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https://hpmuseum.org/forum/showthread.php?mode=threaded&tid=9887&pid=88265 | 1,638,592,566,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362930.53/warc/CC-MAIN-20211204033320-20211204063320-00633.warc.gz | 324,843,546 | 5,575 | Triangular number AND sum of first m factorials
01-09-2018, 04:31 PM
Post: #1
Joe Horn Senior Member Posts: 1,774 Joined: Dec 2013
Triangular number AND sum of first m factorials
153 (my favorite number) is both a triangular number (the sum of the integers 1 through $$n$$; in this case $$n=17$$) as well as the sum of the factorials $$1!$$ through $$m!$$ (in this case $$m=5$$).
The first three natural numbers which have both of those properties are 1, 3 (both trivial) and 153. Find the next number in this sequence. For extra credit, find the mathematical relationship between $$n$$ and $$m$$ for all members of this sequence (which apparently is not yet in OEIS).
<0|ΙΈ|0>
-Joe-
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Messages In This Thread Triangular number AND sum of first m factorials - Joe Horn - 01-09-2018 04:31 PM RE: Triangular number AND sum of first m factorials - Gerson W. Barbosa - 01-09-2018, 08:53 PM RE: Triangular number AND sum of first m factorials - Dieter - 01-09-2018, 10:19 PM RE: Triangular number AND sum of first m factorials - Gerson W. Barbosa - 01-09-2018, 11:00 PM RE: Triangular number AND sum of first m factorials - Valentin Albillo - 01-09-2018, 10:16 PM RE: Triangular number AND sum of first m factorials - John Keith - 01-09-2018, 11:10 PM RE: Triangular number AND sum of first m factorials - Gerson W. Barbosa - 01-10-2018, 04:03 AM RE: Triangular number AND sum of first m factorials - Joe Horn - 01-10-2018, 04:58 AM RE: Triangular number AND sum of first m factorials - Paul Dale - 01-10-2018, 06:35 AM RE: Triangular number AND sum of first m factorials - Joe Horn - 01-11-2018, 03:01 AM RE: Triangular number AND sum of first m factorials - Paul Dale - 01-11-2018, 10:21 AM RE: Triangular number AND sum of first m factorials - Gerson W. Barbosa - 01-11-2018, 06:29 PM RE: Triangular number AND sum of first m factorials - John Keith - 01-11-2018, 10:43 PM RE: Triangular number AND sum of first m factorials - John Keith - 01-11-2018, 10:30 PM RE: Triangular number AND sum of first m factorials - John Cadick - 01-11-2018, 02:22 PM
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https://www.bartleby.com/questions-and-answers/a-block-of-mass-m1-is-placed-on-top-of-a-block-of-mass-m2.-you-push-the-two-blocks-so-that-they-move/4f9bc324-b2b9-4569-9f07-8e48a1d97bb7 | 1,585,789,375,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506477.26/warc/CC-MAIN-20200401223807-20200402013807-00309.warc.gz | 766,271,911 | 30,076 | # A block of mass m1 is placed on top of a block of mass m2. You push the two blocks so that they move up a vertical wall, with a constant force of magnitude P directed at angle theta. The coefficient of friction between m2 and the wall is u. There is no friction between the blocks and no friction between m1 and the wall. Find the force exerted by the lower block on the upper block. The answer the book gives me is (P(cos(theta)-u*sin(theta)))/(1+(m2/m1))I get how to get the numerator but am unsure of how the denominator answer was gotten and where gravity went in the equation
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A block of mass m1 is placed on top of a block of mass m2. You push the two blocks so that they move up a vertical wall, with a constant force of magnitude P directed at angle theta. The coefficient of friction between m2 and the wall is u. There is no friction between the blocks and no friction between m1 and the wall. Find the force exerted by the lower block on the upper block.
The answer the book gives me is (P(cos(theta)-u*sin(theta)))/(1+(m2/m1))
I get how to get the numerator but am unsure of how the denominator answer was gotten and where gravity went in the equation
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Step 1
Free body diagram of the mass m2 is shown below.
Step 2
Let N12 is the normal force between mass m1 and m2. From the free body diagram, write the expression for net force in the vertical direction.
Step 3
From the free body diagram, write the expression ...
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SciencePhysics | 418 | 1,768 | {"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} | 3.421875 | 3 | CC-MAIN-2020-16 | latest | en | 0.925407 |
http://www.instructables.com/id/Simple-Cheap-Motor-Driver-Board-for-Arduino/ | 1,498,482,737,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320736.82/warc/CC-MAIN-20170626115614-20170626135614-00261.warc.gz | 566,350,221 | 18,278 | Many projects with Arduino often need motors, whether it is a car or a vacuum cleaner or anything else. But the problem is that you cannot connect a motor directly to Arduino as the output current is very low. So, you can use an H-bridge circuit to operate the motors. The H-bridge circuit can operate the motors in both directions (clockwise and anti clockwise), but in many projects you don't need a motor to turn in both directions (e.g. fan). In such cases, it is a waste of money to buy an H-bridge circuit. So, in this instructable I'm going to show you how to make a motor driver board using transistors which can power two motors. This motor driver board will be able to operate the motors only in one direction, but would be cheaper than an H-bridge (E.g. For me, H-bridge would cost about Rs 150+ , whereas I could make this board in about Rs 30); and more importantly it's fun to do it.
Step 1: Components and Tools Needed
COMPONENTS:
2* BC-547 Transistor (NPN)
2* 1k Ohm resistor
2* capacitor (between 1 and 5 uF) (I used 2.2 uF)
Perf board (Any size you like, you can cut it later)
Single stranded wire
TOOLS:
Soldering iron
FOR TESTING;
Arduino (Any will do)
9v Battery
Step 2: The Circuit
Now, for the circuit. There are two transistors which do the same thing, have the same connections except the motors they power are two different ones, and their bases are connected to two different pins on Arduino, one is connected to pin 10 and the other 11 (refer to first picture).
When either of the transistor's base goes HIGH, it forms connectivity between the collector an the emitter (refer to the second picture) . As +9v is connected the collector and the motor is connected to the emitter the motor starts spinning (refer to the first picture). The capacitor is connected to protect the Arduino from noise created by the motor.
Now assemble the circuit on your breadboard, as it is not a good idea to do it on the perf board before making sure that your circuit is working (diagram shown in first picture). To make sure that your circuit is working upload the sketch given below. Connect the battery as shown in the first picture.
If your circuit works, then the motor would be stationary at the beginning then will gradually increase its speed then stop again (the same thing would keep repeating).
Step 3: Perf Board It!!
If your circuit works then it's ready for perf boarding. I haven't made it as a shield because it uses very few pins of the Arduino so, it is much harder to make it as a shield. Instead I have made it as a separate board with wires sticking out of it which you have to connect to the Arduino. But if you like, you can make it as a shield (here is a link to help you on how to make shields - https://www.instructables.com/id/How-to-Make-Custom...) .
To create a circuit board, solder all components shown in the fritzing image above to your perf board (except for the Arduino, of course). If you don't know how to solder on perf board watch the video below (This is not my video, of course).
After you have finished soldering on the perf board, test it out and your motor driver is ready!
<p>So I tried this instructables and its not working! :/ damn transistors. I hooked everything up correctly and the transistor either did nothing or started to heat up. I got some new transistors (not radioshacks assorted transistors) and I still cant get the circuit to work..... the npn transistor seems fine when checked with an ohm meter but circuit will not work when I try to dry run (without the arduino). Any help would be great, I might post my question on electrical engineering stack exchange.</p>
<p>really sorry for the late reply.</p><p>Make sure you have wired it all properly. If it all seems right then try connecting the motor to the collector and the emitter directly to gnd .If it still doesn't work just post the picture of the one you made.and I'll try to tell you anything further.</p><p>Thanks</p>
<p>Thanks for the help! The transistor gets hot to the touch and the motor will continue to run when I pull the transistors base up to 9v + and disconnect it. The motor will stop if I disconnect the power supply.</p>
<p>Hope you liked the project. Please tell me if you have any suggestions or ideas.</p><p>Thanks</p>
<p>A trouble shooting section. lol so I guess my discription above means it working?</p>
<p>Sorry, I misunderstood your previous comment . It is quite weird that you need to pull the base to 9v. Try connecting the motor between 9v , the collector and connect the Emitter to Gnd. You will get a better voltage to your motors that way. If that still doesn't work tell me.</p><p>thanks </p>
<p>Another remark: in your circuit you first put your load, and then your transistor or mosfet. Otherwise you won't have 9V over your motors as you can see in the link below </p><p><a href="http://goo.gl/vSLRCM" rel="nofollow">http://goo.gl/vSLRCM</a></p>
<p>Thanks for telling me. I saw the simulation but I did not understand why that happens and how it makes a difference. Could you tell me?</p><p>Thanks. </p>
The voltage at the emitter can maximum be 0.7V below the voltage at the base. So in the case where the load is connected to the emitter, the load only sees 4.3V approx. Over a 1k resistor, you have 4.3mA. The transistor is a current amplifier (x100 approx.) So the current flowing thru the base is 4.3/100 = 43µA. Over the 1k resistor at the base, you have a voltage loss of 43µA*1k=43mV. (so in reality there is only 4.957V at the base)<br><br>When the load is connected to the collector, you can see that the emitter is grounded (0V) So in theory, when you have 4.3V over a 0 ohm resistor, you have infinite ampere. So the resistor at the base counts. There you have 5V-0.7V (from the transisitor) over a 1k resistor. this gives 4.3mA. The transistor amplifies this to maximum 0.43A. With the load of 1k and a voltage of 9V, you need around 9mA. Which is a lot less than the maximum that the transistor can deliver.The full power is released to the load.<br><br>Hope this helps, if not, just ask...
<p>Thanks. It was really help full. If you have any more suggestions please tell me.</p>
<p>easy project. but i think using breadboard is a good idea for prototyping. see how i started from breadboard and shifted to box here <a href="https://www.instructables.com/id/Arduino-Based-Solar-PV-Energy-Meter-With-Xively-Co/">https://www.instructables.com/id/Arduino-Based-Sola...</a></p>
<p>Thanks, I saw your intructable and it is great. It is a really nice idea to use a old router as a case. Please tell me if you have any more suggestiones.</p>
You are welcome :) if you need any help with arduino i will be there :)
<p>Nice project. I like seeing folks still making one-off circuits. Days past so many folks were into making circuits, but it seems computer tech has removed so much of the basic component ideas these days. </p><p>It seem like you are basically using the NPN transistor(s) for on/off switching transistors. If that is all you are doing, a relay would work equally well. But you could use the Arduino to drive the motors through a variable speed circuit using transistors or even MOSFETS. Just a thought. Nice project anyways and glad you are into electronics. A very good field to be into by the way!</p>
<p>I guess MOSFET's would be the better option since they can handle higher current for larger motors.</p>
<p>You could use a MOSFET, but it isn't necessary for small motors like the ones I used in the picture. BTW if you have any suggestions please tell me.</p>
<p>Thank you. Yes of course you can use relays instead of transistors which will work fine. But as I'm not using any heavy duty motors and I'm using a simple 200 mah 9v battery it is not really necessary to use MOSFETS or relays. But if you are using a high power battery or power hungry motors you could use them.</p>
<p>I fully agree.<br>However, though I built many circuits in my days, it seems we have come to a point that at least financially it isnt worth building various things anymore, seeing what the price for e.g. a motordrive module is nowadays.<br>I know some people critcise me for remarks like this, but it is just a fact. Still buying a module doesnt beat the satisfaction of indeed grabbing a few components, soldering them together and see it work.<br>Well done</p>
<p>Just curious: instead of capacitors to reduce noise, could you use diodes between the Arduino and the motors? It seems like that would protect the Arduino from noise and from stray currents.</p>
<p>Yes you can, like like the person above mensioned you can use Fly back diodes instead of a capacitor.</p>
<p>I'd suggest using flyback diodes to prevent current surge from motors</p><p>https://en.wikipedia.org/wiki/Flyback_diode</p>
<p>Thanks. I am pretty new in electronics so please continue giving such suggestions. I checked the Wikipedia page and I think a fly back diode will work</p> | 2,190 | 8,966 | {"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-2017-26 | longest | en | 0.94708 |
https://www.sarthaks.com/1028457/find-hcf-of-188-and-230-by-euclids-game | 1,660,348,502,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571847.45/warc/CC-MAIN-20220812230927-20220813020927-00004.warc.gz | 841,901,685 | 14,740 | # Find HCF of 188 and 230 by Euclid’s game.
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Find HCF of 188 and 230 by Euclid’s game.
## 1 Answer
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By Euclid’s game HCF (a, b) = HCF (a, a – b) if a > b.
Here HCF (188, 230) = HCF (230, – 188) because 230 > 188
= HCF (188, 42) = HCF (146, 42)
= HCF (104, 42) = HCF (62, 42)
= HCF (42, 20) = HCF (22, 20)
= HCF (20,2) = HCF (18, 2) = 2
∴ HCF (230, 188) = 2
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1 answer | 237 | 528 | {"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} | 3.8125 | 4 | CC-MAIN-2022-33 | latest | en | 0.914279 |
https://web2.0calc.com/questions/lowest-common-multiple-of-600-and-108 | 1,548,308,765,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584518983.95/warc/CC-MAIN-20190124035411-20190124061411-00627.warc.gz | 660,462,668 | 6,073 | +0
# lowest common multiple of 600 and 108
0
539
2
lowest common multiple of 600 and 108
Jun 28, 2015
#1
+1667
+10
What you would do is find the multiples of each number first (besides 0). For 600, there is 600, 1200, 1800, 2400, 3000, 3600, 4000, etc.
And for 108: 108, 216, 324, 432, 540, 648, etc.
The lowest number that is included in both sets would be 5,400, therefore that is the LCM.
Jun 28, 2015
#1
+1667
+10
What you would do is find the multiples of each number first (besides 0). For 600, there is 600, 1200, 1800, 2400, 3000, 3600, 4000, etc.
And for 108: 108, 216, 324, 432, 540, 648, etc.
The lowest number that is included in both sets would be 5,400, therefore that is the LCM.
Anonymous4338 Jun 28, 2015
#2
+95369
+5
Excellent work anonymous4338
Here is another way
$${factor}{\left({\mathtt{600}}\right)} = {{\mathtt{2}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{{\mathtt{2}}}$$
$${factor}{\left({\mathtt{108}}\right)} = {{\mathtt{2}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{{\mathtt{3}}}$$
Include all factors but if the factor appears in both only include it once.
so lowest common multiple will be $${{\mathtt{2}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{{\mathtt{2}}} = {\mathtt{5\,400}}$$
.
Jun 30, 2015 | 540 | 1,366 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2019-04 | latest | en | 0.870179 |
https://www.doubtnut.com/question-answer/if-fxacos-x-cos-b-x-x2x0a-n-df04-is-continous-at-x0-then-the-ordered-pair-a-b-is-a-13-b-1-3-c-1-3-d--44185 | 1,627,220,693,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151672.96/warc/CC-MAIN-20210725111913-20210725141913-00510.warc.gz | 774,185,369 | 80,040 | Home
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# If f(x)=(acos x-cos b x)/(x^2),x!=0a n df(0)=4 is continous at x=0, then the ordered pair (a ,b) is a.(+-1,3) b. (1,+-3) c. (-1,-3) d. (-1,3)
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Limits And Derivatives | 555 | 1,353 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2021-31 | latest | en | 0.639108 |
http://www.cplusplus.com/forum/beginner/110573/ | 1,503,155,905,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105455.37/warc/CC-MAIN-20170819143637-20170819163637-00540.warc.gz | 517,462,416 | 8,466 | ### Help with Homework problem
I'm having a problem with this homework problem which I'm sure will end up being a duh moment, but I'm stuck. I know I'm missing the else statement I just haven't been able to figure it out.
The program is a company giving scholarships in the amounts of 1000, 500 and 250. The funds need to be awards
The awards need to be split to go into each amount if theres enough funds to. Ex. 5 \$1000 scholarships, 10 \$500 scholarships, and as many \$250 as they have money left for. With \$15,000 the Fund can award 5 \$1000 scholarships, 10 \$500 scholarships, and 20 \$250 scholarships.
``1234567891011121314151617181920212223242526272829303132333435363738394041`` `````` //included libraries #include #include //main function int main() { int cost1, cost2, cost3, num1, num2, scholarship_amount; printf("How much was in the fund last year?\n"); scanf("%d", &num1); printf("What is the yearly percentage rate?\n"); scanf("%d", &num2); scholarship_amount = (num1 * num2/100); { if (scholarship_amount >= 1000) cost1 = scholarship_amount/1000; printf ("\n%d \$1000 scholarships will be awarded.\n", cost1); if (scholarship_amount >= 500) cost2 = scholarship_amount/500-cost1; printf("%d \$500 scholarships will be awarded.\n", cost2); if (scholarship_amount >= 250) cost3 = scholarship_amount/250; printf ("%d \$250 scholarships will be awarded.\n\n", cost3); } return 0; } ``````
If statements that have more than one function require braces.
The main problem is that you're not subtracting the cost needed for scolarships before going on checking if there is money for more.
Is this heading in the right direction, or am I completely off here
``123456`` `````` if (scholarship_amount >= 1000) cost1 = scholarship_amount/1000; if(scholarship_amount-1000); printf ("\n%d \$1000 scholarships will be awarded.\n",cost1) else if(scholarship_amount =< 1000); printf("0 \$1000 scholarships will be awarded.\n" ,cost1);``````
Sorry, this is only my 3rd code ever so I'm a little confused.
The only correct lines are the first 2.
The problem is that you don't have a graps of what you have to do. Get a sheet of paper and perform the math to solve the problem, then translate that into code.
Try to use better variable names so you don't get confused when using them for operations. Also use better indentation and put brackets after ifs to separate the logic.
I'd do it like this.
``1234567891011121314151617`` ``````const int BEST_SCHOLARSHIP_COST = 1000; const int MEDIUM_SCHOLARSHIP_COST = 500; const int WORST_SCHOLARSHIP_COST = 250; int funds; int bestScholarshipAwarded = 0; int averageScholarshipAwarded = 0; int worstScholarshipAwarded = 0; // ... if(funds >= BEST_SCHOLARSHIP_COST) { bestScholarshipAwarded = funds / BEST_SCHOLARSHIP_COST; // Etc } // bestScholarshipAwarded is initialized to 0, so even if there are less than 1000\$ // and the if is not executed it will still print that there will be 0 scholarships awarded printf ("\n%d \$1000 scholarships will be awarded.\n", bestScholarshipAwarded);``````
Last edited on
I went back and worked on the math and now I think the concept is showing but I'm still missing something.
``12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849`` ``````//main function int main() { const int BEST_SCHOLARSHIP_COST = 1000; const int MEDIUM_SCHOLARSHIP_COST = 500; const int WORST_SCHOLARSHIP_COST = 250; int funds; int bestScholarshipAwarded = 0; int mediumScholarshipAwarded = 0; int worstScholarshipAwarded = 0; int amount, interest, total, remaindingbest, remaindingmedium; printf("How much was in the fund last year?\n"); scanf("%d", &amount); printf("What is the yearly percentage rate?\n"); scanf("%d", &interest); funds = (amount * interest/100); if (funds >= BEST_SCHOLARSHIP_COST) { bestScholarshipAwarded = funds / 5000; printf("%d \$1000 scholarships will be awarded.\n", bestScholarshipAwarded); remaindingbest = funds - bestScholarshipAwarded } if (funds >= MEDIUM_SCHOLARSHIP_COST) { mediumScholarshipAwarded = (funds - remaindingbest)/500; printf("%d \$500 scholarships will be awarded.\n", mediumScholarshipAwarded); remaindingmedium = funds - remaindingbest; } if (funds >= WORST_SCHOLARSHIP_COST) { worstScholarshipAwarded = (funds - remaindingmedium)/250; printf("%d \$250 scholarships will be awarded.\n", worstScholarshipAwarded); } return 0; }``````
You're missing the check for the maximum of scholarships awardable. If you have \$15000 you can award 15 best scholarships, but you want a maximum of 5, so check how many you can affor and if it's more that 5 set the amount to 5
``123456789`` ``````if (funds >= BEST_SCHOLARSHIP_COST) { bestScholarshipAwarded = funds / BEST_SCHOLARSHIP_COST; // or you can write 1000, it's just for the sake of clarity if(bestScholarshipAwarded > 5) { bestScholarshipAwarded = 5; } // ... }``````
I suppose the division on line 26 is a typo since 34 and 43 are correct.
The remainding variables are not needed, because if you award scholarships their cost can be immediately subtracted from the funds. And using them needlessly complicates the logic.
It's also dangerous to use uninitialized variables. If 500 <= funds < 1000 holds true the operations for the best scholarships will not be performed and remaindingbest will not be calculated. This means that, when line 34 executes, its value is an unpredictable random value, making the program fail. Same thing for remaindingmedium
Remember that xxxxScholarshipAwarded variables are the amount of sholarships, so to get the total cost you have to multiply by the cost of a single scholarship.
Okay you lost me on this part...So should I removing the remaindingbest and remaindingmedium?
Also would this program make more since with modules?
It would be easier to code if you just use the funds variable.
`funds = funds - bestScholarshipAwarded * BEST_SCHOLARSHIP_COST;`
Or in the short form
`funds -= bestScholarshipAwarded * BEST_SCHOLARSHIP_COST`
This way you subtract 5000 from the initial 15000, and in the second calculation you start with 10000 and divide that by the cost of a medium scholarship.
Also would this program make more since with modules?
What do you mean?
Okay here is where I'm at now with the suggestion, but I'm still not figuring out how to only have the amount only 5 to 1000 10 to 500 and the rest go to 250.
``12345678910111213`` ``````if (funds = funds - bestScholarshipAwarded * BEST_SCHOLARSHIP_COST); {bestScholarshipAwarded = (funds / BEST_SCHOLARSHIP_COST); printf("%d \$1000 scholarships will be awarded.\n", bestScholarshipAwarded); if (funds >= MEDIUM_SCHOLARSHIP_COST) {mediumScholarshipAwarded = funds / MEDIUM_SCHOLARSHIP_COST; printf("%d \$500 scholarships will be awarded.\n", mediumScholarshipAwarded); funds = funds - mediumScholarshipAwarded * MEDIUM_SCHOLARSHIP_COST;} if (funds >= MEDIUM_SCHOLARSHIP_COST); {worstScholarshipAwarded = funds / WORST_SCHOLARSHIP_COST; printf("%d \$250 scholarships will be awarded.\n", worstScholarshipAwarded);} funds = funds - worstScholarshipAwarded * WORST_SCHOLARSHIP_COST;}``````
Don't put everything inside the first if. The calculations need to be run independently from the first one.
Why did you change the first condition? It was correct before, and it is correct in #2 and #3.
I wrote how to check the max amount of scholarships 4 posts ago.
Here is what I have now... i'm just having problems with the smaller numbers now.
``1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465666768697071727374757677787980818283848586878889909192939495`` ``````//main function int main() { const int BEST_SCHOLARSHIP_COST = 1000; const int MEDIUM_SCHOLARSHIP_COST = 500; const int WORST_SCHOLARSHIP_COST = 250; int funds; int bestScholarshipAwarded = 0; int mediumScholarshipAwarded = 0; int worstScholarshipAwarded = 0; int remainder = 0; int remainingmedium = 0; int amount, interest, remainingbest; printf("How much was in the fund last year?\n"); scanf("%d", &amount); printf("What is the yearly percentage rate?\n"); scanf("%d", &interest); //formula funds = (amount * interest/100); // Checking the if funds are more than or equal to BEST if (funds >= BEST_SCHOLARSHIP_COST) { bestScholarshipAwarded = funds / BEST_SCHOLARSHIP_COST; if(bestScholarshipAwarded > 5){ remainder = bestScholarshipAwarded - 5; bestScholarshipAwarded = 5; remainingbest = BEST_SCHOLARSHIP_COST * remainder; printf("%d \$1000 scholarships will be awarded.\n", bestScholarshipAwarded); }else{ remainder = bestScholarshipAwarded % 1; bestScholarshipAwarded -= remainder; remainingbest = BEST_SCHOLARSHIP_COST * remainder; printf("%d \$1000 scholarships will be awarded.\n", bestScholarshipAwarded); } // taking the remainder of the Math above and checking to see if it is // more or equal to Medium and the else is if it is more or equal to Worst if(remainingbest >= MEDIUM_SCHOLARSHIP_COST){ mediumScholarshipAwarded = remainingbest / MEDIUM_SCHOLARSHIP_COST; if(mediumScholarshipAwarded > 10){ remainder = mediumScholarshipAwarded - 10; mediumScholarshipAwarded = 10; remainingmedium = MEDIUM_SCHOLARSHIP_COST * remainder; printf("%d \$500 scholarships will be awarded.\n", mediumScholarshipAwarded); }else{ remainder = mediumScholarshipAwarded % 1; mediumScholarshipAwarded -= remainder; remainingmedium = MEDIUM_SCHOLARSHIP_COST * remainder; printf("%d \$500 scholarships will be awarded.\n", mediumScholarshipAwarded); } worstScholarshipAwarded = remainingmedium / WORST_SCHOLARSHIP_COST; printf("%d \$250 scholarships will be awarded.\n", worstScholarshipAwarded); }else{ worstScholarshipAwarded = remainingmedium / WORST_SCHOLARSHIP_COST; printf("%d \$250 scholarships will be awarded.\n", worstScholarshipAwarded); } // Checking if funds are more than or equal to MEDIUM but obivously less then BEST }else if(funds >= MEDIUM_SCHOLARSHIP_COST){ mediumScholarshipAwarded = funds / MEDIUM_SCHOLARSHIP_COST; if(mediumScholarshipAwarded > 5){ mediumScholarshipAwarded -= 5; remainder = mediumScholarshipAwarded % 2; mediumScholarshipAwarded -= remainder; remainingmedium = MEDIUM_SCHOLARSHIP_COST * remainder; printf("%d \$500 scholarships will be awarded.\n", mediumScholarshipAwarded); }else{ remainder = mediumScholarshipAwarded % 1; mediumScholarshipAwarded -= remainder; remainingmedium = MEDIUM_SCHOLARSHIP_COST * remainder; printf("%d \$500 scholarships will be awarded.\n", mediumScholarshipAwarded); } // taking the remainder of the Math above and checking to see if it is // more or equal to Worst if (remainingmedium >= WORST_SCHOLARSHIP_COST) { worstScholarshipAwarded = remainingmedium / WORST_SCHOLARSHIP_COST; printf("%d \$250 scholarships will be awarded.\n", worstScholarshipAwarded); } //Lastly seeing if it is more or equal to WORST }else{ worstScholarshipAwarded = funds / WORST_SCHOLARSHIP_COST; printf("%d \$250 scholarships will be awarded.\n", worstScholarshipAwarded); } return 0; }``````
You're still failing to get the logic right, and honestly it's getting a bit hard to explain the mistakes through the forum.
