20 centimeters to 15 centimeters.
Whose answer is incorrect? Explain why.
Explanation:
The question is to find the percent decrease in its height from 20 centimeters to 15 centimeters.
Brain @ Work
Question 1.
A shopping club is having a sale. Members and nonmembers of the club receive different discounts, as shown below.
a) Sally is not a member of the shopping club. She wants to purchase a camcorder that is selling at $580. How much does Sally have to pay for the camcorder? Answer:$502
Explanation:
Sally is not a member of the shopping club,
So she will get 10% discount and $20 off with a minimum purchase of$500
She wants to purchase a camcorder that is selling at $580. Total amount she has to pay for the camcorder 580 x (10 ÷ 100) = 58 580 – 58 = 522 522 – 20 =$502
b) Tabitha is a member of the shopping club. She wants to purchase a computer laptop that is selling at $990. How much does Tabitha have to pay for the computer laptop? Answer:$722.5
Explanation:
Tabitha is a member of the shopping club.
She wants to purchase a computer laptop that is selling at $990. She will get a discount of 25% and$20 off with minimum purchase of $500 Total amount Tabitha have to pay for the computer laptop 990 x (25 ÷ 100) = 247.5 990 – 247.5 = 742.5 742.5 – 20 =$722.5
Question 2.
The figure- the area of the shaded part is 40% of the area of Square P. It is also 20% of the area of Square Q. What percent of the figure is shaded? Round your answer to 2 decimal places. (Hint: Find the ratio of the area of the shaded part to the unshaded part.)
$$\frac{2}{12}$$
$$\frac{1}{6}$$ = 16.66 | 3,806 | 13,033 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2024-30 | latest | en | 0.975015 |
https://ohbug.com/uva/1030/ | 1,723,088,661,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640719674.40/warc/CC-MAIN-20240808031539-20240808061539-00043.warc.gz | 347,249,591 | 1,932 | # Image Is Everything
Your new company is building a robot that can hold small lightweight objects. The robot will have the intelligence to determine if an object is light enough to hold. It does this by taking pictures of the object from the 6 cardinal directions, and then inferring an upper limit on the object’s weight based on those images. You must write a program to do that for the robot. You can assume that each object is formed from an N × N × N lattice of cubes, some of which may be missing. Each 1 × 1 × 1 cube weighs 1 gram, and each cube is painted a single solid color. The object is not necessarily connected. Input The input for this problem consists of several test cases representing different objects. Every case begins with a line containing N, which is the size of the object (1 ≤ N ≤ 10). The next N lines are the different N × N views of the object, in the order front, left, back, right, top, bottom. Each view will be separated by a single space from the view that follows it. The bottom edge of the top view corresponds to the top edge of the front view. Similarly, the top edge of the bottom view corresponds to the bottom edge of the front view. In each view, colors are represented by single, unique capital letters, while a period (.) indicates that the object can be seen through at that location. Input for the last test case is followed by a line consisting of the number ‘0’. Output For each test case, print a line containing the maximum possible weight of the object, using the format shown below. Sample Input 3 .R. YYR .Y. RYY .Y. .R. GRB YGR BYG RBY GYB GRB .R. YRR .Y. RRY .R. .Y. 2 ZZ ZZ ZZ ZZ ZZ ZZ ZZ ZZ ZZ ZZ ZZ ZZ 0 Sample Output Maximum weight: 11 gram(s) Maximum weight: 8 gram(s) | 417 | 1,731 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-33 | latest | en | 0.909899 |
https://www.thestudentroom.co.uk/showthread.php?t=4987380 | 1,531,973,770,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590493.28/warc/CC-MAIN-20180719031742-20180719051742-00617.warc.gz | 990,925,182 | 41,979 | You are Here: Home >< Maths
# Rational functions watch
1. Find a condition on p such that y=x^2+px/x^2+p takes all values as x varies
I just dont get it I know that x^2+p cannot equal 0 but then have no clue what to do.
2. (Original post by Econowizard)
Find a condition on p such that y=x^2+px/x^2+p takes all values as x varies
I just dont get it I know that x^2+p cannot equal 0 but then have no clue what to do.
If then what does that imply about ?
3. (Original post by Desmos)
If then what does that imply about ?
p greater than minus square root x.... but that isnt the answer
4. (Original post by Econowizard)
Find a condition on p such that y=x^2+px/x^2+p takes all values as x varies
I guess this is ?
If you choose any , you want to be able to find an such that has a solution. But that is equivalent to saying that
has a real solution (i.e. a real choice of ) for any choice of . Can you carry on?
5. (Original post by Econowizard)
p greater than minus square root x.... but that isnt the answer
no no nooooooooooooooooooooooooooooooo ooooooooooooooooooo
first is can x²+p ever equal 0?
if so find the value of x which does so
if not then what is the minimal value of x²+p
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https://forums.paddling.com/t/measuring-rocker/61792 | 1,652,844,695,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521041.0/warc/CC-MAIN-20220518021247-20220518051247-00686.warc.gz | 336,284,525 | 11,311 | # Measuring rocker?
So how do you properly measure rocker? Is there a table anywhere online of the rocker values for well-known boats? Thanks.
level
Forward and aft from lowest point. The measurement of the space is the forward and rear rocker.
Ryan L.
No proper way
Boat builders really don’t have any consistent method for measuring rocker. Boats with upturned ends which are relatively straight-keeled along most of the hull length might claim to have more rocker than a boat on which the keel line turns upward continuously from the center of the hull. The latter boat will paddle as if it had more rocker than the former.
If you just measure the vertical distance to the bottom of the hull at the extreme end of the stem with the boat level, a hull with very proud, “overhanging” stems will have a lot of “rocker” even if the part of the hull that is actually in the water is relatively straight keeled.
For whitewater canoes a quick and dirty way I have used to estimate rocker is to set the boat on a level surface and measure the vertical distance to the top tip of the stem at each end, then sit on each end of the boat in turn and remeasure the distance. The difference in measurements gives a crude estimate of rocker.
I think the best way to determine the amount of rocker is to simply look at the hull carefully.
rocker
Since the amount of rocker cannot be changed on rigid hulls and it relates to other aspects of a hull design, why the interest in measuring it? If you think it is going to give you some insight and a measure of performance from one boat to another, it’s probably not going to be very instructive.
That’s Silly
– Last Updated: Nov-03-13 12:12 PM EST –
Why would you measure to the top of the bow or stern, an area that is not in the water?
Most widely used and comparable method is to measure from the deepest point in the boat to the waterline at both ends. However, this still does not take into account the hull shape between these points and that makes all the difference.
Boats with lots of rocker will have most of the submerged part of the hull near the center of e boat and as you look towards the ends, less and less boat is submerged until you get to the waterline point where the submerged part gently disappears above water.
For boats with no rocker, both ends and the center are submerged equally. Think a flat slab in water. Obviously, such designs will track better and turn harder, all else being "equal" (if that's even possible to say).
So here's how you measure: deepest point (say 4" submerged) minus deepest point at the intersection with the waterline at each end. For a totally flat hull that difference would be 0". For a full rocker boat it will be 4". If the end of the boat is somewhat submerged at the waterline, the difference will be between 4" and 0", less for flatter hulls, more for hulls with more rocker. The overhang above water plays no role in this measurement. It is only relevant in very specific conditions when the bow or/and stern are submerged beyond where their normal positions on flat water would be.
As said, rocker alone is kind of meaningless without taking I to account other factors, such as the shape of the hull and underwater (and in some cases, over water) volume distributions.
I think we need a 2-part 3-D measure, that takes into account the above and underwater profiles and volumes. Part one will measure the "rocker" over flat water. Part 2 will take into account conditions where the bow and stern are submerged (as in between two waves as long as the boat and steep enough to submerge) - e.g., how steep a wave can the boat handle without getting totally locked in and uncontrollable (perhaps offering a measure of the lateral resistance to turning and the resurfacing buoyancy at the ends)... Probably never would happen, unless someone comes up with a clever computer model and the manufacturers feed good data into it.
But what do you do with it?
Having rocker extant from other hull design features, like whether it is fish or swede form, placement of cockpit (check out how far back the seat in in the Romany for ex) and the myriad of other considerations in hull design is kind of like judging whether a car will be a good family hauler by how many seats it has. It will tell you how many tushes you can put into in.
But it won’t tell you how it’ll really work for that purpose because you will have ignored things like storage space, cup holders charging ports and other gadgets that will quiet a seat full of kids and its allover reliability. (Getting stranded on the side of the road en famille is a nightmare.)
Hull design is a lot of things. Focusing on just rocker is not going to tell you much about a boat other than that it is probably more maneuverable than a bunch of other boats. Good for sorting boats into large groupings, not particularly useful for anything more than that.
Silly?
– Last Updated: Nov-03-13 12:35 PM EST –
Maybe, but I have seen a lot of rocker measurements by boat makers that could only have been obtained by measuring to a point of the hull that is usually dry.
I'll give an example: the Dagger Prophet whitewater canoe was given a rocker specification by Dagger of 8" at the bow and 7" at the stern. Do you think that boat draws 8 inches amidships with a normal load?
Your method is not bad but the point at which the hull intersects the water is going to be influenced by load as well as a lot of other factors that go into determining the overall buoyancy of the hull, such as beam and prismatic coefficient, that have nothing whatsoever to do with rocker. Likewise, V-bottomed hulls, regardless of rocker, are going to draw a bit more water than shallow arch hulls amidships.
The fact remains that boat makers have not established any uniform method for determining rocker.
Good method. I’d add . . .
. . . that to get the rocker “for you”, you should measure to the waterline point when the hull is burdened with the weight it will carry with you and your gear in it.
Of course, as you and others have stated, this depth measurement does not take into consideration the shape of the underwater keel rocker or the distribution of the underwater hull volume. No problem. A solid understanding of integral calculus and differential equations, plus a few graduate courses in naval architecture and relativistic fluid dynamics, can calculate these things on any quad-core personal computer.
But computing keel rocker is only the beginning. Since most boats are turned when at heel angles, the “side rocker” at all heel angles to the rail must also be calculated and integrated.
Then, one must compare rocker turns to flat hull spin turns. For this, we must calculate the angular momentum and vortex shedding of the so-called planing hull.
All in all, I recommend: (a) putting the boat on a flat surface and literally rocking it; (b) turning the boat over to look at the shape of the keel line and side rocker lines; and, most importantly, © paddling the sucker.