I didn't want to write the code for you because homework is supposed to teach you this stuff, but I'll do it and hopefully you will not just turn it in without reading and understaing it.
``123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384`` ``````int main() { // These constants are not really important. // They make the code more clear but you can manually // write the values every time they are needed. const int BEST_SCHOLARSHIP_COST = 1000; const int MEDIUM_SCHOLARSHIP_COST = 500; const int WORST_SCHOLARSHIP_COST = 250; int amount = 0; printf("How much was in the fund last year?\n"); scanf("%d", &amount); // If interest is integer then interest/100 will be 0 unless interest is greater than 100 float interest = 0.0; printf("What is the yearly percentage rate?\n"); scanf("%f", &interest); int funds = amount * interest / 100; // I'll use these names since they are shorter... int bsa = 0; // bestScholarshipsAwarded int msa = 0; // mediumScholarshipsAwarded int wsa = 0; // worstScholarshipsAwarded // Check if there is enough money for at least one scholarship if(funds >= BEST_SCHOLARSHIP_COST) { // Calculate how many scholarships can be awarded with the current funds bsa = funds / BEST_SCHOLARSHIP_COST; // If the amount is more than we want to award... if(bsa > 5) { // ...set the amount to the maximum bsa = 5; } // Subtract from the funds the total cost of the scholarships we can award funds -= bsa * BEST_SCHOLARSHIP_COST; } // If you put the printf() inside the 'if' it won't print // anything if there areless than \$1000 // It doesn't really matter though printf ("%d \$1000 scholarships will be awarded.\n", bsa); // When you get at this points there are two cases: // 1) scholarships were already awarded so the total funds are diminished // 2) there wasn't enough money to award a higher-class scholarship // so the funds are unchanged // In either case the formula doesn't care about that, it only needs to know // how many funds there are in order to check if there are enough to pay for // its kind of scholarship if(funds >= MEDIUM_SCHOLARSHIP_COST) { msa = funds / MEDIUM_SCHOLARSHIP_COST; if(msa > 10) { msa = 10; } funds -= msa * MEDIUM_SCHOLARSHIP_COST; } printf ("%d \$500 scholarships will be awarded.\n", msa); // As you can see the logic is the same in all the cases, // only with different numbers if(funds >= WORST_SCHOLARSHIP_COST) { wsa = funds / WORST_SCHOLARSHIP_COST; // No maximum in the third case, so no 'if' needed funds -= wsa * WORST_SCHOLARSHIP_COST; } printf ("%d \$250 scholarships will be awarded.\n", wsa); printf("There are \$%d remaining\n", funds); // End the program return 0; }``````
If you don't understand something feel free to ask, but please make an honest effort to understand it.
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Chapter 6 Class 10 Triangles
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### Transcript
Example 4 In figure, if PQ || RS, prove that Δ POQ ~ Δ SOR. Given: PQ II RS To Prove :- Δ POQ ~ Δ SOR Proof: In Δ POQ & Δ SOR ∠ POQ = ∠ SOR (Vertically Opposite Angles) ∠ P = ∠ S (AS PQ || RS, alternate angles) ∠ Q = ∠ R (AS PQ || RS, alternate angles) Using AAA similarity criterion Therefore , Δ POQ ~ Δ SOR Hence proved | 148 | 442 | {"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} | 3.328125 | 3 | CC-MAIN-2023-14 | longest | en | 0.645407 |
https://cboard.cprogramming.com/cplusplus-programming/134646-warning-c4700-help-printable-thread.html | 1,495,863,448,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608773.42/warc/CC-MAIN-20170527040600-20170527060600-00101.warc.gz | 910,077,166 | 3,869 | # warning C4700 HELP
• 02-12-2011
rbduck09
warning C4700 HELP
I am working on the program for class that im taking and its not working this is the algorithm that Im using but I dont know if im doing it right
Note: I am very new to this
Algorithm Name: UPDATE_CHECKBOOK
Problem Input(s): start_balance - the beginning checkbook balance. xact_amount - the amount of the transaction. xact_type - the type of the transaction (credit or debit). Problem Output(s): new_balance -- the ending checkbook balance. Relevant Formula(s): Adding a credit = (starting balance) + (transaction amount) Subtracting a debit = (starting balance) - (transaction amount) Algorithm:
BEGIN
Ask the user for the starting balance.
Input the starting balance (start_balance).
Ask the user for the transaction amount.
Input the amount of the transaction (xact_amount).
Ask the user for the type of transaction.
Input the type of transaction (xact_type).
If (the value of xact_type is "Credit") Then
Compute new_balance = (start_balance + xact_amount).
Else
Compute new_balance = (start_balance - xact_amount).
End If
Output the desired value (new_balance).
END
Heres what I have
float main()
{
float start_balance, xact_amount, xact_type, new_balance, credit, debit;
cout << "Please the starting balance: ";
cin >> start_balance;
cout << "Please enter the transaction amount: ";
cin >> xact_amount;
cout << "Please enter the type of transaction: ";
cin >> xact_type;
if (xact_type = credit) {new_balance = start_balance + xact_amount;}
else if (xact_type = debit) {new_balance = start_balance - xact_amount;}
return 0;
}
• 02-12-2011
nimitzhunter
No such thing as float mAin(). Who taught you that should be shot. Put your code around a code tag. Xact-type is better be bool, I.e 1 for credit and 0 for debit, or any way you want to define it. Then your if statement just becomes
Code:
```if( xact-type) Do something Else Do something else```
• 02-12-2011
rbduck09
so what would put next to main instead of float. and what do you mean by bool. Does you mean boolean and is everything else right
is this line right in my code float start_balance, xact_amount, xact_type, new_balance, credit, debit;
Sorry for all these questions but I am very new to this.. still trying to figure everything out
• 02-12-2011
nimitzhunter
one of the two legal versions of main:
Code:
`int main()`
you don't need "credit", and "debit" becuase the value of xact_type represent them. You can use int to represent the boolean type for now.
Code:
```int xact_type; cout << Enter 1 for credit and 0 for debit: "; << something like this would do. cin >> xact_type;```
• 02-12-2011
rbduck09
This is what I have so far
Code:
```int main() { int start_balance, xact_amount, xact_type, new_balance; cout << "Please the starting balance: "; cin >> start_balance; cout << "Please enter the transaction amount: "; cin >> xact_amount; int xact_type; cout << "Please enter 1 for credit or 0 for debit: "; cin >> xact_type; }```
but it's still giving me errors
error C2086: 'int xact_type' : redefinition
see declaration of 'xact_type'
• 02-14-2011
Salem
I tagged your code, and highlighted the problem.
• 02-15-2011
Elysia
Basically it says "you are trying to define/declare something more than once." And easy fix, and Salem even pointed out where.
Simply put: once is enough. | 874 | 3,341 | {"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} | 2.859375 | 3 | CC-MAIN-2017-22 | longest | en | 0.748422 |
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# SUMPRODUCT within a pivot table
Finden Sie Table. Hier nach relevanten Ergebnissen suchen! Table finden. Erhalten Sie viele Informationen I am trying to define a sumproduct formula in a pivot table. Here is an example of my source data below: Rate per unit Units sold. \$1.00 200 \$2.00 600 \$5.00 300. I am trying to create a formula in my pivot table that creates a weighted average rate per unit. The formula I currently have is Adding an extra line into the Pivot Table Source data is not always the best option. Isn't there a way to put that sum product line into a pivot tables calculated field? My thought process is such: =SUMPRODUCT ('Count', 'Average Time')/SUM ('Count'
i want to use sumproduct formula in pivot table. for example i have table below. i want sumproduct of engg man hrs nad Drafting ManHrs from the pivot table. I have a 7200 row and 19 Department (Civil, Str, Etc) and Three diff Phase (BE, DE, PROC). so please guide me how can i generate the pivot table to show the Sumproduc of Engg & Drafting. A PivotTable is a semi-dynamic, tabular summary of data. It is one of Excel's most flexible tools and can provide results that would take some time to reconstruct with sophisticated uses of functions such as SUMIF, SUMPRODUCT, etc. Excel 2007 has increased PivotTable flexibility significantly from earlier incarnations Supposing you have created a pivot table as below screenshot shown. And I will take the pivot table as example to calculate the weighted average price of each fruit in the pivot table. 1. First of all, add a helper column of Amount in the source data
### AllWealthinfo.com - Tabl
• Re: PowerPivot - SUMPRODUCT (SUMX) within same column This problem is particularly for PowerPivot and since there is no sumproduct function, SUMX works as a substitute. There are tens of items within the data model, therefore it is easier to have it in that way. The example is just simplified version
• The SUMPRODUCT function is my favorite Excel function by a stretch! You can create some powerful calculations with the EXCEL SUMPRODUCT function by creating a criteria for a selected array. For example, you can see how much sales your sales rep did in a particular region and for a particular quarter without having to create a Pivot Table
• Technically, the SUMPRODUCT function in Excel multiplies the numbers in the specified arrays, and returns the sum of those products. The syntax of the SUMPRODUCT function is simple and straightforward: SUMPRODUCT (array1, [array2], [array3],
• SUMPRODUCT function normally multiplies the ranges or arrays together and then returns the sum of products. This SUMPRODUCT is a versatile function where it can be used to count and sum like COUNT IFS and SUMIFS function
• In this post we look at using Excel Pivot Tables versus the SumProduct formula to calculate which month had the highest production scrap within a given year. Being able to get this information can allow management to dig deeper into why a given month had such a high volume of scrap and help address issues that may have caused the spike
• The SUMPRODUCT function works with arrays, but it doesn't require the normal array syntax (Ctrl + Shift + Enter) to enter. The purpose of the SUMPRODUCT function is to multiply, then sum, arrays. If only one array is supplied, SUMPRODUCT will simply sum the items in the array. Up to 30 arrays can be supplied
• Hi All, I'm looking to create a pivot table (or hell, I'll take a straight spreadsheet formula) that utilizes sumproducts, or a variation formula of, that will allow me to filter out data and automatically update my sumproduct to exclude the filtered data. In the chart below, to get the data I need I would want to be able to filter out Product Y and get an updated weighted average based on the.
Hi there, I started to use PowerPivot two weeks ago for a project which grew too large in excel, and I must say it looks absolutely fantastic. Unfortunately, I'm struggling since two days already to find an alternative to the excel formula 'sumproduct'. I have a table with data for every hour of the year, and this for different locations, which are gathered in 'location pools' (they all have a. Sumproduct in a Pivot Table? Hello, What you need to do is calculate the percentages within the pivot table using a formula. Assuming you have Excel 2007 or later, when you select a cell in the pivot table you get a Pivot Table Tools ribbon. One of the sub ribbons is Analyze and it has a selection for formulas
Note: If you're using Excel 2010 or Excel 2007, and a named table, you can use table references to create the names. The SUMPRODUCT Function. With the a simple SUMPRODUCT function, you could sum the amounts for all the North region rows. This works well if the list is not filtered. =SUMPRODUCT((Region=A2)*(Amt) Pivot Table Training. If you use Excel, but don't know how to use Pivot Tables, you're missing out...wasting time trying to do things that a Pivot Table can do for you automatically. Core Pivot is a step-by-step Excel video course that will teach you everything you need to know to use this powerful tool
Quarterly Totals from Monthly Data | Sumproduct and Pivot TableDownload Basic Excel Assignment folder for Practicehttp://bit.ly/2PJs1PRDownload the Assignmen.. Click the insert function button (fx) under the formula toolbar, a dialog box will appear, type the keyword SUMPRODUCT in the search for a function box, SUMPRODUCT function will appear in select a function box. Double click on the SUMPRODUCT function. Popular Course in this categor
Data table or decide to perform the calculations t pivot table calculated field sumproduct to other Items within a specific pivot field SUMPRODUCT. Based on existing fields in a pivot table) ; that divides by. And Sets > calculate Item Analyze and it has a selection for formulas percentages from Download Excel Files:Start File: https://people.highline.edu/mgirvin/YouTubeExcelIsFun/YouTubersLoveExcel9-12.xlsFinished File: https://people.highline.edu/m.. A normal pivot table won't calculate a unique count, either with a calculated fieldor with a Summary. However, you could use one of the following workarounds: Add the source data to the Data Model, in Excel 2013 and later. Use PowerPivot to create the pivot table, and use its functions to create a unique count. See the details below
### Sumproduct formula in pivot table - Microsoft Communit
• g a sensitivity analysis by staying tuned for our next post on Power Pivot in the Blog section. In the meantime, please remember we have training in Power Pivot which you can find out more about here.If you wish to catch up on past articles in the meantime, you can find all of our Past Power.
• Count unique values in pivot table with Value Field Settings in Excel 2013 and later versions. In Excel 2013 and later versions, a new Distinct Count function has been added in the pivot table, you can apply this feature to quickly and easily solve this task.. 1.Select your data range and click Insert > PivotTable, in the Create PivotTable dialog box, choose a new worksheet or existing.
• However Sumproduct can be extended even further through use 2D Ranges together with carefully constructed queries. The examples below are included in the Example File, Excel 2003 Example File. Scenario 1: Lookup a value within a 2D Range matching 2 criteria. You have a table of Dates and Fruit Sold and Number Sold each Da
• SUMPRODUCT treats non-numeric array entries as if they were zeros. For best performance, SUMPRODUCT should not be used with full column references. Consider =SUMPRODUCT(A:A,B:B), here the function will multiply the 1,048,576 cells in column A by the 1,048,576 cells in column B before adding them. Example
• The MATCH section results in a row number that gets applied to the ColumnToIndex. When looking up items with more than one criteria, I like to use an INDEX/SUMPRODUCT formula, replacing the MATCH part of the single criterion formula with SUMPRODUCT array multiplication, as descibed by Chandoo.Very generically that looks like
• About Pivot Table Formulas. Use custom formulas in an Excel pivot table, to create calculated fields and calculated items. There are a few general restriction on using formulas though: Formulas are available only in non-OLAP-based pivot tables ; You can't create formulas that refer to the pivot table totals or subtotals
• e the average salary of employees, the average grade of an exam or the average selling price of a company´s stock list, as can been seen below
### Sumproduct formula inside Pivot Table MrExcel Message Boar
• Connect and share knowledge within a single location that is structured and easy to search. Learn more 'SUMIFS' or SUMPRODUCT is probably the way to go. So those formulas worked to create the data that goes into the pivot table, but something breaks down in the table..
• In Excel pivot table, calculated field is like all other fields of your pivot table, but they don't exist in the source data. But, they are created by using formulas in the pivot table. Follow these simple steps to insert calculated field in a pivot table. First of all, you need a simple pivot table to add a Calculated Field
• Yes. The reasons are obvious, pivot tables are one of the most powerful features in Excel for analyzing data. Here are some not so obvious reasons why pivot tables are awesome: * Pivot tables are the fastest way to analyze data and create powerful..
• Pivot Table is a great tool for summarizing and analyzing data in Excel. We can use a Pivot Table to perform calculations on our data based on certain criteria. For example - Sales per Store, Sales per Year, Average Discount per Region, and many more Here are some of the advantages of using Pivot Tables
• The SUMPRODUCT function is an extremely versatile function and can produce the same result as many built-in functions in Excel and even array formulas! 1a. For example, the COUNTIF function below counts the number of cells that contain exactly star. 1b. The SUMPRODUCT function below produces the exact same result
• If there is a change in the source data, then the formulas will be unchanged until the pivot table is refreshed. Recommended Articles. This has been a guide to Pivot Table Calculated Field. Here we discuss the use of use formulas in the Pivot table using calculated field along with practical examples and downloadable excel template
• Click within your pivot table, head to the Pivot Table Analyze tab within the ribbon, click Field List, and then drag Type to the filters list. 2. With that filter applied, Jason would click inside the pivot table, go back to the Pivot Table Analyze tab in the ribbon, click Options, and then select Show Report.
### Want to Use SumProduct in pivot table - Microsoft Communit
• The SUMPRODUCT formula for Total Payout is: =SUMPRODUCT( (Attainment > [Tier Min]) * (Attainment - [Tier Min]) * [Differential Rate] ) Variables in brackets [] refer to entire column in rate table. The following splits the SUMPRODUCT formula into multiple columns and rows for a clearer visual of how the formula is calculating the total payout
• Types of Filters in a Pivot Table. Here is a demo of the types of filters available in a Pivot Table. Let's look at these filters one by one: Report Filter: This filter allows you to drill down into a subset of the overall dataset. For example, if you have retail sales data, you can analyze data for each region by selecting one or more than regions (yes, it allows multiple selections as well)
• As you have understood from the above snapshot and table, that SUMPRODUCT function multiplies corresponding values and returns the Sum. 1's & 0's in the left most column is the IF logic used with SUMPRODUCT function. Hope this article about how to SUMPRODUCT with IF logic in Excel is explanatory. Find more articles on SUMIF functions here
Also, in my experience, pivot tables tend to have a problem when used for a 200x2000 table (we are now over 400k cells/formulas). - dan Sep 11 '15 at 12:29 Also, I mostly need to really have that spreadsheet with all the information (>400k cells) available at the same time to do my charts I have a table which has the Client names, the number of tickets and (per ticket) if they were solved within contracttime or not 'Binnen SLA' / 'Buiten SLA' (these values are a result of a formula) I want to filter, per client , (in a different table) the times a tickets was solved within contractime and the number of times we were to late as a. In recent blog posts, we've looked at the pivot table Count function, the Average function, and the Min and Max functions.In this article we'll look at the Product function. Product Summary Function. The Product summary function shows the result of multiplying all the underlying values in the Values area.The result is the same as using the PRODUCT function on the worksheet to calculate the.
### Pivotal PivotTables < Thought SumProduct are experts in
The sumproduct formula takes 1 or more arrays of numbers and gets the sum of products of corresponding numbers. The formula can be used to SUMIF with multiple conditions, calculate weighted averages and replace excel array formulas. In this article we explore the SUMPRODUCT formula syntax, usage, examples and tell you how to write a sumproduct formula to calculate sum of values meeting. Answer: I highly recommend a pivot table for this task, however, this article demonstrates a formula combined with an Excel defined table.A pivot table is lightning fast if you have lots of data to work with and is easy to learn. Formula in cell G6 The SUMPRODUCT function considers logic value TRUE as 1 and False as 0. The argument array must be of the same length as the function; Hope this article about how to Return Count if with SUMPRODUCT in Excel is explanatory. Find more articles on SUMPRODUCT functions here. Please share your query below in the comment box. We will assist you
### How to calculate weighted average in an Excel Pivot Table
• And it can be as long as you wish. If you want to look up 20+ different names, all you have to do is provide the list of names. If you would add this to a table, the MATCH formula could reference the entire list of names within the table and you would never have to adjust the formula at all
• I have discovered this amazing function, but my Mac struggles with it. I would like to multiply array 1 by array 2. One of these is a lookup in a table based on array 2. Is this possible? The other problem is with gaps or blanks in arrays, do these disable the SUMPRODUCT function. Repl
• Unfortunately, a pivot table doesn't have a built-in function to calculate a unique count. As a workaround, you could add a column to the source data, then add that field to the pivot table. Add a Field to the Source Data. In this example, we want to count the number of unique Customer who ordered each product
• Excel SUMPRODUCT Formula Examples: If you have a table as shown in the below image: Now if you want to apply SUMPRODUCT formula on these values then your formula should be: =SUMPRODUCT(B2:B11,C2:C11) This will give you an output = 935. So, internally SUMPRODUCT has multiplied the each element of both the arrays and then added it to one another
• The only purpose of this pivot table is to add your source table to data model - Find in menu Data->Existing connections, select here your source table and add it to Excel as new table - you shall receive exact copy of your source table - to check, right click within the table as on screenshot and find edit DAX - now a bit of DAX coding is.
• This is one of that post where filtering the pivot table is extremely easy, not the fact that it was hard (concept) earlier.. but still this trick is fun! Consider this Pivot. 2 insanely simple things here - Sum and Count of Sales. Now let's say I want to filter only the last 3 Customers (Data Tronics, White Associates and MNTL). How should.
### PowerPivot - SUMPRODUCT(SUMX) within same colum
1. SUMPRODUCT is one of Excel's most-powerful worksheet functions. Here, for example, you can use it in one formula to search text in one cell for many items. In How to Add Advanced Filter Capabilities to Excel Tables, I explained how to use a long formula within a Table to simplify complex filtering
2. Hi, I want to use the sumproduct formula to calculate weighted averages in my pivot table. Prices are in column B and quantities are in column M. The problem is that the range of columns B and M are changing every time i refresh the data form the pivot table and the total at the bottom of my pivot table witch i need to see
3. Effectively is calculating the variance (e.g. how far each row varies from the mean), however in DAX, I had to change the sumproduct()'s to sumx()'s, which in turn, modify the context so the 'mean' it is calculating is based upon the single row, opposed to all rows (within the filter)
4. To create the pivot table below, instead of the Product field, add the Date field to the Rows area. The Date field contains many items. 6-Jan, 7-Jan, 8-Jan, 10-Jan, 11-Jan, etc. To group these dates by quarters, execute the following steps. 1. Click any cell inside the column with dates. 2. Right click and click on Group
### Excel SUMPRODUCT Multiple Criteria MyExcelOnlin
There is no SUMPRODUCTIF function in excel, but you can use the SUMPRODUCT function to solve one criteria or multiple criteria questions, in the same column or different columns.. For example, there are data such as the image below. Question 1 - No Criteria. What is the total amount of money earned from the sales transaction above Hi All, How can I do Countifs or Sumproduct of dates in power query, dax/power pivot ? I have multiple columns with 2 dates columns. I need to count the rows that meets the 2 dates criteria of status Excel 2010 - SUMIF not working when cells reference Pivot Table data. There was a hotfix (which is installed) to address this issue but it doesn't seem to fix everything. While I can reference dates on another spreadsheet, its having trouble dealing with dates that get created in a Pivot Table
### Excel SUMPRODUCT function with multiple criteria - formula
1. Select Insert ->Pivot Table, it is on the left-hand side of the toolbar: Select the all the rows and columns (including the salesman name) as shown below: The pivot table dialog box appears to the right-hand side of the spreadsheet. If we click all four fields in the pivot table dialog box (Quarter, Year, Sales, and Salesperson) Excel adds a.
2. It will automatically calculate a separate percentile ranking for each country within its own continent! the world of excel-2013 with power pivot: 1. is it possible to use sumproduct & countif as 'measures' in power pivot? 2. the intent would be implement 'percentile' and/or 'percent rank' so that as the pivot table context.
3. Pivot Table Sort in Excel. To sort any pivot table, there are 2 ways. First, we can click right the pivot table field we want to sort and select the appropriate option from the Sort by list. Also, we can choose More Sort Options from the same list to sort more. Another way is by applying the filter in a Pivot table
4. To calculate the weighted average of a data with a Pivot Table, we can add a column to our source data as an intermediate calculation. This is because Excel doesn't provide a function in the Pivot Table that automatically calculates the weighted average.The steps below will walk through the process
5. Re: Database Functions Vs. Array Formulas Vs. SUMPRODUCT Vs. Pivot Tables. Hi Dave. I found out first hand that lots of SUMPRODUCT formulas slow excel right down (this was for another project I was doing not the one related to this thread), and now, although it does the job I have got to make lots of changes to do away with the SUMPRODUCT (well, majority of it) as users complain of the slowness
6. Using SUMPRODUCT to Count Records. In this example the SUMPRODUCT function is being used to count the number of orders in the list shown below. This list has been formatted as a table providing many benefits. One of the benefits is the ease of referencing the necessary columns within a formula
SUMPRODUCT; The Table in our example is called tblSales, which is referred to throughout the rest of the post. The Structured References used with Tables can also be used as a text string within the INDIRECT function. The structured reference for the 2019-Q3 column of the tblSales Table would be Excel VBA for Pivot Tables. INFO function. Hi there! I have a question about using sumif function in pivot table as below: I have a set of data like this: Date \$ Name. 1 Mar 18 10 Abby. 3 Mar 18 20 Abby. 5 May 18 25 Bob. 7 May 18 15 Carl. 12 May 18 30 Bo The other option is to use a Pivot Table. This would quickly give you a report that showed purchases for each cost centre, broken down by month. You could just easily get a report by month broken down by cost centre. The advantage of a Pivot Table is that you wouldn't need to add a separate column to calculate the month SUMPRODUCT with Criteria. This tutorial assumes you understand the basic functionality of SUMPRODUCT - in that it finds the product of corresponding values in each column/row/array and then sums those calculation results. We will focus on the opportunities to count and sum records based on criteria using SUMPRODUCT @Martin, I'm not sure if you're asking the advantage of Pivot Tables, or GetPivotData, over Sumproduct. If it's the first a short list would be: Subtotals; Automatically generating the universe of items and adjusting the table size when new ones are added; Displaying values as percents of rows; Drilling down; Doesn't break when encountering #NA in data; Filtering at the Report and field level
The steps for creating a pivot table under Solution B are: The Objective is to change the orientation of the base data such that a calculated field formula can be written within a Pivot table instead of a calculated item formula. The process to change the orientation of the base data is described below: 1 4. How to enable Drill Down Detail in Pivot Table in Google Sheets. This is the important step. Just click on the Pivot Table report to activate the Pivot table editor pane on the right. On the Pivot Table editor, click ADD against the Rows and add the Date field again This video shows you how to calculate weighted average cost from a transactional data set using SUMPRODUCT function. It explains things very clearly. If you're working with a very large database transactions the SUMPRODUCT function makes it. It works as text if you do the formatting first and then type the number. We put the number in first and then format as text, then press F2 and ENTER
### SUMPRODUCT with Multiple Criteria Function and Example
A forum for all things Excel. Ask a question and get support for our courses Sum of rows in pivot tables. The QlikView straight table has a choice for its totals between a simple sum of rows and a calculated expression total. The QlikView pivot table lacks this choice. Pivot table totals are always calculated as expression total. This is normally a good thing, since it is a rather rare occasion that a sum of rows total is relevant when the two differ I'm sure that it's (the FLATTEN, Ampersand and SPLIT combo) the simplest way to reverse pivot a table without using Google Apps Script. Note: If your pivot data is very large like 1000+ rows (it may be very rare), this formula may not work as the SPLIT function has normally issue with such a large set of data
### Sumproduct Vs. Excel Pivot Tables: Which Method is Best ..
Post category: PivotTables / SUMPRODUCT Post comments: 0 Comments Let's compare these three ways of creating a summary chart from our raw sales data to see which is more preferable You can see all Calculated Fields within a Pivot Table by following these 2 easy steps: Select Pivot Table. Go to Ribbon > Analyze > Fields, Items & Sets > List Formulas. Alternatively, use the keyboard shortcut Alt, JT, J, L. Excel creates a new worksheet. The worksheet has the following 2 tables: Calculated Fields. Calculated Items Even though I am trying to use the 'Group' function in a separate Pivot Table, excel somehow finds out that as source data is same for both pivots. The same problems happens vise versa as well. i.e. after I use the 'Group' function on source data and then create another pivot table where I want to add 'Calculated Items', Excel.
SUMPRODUCT with Multiple Criteria in excel helps in comparing the different arrays with multiple criteria. The format for SUMPRODUCT with Multiple Criteria in excel will remain the same as of Sum product formula. The only difference is that it will have multiple criteria for to multiple two or more ranges & then adding up those products However, if I wanted to do that in one step, I would use SUMPRODUCT, with the following formula: =SUMPRODUCT(Value,Percentage) And I would get the same result: But what if I need to find out the full number that, when multiplied by the percentage, gave me the value The next step is to click inside your pivot table so that the Pivot Table tools options appear in the ribbon toolbar, as shown here: From there, click Options in Excel 2010 or earlier, or Analyze in Excel 2013. This will show you a range of different options for managing your pivot table Once connected, you can use a single Slicer to filter all the connected Pivot Tables simultaneously. Remember, to connect different Pivot Tables to a Slicer, the Pivot Tables need to share the same Pivot Cache. This means that these are either created using the same data, or one of the Pivot Table has been copied and pasted as a separate Pivot.
### How to use the Excel SUMPRODUCT function Excelje
1. Select your pivot table and go to the Analyze tab in the ribbon. Press the Options button in the PivotTable section to open the options menu.. In the PivotTable Options menu go to the Data tab and uncheck the Enable show details box to disable this feature.. Replace Blank Cells. This pivot table contains blank cells because our source data does not contain any records for those combinations of.
2. Step 1: Inserting Pivot Table. Select any cell within the table. The last way to make a frequency distribution table in Excel is using SUMPRODUCT Function. I have used the same survey data, the same Income (Yearly) column, and the same bins_array to make the frequency distribution table. Take a look at the following image
3. Create a Pivot Table inside another Pivot Table. I'm going to use the following table. Create a pivot table from the table. Select all Pivot Table fields and move them to Rows, so the State is at the top. Press Left Alt (don't hold), then d, and then p to open Pivot Table wizard. Select Another PivotTable report or PivotChart report
4. #5 select any cell in the date column in the existing Pivot table. Right click on it, and select Group from the popup menu list. And the Grouping dialog will open. #6 select Days option from the By list box in the Grouping dialog box. And deselect any other selected options. Then enter number 7 in the Number of days text box. Click Ok button. #7 you would notice that your pivot table has.
5. Figure 2. Data that we will use for the Pivot table creation. Creating the Pivot Table and Finding Unique Values. First, we will create the Pivot table step by step to show important settings for finding unique values. In the new Worksheet, we choose the option to insert a Pivot table on the Insert tab. Furthermore, we enter the range with our.
6. Additionally, there shouldn't be any blank rows or columns within the data range. Excel tables Excel tables are already in list format and are good candidates for PivotTable source data. When you refresh the PivotTable, new and updated data from the Excel table is automatically included in the refresh operation
### Sumproduct in Pivot Tables - can it be done
1. You can create a new Pivot Table that doesn't share the Pivot Cache with a previously existing Pivot Table using the Pivot Table Wizard and following these 8 simple steps: Select a cell within the source data. Use the keyboard shortcut Alt, D, P. Excel displays the Pivot Table Wizard. In Step 1 of 3 of the Pivot Table Wizard, click Next
2. e a couple of ideas for computing income tax in Excel using tax tables. Specifically, we'll use VLOOKUP with a helper column, we'll remove the helper column with SUMPRODUCT, and then we'll use data validation and the INDIRECT function to make it easy to pick the desired tax table, such as single or married filing joint
3. A simple method would be to create an inline Table using the Source field and giving the 'A-' records a value of 'A' and the remaining 3 a value of 'B'. Chose suitable field names to join with your main table. You can then use the new field with the A & B values as your first dimension and use subtotals on this field. Regards. Dav
With Pivot Tables, you can easily evaluate data. Per drag-and-drop you arrange analysis layouts. Within seconds, you'll see your results - without using any formulas. Usually the first obstacle comes up, when you try to create a Pivot Table. There are some rules to regard in order to create Pivot Tables and your data needs a certain structure Pivot Table Options. The PivotTable Options enables to control Layout & Format, Display and various other settings for a pivot table. To access the pivot table options, right-click on any cell within the pivot and select Pivot Table Options. Layout & Format. This section deals with the layout of the Pivot Table using the following =SUMPRODUCT(--(dates>=start_date),--(dates<=end_date),prices) Where: dates refers to the range that contains your dates (Column A in my example below) products refers to the range that contains product names (Column B in my example below) prices refers to the range that you'd like to sum up (Column C in my example below) Example: Suppose I have. Excel's VLOOKUP function allows you to look up data from a table based on criteria that you specify, and has four arguments: lookup_value - This is the data that you want to look for in the first column of the table array.; table array - The table array is a range of two or more columns.; col_index_num - This argument allows you to specify the column within the table array for which you. Choose where you want the pivot table to be placed, new or existing worksheet. Click OK. The new pivot table will be created using the Table as the source data range. Changing the Data Source for an Existing Pivot Table. If you have an existing pivot table that uses a regular range as the source, you can change it to use a Table as the source
### Sumproduct in PowerPivot
1. Figure 1: I'll use this data set to explain why duplicate data may appear within a pivot table.. Excel 2007 and later: As shown in Figure 2, click on cell A1, choose Insert, Table, and then click OK.Click Summarize with Pivot Table from the Design tab, and then click OK. Excel 2003 and earlier: Choose Data, List, Create, and then click OK.Next, choose Data, Pivot Table Wizard, and then click.
2. Click inside the table and go to Insert -> Tables -> Pivot Table. In the Create PivotTable window use the following settings and click OK. In PivotTable Fields on the right side check both positions (Group and Name). Drag fields between areas, as shown on the image below. Now, you should see the following result
3. Like most other functions in Excel, SUMPRODUCT can be entered into a worksheet using the Functions Library found on the Formulas tab. Because the weighting formula in this examples uses SUMPRODUCT in a non-standard way (the function's result is divided by the weight factor), the weighting formula must be typed into a worksheet cell
The SUMPRODUCT function calculates the product of corresponding values and then returns the sum of each multiplication. SUMPRODUCT and nested IF functions I have demonstrated in a previous post how to simplify nested IF functions, in this article I will show you how [ Edit the source data range for your pivot table. In Excel 2007 and 2010, choose Change Data Source from the Data group of options. In Excel 2003, launch the Wizard utility by right-clicking inside the pivot table and choosing Wizard from the pop-up menu. Click the Next button until you see the screen with the source data range
So I wrote some code to create pivot table named ranges. Programming pivot tables is fun. The extensive object model is a VBA wonderland with treats around every turn. There are great web sites out there with excellent pivot table coding samples - Contextures leaps to mind. In terms of identifying PivotFields, DataFields and other pivot table. A Pivot Table is a summary of a large dataset that usually includes the total figures, average, minimum, maximum, etc. let's say you have a sales data for different regions, with a pivot table, you can summarize the data by region and find the average sales per region, the maximum and minimum sale per region, etc. Pivot tables allow us to. Syntax of SUMPRODUCT function. The general syntax of using the SUMPRODUCT function is: SUMPRODUCT(array1, [array2], [array3], ) The first argument is the array1 or range1 that you want to multiply and get the sum. This is required. The second, third and so on arrays are optional. The SUMPRODUCT function treats any non-numeric array item as zero Alter the source data for your pivot table. In Excel 2003, relaunch the pivot table wizard utility by clicking inside the pivot table and choosing Wizard from the pop-up menu. In Excel 2007 or 2010, click the Change Source Data button on the Pivot Tools Options tab
One important part of this is referencing the various ranges within a pivot table by their special VBA range names (which are actually properties of the Pivot Table object). I'll illustrate these special ranges using this simple pivot table, which comes from an example formerly available on the Microsoft web site (I can no longer locate it) Hi, In the first sheet, i have the following data S INR 2000 17.6 P SAR 300 30.6 P USD 100 38.4 P USD 200 76.8 S GBP 100 71.5 S GBP 200 143 S EUR 100 47 P AED 200 20.9 In the second sheet it is to be sorted by formula, not by pivot table i made a formula but it is giving wrong result, if somebody helps i this matter much obliged Introduction. Most people likely have experience with pivot tables in Excel. Pandas provides a similar function called (appropriately enough) pivot_table.While it is exceedingly useful, I frequently find myself struggling to remember how to use the syntax to format the output for my needs
It seems that=SUMPRODUCT(Annual rate column, Current balance column) / SUM This does not seem to work. I am attempting to do the calculation wholly within the pivot table itself, as opposed to a self-written cell. At least on the bright side I am learning better what pivot tables can and cannot do it seems. View entire discussion ( 15 comments =SUMPRODUCT(C3:H8*(C2:H2=Category 2″)*(B3:B8=Level 3″)) Related posts How to sum values with OR operator using SUMPRODUCT with multiple criteria How to sum by week number How to create an Excel summary table using UNIQUE and SUMIFS How to remove blanks from a list How to create a year-by-year average calculato How To: Do 2 pivot tables w/ different date groupings in Excel How To: Use a VLOOKUP function inside a text formula in Excel How To: Use SUMIFS or SUMPRODUCT to get invoice info in Excel How To: Extract numbers from a text string with Excel's LEFT, SEARCH & TRIM function Next, click in the Range Field again > click on Store#2 worksheet > select Data Range in this worksheet and click on the Add button.. 7. Next, select the first data range in 'All Ranges' section and type a Name for this Data Range in 'Field' section.. Note: Type a descriptive Name for Data Range, so as to makes it easy for you to identify the Data Range on the pivot table
The Python Pivot Table. You may be familiar with pivot tables in Excel to generate easy insights into your data. In this post, we'll explore how to create Python pivot tables using the pivot table function available in Pandas. The function itself is quite easy to use, but it's not the most intuitive Use this method if the blanks are in the values area of the pivot table. To set pivot table options for empty cells: Click in the pivot table. Click the PivotTable Tools Analyze tab in the Ribbon. Click Options in the PivotTable group. You can also right-click in the pivot table and select PivotTable Options from the drop-down menu UNPIVOT Example. UNPIVOT carries out almost the reverse operation of PIVOT, by rotating columns into rows.Suppose the table produced in the previous example is stored in the database as pvt, and you want to rotate the column identifiers Emp1, Emp2, Emp3, Emp4, and Emp5 into row values that correspond to a particular vendor. As such, you must identify two additional columns
### Sumproduct in a Pivot Table? - Excel Help Foru
Pivot Table has a way to avoid this. Right click within Pivot Table, choose Pivot Table and then uncheck the box in layout and format that says, Autofill column width on update 10. Display Report filter on Numerous Pages. With Excel Pivot table, it is possible to show Report filter on different sheets in the workbook No, you cant have multiple different graphs from the same pivot. If you create more than one, changing one will change all others. The alternative is to copy and paste values and create many graphs from the value table, or if you want it dynamic, create a parallel table that calls out the values from the pivot table, so every time the pivot changes your parallel table will also change Pivot Table Part 1 - Pivot tables are one of the most underutilized features in Excel, primarily because users often believe there's a long learning curve. Pivot Table Part 2 - Learn advanced Pivot Table techniques to add more interactivity to pivot tables and minimize repetitive tasks The PivotTable is updated to include the additional values. The order you place the fields in each area in the Fields pane affects the look of the PivotTable. You can drag the field values up or down within an area (the Rows area, for example) to adjust which data appears first 2. Insert pivot table. Believe it or not, we're already to the point in the process when you can insert a pivot table into your workbook. To do so, highlight your entire data set (including the column headers), click Insert on the ribbon, and then click the Pivot Table button. 3. Choose where to place your pivot table
### Excel Function Friday: Subtotal and Sumproduct with Filter
Select any cell within the pivot table and in the ribbon, look for an 'Insert' feature. Depending on what version of Excel you have, it is usually on the HOME > Insert > Insert Calculated Field OR you might see an ANALYZE tab appear and it's at ANALYZE > Fields, Items & Sets > Calculated Field Sumproduct & Dates & Multiple similar alpha Data Please help! I am currently working on a workbook that has various part nmbers (2060207-WPI001) on work sheet 7-14-07 in E:E in which I match a 1 to 3 letter designations as in WPI
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• Achilles tendon rupture non surgical rehabilitation protocol. | 8,871 | 40,601 | {"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-2021-25 | latest | en | 0.868396 |
https://socratic.org/questions/how-do-you-solve-the-system-of-equations-by-graphing-and-then-classify-the-syste-51 | 1,575,924,185,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540523790.58/warc/CC-MAIN-20191209201914-20191209225914-00209.warc.gz | 544,086,352 | 6,098 | # How do you solve the system of equations by graphing and then classify the system as consistent or inconsistent 4x + 6y = 0 and x - 2y = 14?
May 6, 2018
$y = - 4$ and $x = 6$
#### Explanation:
show below
$4 x + 6 y = 0$.....(1)
$x - 2 y = 14$......(2)
from equation (2)
$x = 14 + 2 y$.....(3)
$4 \left(14 + 2 y\right) + 6 y = 0$
$56 + 8 y + 6 y = 0$
$14 y + 56 = 0$
$14 y = - 56$
$y = - 4$
$x = 14 + 2 \left(- 4\right) = 14 - 8 = 6$
$x = 6$ | 206 | 457 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "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} | 4.1875 | 4 | CC-MAIN-2019-51 | latest | en | 0.676927 |
https://legoland.com.vn/how-many-ounces-is-3-liters.html | 1,721,494,029,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515300.51/warc/CC-MAIN-20240720144323-20240720174323-00808.warc.gz | 306,641,736 | 52,182 | # 10+ answer : how many ounces is 3 liters most accurate
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You are looking : how many ounces is 3 liters
## 1.How many ounces are in 3 liters? – Quora
• Author: www.quora.com
• Publish: 9 days ago
• Rating: 4(884 Rating)
• Highest rating: 5
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• Descriptions: 3 liters = 101.442 fluid ounces. Formula: multiply the volume value by 33.814.
• More : 3 liters = 101.442 fluid ounces. Formula: multiply the volume value by 33.814.
• Source : https://www.quora.com/How-many-ounces-are-in-3-liters
## 2.3 Liters to Ounces – How Many Ounces in 3 Liters? – Online Calculator
• Author: online-calculator.org
• Publish: 7 days ago
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• Lowest rating: 1
• Descriptions: 3 liters equals 101.44 ounces. 1 liter is roughly 33.814, therefore 3 liters is equivalent to 101.44 oz. 3 Liters Conversion. Liter: …
• More : 3 liters equals 101.44 ounces. 1 liter is roughly 33.814, therefore 3 liters is equivalent to 101.44 oz. 3 Liters Conversion. Liter: …
• Source : https://online-calculator.org/3-liters-to-ounces
## 3.3 Liters to Oz | 3 Liters to Ounces – Liter to Kg
• Author: litertokg.com
• Publish: 11 days ago
• Rating: 2(1242 Rating)
• Highest rating: 3
• Lowest rating: 3
• Descriptions: How many Ounces in 3 Liters · 3 L = 101.44 US customary fluid ounces · 3 L = 100 US food labeling fluid ounces · 3 L = 105.59 Imperial fluid ounces.
• More : How many Ounces in 3 Liters · 3 L = 101.44 US customary fluid ounces · 3 L = 100 US food labeling fluid ounces · 3 L = 105.59 Imperial fluid ounces.
• Source : https://litertokg.com/3-liters-to-oz
## 4.Convert 3 Liters to Ounces – CalculateMe.com
• Author: www.calculateme.com
• Publish: 11 days ago
• Rating: 4(565 Rating)
• Highest rating: 5
• Lowest rating: 1
• Descriptions: Convert 3 Liters to Ounces ; 3.16, 106.85 ; 3.17, 107.19 ; 3.18, 107.53 ; 3.19, 107.87.
• More : Convert 3 Liters to Ounces ; 3.16, 106.85 ; 3.17, 107.19 ; 3.18, 107.53 ; 3.19, 107.87.
• Source : https://www.calculateme.com/volume/liters/to-fluid-ounces/3
## 5.3 Liters to Ounce Conversion Calculator – 3 l to oz – Flightpedia
• Author: www.flightpedia.org
• Publish: 20 days ago
• Rating: 5(1675 Rating)
• Highest rating: 3
• Lowest rating: 2
• Descriptions: Convert 3 Liters to Ounces · 1 l = 35.274 oz · 1 oz = 0.02835 l …
• More : Convert 3 Liters to Ounces · 1 l = 35.274 oz · 1 oz = 0.02835 l …
• Source : https://www.flightpedia.org/convert/3-liters-to-ounces.html
## 6.3 liters to ounces – CoolConversion
• Author: coolconversion.com
• Publish: 28 days ago
• Rating: 2(1355 Rating)
• Highest rating: 5
• Lowest rating: 1
• Descriptions: How many ounces in 3 liters? … To convert any value in liters to ounces, just multiply the value in liters by the conversion factor 33.814022701843. So, 3 …
• More : How many ounces in 3 liters? … To convert any value in liters to ounces, just multiply the value in liters by the conversion factor 33.814022701843. So, 3 …
• Source : https://coolconversion.com/volume/3-liter-to-ounce
## 7.What is 3 Liters in Fluid Ounces? Convert 3 L to fl oz
• Author: whatisconvert.com
• Publish: 2 days ago
• Rating: 3(796 Rating)
• Highest rating: 3
• Lowest rating: 1
• Descriptions: 3 Liters is equivalent to 101.44206767676 Fluid Ounces. How to convert from Liters to Fluid Ounces. The conversion factor from Liters to Fluid Ounces is …
• More : 3 Liters is equivalent to 101.44206767676 Fluid Ounces. How to convert from Liters to Fluid Ounces. The conversion factor from Liters to Fluid Ounces is …
• Source : https://whatisconvert.com/3-liters-in-fluid-ounces
## 8.Convert 3 Litres to Fluid Ounces
• Author: converter.net
• Publish: 0 days ago
• Rating: 4(1085 Rating)
• Highest rating: 5
• Lowest rating: 2
• Descriptions: Result: 3 l is equal to 101.44 fl-oz. OTHER CONVERSIONS …
• More : Result: 3 l is equal to 101.44 fl-oz. OTHER CONVERSIONS …
• Source : https://converter.net/volume/3-litres-to-fluid-ounces
## 9.How Much Is 3 Liters Of Water In Ounces? Liter To Ounce Converter
• Author: waterseer.org
• Publish: 28 days ago
• Rating: 2(1865 Rating)
• Highest rating: 5
• Lowest rating: 2
• Descriptions: A standard water bottle holds about 16.9 oz of water. One 16.9 fl oz bottle contains around 500 mL of water. To get to 3 liters, you need to multiply that by …
• More : A standard water bottle holds about 16.9 oz of water. One 16.9 fl oz bottle contains around 500 mL of water. To get to 3 liters, you need to multiply that by …
• Source : https://waterseer.org/how-much-is-3-liters-of-water/
## 10.How much is 3 liters in fluid ounces? – Unit Converter
• Author: convertoctopus.com
• Publish: 20 days ago
• Rating: 3(363 Rating)
• Highest rating: 5
• Lowest rating: 3
• Descriptions: We can also convert by utilizing the inverse value of the conversion factor. In this case 1 fluid ounce is equal to 0.0098578432291667 × 3 liters. Another way …
• More : We can also convert by utilizing the inverse value of the conversion factor. In this case 1 fluid ounce is equal to 0.0098578432291667 × 3 liters. Another way …
• Source : https://convertoctopus.com/3-liters-to-fluid-ounces | 1,704 | 5,280 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-30 | latest | en | 0.750948 |
https://www.comeforanswers.com/in-april-anthonys-shop-made-a-daily-profit-of-400-in-12-days-and-a-daily-loss-of-300/ | 1,696,309,817,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511053.67/warc/CC-MAIN-20231003024646-20231003054646-00396.warc.gz | 781,384,119 | 8,861 | # In April, Anthony’s shop made a daily profit of \$400 in 12 days and a daily loss of \$300
In April, Anthony’s shop made a daily profit of \$400 in 12 days and a daily loss of \$300 in the rest of the month. It is given that his shop opened every day in April. Find the average daily profit or loss of Anthony’s shop in April.
use directed numbers to solve | 90 | 360 | {"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 | 0.962102 |
http://mathematica.stackexchange.com/questions/46238/scan-vs-map-vs-apply/46248 | 1,467,354,439,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783400572.45/warc/CC-MAIN-20160624155000-00134-ip-10-164-35-72.ec2.internal.warc.gz | 196,524,975 | 23,283 | # Scan vs. Map vs. Apply
I cannot understand the difference between Apply, Scan, and Map.
I have tried to play with these functions using the codes below:
f[x_] := x^2
Map[f, {{a, b}, c}]
Output:
{{a^2, b^2}, c^2}
Apply[f, {{a, b}, c}]
Output:
f[{a, b}, c]
Scan[f, {{a, b}, c}]
There is no output for this one.
The function of Map is understandable. But I cannot get the purpose of Apply and Scan. What are the differences between these three functions?