If you are really interested
– Last Updated: Nov-03-13 2:22 PM EST –
The problem of measuring rocker and an attempt to adopt a uniform method has been discussed a couple of times on another forum. Three boat builders (Richard Guin of Mohawk Canoe, John Kazimierczyk of Millbrook Canoe, and Jeremy Laucks of Blackfly Canoe) and a boat designer (Craig Smerda) chimed in on the discussion with suggestions.
The discussion pertained to canoes and the OP was talking about kayaks, but the water doesn't know whether the boat has a deck or not (until it's upside down).
https://cboats.net/cforum/viewtopic.php?f=2&t=7965353&hilit=rocker
https://www.cboats.net/cforum/viewtopic.php?f=2&t=7967689&p=104852&hilit=measuring+rocker#p104852
Thanks, but those discussions …
Which is: There is no industry standard for measuring rocker, and any measurement protocol is arbitrary and not necessarily informative.
However, an industry standard rocker protocol consistently applied would be better than the current situation.
Which is: Different rocker measurement protocols inconsistently applied.
Clearly
Good info for surf and whitewater boats
– Last Updated: Nov-03-13 4:50 PM EST –
Measured nose and tail rocker gives you a good idea of how a boat or waveski will surf on a wave, and how it will perform in general. Measurement is the height of the tip of the bow or tail off the ground when measured flat. Tail rocker is actually more tricky to measure. Best to look at a side on photo of the boat to judge the tail rocker.
http://sowsa.org/San_Onofre_Wave_Ski_Association/Similar.html
Its the kind of thing
that you recognize when you see. Give is a look from the side and you can pretty much tell what kind of rocker its has and actually it does tell you quite a bit about how the boat will behave.
The Prophet
Sorry for the “silly” remark, that wa silly to say (on my part)!
If there ever was a banana boat, the Prohpet it is:
I would not be surprised if it has 8" of rocker by “my” method of measurement…
Good point…
– Last Updated: Nov-03-13 7:38 PM EST –
I suppose it is a matter of definition, especially where to measure. Many of these surf and short white water kayaks have a flat hull and very upturned noses. It is important to them to have these upturned noses. That makes measuring the rocker necessary the way you say.
But for a long touring/sea kayak type of kayak, such measurement would be meaningless. Because the area above water does not do much most of the time and is not nearly as important for that long boat as the upturned nose on a surf or whitewater kayak is for it.
For a long waterline kayak that is designed to slice rather than to plane, the curvature of the hull below the waterline is much more important.
I think there should be different ways of measuring for surf vs. for sea kayak.
I agree with both of you
There probably is no one protocol for measuring rocker that will make sense for both long boats, intended primarily for flat water, and short boats, intended for whitewater or wave surfing.
On a flat water boat a long, proud, overhanging stem up out of the water usually does nothing but catch wind and measuring rocker at a point above the waterline would not be meaningful in any functional sense.
On the other hand, big, bulbous, upturned bows and sterns become very functional on whitewater and surf boats. I haven’t surfed big ocean waves, but I can well imagine that when surfing down the face of a large wave the keel line of the boat is at an angle well off the horizontal, and an up-swept bow helps prevent the boat from pearling and pitch poling.
Likewise, whitewater boats going over big drops may come down at an angle well above 45 degrees to the horizontal. In this situation, or when plowing through large waves, a big, bent up nose keeps the boat from pitoning, diving deep, or taking on a lot of water and is very functional. Likewise, a highly rockered stern helps the aft end of the boat clear rocky shelves and ledges without hanging up.
The consensus of opinion on the cboats threads I cited seemed to be that comparing rocker figures for different manufacturers boats, even when they are of the same general type, is basically futile, and that the best way to evaluate how much rocker a hull has is to carefully look at it.
It isn’t simple
If you’re looking to compare boats, measuring at the ends doesn’t tell the entire story. You would have to factor in the rate of rise from the center to the ends. For instance, in the recent generation of creekers (Stomper and Recon), you see continuous rocker, with the deviation occurring from end to end, not just near the ends.
Rocker?
Rocker is vertical rise off the keel line at the stems. If we had stable measurements the data would be useful, but only within given hull lengths. I.e. 2" rocker for a 12 ft hull is moderate, 2" for an 18 ft hull is less so.
The symmetry of rocker is also important. Many hulls, think Prospectors, have symmetrical rocker; the same amount at both stems. Many modern hulls have differential rocker, more in the bow than stern, which resists poor paddle skills tendency to turn hulls away from the last stroke, almost universally at less then a vertical angle and always carried behind the body.
Another rocker feature is carry-out. Does the rocker extend towards amidships or is it abrupt? ICF boats have long carry-out, many WW hulls have abruptly rockered stems.
Winters proposed a standard measurement almost two decades ago; vertical rise one foot inboard from the waterline. Nobody uses that, including, maybe, John Winters, and certainly David Yost. Yost claims rocker is a drafting convention unique to individual designers. Just comparing those guys’ 15 ft solo trippers, both with differential rocker, is interesting. DY’s Kee 15 catalogs 2.5" bow rocker, John’s Osprey 1.5", yet Osprey obviously has more rocker into the bow?? So it goes.
SO rocker dimensions are helpful in determining hull performance between similar length hulls and would be more so if standardized. We need someone with a straight twenty foot board, a tape measure and, say, three or four weeks of time free…
My seat of the pants system for
assessing the effect of rocker is to see how it affects a boat’s ability to spin, to turn, and to track. There might be ways of standardizing tests for those, but I think I can usually make a good guess from photos of a hull from different angles.
Of course, rocker (however measured) isn’t the only thing affecting spin, turning, tracking. And so unless other hull characteristics can be held constant while rocker is varied, knowing the rocker doesn’t help a great deal.
Something I noticed when inspecting boats for slalom races was that a lower rocker boat might turn well because the hull was designed to sit lighter on the water. Sometimes people forget that when considering boats.
Hey, I’ve forgotten 5 semesters of
calculus. It was easy! | 3,083 | 13,975 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-21 | latest | en | 0.956851 |
https://iwantto.be/converting-mixed-numbers-to-improper-fractions-worksheet.html | 1,695,907,026,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510412.43/warc/CC-MAIN-20230928130936-20230928160936-00139.warc.gz | 368,920,875 | 5,292 | # Converting Mixed Numbers To Improper Fractions Worksheet
## Mixed Number Fraction To Improper Fraction Worksheet With Answer Key
Converting Mixed Numbers To Improper Fractions Worksheet. Describing a fraction using improper. Web converting improper to mixed fractions.
Web converting improper to mixed fractions. Web mixed numbers & improper fractions subject: Web free mixed number to improper fractions worksheets for 4th grade and 5th grade, 6th grade, 7th grade and middle school Converting improper fractions to mixed numbers can be made easier using a visual method. Below are six versions of our grade 6 math worksheet on rewriting mixed numbers as improper fractions. Web convert between mixed fraction and improper fraction sheet 1 write as mixed fraction: Web what method can help convert improper fractions to mixed numbers? Students will color in correct answers to help them. 12 25= = 7 3 39 49= = 8 11 23 35= = 4 6 44 22= = 9 5 16 31= = 3 2 write as improper. Converting improper to mixed fractions. | 231 | 1,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-40 | latest | en | 0.835546 |
http://stackoverflow.com/questions/15467219/calculate-group-characteristics-without-ddply-and-merge/15477969 | 1,454,954,171,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701153736.68/warc/CC-MAIN-20160205193913-00161-ip-10-236-182-209.ec2.internal.warc.gz | 205,467,143 | 20,575 | # Calculate “group characteristics” without ddply and merge
I wonder whether there is a more straighforward way to calculate a certain type of variables than the approach i normally take....
The example below probably explains it best. I have a dataframe with 2 columns (fruit and whether the fruit is rotten or not). I would like to, for each row, add e.g. the percentage of fruit of the same category that is rotten. For example, there are 4 entries for apples, 2 of them are rotten, so each row for apple should read 0.5. The target values (purely as illustration) are included in the "desired outcome" column.
I have previously approached this problem by * using the "ddply" command on the fruit variable (with sum/lenght as function), creating a new 3*2 dataframe * use the "merge" command to link these values back into the old dataframe.
This feels like a roundabout way, and I was wondering whether there are better/faster way of doing this! ideallly a generic approach, that is easily adjusted if one instead of the percentage needs to determine whether e.g. all fruits are rotten, any fruits are rotten, etc. etc. etc....
Many thanks in advance,
W
`````` Fruit Rotten Desired_Outcome_PercRotten
1 Apple 1 0.5
2 Apple 1 0.5
3 Apple 0 0.5
4 Apple 0 0.5
5 Pear 1 0.75
6 Pear 1 0.75
7 Pear 1 0.75
8 Pear 0 0.75
9 Cherry 0 0
10 Cherry 0 0
11 Cherry 0 0
#create example datagram; desired outcome columns are purely inserted as illustrative of target outcomes
Fruit=c(rep("Apple",4),rep("Pear",4),rep("Cherry",3))
Rotten=c(1,1,0,0,1,1,1,0,0,0,0)
Desired_Outcome_PercRotten=c(0.5,0.5,0.5,0.5,0.75,0.75,0.75,0.75,0,0,0)
df=as.data.frame(cbind(Fruit,Rotten,Desired_Outcome_PercRotten))
df
``````
-
Related discussion on the first part of your question: stackoverflow.com/q/11562656/636656 . Answers below are nicer because they combine the split-apply-combine operation with the merging in a single step. – Ari B. Friedman Mar 18 '13 at 0:59
user1885116, use `df <- data.frame(Fruit, Rotten, Desired_Outcome_PercRotten)` to create a `data.frame` from scratch instead of `as.data.frame` with `cbind`. It gets the column `Rotten` as factor, which is undesirable. – Arun Mar 18 '13 at 13:06
You can do this with just `ddply` and `mutate`:
``````# changed summarise to transform on joran's suggestion
# changed transform to mutate on mnel's suggestion :)
ddply(df, .(Fruit), mutate, Perc = sum(Rotten)/length(Rotten))
# Fruit Rotten Perc
# 1 Apple 1 0.50
# 2 Apple 1 0.50
# 3 Apple 0 0.50
# 4 Apple 0 0.50
# 5 Cherry 0 0.00
# 6 Cherry 0 0.00
# 7 Cherry 0 0.00
# 8 Pear 1 0.75
# 9 Pear 1 0.75
# 10 Pear 1 0.75
# 11 Pear 0 0.75
``````
-
I would also suggest `mutate` (the `plyr` implementation of `transform` which allows you to refer to created columns eg `ddply(df ,.(Fruit), mutate, percR = sum(Rotten) / length(Rotten), pp = Rotten *percR)` compared to `ddply(dd ,.(Fruit), transform, percR = sum(Rotten) / length(Rotten), pp = Rotten *percR)` – mnel Mar 18 '13 at 0:45
`data.table` is super fast as it updates by reference. What about using it?