-
Do not define f when you do this, then it'll be much easier to understand what's going on. ...^2 threads over lists so your results don't reflect only the effect of Map, etc. but also this auto-threading. Take a look here and in the documentation page of each function. There are plenty of examples. For example the documentation clearly explains that Scan is the same as Map except it doesn't return anything. If is useful when f has a side-effect, e.g. f=Print. – Szabolcs Apr 16 '14 at 16:11
In addition to the references given by @Szabolcs: I covered these functions in considerable detail here. – Leonid Shifrin Apr 16 '14 at 16:28
These three functions are similar (speaking commonly), and in some applications any of them could be used, yet they have very different special applications.
Rudimentarily:
(For each example below f should be undefined; start with ClearAll[f].)
## Map
Most basically Map works like this:
f /@ head[1, 2, 3] (* shorthand for Map[f, head[1, 2, 3], {1}] *)
head[f[1], f[2], f[3]]
As with each of the functions under discussion it can operate at different levels:
Map[f, {{1, 2}, {3, 4}}, {0}]
Map[f, {{1, 2}, {3, 4}}, {1}]
Map[f, {{1, 2}, {3, 4}}, {2}]
Map[f, {{1, 2}, {3, 4}}, {0, 2}]
f[{{1, 2}, {3, 4}}]
{f[{1, 2}], f[{3, 4}]}
{{f[1], f[2]}, {f[3], f[4]}}
f[{f[{f[1], f[2]}], f[{f[3], f[4]}]}]
Map is an additive process; it inserts additional heads into an expression at the specified levelspec and then evaluates it.
If the expression is held nothing evaluates. (Held meaning the outermost Head has an Attribute such as HoldAll. Hold has this attribute. Do not confuse the Head with the Attribute.)
Print /@ Hold[a[1, 2], b[3, 4]]
Hold[Print[a[1, 2]], Print[b[3, 4]]]
## Apply
Apply works like Map, except that instead of adding a Head it replaces an existing one, if a head exists at the specified level. Unlike Map it is frequently used at levelspec {0} with the short form @@, which replaces the outermost head:
f @@ head[1, 2, 3] (* short form of Apply[f, head[1, 2, 3], {0}] *)
f[1, 2, 3]
A second short form (@@@) exists for levelspec {1}:
f @@@ {{1, 2}, {3, 4}}
{f[1, 2], f[3, 4]}
If a sub-expression is atomic (has no operable Head) Apply does not modify it:
f @@@ head[1, 2, x + y]
head[1, 2, f[x, y]]
Note that 1 and 2 are unmodified, while x + y (which has a FullForm of Plus[x, y]) had its Head replaced.
As with Map, Apply transforms an expression and then evaluates it in entirety; if it is held evaluation does not continue:
Print @@@ Hold[a[1, 2], b[3, 4]]
Hold[Print[1, 2], Print[3, 4]]
One way to "release" this expression for evaluation is to replace Hold with something else, like List:
List @@ Print @@@ Hold[a[1, 2], b[3, 4]]
12
34
{Null, Null}
Note here that the elements 1, 2, and 3, 4 are printed by Print, but since Print[. . .] itself evaluates to Null the output of the line is {Null, Null}.
## Scan
Scan is also like Map, except that each individual expression which is wrapped in the specified head is evaluated outside of the main expression, and the main expression is never returned. If we Scan a function without side-effects over an expression we get nothing:
Scan[f, head[1, 2, 3]]
(This could also be written f ~Scan~ head[1, 2, 3].) We do not even get an output line because the output is Null and Mathematica does not print a lone Null output line (by convention). If we use a function with side-effects, such as Print, we get a different result:
Scan[Print, head[1, 2, 3]]
1
2
3
Crucially Scan will carry out its evaluation even when an expression is held because it only looks at the subexpressions:
Print ~Scan~ Hold[a[1, 2], b[3, 4]]
a[1,2]
b[3,4]
Its evaluation can also be interrupted because, unlike Map and Apply which transform the entire expression, then evaluate it, Scan works incrementally:
(Print[#]; If[# > 2, Return[]]) & ~Scan~ {1, 2, 3, 4, 5}
1
2
3
Because Scan does not first duplicate and then modify the entire expression it may be more memory efficient than the use of Map or Apply, but it will still unpack packed arrays therefore if memory efficiency is a priority other methods may be preferred.
## Conclusion
I hope these simple examples illustrate that these three functions, while closely related, have unique characteristics that differentiate their use, and each has powerful applications that the others do not.
### Notes on atomic objects
Not all expressions that have the appearance head[arguments] in FullForm are actually standard expressions in that form that can be manipulated with structural tools such as Apply. Instead they have a special internal format and they are merely displayed in the form head[arguments].
From the documentation on Atomic Objects:
All expressions in Mathematica are ultimately made up from a small number of basic or atomic types of objects.
These objects have heads that are symbols that can be thought of as "tagging" their types. The objects contain "raw data", which can usually be accessed only by functions specific to the particular type of object. You can extract the head of the object using Head, but you cannot directly extract any of its other parts.
Standard objects such as strings and integers are atomic, but so are other expressions that may not appear to be:
list = {1, "test", 1/2, 2 + 3 I};
list // FullForm
AtomQ /@ list
List[1, "test", Rational[1, 2], Complex[2, 3]]
{True, True, True, True}
Also atomic are more complex structures such as SparseArray, Graph, BooleanFunction, and (in recent versions) Image.
Atomic objects (or atoms) are handled differently from standard (compound) expression forms.
• Rational[1, 2] is atomic even though it appears otherwise.
• Even though 1 is formatted without an obvious head, Head will return Integer.
(See Is there a summary of answers Head[] can give? for other examples.)
• Apply does not work in the "normal" way on atoms:
new @@@ list
{1, "test", 1/2, 2 + 3 I}
• Replace and kin often work on the apparent FullForm of atoms:
Replace[list, head_[args__] :> {head, args}, {1}]
{1, "test", {Rational, 1, 2}, {Complex, 2, 3}}
• Functions may be overloaded to handle atoms differently; e.g. many functions treat a SparseArray as they would the Normal form expression:
new @@ SparseArray[{1, 2, 3}]
new[1, 2, 3] (* not new[{1, 2, 3}] *)
• Handling may appear somewhat irregular even within functions; Part will "extract" (or return) the head of Rationa[1, 2] but not anything else:
Part[1/2, 0]
Rational
Part[1/2, 1]
Part::partd: Part specification (1/2)[[1]] is longer than depth of object. >>
(1/2)[[1]]
-
Worth to link question I think you should now :P Quite relevant but not closely related. – Kuba Apr 16 '14 at 20:21
It may be worthwhile adding a note about things that are atomic, but not immediately obviously so, like AtomQ@Complex[1, 1] == True vs. AtomQ@Complex[1, a] == False. +1 – rcollyer Apr 17 '14 at 16:08
@Mr.Wizard Could you point me to an explanation for the syntax f ~Scan~ head[1, 2, 3]? – Sascha Apr 18 '14 at 9:12
+1, nice answer. I'd perhaps emphasize more strongly that because Scan does not produce an output, it can be significantly more memory-efficient than Map, particularly when such an output is not needed and in discarded in any case. I've seen cases where this made a real difference. – Leonid Shifrin Apr 22 '14 at 22:49
@Mr.Wizard Unfortunately, due to the nature of Do, the by-element indexing method is slow to the degree that it defeats the purpose. Of course, in such cases for large numerical (packed) arrays there is always Compile, and for non-packed it doesn't matter that much. The good thing would be to systematically enumerate these approaches with their respective domains of applicability, pros and cons (I mean, Scan, Map, chunked scan based on Map, Do loop, Compile). Otherwise it is hard to keep all these things in mind, even for us. – Leonid Shifrin May 26 at 17:13
For a visual, animated, description of basic behavior of such functions, I recommend:
http://reference.wolfram.com/legacy/flash/
(You may want to turn on your computer's sound.)
-
As a rule we try to avoid link-only answers. (Excepting large collections.) Perhaps you could include a screen capture or something? Also, I don't see an animation for Scan. – Mr.Wizard Apr 19 '14 at 17:08
@Mr.Wizard: The whole point is that the link points to animations of how the various operators work! And given that the pages are copyright by Wolfram Research, I'm not sure it's legal to post even a static screenshot (would it be "fair use"?) But the animations are very useful ways to understand what Map, Apply, etc., do -- using a different modality than the usual static examples. – murray Apr 19 '14 at 21:04 | 2,522 | 9,222 | {"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} | 3.25 | 3 | CC-MAIN-2016-26 | latest | en | 0.872496 |
http://www.us-lotteries.com/Arizona/Mega_Millions/Mega_Millions-Full-Wheel.asp | 1,386,411,453,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163053923/warc/CC-MAIN-20131204131733-00076-ip-10-33-133-15.ec2.internal.warc.gz | 744,110,499 | 6,098 | Generate combinations of Arizona lottery Mega Millions numbers. You may fix some numbers, and also see 'what if' scenarios.
# Arizona Mega Millions Combinations Generator (Full Wheel)
Otherwise known as a Lottery Wheel, this utility generates all the lottery combinations of a selected set of Arizona Mega Millions lottery numbers into groups of 5, plus the extra ball, with the option of fixing some selections. Fixed numbers, also known as key numbers, are those lottery numbers that the user wants to appear in all generated lottery plays.
This Arizona Mega Millions lottery wheel is divided into two parts - Preview and Generation, which function independently. The Preview part is intended for analyzing the scenarios and expected outcomes of the combinations of a set of lottery numbers before they are generated. The generation part lists the lottery plays based on the user's selections. If no numbers are fixed, the generated lottery plays will be all the possible combinations of the selected lottery numbers. If some numbers are fixed, then the plays will be generated by combining the fixed numbers with all the possible combinations of the rest of the selected lottery numbers.
If you are a first time user of this tool you may first read the general info and usage instructions provided further below.
Once you have determined how much to spend and selected your Arizona Mega Millions lottery numbers, you can generate the lottery combinations with following tool. Click on a number to select it, click on a selected number to fix it, click on a fixed number to unselect it. Every time you click, the total lottery combinations to be generated will be displayed at the bottom. The 'Generate' button opens a new window with the generated lottery numbers. Becasuse of the time and computer resources involved to generate these combinations, this lottery tool can generate only up to 100,000 lottery combinations.
TOP
## General Information on the Arizona Mega Millions Combinations Generator
You can generate the combinations of a selected set of lottery numbers of the Arizona Mega Millions game with or without fixing some of them. The purpose of fixing some lottery numbers is to reduce the number of plays generated. This results in lower cost, but at the expense of reduces chance of winning. For instance, the combinations of 11 lottery numbers will result in 4158 lottery plays of Arizona Mega Millions; but if 2 of the numbers are fixed, the number of lottery plays becomes 756. This utility allows you to fix up to 4 lottery numbers. To choose which Mega Millions lottery numbers to fix, you may arbitrarily do so, or you may use the results of our Arizona Mega Millions Numbers Analysis as well as our Arizona Mega Millions Pairs Analysis and Arizona Mega MillionsNumber-ball pairs analysis.
A Arizona Mega Millions jackpot is guaranteed if the fixed numbers are all winning numbers, plus if all the other winning lottery numbers are among the selected numbers. If no lottery numbers are fixed, then a jackpot is guaranteed as long as the winning lottery numbers are all within the selected numbers. All other scenarios can be viewed by manipulating the 'what if..' situations in the Preview section as will be demonstrated below.
## Full Wheel Usage Example for Mega Millions
Let us suppose that we want to play all Arizona Mega Millions lottery numbers that end with 0 and 5, that is, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55 (a total of 11). Plus 5, 10, 15, 20, 25, 30, 35, 40, 45 as the Extra Balls (9 of them). To preview the scenarios, select 11 from the drop-down numbers of the regular numbers to play and 9 for the Extra Ball and click on "Show Scenario". The total number of Mega Millions lottery plays to be generated will be shown as 4158. Now, we can manipulate some 'what if' situations to see how the 4158 Mega Millions lottery tickets would fare. For example, if 4 of these lottery numbers are winners, but not the Extra Ball, then click on the "If 4 correct" line. You will see that 63 of our Mega Millions lottery plays(tickets) will match 4-of-5, none will match 3-of-5 + Ball, and so on. If one of the selected Extra Balls is also a winner, click on the "If 4 + Ball correct" line; there will be 7 matches of 4-of-5 + Ball, 56 matches of 4-of-5, and so on. If all of the winning numbers end either with 0 or 5, then it will be celebration time.
Now let us see the effect fixing 2 of our Mega Millions numbers. Go back to the Preview panel and select 2 as the numbers to fix, while leaving the others as they are. In this case there will be only 756 lottery plays to be generated all of which will contain the two fixed lottery numbers. Here, the scenarios will involve how many of the fixed numbers and how many of the other selected lottery numbers are winners, and also if the Extra Ball is a winner. For example, if 1 of the fixed numbers and 3 of the other selected lottery numbers are winners, and if the Ball is also a winner, clicking on the "If 1 fixed + 3 others + Ball correct" line of the options list will reveal that 1 of the 756 Mega Millions lottery tickets will match 4-of-5 + Ball, 8 will match 4-of-5, and so forth.
The generation part of Arizona Mega Millions combinations shouldn't pose any problem. As explained above, a cycle of clicks on a number will select, fix, and unselect it, respectively. The 'Generate' button will list the resulting Mega Millions lottery combinations in a new browser window. Again, note that if you fix the Extra Ball, you cannot have other Ball selections.
At this point, you can select the 11 Mega Millions numbers of our example and see the outcome. Also, you may fix 2 numbers, say, 5 and 10, and click "Generate". Observe that all the generated Mega Millions lottery combinations will contain 5 and 10.
Finally, remember that you can check the past performance of a set of Arizona Mega Millions lottery combination with our Mega Millions Combinations Checker. Also, if the Mega Millions combinations generated by this full lottery wheel are one too many to risk, you may try the Mega Millions abbreviated wheels.
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Arizona lottery Games Data and Analysis Menu
Pick 3 - Pick 3 Game - Winning Numbers - Number Search - Yearly Calendar - Boxes(Combos) - Triples - Digits Analysis - Straights Analysis - Boxes Analysis - Pairs - Sums - Combinations Generator - Game Info Fantasy 5 - Fantasy 5 Game - Winning Numbers - Number Search - Yearly Calendar - Numbers Analysis - Pairs Analysis - Triads Analysis - Check Combinations - Full Wheel - Abbreviated Wheel - Game Info 2By2 - 2By2 Game - Winning Numbers - Number Search - Yearly Calendar - Numbers Analysis - Red Balls Pairs Analysis - White Balls Pairs Analysis - Check Combinations - Full Wheel - Game Info Weekly Winnings - Weekly Winnings Game - Winning Numbers - Number Search - Yearly Calendar - Numbers Analysis - Pairs Analysis - Triads Analysis - Check Combinations - Full Wheel - Abbreviated Wheel - Game Info The Pick - The Pick Game - Winning Numbers - Number Search - Yearly Calendar - Numbers Analysis - Pairs Analysis - Triads Analysis - Check Combinations - Full Wheel - Abbreviated Wheel - Game Info Mega Millions - Mega Millions Game - Winning Numbers - Number Search - Yearly Calendar - Numbers Analysis - Pairs Analysis - Number-Ball Pair Analysis - Check Combinations - Full Wheel - Abbreviated Wheel - Game Info Powerball - Powerball Game - Winning Numbers - Number Search - Yearly Calendar - Numbers Analysis - Pairs Analysis - Number-Ball Pair Analysis - Check Combinations - Full Wheel - Abbreviated Wheel - Game Info
- Arizona Lottery Recent Results and Tools - Compact page for quick check - Arizona Lottery General Information - Arizona Lottery Games - US-Lotteries Home Page
• WE DO NOT SELL ANY LOTTERY, WE DO NOT BUY LOTTERY NUMBERS ON BEHALF OF ANYONE; this is only an information and guide web site. | 1,830 | 7,913 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2013-48 | latest | en | 0.89446 |
https://cs.stackexchange.com/questions/149912/fast-algorithm-for-computing-minimal-closure-of-a-set-of-sets-under-intersection | 1,656,924,542,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104364750.74/warc/CC-MAIN-20220704080332-20220704110332-00305.warc.gz | 229,438,246 | 65,937 | # Fast algorithm for computing minimal closure of a set of sets under intersection?
A step of an algorithm I’ve designed requires computing the minimal closure under intersection of a set of sets of arbitrary size. By the "minimal closure (of a set $$S$$) under intersection", I mean:
Given a set $$S$$ containing sets $$s_1, \cdots , s_k$$, the smallest set $$𝑆′$$ such that $$S\subseteq S'$$ and $$x\cap y \in S'$$ for any two sets $$x, y \in S'$$.
While I can come up with a pretty straightforward naive approach (loop over the sets and store all the intersections in a new set, then update the set as the union of previous step’s set and the new set, and repeat this process until the new and old set are the same), I am looking for a faster method, but haven’t been able to find any papers working on a similar problem. Does anybody know if there are faster existing algorithms to solve this problem before I attempt to potentially reinvent the — or a worse version of — the wheel? Or if not exact solutions, maybe fast approximation algorithms on other data structures (e.g. sets of strings) that translate to this problem setup?
• The result will literally be the set $S'$ containing all sets $s'$ of the form $s'=\bigcap s\in A$ for some $A\subseteq S'$. Since the output can be incredibely large, (up to an exponantial size in the input) - there is no "fast" algorithm for this problem. Mar 15 at 22:59
There is no polynomial-time algorithm. The output can potentially be exponentially large, so any algorithm will have inputs on which it must take exponential time. For instance, if $$s_i=\{1,2,\cdots, k\}\setminus\{i\}$$, then $$S'$$ contains $$2^k$$ elements, so any algorithm must take $$\Omega(2^k)$$ time (it takes that much time even just to output $$S'$$, let alone to calculate it).
One simple approach is a workload algorithm, where you incrementally generate new sets that are in the closure. In particular:
• Set $$W := S$$. (The worklist is initialized to $$S$$.)
• Set $$S' := S$$.
• While $$W \ne \emptyset$$:
• Pop an element from $$W$$, call it $$s$$.
• Let $$T := \{s \cap s' \mid s' \in S', s \cap s' \notin S'\}$$.
• Set $$W := W \cup T$$ and $$S' := S' \cup T$$.
• @JohnL., oh, right! Thank you. Fixed.
– D.W.
Mar 17 at 3:32
• Here is a simpler and faster algorithm. Computes $C_i$, the minimal closure of $\{s_1, s_2,\cdots, s_i\}$ for $i=0,1,2,\cdots,k$. $C_0=\emptyset$. $C_{i+1}=C_i\cup\{s\cap s_{i+1}\mid s\in C_i\}$. Return $C_k$. This algorithm is optimal in the sense that in the worst case, each $s\cap s_{i+1}$ will always be a new set. (I do not want to write another answer). Mar 17 at 3:44
• @JohnL., Thank you for the simpler algorithm! In my algorithm, if $s=s'$, then $s \cap s' \in S'$, and nothing is added to $W$ -- so I don't understand why you say $W$ will never become empty.
– D.W.
Mar 17 at 4:19
• Changing a collection such as a set or a list while looping over it is apparently ambiguous. Mar 17 at 5:02
• @JohnL., "pop" means pick a single element of $W$ (any one is fine) and remove it. It doesn't mean looping over it.
– D.W.
Mar 17 at 5:19 | 902 | 3,110 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2022-27 | longest | en | 0.858059 |
https://www.teacherspayteachers.com/Product/NEW-Problem-Solving-with-Measurement-Task-Cards-48C-57A-4MDA2-5MDA1-2301925 | 1,524,576,551,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946688.88/warc/CC-MAIN-20180424115900-20180424135900-00059.warc.gz | 871,008,091 | 19,408 | Total:
\$0.00
# NEW Problem Solving with Measurement Task Cards 4.8C, 5.7A, 4.MD.A2 & 5.MD.A.1
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Common Core Standards
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Product Description
Measuring lengths, time, liquid volumes, mass and money AND converting AND in word problems using all 4 operations: that is everything all rolled into this fourth grade standard!! It seems to encompass quite a bit of information and a little bit of everything from everywhere.
I have created 32 task cards to help your students practice and master this smorgasbord of a standard! There are a variety of questions to help ease your students into the standard and then challenge them.
Cards 1-6 focus on word problems involving length, time, liquid volumes, mass and money and do NOT require conversions.
Cards 7-24 focus on word problems involving length, time, liquid volumes, mass and money and require students to convert units of measure.
Cards 25-32 require students to use their knowledge of rulers to estimate measurements of objects in both inches and centimeters.
All of these word problems involve a mixture of addition, subtraction, multiplication and division.
There are a mixture of open ended and multiple choice questions.
All you need to do is print, preferably in color, cut, laminate and shuffle cards. Then the options are endless:
•Have your students complete the cards at a center and record their answers on the recording key provided. Or get up and move by placing them around the room and playing a game of SCOOT.
•Use as a ticket out the door, picking one out of the stack to complete as a quick check or spiral throughout the year.
•Assign a sheet for homework for the night.
•Even use as an assessment or intervention.
•Great for fast finishers too!
This mixture of task cards is perfect for centers, independent practice, small group work, intervention or just as a spiral review game. Recording sheet for students provided as well as an answer key for teacher quick check or student self-check.
This document is aligned to Texas: 4.8C (Readiness) and 5.7A (supporting) and Common Core 4.MD.A.2 & 5.MD.A.1
If you like this product, don’t forget to leave feedback and hit that follow button!!! Also, check out my store for more fun math products.
Area and Perimeter Practice Problems
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\$3.25 | 567 | 2,496 | {"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} | 2.859375 | 3 | CC-MAIN-2018-17 | latest | en | 0.939151 |
https://www.edok-burger.ru/solve-math-word-problems-online-460.html | 1,627,885,399,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154304.34/warc/CC-MAIN-20210802043814-20210802073814-00246.warc.gz | 767,241,939 | 6,287 | # Solve Math Word Problems Online
Below is a math problem solver that lets you input a wide variety of math problems and it will provide the final answer for free. The version below will show you the final answer only.
You'll see a button "View steps" and this takes you to the developer's site where you can purchase the full version of the solver (where you can see the steps).
Check There are now 16 boys and 12 girls, so the ratio of boys to girls is 16 : 12 = 4 : 3 At the start of the year there were 20 boys and 10 girls, so the ratio was 20 : 10 = 2 : 1 Consecutive means one after the other.
And they are even, so they could be 2 and 4, or 4 and 6, etc.
You would be expected to understand that this meant that she worked eight hours for each of the four days Monday, Tuesday, Thursday, and Friday; and six hours for each of the two days Wednesday and Saturday.
Suppose you're told that Shelby earns "time and a half" for any hours she works over forty for a given week.
If a problems says "the ratio of Some times, you'll be expected to bring your "real world" knowledge to an exercise.
Probably the greatest source of error, though, is the use of variables without definitions.Sometimes you'll not feel sure of your translation of the English into a mathematical expression or equation. For instance, if you're not sure if you should be dividing or multiplying, try the process each way with regular numbers. For instance, suppose you're not sure if "half of (the unknown amount)" should be represented by multiplying by one-half, or by dividing by one-half. Visual students will be keen on the ability to visualize math in this way, using the virtual blocks to represent word problems.* Use with the whole class using an interactive whiteboard or projector-connected computer.If the former, what does this mean, in practical terms?(And, if you can't think of any meaningful definition, then maybe you need to slow down and think a little more about what's going on in the word problem.) In all cases, don't be shy about using your "real world" knowledge.* Students can take turns working out word problems on the whiteboard while students at their seats work to arrive at a common solution.The hardest thing about doing word problems is using the part where you need to take the English words and translate them into mathematics.Try first to get a feel for the whole problem; try first to see what information you have, and then figure out what you still need. Figure out what you need but don't have, and name things. And make sure you know just exactly what the problem is actually asking for.Pick variables to stand for the unknows, clearly labelling these variables with what they stand for. You need to do this for two reasons: " stands for, so you have to do the whole problem over again.
## One thought on “Solve Math Word Problems Online”
1. No spare time for you to deal with your studies and no lack of ideas, no lack of inspiration for them.
2. Rand believes that altruists lose a sense of their own identity when they dedicate themselves to others as a validity and justification for their existence.
3. The families are run on set rules and the whole family traditionally stays in a large triangular house (Chen 2001, p. The cross-cultural psychology makes an effort to understand individuals of different cultures and how they interact with each other. | 732 | 3,390 | {"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} | 3.609375 | 4 | CC-MAIN-2021-31 | latest | en | 0.956809 |
https://beta.mapleprimes.com/questions/223547-Solve-The-Nonlinear-ODE- | 1,725,981,356,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651255.81/warc/CC-MAIN-20240910125411-20240910155411-00655.warc.gz | 111,946,550 | 24,433 | # Question:Solve the Nonlinear ODE
## Question:Solve the Nonlinear ODE
Maple
hi every body:
i want to solve the below ODE with boundry conditions ics, can anybody help me?
The ODE:
ode := x^3*(diff(f(x), x, x, x, x))+alpha*(x^4*(diff(f(x), x, x))+x^3*(diff(f(x), x, x))-x^2*(diff(f(x), x)))-2*x^2*(diff(f(x), x, x, x))+3*x*(diff(f(x), x, x))-3*(diff(f(x), x))+R*x*f(x)^2-R*x^2*(diff(f(x), x, x))*(diff(f(x), x))-3*R*x*f(x)*(diff(f(x), x, x))+3*R*f(x)*(diff(f(x), x))+x^2*R*f(x)*(diff(f(x), x, x, x))-M^2*(x^3*(diff(f(x), x, x, x))-x^2*(diff(f(x), x))) = 0
The boundry conditions:
ics := f(0) = 0, f(1) = 1, (D(f))(1) = 0, limit(diff((diff(f(x), x))/x, x), x = 0) = 0
M,R,alpha are constants.
with regards...
| 305 | 719 | {"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-2024-38 | latest | en | 0.470748 |
https://www.cnblogs.com/zhouzhendong/p/UOJ348.html | 1,571,401,386,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986682037.37/warc/CC-MAIN-20191018104351-20191018131851-00224.warc.gz | 813,837,451 | 6,553 | 题解
$$f[S] = \frac{1}{(sum[S])^p}\sum_{T\subsetneq S} f[T]g[S-T]$$
代码
#pragma GCC optimize("Ofast","inline")
#include <bits/stdc++.h>
#define clr(x) memset(x,0,sizeof (x))
#define For(i,a,b) for (int i=a;i<=b;i++)
#define Fod(i,b,a) for (int i=b;i>=a;i--)
#define pb(x) push_back(x)
#define mp(x,y) make_pair(x,y)
#define fi first
#define se second
#define _SEED_ ('C'+'L'+'Y'+'A'+'K'+'I'+'O'+'I')
#define outval(x) printf(#x" = %d\n",x)
#define outvec(x) printf("vec "#x" = ");for (auto _v : x)printf("%d ",_v);puts("")
#define outtag(x) puts("----------"#x"----------")
#define outarr(a,L,R) printf(#a"[%d...%d] = ",L,R);\
For(_v2,L,R)printf("%d ",a[_v2]);puts("");
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef vector <int> vi;
LL x=0,f=0;
char ch=getchar();
while (!isdigit(ch))
f|=ch=='-',ch=getchar();
while (isdigit(ch))
x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
return f?-x:x;
}
const int N=23,S=1<<21,mod=998244353;
const ULL Bmod=16ULL*mod*mod;
int Pow(int x,int y){
int ans=1;
for (;y;y>>=1,x=(LL)x*x%mod)
if (y&1)
ans=(LL)ans*x%mod;
return ans;
}
if ((x+=y)>=mod)
x-=mod;
}
void Del(int &x,int y){
if ((x-=y)<0)
x+=mod;
}
int n,m,s,p;
vector <int> e[N];
int w[N];
int cnt1[S],sum[S],f[S];
int g[N][N];
int u[N][S],v[N][S];
int check(int s){
if (!s)
return 0;
clr(vis),clr(in);
int fir=-1;
For(i,0,n-1)
if (s>>i&1){
fir=i;
break;
}
q[++tail]=fir,vis[fir]=1;
for (auto y : e[x])
if (s>>y&1){
in[y]^=1;
if (!vis[y])
vis[y]=1,q[++tail]=y;
}
}
if (tail!=cnt1[s])
return 1;
For(i,0,n-1)
if (in[i])
return 1;
return 0;
}
void FMT(int *a){
For(i,0,n-1)
For(j,0,s-1)
if (j>>i&1)
}
void IFMT(int *a){
For(i,0,n-1)
For(j,0,s-1)
if (j>>i&1)
Del(a[j],a[j^1<<i]);
}
int main(){
s=1<<n;
clr(g);
For(i,1,m){
e[x].pb(y),e[y].pb(x);
}
For(i,0,n-1)
For(i,0,s-1){
For(j,0,n-1)
if (i>>j&1){
cnt1[i]++;
sum[i]+=w[j];
}
f[i]=check(i);
sum[i]=Pow(sum[i],p);
if (f[i])
u[cnt1[i]][i]=sum[i];
}
For(i,0,n)
FMT(u[i]);
v[0][0]=1;
FMT(v[0]);
For(i,1,n){
For(k,0,s-1){
ULL tmp=0;
For(j,0,i-1){
tmp+=(LL)v[j][k]*u[i-j][k];
if (tmp>=Bmod)
tmp-=Bmod;
}
v[i][k]=tmp%mod;
}
IFMT(v[i]);
For(k,0,s-1)
if (cnt1[k]==i)
v[i][k]=(LL)v[i][k]*Pow(sum[k],mod-2)%mod;
else
v[i][k]=0;
FMT(v[i]);
}
IFMT(v[n]);
cout<<v[n][s-1]<<endl;
return 0;
}
posted @ 2019-03-31 21:50 -zhouzhendong- 阅读(...) 评论(...) 编辑 收藏 | 999 | 2,309 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2019-43 | latest | en | 0.115049 |
https://mathoverflow.net/questions/178213/two-definitions-of-non-commutative-lp-space | 1,571,811,700,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987829458.93/warc/CC-MAIN-20191023043257-20191023070757-00371.warc.gz | 561,729,062 | 25,835 | # Two Definitions of Non-commutative $L^p$ space
Throughout, let $(\mathcal{M},\tau)$ be a von Neumann algebra $\mathcal{M}$, acting on a Hilbert space $H$, with normal semifinite faithful trace $\tau$.
In the survey article by Pisier and Xu, the non-commutative $L^p$ space $1\leq p<\infty$ is defined as follows. Let $S_+$ be the set of all positive elements $x\in \mathcal{M}$ whose support projection $\text{supp}(x)$ has finite trace. If $\mathcal{S}$ is the linear span of $S_+$, then the non-commutative $L^p$ space $L^p(\mathcal{M},\tau)$ is just the completion of $\mathcal{S}$ with respect to the norm $$|| x ||_p:=[\tau(|x|^{p})]^\frac{1}{p}.$$
On the other hand, for instance in Nelson, an approach analogous to the construction of the classical $L^p$ spaces in measure theory is used. Briefly, we introduce the concept of a $\tau$-measurable operator. These are, by definition, closed densely defined operators $x$ on $H$ affiliated with $\mathcal{M}$ and satisfy the condition $$\tau(E_\lambda(|x|)<\infty,$$ where $E_\lambda(|x|)$ is the spectral resolution of $|x|$ corresponding to the indicator function of $(\lambda,\infty)$. By placing a specific topology on the set of all $\tau$-measurable operators, we obtain a metrizable topological $*$-algebra, denoted $L^0(\mathcal{M},\tau)$. After extending the trace $\tau$ to $L^0(\mathcal{M},\tau)$, we then define for $0<p<\infty$ $$L^p(\mathcal{M},\tau):=\{x\in L^0(\mathcal{M},\tau)|\ \tau(|x|^p)<\infty\}.$$
I have two questions:
1. Why are these two approaches in defining the non-commutative $L^p$ spaces ($1\leq p<\infty$) equivalent? Are there any references to proof(s) of this fact?
2. The space $\mathcal{S}$ above turns out to be weak$^*$ dense in $\mathcal{M}$. Will the completion of $(\mathcal{S},|| \cdot||_p$) still yield the non-commutative $L^p$ space if we assume that $\mathcal{S}$ is dense in a different operator topology?
• Can't you just show that the map $\mathcal S\to L^0(\mathcal M, \tau)$ is an isometry for the p-norm, or am I missing something? – Yemon Choi Aug 10 '14 at 9:05
• The two norms induced on S coincide by definition and S is dense in both spaces. Finally, both ambient spaces are complete, which proves the desired statement. – Dmitri Pavlov Aug 11 '14 at 18:34
• @DmitriPavlov Thanks, I did not think carefully about this. I have to think harder!...:) – Malcolm King Aug 12 '14 at 13:18 | 747 | 2,402 | {"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} | 2.53125 | 3 | CC-MAIN-2019-43 | latest | en | 0.767846 |
https://planetside.co.uk/forums/index.php/topic,23830.0/prev_next,next.html | 1,571,168,858,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986660231.30/warc/CC-MAIN-20191015182235-20191015205735-00297.warc.gz | 623,334,587 | 10,400 | ## Too Smooth voronoi 3D cell scalar
Started by cyphyr, January 17, 2018, 04:50:11 am
#### cyphyr
##### January 17, 2018, 04:50:11 am
Is there any way to alter the sharpness of a "Smooth voronoi 3D cell scalar"?
It's very useful to have the soft change between voroni cells but sometimes (often) it is too soft.
www.richardfraser.co.uk
/|\
Ryzen 9 3900X @3.79Ghz, 64Gb (TG4 benchmark 6:20)
i7 5930K @3.5Ghz, 32Gb (TG4 benchmark 13.44)
#### Hetzen
##### January 17, 2018, 08:21:02 am #1
Yes I think there is. I was playing with this a while back. I need to get on a machine when I get in the office to check a few things before I give out a clip file. I just need to confirm that cell noise and smooth cell share the same base pattern.
The principle is to extract the difference between the two noises, apply a gain curve then apply that to the hard edged cell noise.
#### cyphyr
##### January 17, 2018, 09:17:43 am #2
Excellent, thankyou.
Yes, cell noise and smooth cell noise do indeed share the same base pattern.
I was trying to "tighten" up the smooth noise and getting nowhere lol
The reverse looks a better way to go.
I've extracted the difference (twice!)
Subtracting the cell noise from the smooth cell noise gives a different result from subtracting the smooth cell noise from the cell noise. So I did both and combined the result ... that's where it all starts to go wrong ... I can't seem to find a way of properly combing them that gives a result I can work with further.
www.richardfraser.co.uk
/|\
Ryzen 9 3900X @3.79Ghz, 64Gb (TG4 benchmark 6:20)
i7 5930K @3.5Ghz, 32Gb (TG4 benchmark 13.44)
#### Dune
##### January 17, 2018, 10:25:12 am #3
That would be cool, Jon! I asked for this a while ago, but never really got any reply. I thought softness was baked in.
Mixing stuff is always hard as you very soon get very narrow differences and thus strange displacements.
#### Hetzen
##### January 17, 2018, 10:28:59 am #4
Well here is the clip file. I wouldn't say it's a success, we still get some horrible stretching at the mid point of the transition. There maybe a better method other than using the Gain scaler.
Tightness of 0.45
[attachimg=1]
Tightness of 0.2
[attachimg=2]
#### Hetzen
##### January 17, 2018, 10:44:19 am #5
This version uses a Bias curve instead. Still getting stretching at the mid point, plus some displacement naughties.
[attachimg=1]
It maybe a case of applying the difference to the Smooth cell, but not as it's set up in these clip files. Needs more thought if it is indeed possible.
#### cyphyr
##### January 17, 2018, 10:48:17 am #6
Thanks, this looks useful, I shall experiment
www.richardfraser.co.uk
/|\
Ryzen 9 3900X @3.79Ghz, 64Gb (TG4 benchmark 6:20)
i7 5930K @3.5Ghz, 32Gb (TG4 benchmark 13.44)
#### cyphyr
##### January 17, 2018, 10:54:50 am #7
I think the sharp edges/stretches may be because of the difference between subtracting the cell noise from the smooth cell noise and subtracting the smooth cell noise from the cell noise.
See below.
I'm trying to combing the result of the two so I can fed it into the Add scalar but all I get is BLACK.
www.richardfraser.co.uk
/|\
Ryzen 9 3900X @3.79Ghz, 64Gb (TG4 benchmark 6:20)
i7 5930K @3.5Ghz, 32Gb (TG4 benchmark 13.44)
#### Dune
##### January 17, 2018, 12:06:56 pm #8
I think Matt needs to take a look at the functions, maybe he can change the whole thing from the start to a smoother voro.
#### cyphyr
##### January 20, 2018, 05:26:17 am #9
... or I could just subtract a Voroni 3D diff scalar ... achieved pretty much the same effect ...
Sometimes I miss the obvious solution hehe
www.richardfraser.co.uk
/|\
Ryzen 9 3900X @3.79Ghz, 64Gb (TG4 benchmark 6:20)
i7 5930K @3.5Ghz, 32Gb (TG4 benchmark 13.44)
#### Hetzen
##### January 20, 2018, 10:13:59 pm #10
The problem with difference voronoi is that the edges always resort to black, so you don't get transitions from neighbour cell grey values. Ie you don't get a slope from one cell's grey value to the next cell's which smooth voronoi does.
Ideally Ulco is right, it would be great to have an extra input on the smooth voronoi to control the steepness between cell grey values. It's a nice noise to use as is, but can look soft as you've found.
I'd be interested to see some of your test renders of the plain noise to see what you've done.
#### cyphyr
##### January 21, 2018, 05:25:38 am #11
Yes you're right, this way each cell comes from a black (zero) base line but you can control the falloff to the baseline very easily with a colour adjust shader.
As Ulco mentions, yes more control within the Voronoi shader would be ideal (as would more noise flavours ... very old request )
I imagine there should be a way of combining the falloff to black with a variable falloff to grey but I can't figure that out but this is good enough (just) for my current needs.
www.richardfraser.co.uk
/|\
Ryzen 9 3900X @3.79Ghz, 64Gb (TG4 benchmark 6:20)
i7 5930K @3.5Ghz, 32Gb (TG4 benchmark 13.44)
#### KlausK
##### January 21, 2018, 08:36:32 am #12
Thank you for this.
Cheers, Klaus.
/ ASUS WS Mainboard / Dual XEON E5-2640v3 / 64GB RAM / NVIDIA GeForce GTX 1070 TI / Win7 Ultimate
#### bobbystahr
##### January 22, 2018, 10:12:36 am #13
Great thread, off to study the results; thanks very much Jon and Richard.
something borrowed,
something Blue.
Ring out the Old.
Bring in the New
Bobby Stahr, Paracosmologist | 1,647 | 5,400 | {"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} | 2.515625 | 3 | CC-MAIN-2019-43 | latest | en | 0.932653 |
https://www.affordablecebu.com/dir/unexperienced/hich_graph_represents_viable_values_for_y_2x_where_x_is_the_number_of_pounds_of_rice_scooped_and_purchased_from_a_bulk_bin_at_the_grocery/9-1-0-3480 | 1,639,013,660,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363641.20/warc/CC-MAIN-20211209000407-20211209030407-00090.warc.gz | 680,722,664 | 16,040 | Home » Questions » Unexperienced [ Ask a new question ]
# hich graph represents viable values for y = 2x, where x is the number of pounds of rice scooped and purchased from a bulk bin at the grocery
hich graph represents viable values for y = 2x, where x is the number of pounds of rice scooped and purchased from a bulk bin at the grocery store and y is the total cost of the rice Asked by: Guest | Views: 88 | 101 | 412 | {"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-2021-49 | latest | en | 0.923839 |
http://life.familyeducation.com/cig/weather/wind-willows.html | 1,464,522,024,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049278887.83/warc/CC-MAIN-20160524002118-00211-ip-10-185-217-139.ec2.internal.warc.gz | 182,248,684 | 11,688 | # Weather
## The Wind in the Willows
When the atmospheric pressure varies between the horizontal points, the wind blows. As the English poet Christina Rossetti once wrote, "Who has seen the wind, neither I nor you." But that hasn't stopped us from measuring it. A wind vane is almost as old as the beginning of time. A simple pennant or flag will point in the direction from which the wind is coming. On airport runways, a cone-shaped bag opened at each end extends horizontally and points in the direction of the wind. This simple wind vane is called a windsock. Pilots can tell the wind direction as they land.
##### Weather-Speak
A pressure plate anemometer is the oldest type of anemometer. It measures wind speed by a single swinging metal plate rather than the cups of a modern anemometer.
##### Weather-Wise
An anemometer measures wind speed, while a weather vane or wind vane measures wind direction.
By the way and very importantly, the wind direction is never the direction the wind is going, but rather, the direction from which the wind is coming. A northeast wind is coming from the northeast. A south, or southerly, wind blows from the south.
What about wind speed? Again, back in the seventeenth century, English physicist Robert Hooke came up with a device consisting of a plate that moved proportionally to the wind speed. The plate swung out farther in a stronger wind. That became known as a pressure plate anemometer.
In more recent times, rotating cups mounted on a shaft have been used for measuring wind speeds (see the figure at the end). This cup anemometer records the wind speed as the cups spin. Those rotating cups might turn some gears or generate an electrical current. The greater current indicates higher wind speeds.
Some anemometers, called aerovanes, look like airplane propellers. The device swings into the wind, and then the propeller blades rotate with the increasing wind speed.
Something that easily beats an old rusty anemometer is the traditional Beaufort scale, which relates wind speeds to the motion of objects such as trees or water. Anemometers become inaccurate because of their constant exposure.
Beaufort NumberWind Speed Miles/hour (Km/hour)Description
0<1 (<1.6)Calm: Still: Smoke will rise vertically.
11-3 (1.6-4.8)Light Air: Rising smoke drifts, weather vane is inactive.
24-7 (6.4-11.3)Light Breeze: Leaves rustle, can feel wind on your face, weather vane is inactive.
38-12 (12.9-19.3)Gentle Breeze: Leaves and twigs move around. Light weight flags extend.
413-18(20.9-29.0)Moderate Breeze: Moves thin branches, raises dust and paper.
519-24 (30.6-38.6)Fresh Breeze: Moves trees sway.
625-31(40.2-50.0)Strong Breeze: Large tree branches move, open wires (such as telegraph wires) begin to "whistle," umbrellas are difficult to keep under control.
732-38 (51.5-61.2)Moderate Gale: Large trees begin to sway, noticeably difficult to walk.
839-46(62.8-74.0)Fresh Gale: Twigs and small branches are broken from trees, walking into the wind is very difficult.
947-54(75.6-86.9)Strong Gale: Slight damage occurs to buildings, shingles are blown off of roofs.
1055-63 (88.5-101.4)Whole Gale: Large trees are uprooted, building damage is considerable.
1164-72 (103.0-115.9)Storm: Extensive widespread damage. These typically occur only at sea, and rarely inland.
12>73 (>115.9)Hurricane: Extreme destruction.
Beaufort scale.
* The Beaufort number is also referred to as a "Force" number, for example, "Force 10 Gale."
* To calculate knots, divide miles/hour by 1.15.
* Small craft advisories are usually issued when force 6 is reached.
Excerpted from The Complete Idiot's Guide to Weather © 2002 by Mel Goldstein, Ph.D.. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with Alpha Books, a member of Penguin Group (USA) Inc.
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Ommm! 5 Meditation & Mindfulness Activities for Families
Family meditation and mindfulness can help reduce anxiety and promote health and happiness. Learn some fun and easy mindfulness activities for kids, and set them on the path to inner peace! | 1,089 | 4,503 | {"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-2016-22 | longest | en | 0.947663 |
https://es.mathworks.com/matlabcentral/cody/problems/5-triangle-numbers/solutions/1696820 | 1,606,870,646,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141685797.79/warc/CC-MAIN-20201201231155-20201202021155-00419.warc.gz | 242,752,850 | 16,976 | Cody
# Problem 5. Triangle Numbers
Solution 1696820
Submitted on 19 Dec 2018 by Dan Moran
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
n = 1; t = 1; assert(isequal(triangle(n),t))
2 Pass
n = 3; t = 6; assert(isequal(triangle(n),t))
3 Pass
n = 5; t = 15; assert(isequal(triangle(n),t))
4 Pass
n = 30; t = 465; assert(isequal(triangle(n),t))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 183 | 602 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-50 | latest | en | 0.723906 |
https://redrat.net/quick-and-easy-methods-how-to-figure-out-percentages | 1,695,366,230,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506339.10/warc/CC-MAIN-20230922070214-20230922100214-00126.warc.gz | 540,319,229 | 49,633 | # Quick and Easy Methods How to Figure Out Percentages
Welcome to the world of percentage calculation! Whether you’re a student, professional, or simply curious about mastering this essential skill, this comprehensive guide will equip you with the knowledge and techniques needed to figure out percentages with ease.
Understanding percentages is crucial in various fields, from finance and economics to mathematics and everyday life. Whether you’re calculating discounts, analyzing data, or determining proportions, having a solid grasp of percentage calculation is essential.
In this article, we will delve into the fundamental methods and concepts behind figuring out percentages. We’ll explore step-by-step procedures, provide practical examples, and offer useful tips and tricks along the way. So, let’s dive in and unlock the secrets of percentage calculation!
Contents
## Understanding Percentages: A Brief Overview
Percentages play a significant role in various aspects of our lives, from understanding discounts during sales to interpreting statistical data. Before we dive into the intricacies of percentage calculations, let’s start with a brief overview of what percentages actually represent.
### What is a Percentage?
A percentage is a way to express a portion or fraction of a whole as a value out of 100. It is denoted by the symbol “%”. For example, 50% represents half of the whole, while 75% represents three-quarters.
### The Relationship between Percentages, Fractions, and Decimals
Percentages, fractions, and decimals are interconnected. Understanding these relationships can simplify percentage calculations. For instance, 50% is equivalent to the fraction 1/2 and the decimal 0.5.
### Common Uses of Percentages
Percentages are used in various real-life scenarios. Some common applications include calculating discounts, determining tax amounts, analyzing statistical data, and interpreting test scores.
## Basic Percentage Calculation Method
Now that we have a brief overview of percentages, let’s explore the basic method for calculating percentages. This method is straightforward and forms the foundation for more complex percentage calculations.
To calculate a percentage, follow these steps:
1. Step 1: Identify the Whole Determine the total value or quantity that represents 100%. This whole will serve as the reference point for calculating the percentage.
2. Step 2: Identify the Part Identify the specific portion or value that you want to calculate the percentage of. This part will be a fraction of the whole.
3. Step 3: Set Up the Proportion Express the part as a fraction or decimal of the whole. For example, if you have 30 out of 100, you can express it as 30/100 or 0.3.
4. Step 4: Calculate the Percentage Multiply the proportion (fraction or decimal) by 100 to obtain the percentage value. This will give you the percentage representation of the part in relation to the whole.
Let’s take an example to illustrate this method. Suppose you have a class of 40 students, and you want to find out what percentage of the class scored above 90% in the final exam. The basic steps would be:
1. Identify the whole: The whole is the total number of students in the class, which is 40.
2. Identify the part: The part is the number of students who scored above 90%.
3. Set up the proportion: Let’s say there are 10 students who scored above 90%. The proportion is 10/40 or 0.25.
4. Calculate the percentage: Multiply the proportion by 100: 0.25 * 100 = 25%. Therefore, 25% of the class scored above 90%.
The basic percentage calculation method provides a solid foundation for understanding more complex percentage calculations. In the next step, we’ll explore an alternative method using proportions.
## Calculating Percentages using Proportions
In addition to the basic percentage calculation method, another approach to calculating percentages is through the use of proportions. This method can be particularly useful when comparing different values or quantities.
To calculate a percentage using proportions, follow these steps:
1. Step 1: Identify the Whole Just like in the basic method, determine the total value or quantity that represents 100%.
2. Step 2: Identify the Part Identify the specific portion or value you want to calculate the percentage of, similar to the basic method.
3. Step 3: Set Up the Proportion Express the part as a fraction or decimal of the whole, just as in the basic method.
4. Step 4: Cross-Multiply and Solve Set up a proportion by cross-multiplying the values. Then solve for the unknown value, which represents the percentage.
Let’s illustrate this method with an example. Suppose you want to determine the percentage of female students in a college with a total student population of 800, and there are 200 female students. The steps would be as follows:
1. Identify the whole: The whole is the total student population, which is 800.
2. Identify the part: The part is the number of female students, which is 200.
3. Set up the proportion: Express the part as a fraction of the whole: 200/800 = 1/4 or 0.25.
4. Cross-multiply and solve: Set up the proportion as (1/4) = (x/100), where x represents the unknown percentage. Cross-multiplying gives 4x = 100. Solving for x, we find that x = 25. Therefore, the percentage of female students in the college is 25%.
## Converting Decimals and Fractions to Percentages
Converting decimals and fractions to percentages is a common task in percentage calculations. It allows us to express values in a more familiar and easily interpretable form. In this step, we will explore how to convert decimals and fractions into percentages.
### Converting Decimals to Percentages
To convert a decimal to a percentage, follow these steps:
1. Step 1: Multiply by 100 Multiply the decimal by 100 to express it as a proportion out of 100.
2. Step 2: Add the Percentage Symbol Add the percentage symbol “%” to the result to indicate that it represents a percentage.
For example, let’s say you have a decimal value of 0.75. To convert it to a percentage:
Step 1: Multiply 0.75 by 100: 0.75 * 100 = 75.
Step 2: Add the percentage symbol: 75%.
Therefore, 0.75 as a decimal is equivalent to 75% as a percentage.
### Converting Fractions to Percentages
To convert a fraction to a percentage, follow these steps:
• Step 1: Divide the Numerator by the Denominator Divide the numerator (top part of the fraction) by the denominator (bottom part of the fraction).
• Step 2: Multiply by 100 Multiply the result by 100 to express it as a proportion out of 100.
• Step 3: Add the Percentage Symbol Add the percentage symbol “%” to the result to indicate that it represents a percentage.