``````library(data.table)
dt=data.table(Fruit,Rotten,Desired_Outcome_PercRotten)
dt[,test:=sum(Rotten)/.N,by="Fruit"]
#dt
# Fruit Rotten Desired_Outcome_PercRotten test
# 1: Apple 1 0.50 0.50
# 2: Apple 1 0.50 0.50
# 3: Apple 0 0.50 0.50
# 4: Apple 0 0.50 0.50
# 5: Pear 1 0.75 0.75
# 6: Pear 1 0.75 0.75
# 7: Pear 1 0.75 0.75
# 8: Pear 0 0.75 0.75
# 9: Cherry 0 0.00 0.00
#10: Cherry 0 0.00 0.00
#11: Cherry 0 0.00 0.00
``````
-
One solution in base R is to use `ave`.
``````within(df, {
## Because of how you've created your data.frame
## Rotten is actually a factor. So, we need to
## convert it to numeric before we can use mean
Rotten <- as.numeric(as.character(Rotten))
NewCol <- ave(Rotten, Fruit)
})
Fruit Rotten Desired_Outcome_PercRotten NewCol
1 Apple 1 0.5 0.50
2 Apple 1 0.5 0.50
3 Apple 0 0.5 0.50
4 Apple 0 0.5 0.50
5 Pear 1 0.75 0.75
6 Pear 1 0.75 0.75
7 Pear 1 0.75 0.75
8 Pear 0 0.75 0.75
9 Cherry 0 0 0.00
10 Cherry 0 0 0.00
``````
or shorter:
``````transform(df, desired = ave(Rotten == 1, Fruit))
``````
The default function applied with `ave` is `mean`, hence I have not included it here. However, you could specify a different function by appending `FUN = some-function-here` if you wanted to do something different.
-
As `ave` is already out, let me add one solution using my base R function of choice: `aggregate`.
You can get the desired data simply with:
``````aggregate(as.numeric(as.character(Rotten)) ~ Fruit, df, mean)
``````
However, you will need to still `merge` it afterwards (or in one piece):
``````merge(df, aggregate(as.numeric(as.character(Rotten)) ~ Fruit, df, mean))
``````
- | 1,714 | 5,739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2016-07 | latest | en | 0.857047 |
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A302347 a(n) = Sum of (Y(2,p)^2) over the partitions p of n, Y(2,p)= number of part sizes with multiplicity 2 or greater in p 3
0, 0, 1, 1, 3, 4, 10, 13, 25, 34, 59, 80, 127, 172, 260, 349, 505, 673, 946, 1248, 1711, 2238, 3010, 3902, 5162, 6637, 8663, 11051, 14253, 18051, 23047, 28988, 36677, 45840, 57538, 71485, 89082, 110062, 136269, 167487, 206138, 252132, 308640, 375777, 457698 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,5 COMMENTS This sequence is part of the contribution to the b^2 term of C_{1-b,2}(q) for(1-b,2)-colored partitions - partitions in which we can label parts any of an indeterminate 1-b colors, but are restricted to using only 2 of the colors per part size. This formula is known to match the Han/Nekrasov-Okounkov hooklength formula truncated at hooks of size two up to the linear term in b. It is of interest to enumerate and determine specific characteristics of partitions of n, considering each partition individually. LINKS Alois P. Heinz, Table of n, a(n) for n = 0..2000 Guo-Niu Han, The Nekrasov-Okounkov hook length formula: refinement, elementary proof, extension and applications, arXiv:0805.1398 [math.CO], 2008. Guo-Niu Han, The Nekrasov-Okounkov hook length formula: refinement, elementary proof, extension and applications, Annales de l'institut Fourier, Tome 60 (2010) no. 1, pp. 1-29. W. J. Keith, Restricted k-color partitions, Ramanujan Journal (2016) 40: 71. FORMULA a(n) = Sum_{p in P(n)} (H(2,p)^2 + 2*A024786 - 2*A024788), where P(n) is the set of partitions of n, and H(2,p) is the hooks of length 2 in partition p. G.f: (q^2*(1+q^4))/((1-q^2)*(1-q^4))*Product_{j>=1} 1/(1-q^j). a(n) ~ sqrt(3) * exp(Pi*sqrt(2*n/3)) / (8*Pi^2). - Vaclav Kotesovec, May 22 2018 EXAMPLE For a(6), we sum over partitions of six. For each partition, we count 1 for each part which appears more than once, then square the total in each partition. 6............0^2 = 0 5,1..........0^2 = 0 4,2..........0^2 = 0 4,1,1........1^2 = 1 3,3..........1^2 = 1 3,2,1........0^2 = 0 3,1,1,1......1^2 = 1 2,2,2........1^2 = 1 2,2,1,1......2^2 = 4 2,1,1,1,1....1^2 = 1 1,1,1,1,1,1..1^2 = 1 -------------------- Total.............10 MAPLE b:= proc(n, i, p) option remember; `if`(n=0 or i=1, ( `if`(n>1, 1, 0)+p)^2, add(b(n-i*j, i-1, `if`(j>1, 1, 0)+p), j=0..n/i)) end: a:= n-> b(n\$2, 0): seq(a(n), n=0..60); # Alois P. Heinz, Apr 05 2018 MATHEMATICA Array[Total[Count[Split@ #, _?(Length@ # > 1 &)]^2 & /@ IntegerPartitions[#]] &, 44] (* Michael De Vlieger, Apr 07 2018 *) PROG def sum_square_freqs_greater_one(freq_list): tot = 0 for f in freq_list: count = 0 for i in f: if i > 1: count += 1 tot += count*count return tot def frequencies(partition, n): tot = 0 freq_list = [] i = 0 for p in partition: freq = [0 for i in range(n+1)] for i in p: freq[i] += 1 for f in freq: if f == 0: tot += 1 freq_list.append(freq) return freq_list CROSSREFS Cf. A024786, A024788, A302300. Sequence in context: A031367 A073443 A257494 * A092119 A143372 A035594 Adjacent sequences: A302344 A302345 A302346 * A302348 A302349 A302350 KEYWORD nonn AUTHOR Emily Anible, Apr 05 2018 STATUS approved
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Venn Diagram Examples for Problem Solving. Computer Science. Chomsky Hierarchy
A Venn diagram, sometimes referred to as a set diagram, is a diagramming style used to show all the possible logical relations between a finite amount of sets. In mathematical terms, a set is a collection of distinct objects gathered together into a group, which can then itself be termed as a single object. Venn diagrams represent these objects on a page as circles or ellipses, and their placement in relation to each other describes the relationships between them.
The Venn diagram example below visualizes the the class of language inclusions described by the Chomsky hierarchy.
Basic Flowchart Symbols and Meaning
Flowchart Symbols and Meaning - Provides a visual representation of basic flowchart symbols and their proposed use in professional workflow diagram, standard process flow diagram and communicating the structure of a well-developed web site, as well as their correlation in developing on-line instructional projects. See flowchart's symbols by specifics of process flow diagram symbols and workflow diagram symbols.
How to Build a Flowchart
Example of DFD for Online Store (Data Flow Diagram)
Data flow diagrams (DFDs) reveal relationships among and between the various components in a program or system. DFDs are an important technique for modeling a system’s high-level detail by showing how input data is transformed to output results through a sequence of functional transformations.
Example of DFD for Online Store shows the Data Flow Diagram for online store and interactions between the Visitors, Customers and Sellers, as well as Website Information and User databases.
Steps to Creating a Sales Process Flow Chart
Venn Diagram Examples for Problem Solving
In the Venn Diagrams solution, there are the pre-made examples that can be always used for making the unique, great looking diagrams, such as the 2-set Venn ones of any needed colour, the 3-set one, the 4-set ones and the 5-set ones. Having the already previously created samples of the Venn diagrams can help any ConceptDraw DIAGRAM user make it possible to make the needed drawing within only a few minutes by editing the existing ones.
Scientific Symbols Chart
ConceptDraw DIAGRAM is the beautiful design software that provides many vector stencils, examples and templates for drawing different types of illustrations and diagrams.
Mathematics Solution from the Science and Education area of ConceptDraw Solution Park includes a few shape libraries of plane, solid geometric figures, trigonometrical functions and greek letters to help you create different professional looking mathematic illustrations for science and education.
Data Flow Diagrams
ConceptDraw DIAGRAM software enables you to quickly create data flow diagrams that include data storage, external entities, functional transforms, data flows, and control transforms and signals.
Process Flow Diagram Symbols
Chemical and Process Engineering solution contains variety predesigned process flow diagram elements relating to instrumentation, containers, piping and distribution necessary for chemical engineering, and can be used to map out chemical processes or easy creating various Chemical and Process Flow Diagrams in ConceptDraw DIAGRAM.
Health Sciences
Health Sciences solution including professionally designed samples and wide range of high-quality vector icons and pictograms of basic medical sciences, surgery specializations, internal medicine, medical diagnostics, interdisciplinary fields, mental health, obstetrics and gynecology, public health, allied health, and other medical specialties, is a perfect assistant in drawing Health diagrams, schematics, illustrations, infographics and slides for presentations on the healthcare, medical and health sciences thematic, in development knowledge in health science, in studying and representing the physical, mental and social aspects of human health, in determining the role of different health sciences at this process, finding and explaining new methods of treatment different diseases, making an overview and describing the main characteristics of health sciences in a visual form.
Diagram Flow Chart
ConceptDraw DIAGRAM is a software for producing flow charts. The software delivers built-in object libraries with vector stencils that allows you to use RapidDraw technology. By clicking on direction arrows one can add a new object to flowchart. Users can start drawing their own flowchart diagrams in fast and simple way. | 852 | 4,649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2022-33 | latest | en | 0.915294 |
http://www.chegg.com/homework-help/structural-analysis-8th-edition-chapter-2-solutions-9780133002355 | 1,469,948,585,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469258950570.93/warc/CC-MAIN-20160723072910-00014-ip-10-185-27-174.ec2.internal.warc.gz | 347,615,999 | 17,343 | View more editions
# TEXTBOOK SOLUTIONS FOR Structural Analysis 8th Edition
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Chapter: Problem:
100% (3 ratings)
Determine the horizontal and vertical components of reaction at the pins A, B, and C.
SAMPLE SOLUTION
Chapter: Problem:
100% (3 ratings)
• Step 1 of 6
Draw free body diagram of the whole system.
• Step 2 of 6
Apply equilibrium conditions.
Therefore, the horizontal reaction at C is .
• Step 3 of 6
Consider forces along horizontal direction.
Therefore, the horizontal reaction at A is
• Step 4 of 6
Consider the member AB.
• Step 5 of 6
Therefore, the vertical reaction at A is .
Consider forces along vertical direction.
Therefore, the vertical reaction at B is .
• Step 6 of 6
From the whole truss, calculate the value of support reaction at C.
Therefore, the vertical reaction at C is .
Calculate the angle at B.
Calculate force in the member BC. | 263 | 1,078 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2016-30 | longest | en | 0.856347 |
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- - transonic compressor Convrgce pb with transient .. (http://www.cfd-online.com/Forums/cfx/26254-transonic-compressor-convrgce-pb-transient.html)
Noureddine August 13, 2008 15:23
transonic compressor Convrgce pb with transient ..
Hi,
I would excute a transient simulation of an isolated transonic Rotor 37 using ANSYS CFX 11 with rotation along one pitch, i did not have any problem to get results for steady fluid flow but i could not converge with transient any way.
However, I noticed that the value of pitches displayed by the solver doesnt correspond to that i specified.
my setup is as fellow :
- Rotor 37 design speed = 17188.7 rpm = 1800 [rad/s]
- 36 blades ===> 1 pitch = 2*pi/36 = 2*3.1416/36 = 0.1745333 [rad]
- total time to rotate along one pitch = 0.1745333/1800 = 9.7e-5 [s]
- if i choose 100 time steps ===> DT = 9.7e-5/100 = 9.7e-7 [s].
I should see in each time step an increase of pitches by 0.01, but it is not the case !!!!!!
what it can be the problem ?????
Regards,
sfallah November 10, 2014 11:58
Dear Noureddine
I can not simulate steady case of rotor NASA 37, all my simulation leads to overflow. Can you give me your boundary condition containing: total inlet pressure, mass flow rate or static outlet pressure for steady simulation of one blade(with periodic condition) of NASA 37 in CFX??? I guess that overflow arisen from fault boundary condition. :confused:
ghorrocks November 10, 2014 17:05
sfallah November 11, 2014 03:03
Quote:
Originally Posted by ghorrocks (Post 518407) Have you read the FAQ on overflow error? http://www.cfd-online.com/Wiki/Ansys...do_about_it.3F
Yes, I read it but my sticky problem is not solved. My operating condition is selected according to suggestion of AIAA paper:"Fully Coupled Fluid-Structural Interaction of a Transonic Rotor at Near-Stall Conditions Using Detached Eddy Simulation":
Total inlet pressure=17.7 (psi)
Outlet Mass flow rate= 20.19(kg/s)
Total inlet temp=519 (R)
My grid contain about 600000 element for on blade passage, physical time step in steady simulation assumed 0.0001s,
I used Geometry of NASA 37 which exist in turbogrid tutorial. I test it in different inlet and outlet domain length(by extending original geometry in Bladegen)
In solution procedure, first, Mach Number increased gradually, then Notice:"a wall hase been placed at portion of an outlet..." appears in monitor screen and finally :Overflow!!!!!!
ghorrocks November 11, 2014 17:55
Well, that's your problem. If you are running near stall conditions you are unlikely to have a steady state solution. You will probably need to run it transient.
sfallah November 13, 2014 07:34
Quote:
Originally Posted by ghorrocks (Post 518612) Well, that's your problem. If you are running near stall conditions you are unlikely to have a steady state solution. You will probably need to run it transient.
ghorrocks
Thank you ghorrocks
Very useful comment.
1 technical question: Does the length of the computational domain in inlet and outlet is important in turbomachinery?
Simulation of Original geometry of NASA67 as exist in turbogrid tutorial(Small inlet and outlet domain) leads to smooth but low slop convergence curve, in the other hand, using geometry with extended inlet and outlet length leads to steep and oscillatory convergence curve. Which of them is correct and optimum?????
ghorrocks November 13, 2014 18:28
Quote:
Does the length of the computational domain in inlet and outlet is important in turbomachinery?
Yes, it does. This is one of the normal things to check with a sensitivity analysis. A longer domain is more accurate, but will result in a larger model. If the smaller model is converging better then I would be suspicious it is artificially damping the result causing inaccuracy.
sfallah December 23, 2014 12:48
Outflow boundary condition NASA37
Dear All
What is the best outlet boundary condition for transonic(subsonic inlet and outlet but transonic passage) compressor and in general transonic turbomachines? why?
I would like to have specified inlet mass flow rate. I use total pressure(because of more stable and better convergence behavior than inlet mass flow rate) at inlet but by applying static pressure at outlet, desired mass flow rate is not be obtained.
My case is Nasa 37 rotor which outlet length is short. using k-omega sst and steady-state option, I have not converged results but using k-epsilon convergence attainment is easy. I guess that its reason is static outlet boundary condition which forced at non-uniform flow location (outlet). | 1,154 | 4,682 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2016-44 | latest | en | 0.859942 |
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voteerto55 2023-02-14 | 637 | 1,687 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 11, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-38 | latest | en | 0.659503 |
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A143028 A sequence of asymptotic density zeta(2) - 1, where zeta is the Riemann zeta function. 10
1, 2, 4, 5, 6, 8, 9, 10, 12, 13, 16, 17, 18, 20, 21, 22, 24, 25, 26, 28, 29, 30, 33, 34, 36, 37, 38, 40, 41, 42, 44, 45, 46, 48, 49, 52, 53, 54, 56, 57, 58, 60, 61, 62, 64, 65, 66, 69, 70, 72, 73, 74, 77, 80, 81, 82, 84, 85, 88, 89, 90, 92, 93, 94, 96, 97, 98, 100, 101, 102, 105 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS x is an element of this sequence if when m>1 is the least natural number such that the least positive residue of x mod m! is no more than (m-2)!, floor[x/(m!)] is not congruent to m-1 mod m. The sequence is made up of the residue classes 1 mod 4; 2 and 8 mod 18; 4, 6, 28, 30, 52 and 54 mod 96, etc. A set of such sequences with entries for each zeta(k) - 1 partitions the integers. See the linked paper for their construction. A161189(n) = 2 if n is a term of this sequence. Similarly A161189(n) = 3, 4, 5, ... if n is in A143029, A143030...; such that the number system is partitioned into relative densities tending to (zeta(2) - 1), (zeta(3) - 1), ... such that Sum_{k>=2}: (zeta(k) - 1) = 1.0. This implies that the density of 2's in A161189 tends to (zeta(2) - 1) = (Pi^2/6 - 1) = 0.644934... . - Gary W. Adamson, Jun 07 2009 LINKS Amiram Eldar, Table of n, a(n) for n = 1..10000 William J. Keith, Sequences of Density zeta(K) - 1, INTEGERS, Vol. 10 (2010), Article #A19, pp. 233-241. Also arXiv preprint, arXiv:0905.3765 [math.NT], 2009 and author's copy. MATHEMATICA f[n_] := Module[{k = n - 1, m = 2, r}, While[{k, r} = QuotientRemainder[k, m]; r != 0, m++]; IntegerExponent[k + 1, m] + 2]; Select[Range[100], f[#] == 2 &] (* Amiram Eldar, Feb 15 2021 after Kevin Ryde at A161189 *) CROSSREFS Cf. A143029, A143030, A143031, A143032, A143033, A143034, A143035, A143036, A339013. Cf. A161189. - Gary W. Adamson, Jun 07 2009 Sequence in context: A006594 A172276 A260483 * A277018 A277008 A091529 Adjacent sequences: A143025 A143026 A143027 * A143029 A143030 A143031 KEYWORD nonn AUTHOR William J. Keith, Jul 17 2008, Jul 18 2008 STATUS approved
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Last modified June 21 03:51 EDT 2021. Contains 345354 sequences. (Running on oeis4.) | 1,011 | 2,626 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2021-25 | latest | en | 0.77147 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=14&t=35168 | 1,610,870,160,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703509973.34/warc/CC-MAIN-20210117051021-20210117081021-00373.warc.gz | 431,556,905 | 11,267 | ## Problem A15
$c=\lambda v$
Jasmin Argueta 1K
Posts: 69
Joined: Fri Sep 28, 2018 12:16 am
### Problem A15
Can someone help me solve problem A15? I've been stuck on it for about 20 minutes. I tried using the equation deltaE=-hR(1/finialn^2---1/initialn^2). Then it didn't work so I used v=R(1/finialn^2---1/initialn^2) thats shown in the book. I understand n initial is 1 because Lyman but when I work the problems out I can't seem to get the correct answer. Using the books equation I get 2.914x10^15=(3.29x10^15)(1/finialn^2----1). After this I can't get the answer as n final equals 3.
Andrea Zheng 1H
Posts: 61
Joined: Fri Sep 28, 2018 12:26 am
### Re: Problem A15
You got the correct wavelength, however, you just switched n1 and n2 in the equation. The equation should be: wavelength = Rydberg constant *(($\frac{1}{{{n_{1}}^2}}$) - ($\frac{1}{{n_{2}}^2}$)). If you use this, you should get the correct answer of $n_{2}$ = 3.