For example, let’s convert the fraction 3/5 into a percentage:
Step 1: Divide 3 by 5: 3 ÷ 5 = 0.6.
Step 2: Multiply by 100: 0.6 * 100 = 60.
Step 3: Add the percentage symbol: 60%.
Therefore, the fraction 3/5 is equivalent to 60% as a percentage.
## Finding the Percentage Increase or Decrease
Calculating the percentage increase or decrease is a useful skill when analyzing changes in values over time or comparing different quantities. In this step, we will explore the method to find the percentage increase or decrease between two values.
### Percentage Increase
To calculate the percentage increase between two values, follow these steps:
• Step 1: Determine the Difference Find the difference between the new value and the original value. If the new value is greater than the original value, the difference will be positive.
• Step 2: Divide the Difference by the Original Value Divide the difference obtained in Step 1 by the original value.
• Step 3: Multiply by 100 Multiply the result from Step 2 by 100 to express it as a percentage.
Let’s illustrate this with an example. Suppose the price of a product increased from \$50 to \$70. To calculate the percentage increase:
Step 1: Determine the difference: \$70 – \$50 = \$20.
Step 2: Divide the difference by the original value: \$20 / \$50 = 0.4.
Step 3: Multiply by 100: 0.4 * 100 = 40%.
Therefore, the price increase is 40%.
### Percentage Decrease
To calculate the percentage decrease between two values, follow the same steps as for the percentage increase. The only difference is that the difference obtained in Step 1 will be negative since the new value is lower than the original value. The negative sign will indicate a decrease instead of an increase.
Let’s consider an example. If the price of a product decreased from \$80 to \$60:
Step 1: Determine the difference: \$60 – \$80 = -\$20.
Step 2: Divide the difference by the original value: -\$20 / \$80 = -0.25.
Step 3: Multiply by 100: -0.25 * 100 = -25%.
Therefore, the price decrease is 25%.
## Conclusion
Congratulations! You have now learned the essential methods and concepts for figuring out percentages. We covered various topics, including understanding percentages, basic percentage calculation methods, calculating percentages using proportions, converting decimals and fractions to percentages, and finding the percentage increase or decrease.
By following the step-by-step instructions provided in this guide, you can confidently calculate percentages in different scenarios. Whether you need to determine discounts, analyze data, or interpret proportions, you now have the knowledge and tools to do so effectively.
Remember, percentages are widely used in many fields, from finance and economics to everyday life situations. Having a solid understanding of percentage calculations will empower you to make informed decisions and better interpret numerical information.
If you ever need a refresher or encounter more complex percentage calculations, don’t hesitate to revisit this guide or seek additional resources. Practice is key to mastering this skill, so keep applying your knowledge in various scenarios to become even more proficient.
Thank you for joining us on this percentage calculation journey. We hope this guide has provided you with valuable insights and practical techniques. Feel free to explore other related topics and continue expanding your mathematical abilities. Happy calculating!
## FAQs
Q: What are some common real-life scenarios where percentage calculations are useful?
A: Percentage calculations are useful in scenarios such as calculating discounts during sales, determining tax amounts, analyzing statistical data, and interpreting test scores.
Q: How do I calculate a percentage increase between two values?
A: To calculate a percentage increase, find the difference between the new value and the original value, divide it by the original value, and multiply by 100.
Q: Can you provide an example of converting a decimal to a percentage?
A: Certainly! For instance, to convert the decimal 0.75 to a percentage, you would multiply it by 100 and add the percentage symbol, resulting in 75%.
Q: How can I convert a fraction into a percentage?
A: To convert a fraction to a percentage, divide the numerator by the denominator, multiply by 100, and add the percentage symbol. For example, 3/5 is equivalent to 60% as a percentage.
Q: What is the difference between a percentage and a proportion?
A: A percentage is a way to express a portion or fraction of a whole as a value out of 100, while a proportion compares two quantities or values.
Q: How can I calculate the percentage of a whole?
A: To calculate the percentage of a whole, divide the part by the whole, and multiply by 100. This will give you the percentage representation of the part in relation to the whole.
Q: Are there alternative methods to calculate percentages other than the basic method?
A: Yes, another method is using proportions. By expressing the part as a fraction or decimal of the whole and setting up a proportion, you can find the percentage.
Q: Can you explain how to calculate a percentage decrease?
A: Sure! To calculate a percentage decrease, follow the same steps as for a percentage increase, but the difference will be negative since the new value is lower than the original value. The negative sign indicates a decrease. | 2,571 | 12,119 | {"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} | 4.625 | 5 | CC-MAIN-2023-40 | latest | en | 0.857766 |
http://mathoverflow.net/questions/79524/how-do-you-solve-linear-systems-whose-solutions-decay-exponentially?sort=oldest | 1,469,494,236,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824499.16/warc/CC-MAIN-20160723071024-00318-ip-10-185-27-174.ec2.internal.warc.gz | 164,462,782 | 17,159 | # How do you solve linear systems whose solutions decay exponentially?
Consider the heat equation
$$\dot{u} = \Delta u$$
with initial conditions
$$u_0 = \delta(x)$$
for some point $x$ in the domain $\Omega$ of the problem. If $\Omega$ is $\mathbb{R}^n$, then this problem has a closed-form solution given by the Euclidean heat kernel
$$k_t(x,y) = \frac{1}{(4\pi t)^{n/2}}e^{-|x-y|^2/4t};$$
in general solutions to this problem will exhibit this same type of behavior: the magnitude of $u$ decays roughly exponentially as you travel away from the "source" $x$.
Now consider solving this problem numerically by constructing a finite-dimensional linear system whose solution approximates the solution of the original PDE (e.g., using a Galerkin method). Typical numerical linear algebra systems (which work with fixed-precision floating-point arithmetic) guarantee that the maximum error in any component of the solution will be no larger than some small fraction $\epsilon > 0$ of the largest element of the right-hand side.
For instance, in the problem outlined above the right-hand side will look like a Kronecker delta, so the absolute error will be no worse than $\epsilon$. Unfortunately, since the magnitude of the solution decays exponentially, the error $\epsilon$ may be much larger than the smallest entry of the true solution.
Question: using floating-point computations only, can one obtain a solution to a linear system that obtains a desired accuracy relative to the magnitude of each element of the solution vector, rather than the right hand side?
The fundamental problem in achieving better accuracy seems to stem from cancellation effects, i.e., if you add two numbers of very different magnitude in floating point, the smaller one is essentially ignored. I am aware of algorithms that use alternative numerical representations (e.g,. Dixon's method and so on), but this kind of answer is not particularly interesting to me due to considerations of efficiency.
-
My apologies if this question is "too numerical" for MO -- the Numerical Modeling & Simulation Stack Exchange site does not yet exist! – TerronaBell Oct 30 '11 at 15:02
Recent version of LAPACK include routines that compute solutions with componentwise error bounds. It's expensive to do this, and might not be worth it in practice, but see the LAPACK documentation. – Brian Borchers Oct 30 '11 at 17:11
@fuzzytron: on the contrary, I am a "numerical mathematician" and I find that there are too little numerical questions on MO. One could argue on whether numerical linear algebra belongs here rather than in a hypothetical AppliedMathOverflow site, but as long as the pure math people do not get too snobbish I think that both groups can benefit of hanging around in the same environment. – Federico Poloni Nov 3 '11 at 14:32
This subject has been studied in literature --- it turns out that for some class of matrices (such as M-matrices) you can obtain componentwise accurate solutions. A good starting point is the section on "accurate floating point computation" on Jim Demmel's home page; other good search terms are "accurate linear algebra" and "componentwise error analysis". Recent literature has focused more on accurate eigenvalue computation rather than linear systems, but there are results also for them.
EDIT: And let me add that discretizing $\Delta$ with the usual finite-difference scheme results in an M-matrix.
-
Thanks Federico -- I ran across some of Jim's work earlier, but I'll take a closer look (seems like the right idea). – TerronaBell Nov 4 '11 at 16:26
I assume that you're interested in solving the boundary value problem in a situation where the boundaries are far away from the source, so that the solution is close to the analytical solution on an unbounded domain.
One option here would be to write the solution to your problem as the sum of analytical solution to the problem on an unbounded domain plus a correction term. That is,
$u(x,y,t)=k_{t}(x,y)+v(x,y,t)$
Since the PDE is linear, your correction term $v(x,y,t)$ would also have to satisfy the heat equation. You could easily derive boundary conditions for $v$ in terms of the boundary conditions for $u$ and the values of $k_{t}(x,y)$ at the boundaries.
-
Thanks Brian. Actually I'm interested in geometries for which there is no known analytical solution for (k_t). Often these are compact spaces without boundary. – TerronaBell Oct 30 '11 at 16:11
On compact riemannian manifolds, you have an asymptotic heat kernel, which is basically the euclidean heat kernel with correction terms coming from the metric and its derivatives. – Matthias Ludewig Oct 30 '11 at 18:09 | 1,054 | 4,663 | {"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} | 3.359375 | 3 | CC-MAIN-2016-30 | latest | en | 0.876107 |
https://imagine.gsfc.nasa.gov/features/yba/M31_velocity/ | 1,579,441,795,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594603.8/warc/CC-MAIN-20200119122744-20200119150744-00166.warc.gz | 477,642,882 | 5,397 | # What is the Velocity of M31?
A note from your astronomy professor.
Galex image of M31. (Credit: NASA/JPL-Caltech)
You walk into astronomy class one day and find the following question on the board: "What is the radial velocity of the galaxy M31 with respect to our galaxy?" You had already learned that radial velocity means the velocity in a straight line toward or away from something, so the challenge is to find out how fast M31, also known as the Andromeda galaxy, is moving toward or away from our home galaxy, the Milky Way. Once everyone is in class, your professor says that the first person to solve this question using astronomical experiments and data (instead of looking up the answer on the Internet or in a book) will be excused from exams for the rest of the year.
Your professor tells you that the resources you can use include the University's intro astronomy equipment (for example, an optical telescope), as well as astronomical data available in print and on-line. Use your understanding of the laws of physics to select an experiment which will help you to find the answer.
You immediately think of three possible approaches:
1. Use a lightcurve from M31 and the 1/r2 relationship to determine M31's distance. You think that if the distance to M31 is changing over time, you could use 1/r2 at two different times to determine the change in distance. From that changing distance and the time interval between those measurements, you can determine the velocity.
2. Use a spectrum from M31 and the Doppler shift of emission lines. You remember learning that the observed wavelength of emission lines is shifted when the source is moving with respect to the observer, so maybe you can use this to measure the velocity of M31.
3. Use Hubble's Law. You learned about Hubble's Law earlier in astronomy class, which states that galaxies are all moving away from each other, and the further a galaxy is, the faster it is moving away. You think that you could maybe use this law to determine the velocity of M31.
At least one of the above methods will give you the correct answer. Which one do you want to try?
A service of the High Energy Astrophysics Science Archive Research Center (HEASARC), Dr. Alan Smale (Director), within the Astrophysics Science Division (ASD) at NASA/GSFC | 489 | 2,305 | {"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} | 3.375 | 3 | CC-MAIN-2020-05 | longest | en | 0.952517 |
https://lynniezulu.com/what-size-is-dash-10-hose/ | 1,670,452,775,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711221.94/warc/CC-MAIN-20221207221727-20221208011727-00226.warc.gz | 398,509,705 | 11,056 | # What size is Dash 10 hose?
## What size is Dash 10 hose?
JIC 37° & UNO (O Ring Boss)
TUBE OD HOSE ID TUBE or HOSE DASH SIZE THREAD SIZE
3/8 -06 9/16-18
1/2 -08 3/4-16
5/8 -10 7/8-14
3/4 -12 1.1/16-12
### What is Dash size in hose?
Simply put, a dash size refers to an I.D. of a hydraulic hose or fitting in 1/16″ increments. For example, -6 hose would indicate a hose with I.D. of 6/16″ – or 3/8″. Dash numbers, the industry standard, are typically used to describe hose and coupling size.
#### Are hoses measured by ID or OD?
Hoses are measured by inside diameter, unlike tubes which are measured by outside diameter. As a matter of fact, that is how the hoses and tubes are differentiated.
Simply put, a dash size refers to a diameter of a hose, tube or fitting in 1/16″ increments. Each dash size corresponds to a progressively larger diameter, measured in sixteenths of an inch. “-1” is 1/16, “-2” is 2/16, and so on. In this case, -6 hose would indicate a hose with an inside diameter of 6/16”, or 3/8″.
How do you measure a hose ID?
Measure the distance across the opening in the paper tube with a ruler. This measurement equals the diameter of the inside of your garden hose.
## How do you know what size an fitting I have?
You can always identify AN fittings based on the male thread size (outside diameter). These sizes are constant regardless of brand, hose type, or fitting configuration. If your fitting has a convex 37 degree flare at the end, the threads will tell you the AN size (and vice-versa).
### What is SAE 10 thread?
SAE 45° Flare (SAE J512)
Inch size Dash size Male Thread O.D. (in)
5⁄8 -10 0.88
3⁄4 -12 1.06
7⁄8 -14 1.25
1 -16 1.38
#### Is SAE thread the same as UNF?
The Unified Thread Standard uses the term UNC (Unified Coarse) to describe a fastener that previously would have been designated USS and the Unified Thread Standard uses the term UNF (Unified Fine) to describe a fastener that would have previously been designated SAE.
What size is 10 an fitting?
Plumbing Basics – AN Fittings & Hoses
AN Size Hose i.d. – Tube o.d. SAE Thread Size
-8 1/2 in. 3/4 in.-16
-10 5/8 in. 7/8 in.-14
-12 3/4 in. 1 1/16 in.-12
-16 1 in. 1 5/16 in.-12
What does dash size mean on a hose?
For example, nylon tubing is very flexible, and flexible metal hoses are (relatively) very rigid. So remember: dash size is the size of hose, tube or fittings expressed in sixteenths of an inch. Hoses are measured by inside diameter, and tubes are measured in outside diameter. That’s all there is to it!
## What does-10 mean on a hose?
In another example, a -10 tube would indicate a tube with an outside diameter of 10/16″, or 5/8″. Fittings are a little different and slightly more complex. In the part number 10343-8-6, for example, -8 is the size of the fitting end connection, and -6 is the hose size.
### What does-8 mean on hose part numbers?
In the part number 10343-8-6, for example, -8 is the size of the fitting end connection, and -6 is the hose size. Did you notice that for hose part numbers, we talked about inside diameter, and for tube part numbers, we referred to an outside diameter? This is not an accident.
#### What does-6 mean on a hose?
In this case, -6 hose would indicate a hose with an inside diameter of 6/16”, or 3/8″. In another example, a -10 tube would indicate a tube with an outside diameter of 10/16″, or 5/8″. | 974 | 3,381 | {"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-2022-49 | latest | en | 0.921051 |
https://www.mathworks.com/matlabcentral/answers/362378-why-does-my-legend-only-display-50-entries | 1,637,967,480,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358074.14/warc/CC-MAIN-20211126224056-20211127014056-00314.warc.gz | 928,254,678 | 31,466 | # Why does my legend only display 50 entries?
100 views (last 30 days)
MathWorks Support Team on 20 Oct 2017
Commented: Dev-iL on 24 Dec 2020
I have a plot with more than 50 items and I would like to add a legend for all of them. The legend gets truncated at 50.
p = plot(magic(100));
labels = cellstr(num2str((1:100)'));
legend(p,labels)
MathWorks Support Team on 20 Oct 2017
Legends are currently limited to no more than 50 entries. Usually in plots with more than 50 features, the plot is so cluttered and the legend is so large that it is more advisable to select just a few key items to display in the legend.
If you do require more than 50 entries, the workaround is to create an additional legend. This requires creating an additional hidden axes.
First plot the original data.
p = plot(magic(100));
labels = cellstr(num2str((1:100)'));
legend(p(1:50),labels(1:50),'location','westoutside')
Next create an additional axes so that the figure can have two legends. Because the new legend needs information about the color of each line, it is necessary to copy the original lines to this new axes.
newAx = axes;
newP = copyobj(p,newAx);
legend(newP(51:100),labels(51:100),'location','eastoutside')
Now make the new axes and copied lines invisible. Normally, this would be done by setting the Visible property of each line to 'off', however this would gray out the corresponding entries in the legend. Instead, set the XData and YData properties of each line to NaN.
axis(newAx,'off')
set(newP,'XData',NaN,'YData',NaN)
Eric Sargent on 9 Dec 2020
Legend will cap the number of entries at 50 if no handles are specified. To have all objects in an axes show up, pass in their handles to the legend command.
p = plot(magic(100));
legend(p);
Richard Mittleman on 6 Sep 2019
This is an amazingly stupid way to have to do this
Walter Roberson on 16 Sep 2020
You are right, I made a multiplication mistake.
Eric Sargent on 9 Dec 2020
Legend will cap the number of entries at 50 if no handles are specified. To have all objects in an axes show up, pass in their handles to the legend command.
p = plot(magic(100));
legend(p);
Dev-iL on 24 Dec 2020
Thank you for pointing this out! This indicates a different problem, where the legend(target, ___) syntax has side effects that change the outcome of the funciton significantly. At the very least it's a documentation bug, but more likely a sign that the legend implementation is somewhat flawed.
Dev-iL on 24 Dec 2020
I suggest using the following undocumented feature for this:
hF = figure();
hAx = axes(hF);
plot(hAx, magic(100));
hL = legend(hAx, '-DynamicLegend');
set(hL, 'LimitMaxLegendEntries', false); % << Solution
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### Community Treasure Hunt
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Start Hunting! | 721 | 2,838 | {"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-2021-49 | latest | en | 0.87685 |
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! Check to ensure mask is calculated first in WHERE ! statements. program where_18 integer :: a(4) integer :: b(3) integer :: c(3) equivalence (a(1), b(1)), (a(2), c(1)) a = (/1, 1, 1, 1/) where (b .eq. 1) c = 2 elsewhere (b .eq. 2) c = 3 endwhere if (any (a .ne. (/1, 2, 2, 2/))) & STOP 1 a = (/1, 1, 1, 1/) where (c .eq. 1) b = 2 elsewhere (b .eq. 2) b = 3 endwhere if (any (a .ne. (/2, 2, 2, 1/))) & STOP 2 end program | 228 | 490 | {"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-2023-23 | latest | en | 0.573991 |
https://python-control.readthedocs.io/en/0.9.0/generated/control.matlab.care.html | 1,618,254,467,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038069133.25/warc/CC-MAIN-20210412175257-20210412205257-00241.warc.gz | 569,229,029 | 5,423 | control.matlab.care¶
`control.matlab.``care`(A, B, Q, R=None, S=None, E=None, stabilizing=True)
(X, L, G) = care(A, B, Q, R=None) solves the continuous-time algebraic Riccati equation
where A and Q are square matrices of the same dimension. Further, Q and R are a symmetric matrices. If R is None, it is set to the identity matrix. The function returns the solution X, the gain matrix G = B^T X and the closed loop eigenvalues L, i.e., the eigenvalues of A - B G.
(X, L, G) = care(A, B, Q, R, S, E) solves the generalized continuous-time algebraic Riccati equation
where A, Q and E are square matrices of the same dimension. Further, Q and R are symmetric matrices. If R is None, it is set to the identity matrix. The function returns the solution X, the gain matrix G = R^-1 (B^T X E + S^T) and the closed loop eigenvalues L, i.e., the eigenvalues of A - B G , E.
Parameters
• A (2D arrays) – Input matrices for the Riccati equation
• B (2D arrays) – Input matrices for the Riccati equation
• Q (2D arrays) – Input matrices for the Riccati equation
• R (2D arrays, optional) – Input matrices for generalized Riccati equation
• S (2D arrays, optional) – Input matrices for generalized Riccati equation
• E (2D arrays, optional) – Input matrices for generalized Riccati equation
Returns
• X (2D array (or matrix)) – Solution to the Ricatti equation
• L (1D array) – Closed loop eigenvalues
• G (2D array (or matrix)) – Gain matrix
Notes
The return type for 2D arrays depends on the default class set for state space operations. See `use_numpy_matrix()`. | 435 | 1,569 | {"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-2021-17 | latest | en | 0.758785 |
https://www.schoolphysics.co.uk/age16-19/Mechanics/Kinematics/text/Particle_velocity/index.html | 1,624,068,715,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487643380.40/warc/CC-MAIN-20210619020602-20210619050602-00070.warc.gz | 914,916,470 | 2,405 | # Particle velocity
## Question:
The velocity of a particle moving along positive x direction is v = (40 -10t) m/s. Here t is in seconds. At time t = 0, the x coordinate of the particle and initial velocity both are zero.
Find the time when the particle is at a distance of 60 m from origin.
v = (40 – 10t)
x = 0 and v = 0 when t = 0, and so v = at and Distance (s) = ½ at2.
Substituting for t gives a = (40 – 10t)/t and then s = (20t - 5t2)
For x = +60 m there is no solution but for – 60m we get t = 6s. This means that the object is the other side of the origin.
A VERSION IN WORD IS AVAILABLE ON THE SCHOOLPHYSICS USB | 192 | 628 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-25 | longest | en | 0.91396 |
http://www.docstoc.com/docs/82252212/3890sg2 | 1,438,375,217,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042988311.72/warc/CC-MAIN-20150728002308-00062-ip-10-236-191-2.ec2.internal.warc.gz | 406,891,872 | 38,774 | # 3890sg2 by mudoc123
VIEWS: 46 PAGES: 2
• pg 1
``` ECN 3890
Spring 2007
Review Questions for Exam 2
Make sure that you study old homework problems as well.
Chapter 4
Questions from chapter: 1, 2(a-c), 3, 5, 6
Graph a short-run production function (total product curve) and demonstrate the relationship between total product and
marginal product.
Explain how marginal revenue product is derived. Why is the MRP curve the firm’s short-run demand for labor? Explain
how and why the labor demand curves of a perfectly competitive seller and an imperfectly competitive seller differ.
The Whatsa Widget Company has a monopoly in the sale of widgets. The following table contains production and product
demand data for this firm:
L Q MP P TR MR MRP
0 0 \$4.50
1 10 3.00
2 18 2.50
3 24 2.00
4 28 1.50
5 30 1.00
a.) Fill in the table above.
b.) Suppose that the firm hires labor in a perfectly competitive market at a wage of \$9. How many workers
should the firm hire to maximize profit? Why?
Suppose that labor demand is given by the equation: L = 50 – 2W, where L = number of workers demanded and W =
wage rate.
(a) Calculate the elasticity of demand at a wage of \$5. How would your answer be different for a wage of
\$20?
(b) What does your answer in (a) imply about how the wage elasticity varies along the demand curve?
(c) Calculate the change in the firm’s wage bill as the wage changes from \$5 to \$6. Do the same for the
change in wage from \$20 to \$21. What relationship do you find between the firm’s wage bill and the
wage elasticity of labor demand?
Why is the short-run demand curve for labor downward sloping?
The marginal revenue product of labor at a local sawmill is MRPL = 20 – 0.5L, where L = number of workers. If the wage
of workers is \$10 (and labor is hired in a perfectly competitive labor market), how many workers should the sawmill hire to
maximize profit?
Chapter 5
Questions from end of chapter: 1, 2, 3, 7
Briefly explain how the following programs would affect the elasticity of demand for workers in the steel industry:
(a) an increased tariff on imported steel
(b) a law making it illegal for a firm to lay off workers for economic reasons (recession)
(c) a boom in the machinery industry (which uses steel as an input)
Suppose that the wage increases. Show (using an isoquant/isocost diagram) that in the long run the firm will hire fewer
workers. Decompose the employment change into substitution and scale effects. Make sure that you can explain what
each effect measures.
What factors determine the elasticity of labor demand?
What happens to the firm’s long-run demand for labor if the demand for the firm’s output increases? What happens to the
firm’s long-run demand for labor if the price of capital increases? (Make sure that you can decompose these changes into
scale and substitution effects.)
Referring to scale and substitution effects, explain why an increase in the wage rate for autoworkers will generate more of
a negative employment response in the long run than in the short run.
Explain why isoquants for inputs that are substitutes in production are (a) negatively sloped, (b) convex, and (c) never
intersect.
Explain how each of the following would shift an isocost line:
(a) a decrease in the price of labor
(b) a simultaneous and proportionate increase in the prices of labor and capital
(c) a decrease in total expenditure from \$150 to \$120
Explain graphically how the isocost/isoquant diagram can be used to derive both the short-run and long-run demand
curves for labor. Make sure that you distinguish between scale and substitution effects.
Indicate whether you believe that the demand for labor is relatively elastic or relatively inelastic for each of the following
groups of workers. Make sure that you can use the laws of derived demand to explain your answer.