Sana_Mian_3G
Posts: 30
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### Re: Problem A15
Another equation that can be used is En= -hR/n^2. Use this equation with both n values and then subtract them from each other to get the answer. | 392 | 1,163 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2021-04 | latest | en | 0.867224 |
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1 CHAPTER 25 Solved Problems P The following data are furnished by the Hypothetical Leasing Ltd (HLL): Investment cost Rs 500 lakh Primary lease term 5 years Estimated residual value after the primary period Nil Pre-tax required rate of return 24 per cent The HLL seeks your help in determining the annual lease rentals under the following rental structures: (a) Equated, (b) Stepped (an annual increase of 15 per cent), (c) Ballooned (annual rental of Rs 80 lakh for years 1 4) and (d) Deferred (2 years deferment period). You are required to compute the relevant annual rentals. (a) Equated annual lease rentals, Y: Y = Investment cost/pvifa (24, 5 years) = Rs 500 lakh/2.745 = Rs lakh (b) Stepped lease rental (assuming annual increase of 15 per cent annually), Y: Y PVIF(24, 1) + (1.15)Y PVIF(24, 2) + (1.15) 2 Y PVIF(24, 3) + (1.15) 3 Y PVIF(24, 4) + (1.15) 4 Y PVIF(24, 5) = Rs 500 lakh. Or 0.806Y Y Y Y Y = Rs 500 lakh Or Y = Rs 500 lakh or Y = Rs 500 lakh/ = Rs lakh Lease rentals (year-wise) (in lakh of rupees) Year Lease rent (c) Ballooned lease rental (Rs 80 lakh for years, 1 4) Rs 80 lakh PVIFA(24, 4), + Y PVIF (24, 5) = Rs 500 lakh Rs 80 lakh Y = Rs 500 lakh 0.341Y = Rs 500 lakh Rs lakh = Rs lakh or Y = Rs /0.341 = Rs lakh (ballooned payment) (d) Deferred lease rental (deferment of 2 years) Denoting Y as the equated annual rental to be charged between years 3 5, Y PVIF (24, 3) + Y PVIF (24, 4) + Y PVIF (24,5) = Rs 500 lakh Y Y Y = Rs 500 lakh Y = Rs 500 lakh/1.288 = Rs lakh. P Mr X, the Finance Manager of ABC Ltd, had almost decided to finance the purchase of Rs 20 lakh in new computer equipment with 16 % long-term debt when he was contacted by First Leasing Company Ltd. The manager of the leasing company tried to convince Mr X that leasing the equipment would be more beneficial to ABC Ltd. If ABC borrowed, the firm would be required to pay 16 per cent interest on the borrowed funds plus an annual sinking fund payment of Rs 2,00,000. The equipment has an expected life of 10 years, with an anticipated salvage value of Rs 4,00,000. The firm uses the straight line method of depreciation, and is in the 50 per cent tax bracket. The leasing company is willing to lease the equipment for Rs 3,80,000 per year. Further, it was stressed that the lease payments were fully tax deductible, while debt repayment was not. Mr X seeks your advice before committing to lease the computer equipment. What advise would you, as a financial consultant, give to the finance manager of ABC Ltd? PV of cash outflows under leasing alternative Year end Lease payment after PV factor Total PV taxes (L) (1 0.5) (0.08) 1 10 Rs 1,90, Rs 12,74,900
3 on interest depreciation (Col 2 tax cost alternative payment (Col 3 + 4) of debt (0.07) Rs 1,00,000 Rs 24,031 Rs 34,330 Rs 41, Rs 38, ,00,000 20,395 34,330 45, , ,00,000 16,250 34,330 49, , ,00,000 11,526 34,330 54, , ,00,000 6,147 34,330 59, ,440 m Total 2,02,536 Recommendation Since the PV of cash outflows for buying/borrowing (Rs 2,02,536) is lower than that of leasing (Rs 2,46,000), the buying alternative is preferred. (b) PV of cash outflows under leasing alternative, when lease rental is paid in advance Year-end Lease payment Tax shield Cash outflows PV factor Total PV after taxes (0.07) Rs 1,20,000 Rs 1,20, Rs 1,20, ,20,000 Rs 60,000 60, ,03, ,000 (60,000) (42,780) 2,80,440 Recommendation Buying alternative is better. P For the Hypothetical Ltd in P.25.3 assume, (i) The company follows written down value method of depreciation, the depreciation rate being 25 per cent. There is no other asset in this asset block; (ii) The corporate tax rate is 35 per cent; (iii) Post-tax marginal cost of capital is 10 per cent; (iv) Salvage value, Rs 40,000 at the end of 5th year. Compute the NAL to the lessee if lease rentals are paid (a) at the end of the year (b) in advance. (a) Computation of NAL (lease rentals are paid in arrear, that is, at the year-end) Benefits from leasing: Cost of the machine Rs 3,43,300 PV of tax shield on lease rentals (working note 2) 1,59,222 Total 5,02,522 Cost of leasing: PV of lease rentals (1) 4,11,960 PV of tax shield foregone on depreciation (3) 67,259 PV of interest tax shield foregone on debt (4) 43,810 PV of salvage proceeds (Rs 40, ) 24,840 PV of tax shield on short-term capital loss (5) 24,018 Total 5,71,887 NAL (69,365) Recommendation Leasing is not financially viable. (1) PV of lease rentals: Lease rentals PVIFA (14,5) = Rs 1,20, = Rs 4,11,960 (2) PV of tax shield on lease rentals: Rs 1,20, = Rs 1,59,222 (3) PV of shield foregone on depreciation Year Depreciation* Tax shield PV factor (at 0.10) Total PV 1 Rs 85,825 Rs 30, Rs 27, ,369 22, , ,277 16, , ,207 12, ,655 67,259 *No depreciation is to be charged in 5 th year as the block of assets ceases to exist.
4 (4) PV of interest tax shield Year Interest Tax shield PV factor (at 0.10) Total PV 1 Rs 48,062 Rs 16, Rs 15, ,791 14, , ,501 11, , ,052 8, , ,294 4, ,672 43,810 (5) PV of tax shield on short-term capital loss: (Cost of machine Accumulated depreciation Salvage value) t = (Rs 3,43,000 Rs 2,34,678 Rs 40,000) = Rs 68, = Rs 24,018. (b) Computation of NAL (lease rentals are paid in advance) Benefits from leasing: Cost of the machine Rs 3,43,300 PV of tax shield on lease rentals 1,59,222 Total 5,02,522 (Contd.) Cost of leasing: PV of lease rentals (1) 4,69,680 PV of tax shield foregone on depreciation 67,259 PV of interest tax shield foregone on debt 43,810 PV of salvage proceeds 24,840 PV of tax shield on short-term capital loss 24,018 Total 6,29,607 NAL (1,27,085) Recommendation Leasing is not financially viable. (1) PV of lease rentals Year Lease payment PV factor (at 0.14) Total PV 0 Rs 1,20, Rs 1,20, ,20, ,49,680 4,69,680 P For the facts in P.25.19, determine the break even lease rental (BELR) for the lessee in both the situations. (a) Computation of BELR (lease rents are paid at the end of the year) Benefits from leasing: Cost of the machine Rs 3,43,300 PV of tax shield on lease rentals (2) L Cost of leasing: PV of lease rentals (1) 3.433L PV of tax shield foregone on depreciation Rs 67,259 PV of interest tax shield foregone on debt 43,810 PV of salvage proceeds 24,840 PV of tax shield on short-term capital loss 24,018 1,59,927 BELR (L) = Rs 3,43, L = 3.433L + Rs 1,59, L = Rs 1,83,373 L = Rs 82,177 (1) PV of lease rentals: L PVIFA (14,5) = L = 3.433L (2) PV of tax shield on lease rentals: 3.433L tax rate = 3.433L 0.35 = L Benefits from leasing (b) BELR (lease rents paid in advance)
5 Cost of the machine Rs 3,43,300 PV of tax shield on lease rentals (2) L Cost of leasing PV of lease rentals (1) 3.914L Other costs (already computed) 1,59,927 BELR(L) = Rs 3,43, L = 3.914L + Rs 1,59, L = Rs 1,83,373 L = Rs 1,83,373/ = Rs 72,078 (1) PV of lease rentals = L = 3.914L, PVIFA = (years, 1 4) + 1 (year 0) = (2) PV of tax shield on lease rentals: 3.914L 0.35 = L P Agrani Ltd. is in the business of manufacturing bearings. Some more product lines are being planned to be added to the existing system. The machinery required may be bought or may be taken on lease. The cost of machine is Rs 40,00,000 having a useful life of 5 years with the salvage value of Rs 8,00,000. The full purchase value of machine can be financed by 20 per cent loan repayable in five equal instalments falling due at the end of each year. Alternatively, the machine can be procured on a 5 years lease, year-end lease rentals being Rs 12,00,000 per annum. The Company follows the written down value method of depreciation at the rate of 25 per cent. Company s tax rate is 35 per cent and cost of capital is 16 per cent. (i) Advise the company which option it should choose lease or borrow. (ii) Assess the proposal from the lessor s point of view examining whether leasing the machine is financially viable at 14 per cent cost of capital (Detailed working notes should be given). (i) PV of cash outflows under leasing alternative Year-end Lease rent after taxes PVIFA at 13 per cent Total PV [LR (1-t)] [Rs 12,00,000 (1 0.35)] [20% (1 0.35)] 1-5 Rs 7,80, Rs 27,43,260 (ii) Borrowing/Buying option Equivalent annual loan instalment = Rs 40,00,000/2.991 (PVIFA for 5 years at 20 per cent) = Rs 13,37,345. PV of cash outflows under buying alternative Year- Loan Tax advantage on Net cash outflows PVIF Total PV end instalment on Interest Depreciation at 13% (I 0.35) (D 0.35) (Col. 2 Col ) Rs 13,37,345 Rs 2,80,000 Rs 3,50,000 Rs 7,07, Rs 6,26, ,37,345 2,42,386 2,62,500 8,32, ,51, ,37,345 1,97,249 1,96,875 9,43, ,53, ,37,345 1,43,084 1,47,656 10,46, ,41, ,37,345 77,635 1,10,742 11,48, ,23,890 Total PV of cash outflows 31,96,926 Less: PV of salvage value (Rs 8,00, ) 4,34,400 Less: PV of tax savings on short-term capital loss (9,49,279 8,00,000) 0.35 = (52, ) 28,358 NPV of cash outflows 27,34,168 Schedule of debt payment Year- Loan Loan at the Payments Loan outstanding end instalment beginning of Interest Principal at the year the year (Col. 3 20%) repayment (Col. 3 Col. 5) Rs 13,37,345 Rs 40,00,000 Rs 8,00,000 Rs 5,37,345 Rs 34,62, ,37,345 34,62,655 6,92,531 6,44,814 28,17, ,37,345 28,17,841 5,63,568 7,73,777 20,44,064
6 4 13,37,345 20,44,064 4,08,813 9,28,532 11,15, ,37,345 11,15,532 2,21,813* 11,15,532 *Difference between loan instalment and loan outstanding. Schedule of Depreciation Year Depreciation Balance at the end of the year 1 Rs 40,00, = Rs10,00,000 Rs 30,00, ,00, = 7,50,000 22,50, ,50, = 5,62,500 16,87, ,87, = 4,21,875 12,65, ,65, = 3,16,406 9,49,219 (ii) Recommendation The Company is advised to go for borrowing as the PV of cash outflows under borrowing option is lower than under leasing alternative. Assumption The machine is sold after the expiry of its useful of 5 years; for this reason, the depreciation is charged in 5th year and there is no other asset in this block. Determination of NPV of cash inflows Particulars Years Lease rent Rs 12,00,000 Rs 12,00,000 Rs 12,00,000 Rs 12,00,000 Rs 12,00,000 Less: Depreciation 10,00,000 7,50,000 5,62,500 4,21,875 3,16,406 Earnings before taxes 2,00,000 4,50,000 6,37,500 7,78,125 8,83,594 Less: Taxes (0.35) 70,000 1,57,500 2,23,125 2,72,344 3,09,258 Earnings after taxes 1,30,000 2,92,500 4,14,375 5,05,781 5,74,336 Cash inflows after taxes 11,30,000 10,42,500 9,76,875 9,27,656 8,90,742 (x) PV factor at (0.14) Present value 9,91,010 8,01,682 6,59,391 5,49,172 4,62,295 Total PV of operating CFAT 34,63,550 Add: PV of salvage value of machine (8,00, ) 4,15,200 Add: PV of tax savings on short-term capital loss (52, ) 27,105 Gross PV 39,05,855 Less: Cost of machine 40,00,000 NPV (94,145) Recommendation It is not financially profitable to let out the machine on lease by the leasing Company, as NPV is negative. Assumption The machine is to be sold after the expiry of 5 years. There is no other asset in the block of 25 per cent of the lessee. P The Hypothetical Equipments Ltd (HEL) has recently leased assets worth Rs 2,500 lakh from the Hypothetical Leasing Ltd (HLL). The following facts are available: (1) Lease period, 9 years, of which the first 6 years constitute the lease term; (2) Annual lease rates: First 6 years, Rs 360/Rs 1,000; Next 3 years, Rs 15/Rs 1,000; (3) Incremental borrowing rates for HEL, 22 per cent. (a) Assuming 14 years as the average economic life of the equipment, is the lease a finance lease or an operating lease (b) Assuming further (i) physical life of 14 years, (ii) technological life of 9 years and (iii) productmarket life of 11 years, how will you classify the lease? A lease is finance lease if one of the following two conditions is satisfied: (i) The lease term exceeds 75 per cent of the useful life of the equipment (the minimum of physical useful life, technological life and product market life). (ii) The PV of lease payments exceeds 90 per cent of the fair market value of the equipment (cost of equipment), the discount rate being incremental borrowing rate in the case of lessee and cost of capital in the case of lessor.