(a) computer programmers
(b) economics professors
(c) airline pilots
(d) surgical nurses | 980 | 4,191 | {"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} | 3.40625 | 3 | CC-MAIN-2015-32 | longest | en | 0.931324 |
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05-29-2023, 12:41 AM
Post: #1
Eddie W. Shore Senior Member Posts: 1,585 Joined: Dec 2013
(12C) Log-Normal Distribution Parameter Conversions
The log-normal distribution is transformation of a standard normal variable, where for a standard normal variable t, then a random variable x follows a log-normal distribution, with the form:
x = e^(μ + t * σ)
where:
μ = mean
σ = standard deviation (sample)
The distribution takes the positive values of x. The cumulative distributive function of the log-normal distribution (the area between 0 and x) is:
pdf = 1/2 * (1 + erf((ln x - μ) ÷ (σ * √2)) )
erf is the error function.
erf(θ) = 2 ÷ √(π) * ∫(e^(-s^2) ds, s = 0 to s = θ)
This program on today's blog focuses on the relationship between the distribution mean (μ), standard deviation (σ), the arithmetic expected value (E[x]), and the arithmetic variance (Var[x]):
E[x] =e^(μ + σ^2 ÷ 2)
Var[x] = (e^(σ^2) - 1) * e^(2 * μ + σ^2)
μ = ln( E[x]^2 ÷ √(Var[x] + E[x]^2) )
σ = √( ln (1 + Var[x] ÷ E[x]^2 ) )
Calculate E[x] and Var[x] from μ and σ
Instructions:
To find E[x] and Var[x]:
1. Store μ in memory register 1
2. Store σ in memory register 2
3. Run the program. E[x] is shown in the X stack and is stored in memory register 3. Var[x] is shown in the Y stack in memory register 4.
Code:
(Step: Key Code: Key)
(assume program starts with step 00)
Code:
01: 45, 2: RCL 2 02: 2: 2 03: 21: y^x 04: 44, 0: STO 0 05: 43, 22: e^x 06: 1: 1 07: 30: - 08: 2: 2 09: 45, 1: RCL 1 10: 20: × 11: 45, 0: RCL 0 12: 40: + 13: 43, 22: e^x 14: 20: × 15: 44, 4: STO 4 16: 45, 0: RCL 0 17: 2: 2 18: 10: ÷ 19: 45, 1: RCL 1 20: 40: + 21: 43, 22: e^x 22: 44, 3: STO 3 23: 44, 33, 00: GTO 00
Lines 01 to 03: Store σ^2 in memory register 0
Examples (answers are rounded to four decimal places):
Example 1
Inputs: μ = 1, σ = 0.5
Results: E[x] = 3.0802, Var[x] = 2.6948
Example 2
Inputs: μ = 0, σ = 1
Results: E[x] = 1.6487, Var[x] = 4.6708
Calculate μ and σ from E[x] and Var[x]
Instructions
To find μ and σ:
1. Store E[x] in memory register 3
2. Store Var[x] in memory register 4
3. Run the program. μ is shown in the X stack and is stored in memory register 1. σ is shown in the Y stack in memory register 2.
Code:
(Step: Key Code: Key)
(assume program starts with step 00)
Code:
01: 45, 4: RCL 4 02: 45, 3: RCL 3 03: 2: 2 04: 21: y^x 05: 44, 0: STO 0 06: 10: ÷ 07: 1: 1 08: 40: + 09: 43, 23: LN 10: 43, 21: √ 11: 44, 2: STO 2 12: 45, 0: RCL 0 13: 45, 0: RCL 0 14: 45, 4: RCL 4 15: 40: + 16: 43, 21: √ 17: 10: ÷ 18: 43, 23: LN 19: 44, 1: STO 1 20: 43, 33, 00: GTO 00
Lines 01 to 03: Store E[x]^2 in memory register 0
Lines 12 to 13: Put two copies of memory register 0 on to the stack
Examples (answers are rounded to four decimal places):
Example 1
Inputs: E[x] = 1.84, Var[x] = 0.36
Results: μ = 0.5592, σ = 0.3180
Example 2
Inputs: E[x] = 5.03, Var[x] = 1.72
Results: μ = 1.5825, σ = 0.2565
"Log-normal distribution" Wikipedia. Last Edited May 18, 2023 and retrieved May 24, 2023. https://en.wikipedia.org/wiki/Log-normal_distribution
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https://www.doubtnut.com/qna/646339933 | 1,718,359,060,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861545.42/warc/CC-MAIN-20240614075213-20240614105213-00545.warc.gz | 681,424,252 | 43,474 | # Which of the following matrices can be represented by the notation B=[bij]3×3?
A
b11b12b13b21b22b23b31b32b33
B
b11b12b21b22b31b32
C
[b11b12b13b21b22b23]
D
None of these
Text Solution
Verified by Experts
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Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## Statement-1 (Assertion and Statement- 2 (Reason) Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice as given below. Statement - 1 If mateix A=[aij]3×3,B=[bij]3×3, where aij+aji=0andbij−bji=0 then A4B5 is non-singular matrix. Statement-2 If A is non-singular matrix, then |A|≠0.
AStatement- is true, Statement -2 is true, Statement-2
is a correct explanation for Statement-1
BStatement-1 is true, Statement-2 is true, Sttatement - 2
is not a correct explanation for Stamtement-1
CStatement 1 is true, Statement - 2 is false
DStatement-1 is false, Statement-2 is true
• Question 2 - Select One
## Consider an arbitarary 3×3 non-singular matrix A[aij]. A maxtrix B=[bij] is formed such that bij is the sum of all the elements except aij in the ith row of A. Answer the following questions : If there exists a matrix X with constant elemts such that AX=B, then X is
Bnull matrix
Cdiagonal matrix
Dnone of these
• Question 3 - Select One
## Consider a matrix A=[aij[_(3×3) where, aij={i+2jij=even2i−3jij=odd. If bij is the cafactor of aij in matrix A and Cij=Σ3r=1airbjr, then [Cij]3×3 is
A100010001
B157468393
C880008800088
D231162152
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Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation | 746 | 2,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-26 | latest | en | 0.82012 |
https://gazebosim.org/api/math/6.9/classignition_1_1math_1_1Line3.html | 1,679,841,518,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945473.69/warc/CC-MAIN-20230326142035-20230326172035-00467.warc.gz | 316,677,436 | 7,723 | # Ignition Math
## API Reference
6.9.3~pre2
Line3< T > Class Template Reference
A three dimensional line segment. The line is defined by a start and end point. More...
#include <ignition/math/Line3.hh>
## Public Member Functions
Line3 ()=default
Line Constructor. More...
Line3 (const Line3< T > &_line)
Copy constructor. More...
Line3 (const math::Vector3< T > &_ptA, const math::Vector3< T > &_ptB)
Constructor. More...
Line3 (const double _x1, const double _y1, const double _x2, const double _y2)
2D Constructor where Z coordinates are 0 More...
Line3 (const double _x1, const double _y1, const double _z1, const double _x2, const double _y2, const double _z2)
Constructor. More...
bool Coplanar (const Line3< T > &_line, const double _epsilon=1e-6) const
Test if this line and the given line are coplanar. More...
math::Vector3< T > Direction () const
Get the direction of the line. More...
bool Distance (const Line3< T > &_line, Line3< T > &_result, const double _epsilon=1e-6) const
Get the shortest line between this line and the provided line. More...
Distance (const Vector3< T > &_pt)
Calculate shortest distance between line and point. More...
bool Intersect (const Line3< T > &_line, double _epsilon=1e-6) const
Check if this line intersects the given line segment. More...
bool Intersect (const Line3< T > &_line, math::Vector3< T > &_pt, double _epsilon=1e-6) const
Check if this line intersects the given line segment. The point of intersection is returned in the _pt parameter. More...
Length () const
Get the length of the line. More...
bool operator!= (const Line3< T > &_line) const
Inequality operator. More...
Line3operator= (const Line3< T > &_line)
Assignment operator. More...
bool operator== (const Line3< T > &_line) const
Equality operator. More...
math::Vector3< T > operator[] (const size_t _index) const
Get the start or end point. More...
bool Parallel (const Line3< T > &_line, const double _epsilon=1e-6) const
Test if this line and the given line are parallel. More...
void Set (const math::Vector3< T > &_ptA, const math::Vector3< T > &_ptB)
Set the start and end point of the line segment. More...
void Set (const double _x1, const double _y1, const double _x2, const double _y2, const double _z=0)
Set the start and end point of the line segment, assuming that both points have the same height. More...
void Set (const double _x1, const double _y1, const double _z1, const double _x2, const double _y2, const double _z2)
Set the start and end point of the line segment. More...
void SetA (const math::Vector3< T > &_ptA)
Set the start point of the line segment. More...
void SetB (const math::Vector3< T > &_ptB)
Set the end point of the line segment. More...
bool Within (const math::Vector3< T > &_pt, double _epsilon=1e-6) const
Check if the given point is between the start and end points of the line segment. More...
## Friends
std::ostreamoperator<< (std::ostream &_out, const Line3< T > &_line)
Stream extraction operator. More...
## Detailed Description
### template<typename T> class ignition::math::Line3< T >
A three dimensional line segment. The line is defined by a start and end point.
## ◆ Line3() [1/5]
Line3 ( )
default
Line Constructor.
## ◆ Line3() [2/5]
Line3 ( const Line3< T > & _line )
inline
Copy constructor.
Parameters
[in] _line a line object
## ◆ Line3() [3/5]
Line3 ( const math::Vector3< T > & _ptA, const math::Vector3< T > & _ptB )
inline
Constructor.
Parameters
[in] _ptA Start point of the line segment [in] _ptB End point of the line segment
References Line3< T >::Set().
## ◆ Line3() [4/5]
Line3 ( const double _x1, const double _y1, const double _x2, const double _y2 )
inline
2D Constructor where Z coordinates are 0
Parameters
[in] _x1 X coordinate of the start point. [in] _y1 Y coordinate of the start point. [in] _x2 X coordinate of the end point. [in] _y2 Y coordinate of the end point.
References Line3< T >::Set().
## ◆ Line3() [5/5]
Line3 ( const double _x1, const double _y1, const double _z1, const double _x2, const double _y2, const double _z2 )
inline
Constructor.
Parameters
[in] _x1 X coordinate of the start point. [in] _y1 Y coordinate of the start point. [in] _z1 Z coordinate of the start point. [in] _x2 X coordinate of the end point. [in] _y2 Y coordinate of the end point. [in] _z2 Z coordinate of the end point.
References Line3< T >::Set().
## ◆ Coplanar()
bool Coplanar ( const Line3< T > & _line, const double _epsilon = 1e-6 ) const
inline
Test if this line and the given line are coplanar.
Parameters
[in] _line Line to check against. [in] _epsilon The error bounds within which the check will return true.
Returns
True if the two lines are coplanar.
## ◆ Direction()
math::Vector3 Direction ( ) const
inline
Get the direction of the line.
Returns
The direction vector
## ◆ Distance() [1/2]
bool Distance ( const Line3< T > & _line, Line3< T > & _result, const double _epsilon = 1e-6 ) const
inline
Get the shortest line between this line and the provided line.
In the case when the two lines are parallel, we choose the first point of this line and the closest point in the provided line.
Parameters
[in] _line Line to compare against this. [out] _result The shortest line between _line and this. [in] _epsilon Error tolerance.
Returns
True if a solution was found. False if a solution is not possible.
Referenced by Line3< T >::Intersect().
## ◆ Distance() [2/2]
T Distance ( const Vector3< T > & _pt )
inline
Calculate shortest distance between line and point.
Parameters
[in] _pt Point which we are measuring distance to.
Returns
Distance from point to line.
References Vector3< T >::Length().
## ◆ Intersect() [1/2]
bool Intersect ( const Line3< T > & _line, double _epsilon = 1e-6 ) const
inline
Check if this line intersects the given line segment.
Parameters
[in] _line The line to check for intersection. [in] _epsilon The error bounds within which the intersection check will return true.
Returns
True if an intersection was found.
Referenced by Triangle3< T >::Intersects().
## ◆ Intersect() [2/2]
bool Intersect ( const Line3< T > & _line, math::Vector3< T > & _pt, double _epsilon = 1e-6 ) const
inline
Check if this line intersects the given line segment. The point of intersection is returned in the _pt parameter.
Parameters
[in] _line The line to check for intersection. [out] _pt The point of intersection. This value is only valid if the return value is true. [in] _epsilon The error bounds within which the intersection check will return true.
Returns
True if an intersection was found.
## ◆ Length()
T Length ( ) const
inline
Get the length of the line.
Returns
The length of the line.
Referenced by Line3< T >::Intersect(), and Triangle3< T >::Intersects().
## ◆ operator!=()
bool operator!= ( const Line3< T > & _line ) const
inline
Inequality operator.
Parameters
[in] _line Line to compare for inequality.
Returns
True if the given line is not to this line
## ◆ operator=()
Line3& operator= ( const Line3< T > & _line )
inline
Assignment operator.
Parameters
[in] _line a new value
Returns
this
## ◆ operator==()
bool operator== ( const Line3< T > & _line ) const
inline
Equality operator.
Parameters
[in] _line Line to compare for equality.
Returns
True if the given line is equal to this line
## ◆ operator[]()
math::Vector3 operator[] ( const size_t _index ) const
inline
Get the start or end point.
Parameters
[in] _index 0 = start point, 1 = end point. The _index parameter is clamped to the range [0, 1].
## ◆ Parallel()
bool Parallel ( const Line3< T > & _line, const double _epsilon = 1e-6 ) const
inline
Test if this line and the given line are parallel.
Parameters
[in] _line Line to check against. [in] _epsilon The error bounds within which the check will return true.
Returns
True if the two lines are parallel.
Referenced by Line3< T >::Intersect().
## ◆ Set() [1/3]
void Set ( const math::Vector3< T > & _ptA, const math::Vector3< T > & _ptB )
inline
Set the start and end point of the line segment.
Parameters
[in] _ptA Start point of the line segment [in] _ptB End point of the line segment
Referenced by Line3< T >::Distance(), and Line3< T >::Line3().
## ◆ Set() [2/3]
void Set ( const double _x1, const double _y1, const double _x2, const double _y2, const double _z = 0 )
inline
Set the start and end point of the line segment, assuming that both points have the same height.
Parameters
[in] _x1 X coordinate of the start point. [in] _y1 Y coordinate of the start point. [in] _x2 X coordinate of the end point. [in] _y2 Y coordinate of the end point. [in] _z Z coordinate of both points, by default _z is set to 0.
## ◆ Set() [3/3]
void Set ( const double _x1, const double _y1, const double _z1, const double _x2, const double _y2, const double _z2 )
inline
Set the start and end point of the line segment.
Parameters
[in] _x1 X coordinate of the start point. [in] _y1 Y coordinate of the start point. [in] _z1 Z coordinate of the start point. [in] _x2 X coordinate of the end point. [in] _y2 Y coordinate of the end point. [in] _z2 Z coordinate of the end point.
## ◆ SetA()
void SetA ( const math::Vector3< T > & _ptA )
inline
Set the start point of the line segment.
Parameters
[in] _ptA Start point of the line segment
Referenced by Line3< T >::Distance().
## ◆ SetB()
void SetB ( const math::Vector3< T > & _ptB )
inline
Set the end point of the line segment.
Parameters
[in] _ptB End point of the line segment
Referenced by Line3< T >::Distance().
## ◆ Within()
bool Within ( const math::Vector3< T > & _pt, double _epsilon = 1e-6 ) const
inline
Check if the given point is between the start and end points of the line segment.
Parameters
[in] _pt Point to check. [in] _epsilon The error bounds within which the within check will return true.
Returns
True if the point is on the segement.
References std::max(), std::min(), Vector3< T >::X(), Vector3< T >::Y(), and Vector3< T >::Z().
Referenced by Line3< T >::Intersect().
## ◆ operator<<
std::ostream& operator<< ( std::ostream & _out, const Line3< T > & _line )
friend
Stream extraction operator.
Parameters
[in] _out output stream [in] _line Line3 to output
Returns
The stream
The documentation for this class was generated from the following file: | 2,920 | 10,348 | {"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} | 2.90625 | 3 | CC-MAIN-2023-14 | latest | en | 0.567395 |
https://im.kendallhunt.com/k5_es/teachers/kindergarten/unit-6/lesson-5/preparation.html | 1,725,722,110,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650883.10/warc/CC-MAIN-20240907131200-20240907161200-00780.warc.gz | 291,088,972 | 26,803 | # Lesson 5
¿Cuántos dedos? ¿Cuántos puntos?
### Lesson Purpose
The purpose of this lesson is for students to count to answer “how many” questions about groups of up to 19 images.
### Lesson Narrative
In this lesson students see 11–19 images displayed on fingers and 10-frames. Both representations highlight a group of 10 ones in teen numbers. Students begin to understand that each of these numbers have 10 ones and some more ones (MP7). Although some students may begin to recognize the group of 10 ones and determine how many there are by counting on from 10, this is not an expectation of students in kindergarten. Make sure that numbers 1–20 are posted in the classroom so that students can count from 1 to find the number 17 if they are unsure what the written number 17 looks like. Throughout the section, students have access to a reference sheet that shows numbers 11–20 with dots in 10-frames that they can use to identify written numbers.
• Action and Expression
• MLR8
### Learning Goals
Teacher Facing
• Answer “how many” questions about groups of up to 19 images.
### Student Facing
• Descifremos cuántas cosas hay.
### Required Materials
Materials to Gather
Materials to Copy
• Grab and Count Stage 1 Recording Sheet, Spanish
### Required Preparation
Activity 1:
• Make sure numbers 1–20 are posted in the classroom.
Activity 2:
• Make sure numbers 1–20 are posted in the classroom.
Activity 3:
• Each group of 2 needs around 20 pattern blocks.
• Gather a group of 18 pattern blocks for the activity synthesis.
• Gather materials from:
• Find the Pair, Stage 1
• Tower Build, Stages 1 and 2
Building Towards
### Lesson Timeline
Warm-up 10 min Activity 1 10 min Activity 2 10 min Activity 3 20 min Lesson Synthesis 5 min Cool-down 5 min
### Teacher Reflection Questions
In upcoming lessons, students will compose and decompose numbers 11–19 using 10 ones and some more ones. How does the work of this lesson help build students’ understanding numbers 11–19 as 10 ones and some more ones? | 483 | 2,029 | {"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} | 3.984375 | 4 | CC-MAIN-2024-38 | latest | en | 0.865077 |
https://dsp.stackexchange.com/questions/54902/is-my-kalman-filter-model-reasonable-using-for-a-3-wheels-rover-platform | 1,723,612,559,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641095791.86/warc/CC-MAIN-20240814030405-20240814060405-00418.warc.gz | 162,924,662 | 40,476 | # Is my Kalman Filter model reasonable? Using for a 3 wheels rover platform
Background: I'm building a 3 omniwheel rover platform that looks something like this:
It has 1 IMU sensors on each of the wheels (3 in total). So in theory, I can get gyroscope and accelerometer data from each of the wheels and use that information to calculate the current speed of the wheels.
So let's say this is my model:
$$x_k = \begin{bmatrix} v_k\\a_k \end{bmatrix} = Ax_{k-1} + Bu_{k-1}$$
where: $$A = \begin{bmatrix} 0&\Delta t \\ 0&0 \end{bmatrix}$$ and $$B = \begin{bmatrix} 1&0\\0&0 \end{bmatrix}$$ and $$u_{k-1} = \begin{bmatrix} u_{velocity}\\0 \end{bmatrix}$$
I can "control" the speed of each wheels, but of course that's not accurate. But I cannot control the acceleration of each wheels, so I let $$u_{acceleration}$$ in matrix $$u_{k-1}$$ to be 0.
$$z_k = \begin{bmatrix} v_{measure}\\a_{measure} \end{bmatrix} = H\begin{bmatrix} \omega_{measure}\\a_{measure} \end{bmatrix}$$ with $$H = \begin{bmatrix} R&0\\0&1 \end{bmatrix}$$
$$\omega_{measure}$$ is the data from gyroscope and $$a_{measure}$$ is the data from accelerometer (of each wheels).
Predict:
$$x_k = Ax_{k-1} + Bu_{k-1}$$
$$P_k = AP_{k-1}A^T + Q$$
I haven't really make up my mind about what Q should be.
Update:
$$K_k = P_kH^t(HP_kH^T + R)^{-1}$$
$$x_k = x_k + K_k(z_k - Hx_k)$$
$$P_k = (I - K_kH)P_k$$
I'm also aware that I dropped all the negative sign and the hat sign but hopefully it won't bother the readers too much.
What do you think about this module? Should I change something? Is something is not correct? This is my first time trying out Kalman filter (or any filter at all) after spending a full day learning about it.
In the future I would also like to add an encoder on each wheels to get a better reading.
• Although I can tell each of the wheel to operate at a certain speed, there's nothing guarantee that it will operate at that speed. Should this be consider as a control input? Should I leave B = 0 and let A = [1 t][0 0] ? Commented Jan 18, 2019 at 4:10
• What is the objective? To keep the robot on a route? To regulate the speed?
– A_A
Commented Jan 18, 2019 at 10:08
• @A_A Basically it's to get the current position of the robot while it is moving on a path. But I have another algorithm to calculate the position that is quite accurate, but it is based on the current speed of 3 wheels (without sensor feedback, using stepper motors give reasonable result but I'm working on making it more accurate). What do you think? Commented Jan 18, 2019 at 11:39
• So, the motors on the wheels are stepper motors?
– A_A
Commented Jan 18, 2019 at 13:48
• @A_A yes, it is stepper motors. Commented Jan 19, 2019 at 5:06 | 827 | 2,710 | {"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": 15, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-33 | latest | en | 0.871469 |
https://www.scribd.com/document/167199365/Unit-9-Lecture-Day-4-Colligative-Properties | 1,566,577,695,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318894.83/warc/CC-MAIN-20190823150804-20190823172804-00215.warc.gz | 960,781,629 | 63,328 | You are on page 1of 41
# COLLIGATIVE
PROPERTIES
Unit 9 Chapter 12
Definition:
The properties of a solution that are dependant
only on the number of solute particles in
solution.
Vapor pressure lowering
Boiling point elevation
Freezing point depression
Osmotic pressure
COLLIGATIVE PROPERTIES
The equilibrium vapor pressure of a solvent with a
nonvolatile solute is shown in (a).
The equilibrium vapor pressure of a pure solvent is shown
in (b).
VAPOR PRESSURE
LOWERING
Which equilibrium has a higher vapor
pressure?
VAPOR PRESSURE
LOWERING
The presence of solute particles results in
fewer solvent molecules sitting at the surface
to escape.
VAPOR PRESSURE
LOWERING
According to Raoults Law, the vapor pressure
of a solution with a non-volatile solute is equal
to the vapor pressure of the pure solvent at
that temperature multiplied by its mole fraction.
P
solv
= P
solv
X
solv
VAPOR PRESSURE
LOWERING
The vapor pressure of the solvent is
proportional to the relative number of solvent
molecules at the surface the mole fraction
of the solvent.
When the solute-solvent attractions are stronger than
the original solvent-solvent attractions, the solution
vapor pressure will be lower. This is typically the case
when the solute is nonvolatile.
VAPOR PRESSURE
LOWERING
Definition:
When the vapor pressure of a solvent
containing a solute is compared to the vapor
pressure of the pure solvent, the solutions
vapor pressure is found to be lower.
P
solv
P
solv
= P
VAPOR PRESSURE
LOWERING
Vapor pressure of
the pure solvent
Vapor pressure of
the solvent in
solution
Amount by which
the solvents vapor
pressure changed
Eugenol, C
10
H
12
O
2
, is the chief constituent of oil of clove. It is a
pale yellow liquid that dissolves in ethanol, C
2
H
5
OH; it has a
boiling point of 255C (thus, it has a relatively low vapor
pressure at room temperature). What is the vapor pressure
lowering at 20.0C of ethanol containing 8.56 g of eugenol in
50.0 g of ethanol? The vapor pressure of ethanol at 20.0C is
44.6 mmHg.
VAPOR PRESSURE
LOWERING
P
solv
= P
solv
X
solv
P
solv
P
solv
= P
Solute
Solvent
P
P
solv
= 44.6 mm Hg
P
solv
P
solv
= P
P
solv
P
solv
X
solv
=
P
Eugenol, C
10
H
12
O
2
, is the chief constituent of oil of clove. It is a
pale yellow liquid that dissolves in ethanol, C
2
H
5
OH; it has a
boiling point of 255C (thus, it has a relatively low vapor
pressure at room temperature). What is the vapor pressure
lowering at 20.0C of ethanol containing 8.56 g of eugenol in
50.0 g of ethanol? The vapor pressure of ethanol at 20.0C is
44.6 mmHg.
VAPOR PRESSURE
LOWERING
P
solv
= P
solv
X
solv
P
solv
P
solv
= P
Solute
P
solv
= 44.6 mm Hg
P
solv
P
solv
X
solv
=
P
(44.6) (44.6) X
solv
=
P
Solvent
8.56gEugenol 1mol
164.22g
= 0.0521mol
50.0gEthanol 1mol
46..07g
=1.086mol
Eugenol, C
10
H
12
O
2
, is the chief constituent of oil of clove. It is a
pale yellow liquid that dissolves in ethanol, C
2
H
5
OH; it has a
boiling point of 255C (thus, it has a relatively low vapor
pressure at room temperature). What is the vapor pressure
lowering at 20.0C of ethanol containing 8.56 g of eugenol in
50.0 g of ethanol? The vapor pressure of ethanol at 20.0C is
44.6 mmHg.