7 (a) (i) Term of lease is 9/14 years = 64 per cent. (ii) Determination of PV of lease payment (Rs in lakh) Year Lease payment Discount factor (0.22) Total PV Rs 2, * 23 2,873 *( ) The lease is finance lease as the PV of lease payment exceeds the cost of asset. (b) Finance lease as term of lease is 9/9 = 100 per cent. P From the given facts relating to the Hypothetical Leasing Ltd, calculate the annual rentals under the following rental structure for the 6-year period; (a) Equated, (b) Stepped (annual increase of 12 per cent), (c) Ballooned (annual rental of Rs 15 lakh for year 1 and 2) (d) Deferred (deferment period of 1 year). Investment cost Rs 96 lakh Primary lease term 3 years Residual value Nil Pre-tax required rate of annual return 22 per cent Assume that the lease can be renewed for an additional period of 3 years (secondary lease period). The lease rental for the secondary period will be 5 per cent of the rental charged during the primary period. (a) Equated annual lease rentals, Y Y PVIFA (22, 3) Y PVIFA (22, 4 6) = Rs 96 lakh 2.042Y Y = Rs 96 lakh Y = Rs 96 lakh/ = Rs lakh (primary lease period); Rs 2.29 lakh (secondary lease period). (b) Stepped lease rentals (annual increase of 12 per cent) Y PVIF (22, 1) Y PVIF (22, 2) + (1.12) 2 Y PVIF (22, 3) + (1.12) 3 Y PVIF (22, 4) + (1.12) 4 Y PVIF (22, 5) + (1.12) 5 Y PVIF (22, 6) = Rs 96 lakh Or 0.820Y Y Y Y Y Y = Rs 96 lakh Or Y = Rs 96 lakh/ = Rs lakh (c) Ballooned lease rentals (Rs 15 lakh for years 1 2) Rs 15 lakh PVIFA (22, 2) + Y PVIF (22, 3) Y PVIFA (22, 4 6) = Rs 96 lakh Rs lakh Y Y = Rs 96 lakh Y = Rs 96 lakh Rs lakh/ = Rs lakh (d) Deferred lease rentals (deferment of 1 year), Y Y PVIF (22, 2) + Y PVIF (22, 3) Y PVIF (22, 4 6) = Rs 96 lakh 0.672Y Y Y = Rs 96 lakh Y = Rs 96 lakh/ = Rs lakh P Hypothetical Ltd is expanding its facilities. In the coming year, the company will either purchase or lease equipment which it plans to use for 4 years and then replace it with a new one. Its current tax bracket is 50 per cent. The other data are as follows: Purchase: (i) The purchase price of the equipment is Rs 40,00,000, (ii) The expected salvage value after 4 years is Rs 10,00,000, (iii) The equipment is subject to the straight line method of depreciation, (iv) Funds to finance the equipment can be obtained at 16 per cent, (v) The loan is to be repaid in four equal annual instalments due at the end of each year, (vi) The equipment will increase the annual revenues by Rs 30,00,000, and increase annual cash operating costs by Rs 20,00,000. Leasing: (i) The annual lease is Rs 10,00,000, (ii) The lease rent is payable at the end of each year for 4 years, (iii) The equipment will increase annual revenues by Rs 30,00,000 and increase annual non-depreciation operating costs by Rs 19,00,000, as the lessor will pay Rs 1,00,000 for the maintenance costs associated with the equipment. Determine whether the company should purchase or lease the equipment. PV of cash outflows under leasing alternative
8 Year- Effective lease payment PV factor Total end Gross Savings Net Tax Cash (0.08) PV in main- (Col 2 shield outflows tenance costs Col 3) (Col ) after taxes Rs 10,00,000 Rs 1,00,000 Rs 9,00,000 Rs 4,50,000 Rs 4,50, Rs 14,90,400 Determination of interest and principal components of loan instalment Year- Loan Loan at Payment of Principal end instalment the beginning out-standing Interest Principal at the end (Col ) (Col 2 Col 4) of the year Rs 14,29,593* Rs 40,00,000 Rs 6,40,000 Rs 7,89,593 Rs 32,10, ,29,593 32,10,407 5,13,665 9,15,928 22,94, ,29,593 22,94,479 3,67,117 10,62,476 12,32, ,29,593 12,32,003 1,97,590 12,32,003 *Rs 40,00, that is, PV annuity factor of 4 years at 16 per cent. PV of cash outflows under buying alternative Year Loan Interest Depreciation Cash outflows PV factor Total PV instalment (I t) (D t) after taxes (0.08) [Col 2 (Col 3 + Col 4)] Rs 14,29,593 Rs 3,20,000 Rs 3,75,000 Rs 7,34, Rs 6,80, ,29,593 2,56,832 3,75,000 7,97, ,83, ,29,593 1,83,558 3,75,000 8,71, ,91, ,29,593 98,795 3,75,000 9,55, ,02,512 4 Salvage value (10,00,000) (7,35,000) 20,23,028 Recommendation The lease alternative is better, as it is a cheaper source of finance than debt in terms of the NPV of the cash outflows. Review Questions Beta Limited is considering the acquisition of a personal computer costing Rs 50,000. The effective life of the computer is expected to be 5 years. The company plans to acquire the same either by borrowing Rs 50,000 from its banker at 15 per cent interest per annum or by lease. The company wishes to know the lease rentals to be paid annually which will match the loan option. The following further information is provided to you. (a) The principal amount of the loan will be paid in 5 annual equal instalments. (b) Interest, lease rentals, principal repayments are to be paid on the last day of each year. (c) The full cost of the computer will be written off over the effective life of computer on a straight-line basis and the same will be allowed for tax purposes. (d) The company s effective tax rate is 40 per cent and the after-tax cost of capital is 9 per cent. (e) The computer will be sold for Rs 1,700 at the end of the fifth year. The commission on such sales is 9 per cent on the sale value and the same will be paid. You are required to compute the annual lease rentals payable by Beta Ltd. which will result in indifference to the loan option Welsh Limited is faced with a decision to purchase or acquire on lease a mini car. The cost of the mini car is Rs 1,26,965. It has a life of 5 years. The mini car can be obtained on lease by paying in advance equal lease rentals annually. The leasing company desires a return of 10 per cent on the gross value of the asset. Welsh Limited can also obtain 100 per cent finance from its regular banking channel. The annual
9 rate of interest will be 15 per cent and the loan will be paid in 5 annual equal instalments, inclusive of interest, each instalment becoming due at the beginning of the year. The effective tax rate of the company is 40 per cent. For the purpose of taxation, it is to be assumed that the asset will be written off over a period of 5 years on a straight line basis. (a) Advise Welsh Limited about the method of acquiring the car. (b) What should be the annual lease rental to be charged by the leasing company to match the loan option? Computeronics Ltd sells computer services to its clients. The company has recently completed a feasibility study and decided to acquire an additional computer the details of which are as follows: 1. The purchase price of the computer is Rs 2,30,000; maintenance, property taxes and insurance will be Rs 20,000 per year. The additional annual expenses to operate the computer are estimated at Rs 80,000. If the computer is rented, the annual rent will be Rs 85,000, plus 5 per cent of annual billings. The rent is due on the last day of each year. Maintenance expenses are to be borne by the lessor. 2. Due to competitive conditions, the company feels it will be necessary to replace the computer at the end of of 3 years with a more advanced model. The resale value is estimated at Rs 1,10, The appropriate income tax rate is 50 per cent, and the straight-line method of depreciation is used. 4. The estimated annual billing for the services of the new computer will be Rs 2,20,000 during the first year, and Rs 2,60,000 during the subsequent two years. 5. If the computer is purchased, the company will borrow to finance the purchase from a bank with interest at 16 per cent. The interest will be paid regularly, and the principal will be returned in one lumpsum at the end of year 3. Assuming cost of capital at 12 per cent, you are required to analyse the financial viability of the proposal from the viewpoint of the leasing company as well as the Lessor. Answers Rs 14, (a) Leasing option is better; PV of cash outflows under lease is Rs 81,719; PV of cash outflows under borrowing and buy option is Rs 87,442, (b) Rs 34, (a) Computeronics should buy the computer (PV of cash outflows under leasing alternative is Rs 1.25 lakh and under buying alternative is Rs 1.17 lakh) (b) Proposal is financially unsound from the point of leasing company ( NPV of Rs 0.11 lakh).
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EMBA in Management & Finance Corporate Finance EMBA in Management & Finance Lecture 5: Capital Budgeting For the Levered Firm Prospectus Recall that there are three questions in corporate finance. The | 12,285 | 47,113 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2019-35 | latest | en | 0.891201 |
https://ask.sagemath.org/question/72581/extracting-specific-rows-of-a-matrix/?answer=72656 | 1,721,505,576,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517515.18/warc/CC-MAIN-20240720174732-20240720204732-00360.warc.gz | 100,482,948 | 14,845 | # Extracting specific rows of a matrix
Suppose I have a matrix A which is m x n in order. How do I extract a submatrix of specified rows and colums from A. Like I want a p x q submatrix of A. Will the expanded_submatrix method help in this regard? Thanks beforehand.
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The method matrix_from_rows_and_columns will do just this; it is especially helpful if you don't want to use consecutive rows and/or columns:
sage: M = matrix(4, 4, range(16))
sage: M
[ 0 1 2 3]
[ 4 5 6 7]
[ 8 9 10 11]
[12 13 14 15]
sage: M.matrix_from_rows_and_columns([0,3], [1])
[ 1]
[13]
sage: M.matrix_from_rows_and_columns([0,3], [1,2])
[ 1 2]
[13 14]
The arguments are a pair: a list of rows to use and a list of columns to use. The submatrix method can be used (in addition to the answers already provided) if you want consecutive rows and columns.
more
Slice indexing works for both dense and sparse matrices :
sage: M=matrix(SR, 5, sparse=True) ; M[2, 1]=3 ; M[3,2]=5 ; M
[0 0 0 0 0]
[0 0 0 0 0]
[0 3 0 0 0]
[0 0 5 0 0]
[0 0 0 0 0]
sage: M[2:5, 1:4]
[3 0 0]
[0 5 0]
[0 0 0]
HTH,
more
The syntax was unexpectedly easy. It is similar to the one used in Matlab/Octave. We just use the syntax A[i-1:p,j-1:q], wherei,j are the positions from which we want to extract the p x q matrix, to get the desired answer. So, basically matrix is just storing a list of list as an array data structure.
more
That's true for dense matrix. Try this on a sparse matrix...
( 2023-08-17 12:10:35 +0200 )edit | 525 | 1,554 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-30 | latest | en | 0.837678 |
http://stackoverflow.com/questions/14495636/strange-multiplication-behavior-in-guile-scheme-interpreter/14496685 | 1,448,929,249,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398464386.98/warc/CC-MAIN-20151124205424-00169-ip-10-71-132-137.ec2.internal.warc.gz | 217,210,572 | 25,252 | # Strange multiplication behavior in Guile Scheme interpreter
I was practicing Scheme in `Guile 1.8.8` interpreter on OS X. I noticed something interesting.
Here's `expt` function which is basically does exponentiation `expt(b,n) = b^n` :
`````` (define (square x) (* x x))
(define (even? x) (= (remainder x 2) 0))
(define (expt b n)
(cond ((= n 0) 1)
((even? n) (square (expt b (/ n 2))))
(else (* b (expt b (- n 1))))
))
``````
If I try it with some inputs
`````` > (expt 2 10)
1024
> (expt 2 63)
9223372036854775808
``````
Here comes the strange part:
`````` > (expt 2 64)
0
``````
More strangely, until `n=488` it stays at `0`:
`````` > (expt 2 487)
0
> (expt 2 488)
79916762888089401123.....
> (expt 2 1000)
1071508607186267320948425049060....
> (expt 2 10000)
0
``````
When I try this code with repl.it online interpreter, it works as expected. So what the hell is wrong with Guile?
(Note: On some dialects, `remainder` function is called as `mod`.)
-
How come you had (expt 2 64) two times and the first time it was 0, and then it wasn't (79916762888089401123.....) – malkia Jan 24 '13 at 8:16
Try renaming your `expt` to `my-expt`. Just to rule out any confusion over whether the problem is your `expt` or the builtin `expt`. – soegaard Jan 24 '13 at 8:23
Normally `remainder` and `modulo` works differently for negative numbers. – soegaard Jan 24 '13 at 8:23
Your code computes (expt 2 488) as the square of (expt 2 244). Since (expt 2 488) isn't being reported as zero, I'd bet at pretty hefty odds that what you're seeing is an oddity in display rather than in computation. What happens if you ask for something like (zerop (expt 2 100))? – Gareth McCaughan Jan 24 '13 at 9:36
Further information in case it's useful: I just tried this in guile 1.8.8 on an x64 box running FreeBSD and everything worked just fine. ahmet alp balkan, is your machine 64-bit? (Most recent Macs are.) If so, maybe the problem is somehow specific to OS X. – Gareth McCaughan Jan 24 '13 at 9:41
I recently fixed this bug in Guile 2.0. The bug came into existence when C compilers started optimizing out overflow checks, on the theory that if a signed integer overflow occurs then the behavior is unspecified and thus the compiler can do whatever it likes.
-
Mark, just to clarify, do you mean you've verified that that bug was the cause of the problem here? – Gareth McCaughan Jan 24 '13 at 10:03
The reason I ask is that it seems like a bug in Guile's multiplication couldn't actually cause these symptoms (unless it depended on the depth of the call stack, or involved uninitialized memory, or something like that -- which isn't the case for the bug you fixed). Because (expt 2 488) is calculated by doing (expt 2 244) and squaring the result -- so if (expt 2 244) is truly returning zero, how can (expt 2 488) give the right (bignum) answer? – Gareth McCaughan Jan 24 '13 at 10:07
Do we know for sure that (expt 2 244) returned 0? Could someone please confirm this? If it were the case, then I would agree with Gareth that the bug is most likely in number->string. For bignums Guile uses GNU MP's mpz_get_str, so if it's a bignum display bug, I'd be curious to see the results of "make check" for the version of GNU MP on Homebrew. It would also be worthwhile to try compiling both GNU MP and Guile with -fwrapv. – Mark H Weaver Jan 25 '13 at 3:05
Here comes the hero! Thanks! – ahmet alp balkan Jan 25 '13 at 5:46
I observed this bad behavior in a very simple statement like `(* 10000 100000000000000000000000)`. – ahmet alp balkan Jan 25 '13 at 6:22
I could reproduce the problem with guile 2.0.6 on OS X. It boils down to:
``````> (* 4294967296 4294967296)
\$1 = 0
``````
My guess is that guile uses the native int type to store small numbers, and then switches to a bignums, backed by GNU MP when the native ints are too small. Maybe in that particular case, the check fails, and the computation overflows the native int.
Interestingly, the following loop shows that squaring powers of two between 2^32 and 2^60 results in 0:
``````(let loop
((x 1)
(exp 0))
(format #t "(2^~s) ^ 2 = ~s\n" exp (* x x))
(if (< exp 100)
(loop (* 2 x) (+ 1 exp))))
``````
Results in:
``````(2^0) ^ 2 = 1
(2^1) ^ 2 = 4
(2^2) ^ 2 = 16
(2^3) ^ 2 = 64
(2^4) ^ 2 = 256
(2^5) ^ 2 = 1024
(2^6) ^ 2 = 4096
(2^7) ^ 2 = 16384
(2^8) ^ 2 = 65536
(2^9) ^ 2 = 262144
(2^10) ^ 2 = 1048576
(2^11) ^ 2 = 4194304
(2^12) ^ 2 = 16777216
(2^13) ^ 2 = 67108864
(2^14) ^ 2 = 268435456
(2^15) ^ 2 = 1073741824
(2^16) ^ 2 = 4294967296
(2^17) ^ 2 = 17179869184
(2^18) ^ 2 = 68719476736
(2^19) ^ 2 = 274877906944
(2^20) ^ 2 = 1099511627776
(2^21) ^ 2 = 4398046511104
(2^22) ^ 2 = 17592186044416
(2^23) ^ 2 = 70368744177664
(2^24) ^ 2 = 281474976710656
(2^25) ^ 2 = 1125899906842624
(2^26) ^ 2 = 4503599627370496
(2^27) ^ 2 = 18014398509481984
(2^28) ^ 2 = 72057594037927936
(2^29) ^ 2 = 288230376151711744
(2^30) ^ 2 = 1152921504606846976
(2^31) ^ 2 = 4611686018427387904
(2^32) ^ 2 = 0
(2^33) ^ 2 = 0
(2^34) ^ 2 = 0
(2^35) ^ 2 = 0
(2^36) ^ 2 = 0
(2^37) ^ 2 = 0
(2^38) ^ 2 = 0
(2^39) ^ 2 = 0
(2^40) ^ 2 = 0
(2^41) ^ 2 = 0
(2^42) ^ 2 = 0
(2^43) ^ 2 = 0
(2^44) ^ 2 = 0
(2^45) ^ 2 = 0
(2^46) ^ 2 = 0
(2^47) ^ 2 = 0
(2^48) ^ 2 = 0
(2^49) ^ 2 = 0
(2^50) ^ 2 = 0
(2^51) ^ 2 = 0
(2^52) ^ 2 = 0
(2^53) ^ 2 = 0
(2^54) ^ 2 = 0
(2^55) ^ 2 = 0
(2^56) ^ 2 = 0
(2^57) ^ 2 = 0
(2^58) ^ 2 = 0
(2^59) ^ 2 = 0
(2^60) ^ 2 = 0
(2^61) ^ 2 = 5316911983139663491615228241121378304
(2^62) ^ 2 = 21267647932558653966460912964485513216
(2^63) ^ 2 = 85070591730234615865843651857942052864
(2^64) ^ 2 = 340282366920938463463374607431768211456
(2^65) ^ 2 = 1361129467683753853853498429727072845824
(2^66) ^ 2 = 5444517870735015415413993718908291383296
(2^67) ^ 2 = 21778071482940061661655974875633165533184
(2^68) ^ 2 = 87112285931760246646623899502532662132736
(2^69) ^ 2 = 348449143727040986586495598010130648530944
(2^70) ^ 2 = 1393796574908163946345982392040522594123776
(2^71) ^ 2 = 5575186299632655785383929568162090376495104
(2^72) ^ 2 = 22300745198530623141535718272648361505980416
(2^73) ^ 2 = 89202980794122492566142873090593446023921664
(2^74) ^ 2 = 356811923176489970264571492362373784095686656
(2^75) ^ 2 = 1427247692705959881058285969449495136382746624
(2^76) ^ 2 = 5708990770823839524233143877797980545530986496
(2^77) ^ 2 = 22835963083295358096932575511191922182123945984
(2^78) ^ 2 = 91343852333181432387730302044767688728495783936
(2^79) ^ 2 = 365375409332725729550921208179070754913983135744
(2^80) ^ 2 = 1461501637330902918203684832716283019655932542976
(2^81) ^ 2 = 5846006549323611672814739330865132078623730171904
(2^82) ^ 2 = 23384026197294446691258957323460528314494920687616
(2^83) ^ 2 = 93536104789177786765035829293842113257979682750464
(2^84) ^ 2 = 374144419156711147060143317175368453031918731001856
(2^85) ^ 2 = 1496577676626844588240573268701473812127674924007424
(2^86) ^ 2 = 5986310706507378352962293074805895248510699696029696
(2^87) ^ 2 = 23945242826029513411849172299223580994042798784118784
(2^88) ^ 2 = 95780971304118053647396689196894323976171195136475136
(2^89) ^ 2 = 383123885216472214589586756787577295904684780545900544
(2^90) ^ 2 = 1532495540865888858358347027150309183618739122183602176
(2^91) ^ 2 = 6129982163463555433433388108601236734474956488734408704
(2^92) ^ 2 = 24519928653854221733733552434404946937899825954937634816
(2^93) ^ 2 = 98079714615416886934934209737619787751599303819750539264
(2^94) ^ 2 = 392318858461667547739736838950479151006397215279002157056
(2^95) ^ 2 = 1569275433846670190958947355801916604025588861116008628224
(2^96) ^ 2 = 6277101735386680763835789423207666416102355444464034512896
(2^97) ^ 2 = 25108406941546723055343157692830665664409421777856138051584
(2^98) ^ 2 = 100433627766186892221372630771322662657637687111424552206336
(2^99) ^ 2 = 401734511064747568885490523085290650630550748445698208825344
(2^100) ^ 2 = 1606938044258990275541962092341162602522202993782792835301376
``````
-
Then why does 2^488 work correctly but nothing in 2^(64..487) range? 2^65 does not work as well, that could be because 2^64 is interpreted as 0, but what does change at 2^488? – ahmet alp balkan Jan 24 '13 at 8:41
Also, `(* 4294967297 4294967295)` gives -1, `(* 4294967296 4294967295)` gives -4294967296 and `(* 4294967297 4294967297)` gives 8589934593. That's on Guile 1.8.8 from Homebrew. – Lloeki Jan 24 '13 at 9:26
I wasn't able to reproduce your results running Arch.
Here is a log of my terminal session:
``````\$ uname -r
3.6.10-1-ARCH
\$ guile --version
Guile 1.8.8
Copyright (c) 1995, 1996, 1997, 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008 Free Software Foundation
Guile may be distributed under the terms of the GNU General Public Licence;
certain other uses are permitted as well. For details, see the file
`COPYING', which is included in the Guile distribution.
There is no warranty, to the extent permitted by law.
\$ guile
guile> (define (square x) (* x x))
guile> (define (even? x) (= (remainder x 2) 0))
guile> (define (expt b n)
(cond ((= n 0) 1)
((even? n) (square (expt b (/ n 2))))
(else (* b (expt b (- n 1))))))
guile> (expt 2 10)
1024
guile> (expt 2 64)
18446744073709551616
guile> (expt 2 487)
399583814440447005616844445413525287135820562261116307309972090832047582568929999375399181192126972308457847183540047730617340886948900519205142528
guile> (expt 2 488)
799167628880894011233688890827050574271641124522232614619944181664095165137859998750798362384253944616915694367080095461234681773897801038410285056
guile> (expt 2 1000)
10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652624386837205668069376
guile> (expt 2 10000)
19950631168807583848837421626835850838234968318861924548520089498529438830221946631919961684036194597899331129423209124271556491349413781117593785932096323957855730046793794526765246551266059895520550086918193311542508608460618104685509074866089624888090489894838009253941633257850621568309473902556912388065225096643874441046759871626985453222868538161694315775629640762836880760732228535091641476183956381458969463899410840960536267821064621427333394036525565649530603142680234969400335934316651459297773279665775606172582031407994198179607378245683762280037302885487251900834464581454650557929601414833921615734588139257095379769119277800826957735674444123062018757836325502728323789270710373802866393031428133241401624195671690574061419654342324638801248856147305207431992259611796250130992860241708340807605932320161268492288496255841312844061536738951487114256315111089745514203313820202931640957596464756010405845841566072044962867016515061920631004186422275908670900574606417856951911456055068251250406007519842261898059237118054444788072906395242548339221982707404473162376760846613033778706039803413197133493654622700563169937455508241780972810983291314403571877524768509857276937926433221599399876886660808368837838027643282775172273657572744784112294389733810861607423253291974813120197604178281965697475898164531258434135959862784130128185406283476649088690521047580882615823961985770122407044330583075869039319604603404973156583208672105913300903752823415539745394397715257455290510212310947321610753474825740775273986348298498340756937955646638621874569499279016572103701364433135817214311791398222983845847334440270964182851005072927748364550578634501100852987812389473928699540834346158807043959118985815145779177143619698728131459483783202081474982171858011389071228250905826817436220577475921417653715687725614904582904992461028630081535583308130101987675856234343538955409175623400844887526162643568648833519463720377293240094456246923254350400678027273837755376406726898636241037491410966718557050759098100246789880178271925953381282421954028302759408448955014676668389697996886241636313376393903373455801407636741877711055384225739499110186468219696581651485130494222369947714763069155468217682876200362777257723781365331611196811280792669481887201298643660768551639860534602297871557517947385246369446923087894265948217008051120322365496288169035739121368338393591756418733850510970271613915439590991598154654417336311656936031122249937969999226781732358023111862644575299135758175008199839236284615249881088960232244362173771618086357015468484058622329792853875623486556440536962622018963571028812361567512543338303270029097668650568557157505516727518899194129711337690149916181315171544007728650573189557450920330185304847113818315407324053319038462084036421763703911550639789000742853672196280903477974533320468368795868580237952218629120080742819551317948157624448298518461509704888027274721574688131594750409732115080498190455803416826949787141316063210686391511681774304792596709376
guile> (exit)
``````
- | 4,929 | 12,831 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2015-48 | latest | en | 0.852575 |
https://it.mathworks.com/matlabcentral/answers/495601-error-using-plot-vectors-must-be-same-length | 1,713,128,846,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.9/warc/CC-MAIN-20240414192536-20240414222536-00400.warc.gz | 298,687,253 | 27,498 | # Error using plot, vectors must be same length
4 visualizzazioni (ultimi 30 giorni)
Saad Khan il 9 Dic 2019
Risposto: Saad Khan il 9 Dic 2019
Dear MATLAB experts,
sorry for the dumb question but im in serious need of help.
Im doing alot of different calculations and some of the results that i will be using is placed in two vectors, c and H.
the size for them is: c: 64x4double and H:69x1double.
Im trying to do a plot of H and c where H is the x-axis and c being y-axis.
I have tried changing the code but it messes up my calculations.
If anyone knows a solution to avoid this it would be much appreciated.
I attached the code in the file: optim_o2_co_vs_height.m
p.s im still learning how to use MATLAB
Best Regards
##### 0 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti
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### Risposta accettata
Saad Khan il 9 Dic 2019
Hi,
that's what im looking for, some way to force it to plot identical size of the "c".
Is it possible to crop H?
its a concentration vs height plot and a concentration vs temperature that im trying to do.
Best Regards
##### 1 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
Nicolas B. il 9 Dic 2019
To crop H, you need to decide which samples of H are not relevant to plot data. E.g:
H(5:end)
or
H(1:end-5)
Accedi per commentare.
### Più risposte (2)
Nicolas B. il 9 Dic 2019
Modificato: Nicolas B. il 9 Dic 2019
I have a question of logic about what you want to plot. Because you want t plot with:
• x-axis: H, size 69x1
• y-axis: c(:, 1), size 64x1
So there are 5 missing samples in the y-axis!
You should define whether you want to crop your H-vector while plotting or if you need to add some values to c in order to plot.
Edit: you will have the same problem at line 166 where Temperature's size is also 69x1.
##### 0 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti
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Saad Khan il 9 Dic 2019
I just tried to change the ODE solver and I get different sizes of the different vectors. I assume its because of the limits each ODE has.
For the H, i just want values until 3 is reached or 3+ 0.09.
Should i just implement it in the plot commando e.g:
plot(H(1:end-5),c(1:etc....)?
Best Regards
##### 0 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti
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Translated by | 733 | 2,575 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-18 | latest | en | 0.794936 |
https://dodgerocksgasmonkey.com/forms-of-quadratic-expression-76 | 1,674,991,399,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499713.50/warc/CC-MAIN-20230129112153-20230129142153-00670.warc.gz | 240,616,269 | 4,609 | # Forms of quadratic equation
Standard Form of Quadratic Equation: y=ax^2+bx+c y = ax2 + bx+ c The benefits of standard form include quickly identifying the end behavior of a function and identifying the values of a, b, a,b, and c c. The end behavior of a
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## Forms of Quadratic Functions: Standard, Vertex & Factored
Quadratic Equation in Standard Form: ax 2 + bx + c = 0. Quadratic Equations can be factored. Quadratic Formula: x = −b ± √ (b2 − 4ac) 2a. When the Discriminant ( b2−4ac) is: positive, there
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