VAPOR PRESSURE
LOWERING
P
solv
= P
solv
X
solv
P
solv
P
solv
= P
Solute
P
solv
= 44.6 mm Hg
P
solv
P
solv
X
solv
=
P
(44.6) (44.6) X
solv
=
P
Solvent
X
solv
=
1.086mol
1.086mol + 0.0521mol
X
solv
= 0.954
Eugenol, C
10
H
12
O
2
, is the chief constituent of oil of clove. It is a
pale yellow liquid that dissolves in ethanol, C
2
H
5
OH; it has a
boiling point of 255C (thus, it has a relatively low vapor
pressure at room temperature). What is the vapor pressure
lowering at 20.0C of ethanol containing 8.56 g of eugenol in
50.0 g of ethanol? The vapor pressure of ethanol at 20.0C is
44.6 mmHg.
VAPOR PRESSURE
LOWERING
P
solv
= P
solv
X
solv
P
solv
P
solv
= P
Solute
P
solv
= 44.6 mm Hg
P
solv
P
solv
X
solv
=
P
(44.6) (44.6)(0.954) = P
Solvent
2.05 mm Hg = P
Remember that for a gaseous solution
(homogeneous mixture), the total pressure of
the solution is the sum of the partial pressures.
(Whos law is that?)
Therefore
P
total
= P
solvent
+ P
solute
VAPOR PRESSURE
LOWERING
This means that Raoults law can also be
applied to the vapor pressure of a solution with
a volatile solute. The total vapor pressure of
the mixture would be equal to the vapor
pressure created by the solvent added to the
vapor pressure created by the volatile solute.
P
total
= P
solv
X
solv
+ P
solute
X
solute
VAPOR PRESSURE
LOWERING
WHAT ELSE DOES VAPOR
PRESSURE LOWERING
AFFECT?
In the diagram notice
that the vapor pressure
lowers when a solvent
is mixed with a
nonvolatile solute.
The P
solv
increases
as the temperature
increases; similarly, the
P
solv
increases as the
temperature increases.
This means that the
lowering of the vapor
higher boiling point of
the solution.
Vapor Pressure
(mm Hg)
Temperature (C)
Solvent + Solute
Solvent
P
Definition:
The boiling point of a solution is greater than
the boiling point of the pure solvent because
the solution (which has a lower vapor
pressure) will need to be heated to a higher
temperature in order for the vapor pressure to
become equal to the external pressure (i.e.,
the boiling point).
BOILING POINT ELEVATION
The boiling point elevation, T
b
, is the
difference between the higher boiling point of
the solution and the boiling point of the pure
solvent.
T
b
= T
b
T
b
BOILING POINT ELEVATION
Boiling point
elevation
Boiling point
of solution
Boiling point of
pure solvent
The boiling point elevation, T
b
, is directly
proportional to the molality of the solute.
T
b
= k
b
m
solute
BOILING POINT ELEVATION
Boiling point
elevation
molal boiling point
elevation constant
molality of
the solute
MOLAL BOILING POINT
ELEVATION CONSTANT
Each solvent has a
different boiling
point constant.
Solvent
T
b
of
pure
solvent
(C)
k
b
(C/m)
Water 100.00 +0.5121
Benzene 80.10 +2.53
Camphor 207.4 +5.611
Chloroform 61.70 +3.63
A solution was made up of eugenol, C
10
H
12
O
2
, in diethyl ether. If
the solution was 0.575 m eugenol in ether, what was the boiling
point of the solution? The boiling point of pure ether is 34.6C
and the boiling-point-elevation constant is 2.02 C/m.
BOILING POINT ELEVATION
T
b
= k
b
m
solute
T
b
= T
b
T
b
Solute
Solve
nt
T
b
T
b
= 34.6C
k
b
= 2.02 C/m
m = 0.575 m
T
b
= ??
k
b
m
solute
= T
b
= T
b
T
b
(2.02 C/m)(0.575 m) = T
b
(34.6C)
35.8C = T
b
antifreeze, cocoa, gasoline and ethanol,
salted pasta water (just kidding)
BOILING POINT ELEVATION
Addition of a nonvolatile solute to the volatile
solvent increases the attractions (and lowers
the vapor pressure). This means that the
particles are closer together so a lower
temperature allows them to freeze.
FREEZING POINT
DEPRESSION
The freezing point depression, T
f
, is the
difference between the higher freezing point of
the pure solvent and the freezing point of the
solution.
T
f
= T
f
T
f
FREEZING POINT
DEPRESSION
Freezing point
depression
Freezing point
of pure solvent
Freezing point
of solution
The freezing point depression, T
f
, is directly
proportional to the molality of the solute.
T
f
= k
f
m
solute
FREEZING POINT
DEPRESSION
Freezing point
depression
molal freezing
point depression
constant
molality of
the solute
MOLAL FREEZING POINT
DEPRESSION CONSTANT
Each solvent has a
different freezing
point constant.
Solvent
T
f
of
pure
solvent
(C)
k
b
(C/m)
Water 0.0 +1.86
Benzene 5.50 +5.12
Camphor 179.95 +39.7
A solution was made up of 0.575 m eugenol, C
10
H
12
O
2
, in diethyl
ether. What was the freezing point of the solution? The freezing
point of pure ether is -116.3C and the freezing-point-depression
constant is
1.79 C/m.
FREEZING POINT
DEPRESSION
T
f
= k
f
m
solute
T
f
= T
f
T
f
Solute
Solve
nt
T
f
T
f
= 116.3C
k
f
= 1.79C/m
m = 0.575 m
T
f
= ??
k
f
m
solute
= T
f
= T
f
T
f
(1.79C/m)(0.575 m) = (116.3C)
T
f
117.3C = T
b
antifreeze (HEAT), salt on icy roads,
making ice cream, Crystal Lake freezes
before the pond in Veterans Acres
FREEZING POINT
DEPRESSION
COLLIGATIVE PROPERTY
ANIMATION
Osmosis is the diffusion of small molecules
through a semi-permeable membrane.
Usually, osmosis is seen in the net movement
of the solvent from the pure solvent (low solute
concentration) to solution (high solute
concentration).
OSMOTIC PRESSURE
The membrane is
termed semi-
permeable because
small molecules such
as water or small ions
(Na
+
or K
+
) may pass
in either direction
through the
membrane.
OSMOTIC PRESSURE
The solute reduces the mole fraction of the
solvent in a solution. The lower the mole
fraction of solvent, the greater the net flow
of solvent molecules into the solution (and
greater osmotic pressure).
OSMOSIS
OSMOTIC PRESSURE
Osmotic pressure is the pressure necessary to just
stop osmosis. This is done by pressing on the
solution side to increase the movement of solvent
particles from the solution back into the pure
solvent.
Osmotic pressure is proportional to the
molar concentration of the solute.
OSMOTIC PRESSURE
H ( )
H= MRT
PV = nRT
P =
n
V
RT
Notice how this equation is
based on the ideal gas
law:
Osmotic
pressure
Molarity of
the solute
Dextran, a polymer of glucose units, is produced by bacteria
growing in sucrose solutions. Solutions of dextran in water have
been used as a blood plasma substitute. What is the osmotic
pressure (in mmHg) at 21C of a solution containing 1.50 g of
dextran dissolved in 100.0 mL of aqueous solution, if the average
molecular weight of the dextran is 4.0 x 10
4
amu?
OSMOTIC PRESSURE
Solute
Solvent =
H
2
O
H
H= MRT
H= MRT
1.50g 1mol
4.0x10
4
g
= 3.75x10
5
mol
M =
3.75x10
5
mol
0.1000L
= 3.75x10
4
M
H = 3.75x10
4
M
( )
0.0821
atm-L
mol-L
( )
294K ( )
H = 0.00905atm = 6.9mmHg
reverse osmosis, vinegar and eggs, water
transportation in plants, blood pressure
(normal osmotic pressure when compared
to pure water), dialysis of cell walls,
desalinating ocean water
OSMOTIC PRESSURE
When a nonelectrolyte solute dissolves in a
solvent, it dissolves without separating into
ions.
Like methanol dissolving in water
CH
3
OH(l) CH
3
OH(aq)
Nonelectrolyte Solutes
## When an electrolyte solute dissolves in a
solvent, it does separate into ions.
Like sodium sulfate dissolving in water
Na
2
SO
4
(s) 2 Na
+
(aq) + 1 SO
4
2
(aq)
Electrolyte Solutes
## Since colligative properties depend on the
number of solute particles, not the identity
of the solute, an electrolyte influences those
properties more than a nonelectrolyte.
A 1.0 M sodium sulfate solution creates how many
particles?
Na
2
SO
4
(s) 2 Na
+
(aq) + 1 SO
4
2
(aq)
Electrolyte Solutes
2 + 1 = 3 particles
A strong electrolyte, such as Na
2
SO
4
, will
dissolve 100 %, creating one particle for
each dissociated ion.
These particles are called the vant Hoff
factor, i.
What is i for potassium phosphate?
Electrolyte Solutes
K
3
PO
4
(s) 3 K
+
(aq) + 1 PO
4
3
(aq)
4
A weak electrolyte, such as Ag
3
PO
4
, will
NOT dissolve 100 %. This means that
fewer particles will be able to influence the
colligative properties.
What is i for silver phosphate?
Electrolyte Solutes
Ag
3
PO
4
(s) 3 Ag
+
(aq) + 1 PO
4
3
(aq)
<4
How does this change the colligative
properties?
Electrolyte Solutes
Vapor Pressure
Lowering
Boiling Point Elevation
Freezing Point
Depression
Osmotic Pressure
P
1
= X
1
P
1
o
=
moleSolvent
moleSolvent + (i)moleSolute
|
\
|
.
|
P
1
o
AT
b
= k
b
- m- i = T
b
T
b
o
AT
f
= k
f
- m- i = T
f
o
T
f
H= iMRT | 3,903 | 11,385 | {"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-2019-35 | latest | en | 0.847048 |
http://www.convertit.com/Go/SmartPages/Measurement/Converter.ASP?From=nail&To=depth | 1,618,277,065,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038071212.27/warc/CC-MAIN-20210413000853-20210413030853-00531.warc.gz | 125,788,613 | 3,718 | New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```cloth nail = 0.05715 length (length) ``` Related Measurements: Try converting from "nail" to archin (Russian archin), arpentcan, barleycorn, city block (informal), cloth quarter, digitus (Roman digitus), ell, engineers chain, fathom, furlong (surveyors furlong), gradus (Roman gradus), Greek cubit, Greek palm, mile, nautical mile, pace, Roman cubit, Roman mile, sun (Japanese sun), UK mile (British mile), or any combination of units which equate to "length" and represent depth, fl head, height, length, wavelength, or width. Sample Conversions: nail = .00161082 actus (Roman actus), 3.82E-13 astronomical unit, .125 Biblical cubit, .0015625 bolt (of cloth), .25 cloth quarter, 1.49E-10 earth to moon (mean distance earth to moon), .001875 engineers chain, .07716049 gradus (Roman gradus), .5625 hand, 2.25 inch, 57,150 micron, .00003551 mile, .00208333 naval shot, .075 pace, 1.85E-18 parsec, .1930999 Roman foot, .18860017 shaku (Japanese shaku), .00052083 skein, .00030941 stadium (Roman stadium), .06820249 vara (Mexican vara).
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 452 | 1,640 | {"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-2021-17 | latest | en | 0.680012 |
https://physics.stackexchange.com/questions/146184/whats-the-largest-mushroom-cloud-possible-from-a-coffee-cup-grenade-sized-nucle | 1,701,487,946,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100309.57/warc/CC-MAIN-20231202010506-20231202040506-00706.warc.gz | 511,569,151 | 43,179 | # What's the largest mushroom cloud possible from a coffee cup/grenade sized nuclear bomb?
Assuming the coffee cup is $16$oz = $1$lb = $0.4536$kg or $\sim 450$mL
I did a quick comparison to the W54 assuming there was a linear ratio (wishful thinking perhaps), and got it to be around 8m high.
Height of mushroom cloud: $1450$ ft
Weight of bomb: $58.1$ lb
Volume of bomb: $5.85$ inches in diameter, $15$ inches long
Source, see Socorro test
Assuming comparable height-to-weight ratio:
Mushroom cloud height of $16$ oz ($1$ lb) bomb = $1450$ft / $58.1$ lb $\times$ $1$lb $\sim 7.6$m
Assuming comparable height-to-volume ratio:
Cloud height of $16$ oz ($454$ mL) bomb = $1450$ft / $(\pi 5.85”^2 \times 15”) \times 454$ mL $\sim 8.3$m
But I'm curious what a more realistic value might be, assuming it was designed for that size and assuming maximum efficiency. Thanks in advance.
• ....Are you making a suitcase bomb? Nov 11, 2014 at 21:20
• Haha, much more trivial. Proposing an awareness campaign for the Comprehensive Test Ban Treaty by putting trivia on coffee sleeves and a mushroom cloud model. Nov 11, 2014 at 21:23
• Are we allowed to expend arbitrary amounts of energy to briefly compress a mass to cup size? Nov 11, 2014 at 23:05
• @crclayton I'm almost certain that $1450 ft$ is just the height of the bomb when it went off!
– user12029
Nov 12, 2014 at 0:05
• @NeuroFuzzy Oh, damn, maybe you're right. Nov 12, 2014 at 3:26
## 2 Answers
Of all the common nuclear fuels, Pu-239 has the smallest critical mass. A spherical untampered critical mass is about 11 kg (24.2 lbs),1 10.2 cm (4") in diameter. Using appropriate triggers, neutron reflectors, implosion geometry and tampers, this critical mass can be reduced by more than twofold. This optimization usually requires a large nuclear development organization supported by a sovereign nation.
I measured my coffee cup, and it is clearly smaller in volume than the volume of a spherical critical mass of plutonium. Using all the other fancy hardware described in the second sentence of the quote would reduce the amount of plutonium needed, but would presumably increase the total size of the bomb.
Therefore I think the answer is that the largest possible explosion from a holy nuclear hand grenade is zero, for any common nuclear fuel.
I believe I heard somewhere that according to rough estimates, substances from the nuclear island of stability might have critical masses as small as a pencil eraser. Luckily for our civilization, there is probably no practical way to make bulk quantities of these atoms.
Hirosima was 15kt and had a volume of $1.5m^3$, it produced a mushroom cloud of about $17000m$
if things scaled linearly the coffee mug would produce a cloud of $0.0005m$ in height
However I really doubt it would be linear, even on the site that you linked there doesn't seem to be much of a correlation between kt and mushroom cloud height, not to mention there are multiple types of clouds each of which has varying heights.
for more information concerning the hypothetical you could check out:
http://nuclearsecrecy.com/nukemap/
http://upload.wikimedia.org/wikipedia/commons/6/66/Nukecloud.png
• Yeah, I did similar comparisons to a bunch of bombs like Tsar Bomba and Ivy Mike, all with underwhelming results. Note that Hiroshima wasn't 15kT though, it had a 15kT TNT equivalent. Now I'm more curious about the largest theoretical yield assuming 100% efficiency of the fissile material and such, which is beyond my physics level. Nov 11, 2014 at 21:54
• @crclayton Peak observed efficiency for fission of fissile atoms in nuclear weapons is worse than typical thermodynamic efficiencies for heat engines. 100% fission cannot be reached before the assembly becomes subcritical. Nov 13, 2014 at 0:12
• I doubt that the Hirosima bomb contained 1.5 cubic meters of fissile material. Jun 26, 2020 at 14:55 | 996 | 3,894 | {"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} | 3.390625 | 3 | CC-MAIN-2023-50 | longest | en | 0.884186 |
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A188283 Numbers k such that iterations for the map r -> A061602(r) starting with k ends with a fixed point (factorions A014080). 1
0, 1, 2, 10, 11, 145, 154, 223, 232, 322, 405, 415, 450, 451, 504, 514, 540, 541, 569, 596, 659, 695, 956, 965, 1023, 1032, 1123, 1132, 1203, 1213, 1223, 1230, 1231, 1232, 1302, 1312, 1320, 1321, 1322, 1449, 1494, 1569, 1596, 1659, 1695, 1944, 1956, 1965, 2003 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 COMMENTS If k is a term, then (10^k - 1)/9 is also a term. - Jinyuan Wang, Nov 07 2020 LINKS EXAMPLE Number 405 is in sequence because 405 -> 145 -> 145 -> ... PROG (PARI) is(k) = {my(t=k, v=List([k])); while(t=sum(i=1, #d=digits(t), d[i]!), if(t==v[#v], return(1), if(sum(i=1, #v-1, t==v[i]), return(0))); listput(v, t)); } \\ Jinyuan Wang, Nov 07 2020 CROSSREFS Supersequence of A014080 (factorions). Cf. A061602, A173447, A188284. Sequence in context: A265747 A104459 A008560 * A174703 A258081 A300907 Adjacent sequences: A188280 A188281 A188282 * A188284 A188285 A188286 KEYWORD nonn,base AUTHOR Jaroslav Krizek, Mar 26 2011 EXTENSIONS Missing terms a(1) and a(8)-a(10) aded by Jaroslav Krizek, Jan 28 2012 Name corrected and more terms from Jinyuan Wang, Nov 07 2020 STATUS approved
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Last modified January 15 18:43 EST 2021. Contains 340188 sequences. (Running on oeis4.) | 675 | 1,919 | {"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} | 3.40625 | 3 | CC-MAIN-2021-04 | latest | en | 0.691923 |
https://myyachtguardian.com/how-much-was-a-pound-worth-in-1960-new/ | 1,680,358,081,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950030.57/warc/CC-MAIN-20230401125552-20230401155552-00101.warc.gz | 469,863,742 | 36,629 | # How Much Was A Pound Worth In 1960? New
Let’s discuss the question: how much was a pound worth in 1960. We summarize all relevant answers in section Q&A of website Myyachtguardian.com in category: Blog MMO. See more related questions in the comments below.
Table of Contents
## What was a pound worth in 1960s?
The British pound has lost 96% its value since 1960
£100 in 1960 is equivalent in purchasing power to about £2,453.92 today, an increase of £2,353.92 over 62 years. The pound had an average inflation rate of 5.30% per year between 1960 and today, producing a cumulative price increase of 2,353.92%.
## How much was 100 pounds worth in 1960s?
This means that 100 pounds in 1960 are equivalent to 2,068.1 pounds in 2022. In other words, the purchasing power of £100 in 1960 equals £2,068.1 today. The average annual inflation rate between these periods has been 5.01%.
### UK pounds of the 1960-70’s
UK pounds of the 1960-70’s
UK pounds of the 1960-70’s
See also How Long Can Clam Chowder Stay In The Fridge? Update New
## How much would one pound in 1965 be worth today?
£1 in 1965 is equivalent in purchasing power to about £18.41 in 2017, an increase of £17.41 over 52 years. The pound had an average inflation rate of 5.76% per year between 1965 and 2017, producing a cumulative price increase of 1,740.58%.
## How much was 10 pounds 1960?
£10 in 1960 is equivalent in purchasing power to about £211.34 in 2016, an increase of £201.34 over 56 years. The pound had an average inflation rate of 5.60% per year between 1960 and 2016, producing a cumulative price increase of 2,013.44%.
## What things cost in 1960 UK?
People were a lot poorer in the 1960s. The average weekly pay packet was less than £10 per week. Allowing for inflation that is £150 in today’s money. Today average weekly earnings are more than £500.
Housing.
Year Cost In today’s money
1960 (first quarter) £2,189 £33,000
1969 (last quarter £4,312 £47,500
## What was money worth in 1964?
\$1 in 1964 is equivalent in purchasing power to about \$9.15 today, an increase of \$8.15 over 58 years. The dollar had an average inflation rate of 3.89% per year between 1964 and today, producing a cumulative price increase of 815.21%.
## How much did a house cost in 1960?
In 1960, the median home value in the U.S. was \$11,900, which is the equivalent of around \$98,000 in today’s dollars, and in 2000, SLH notes, it rose to over \$170,000. And it has only kept rising.
## How much was 1000 pounds 1960?
Value of \$1,000 from 1960 to 2022
\$1,000 in 1960 is equivalent in purchasing power to about \$9,712.97 today, an increase of \$8,712.97 over 62 years.
## How much was 100 pounds 1950?
£100 in 1950 is equivalent in purchasing power to about £3,651.14 today, an increase of £3,551.14 over 72 years. The pound had an average inflation rate of 5.12% per year between 1950 and today, producing a cumulative price increase of 3,551.14%.
## How much was a British pound worth in 1850?
£1 in 1850 is equivalent in purchasing power to about £127.96 in 2017, an increase of £126.96 over 167 years. The pound had an average inflation rate of 2.95% per year between 1850 and 2017, producing a cumulative price increase of 12,696.43%.
See also How To Keep T Shirt Print From Cracking? New
## How much was 200 pounds 1919?
For comparison, in the UK £200.00 in 1919 would be equivalent to £11,003.42 in 2022, an absolute change of £10,803.42 and a cumulative change of 5,401.71%.
## How much is a pound worth in 1920?
£1 in 1920 is equivalent in purchasing power to about £42.49 in 2017, an increase of £41.49 over 97 years. The pound had an average inflation rate of 3.94% per year between 1920 and 2017, producing a cumulative price increase of 4,148.62%.
### A guide to old British Money! Pounds, shillings and pence!
A guide to old British Money! Pounds, shillings and pence!
A guide to old British Money! Pounds, shillings and pence!
## How much was 5k in 1960?
\$5,000 in 1960 is equivalent in purchasing power to about \$48,564.86 today, an increase of \$43,564.86 over 62 years.
Value of \$5,000 from 1960 to 2022.
Cumulative price change 871.30%
Average inflation rate 3.73%
Converted amount (\$5,000 base) \$48,564.86
Price difference (\$5,000 base) \$43,564.86
CPI in 1960 29.600
## What would 1000 pounds in 1970 be worth today?
Buying power of £100 in 1970
Initial value Equivalent value
£1,000 pounds in 1970 £16,482.55 pounds today
£5,000 pounds in 1970 £82,412.76 pounds today
£10,000 pounds in 1970 £164,825.52 pounds today
£50,000 pounds in 1970 £824,127.61 pounds today
## What is 6d in today’s money?
Coins and units of money
Coin Amount
6d Sixpence, or ‘tanner’
1s Shilling, or ‘bob’
2s Two shillings, or ‘florin’
2s 6d Half a crown
## What did a loaf of bread cost in 1960?
The Price of Bread
YEAR Cost of 1 lb. of Bread Federal Minimum Wage
1930 \$0.09 None
1940 \$0.10 \$0.30
1950 \$0.12 \$0.75
1960 \$0.23 \$1.00
## How much was a pair of shoes in 1960?
This rate of change indicates significant inflation. In other words, footwear costing \$20 in the year 1960 would cost \$70.06 in 2022 for an equivalent purchase. Compared to the overall inflation rate of 3.72% during this same period, inflation for footwear was lower.
## How much did a car cost in 1960?
In 1960 the average new car costs about 2,752 dollars, and a gallon was gas was around 31. cent.
## What was 10000 dollars worth in 1964?
\$10,000 in 1964 is equivalent in purchasing power to about \$91,521.29 today, an increase of \$81,521.29 over 58 years. The dollar had an average inflation rate of 3.89% per year between 1964 and today, producing a cumulative price increase of 815.21%.
See also How Tall Is 48 Cm? New Update
## What could you buy for a dollar in 1964?
1964: 3 cans of Chicken of the Sea tuna
You could buy three 6.5-ounce cans of Chicken of the Sea tuna in 1964 for a dollar.
## How much was \$50 1964?
Value of \$50 from 1964 to 2022
Cumulative price change 827.43%
Average inflation rate 3.91%
Converted amount (\$50 base) \$463.72
Price difference (\$50 base) \$413.72
CPI in 1964 31.000
## How much did a TV cost in 1960?
Buying power of \$1,000.00 since 1950
Year USD Value Inflation Rate
1958 \$780.13 1.48%
1959 \$790.88 1.38%
1960 \$795.50 0.58%
1961 \$771.75 -2.99%
### PREMIER 1960 Lincoln Pennies Found In Change! – That Sold For INCREDIBLE Money!
PREMIER 1960 Lincoln Pennies Found In Change! – That Sold For INCREDIBLE Money!
PREMIER 1960 Lincoln Pennies Found In Change! – That Sold For INCREDIBLE Money!
## How much was a dress in 1960?
Buying power of \$20.00 since 1935
Year USD Value Inflation Rate
1960 \$34.00 0.20%
1961 \$34.14 0.40%
1962 \$34.22 0.24%
1963 \$34.46 0.71%
## How much did a gallon of milk cost in 1960?
1960: \$1 per gallon.
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You have just come across an article on the topic how much was a pound worth in 1960. If you found this article useful, please share it. Thank you very much. | 2,379 | 7,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-14 | latest | en | 0.964103 |
acosta4.wordpress.com | 1,685,895,793,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650201.19/warc/CC-MAIN-20230604161111-20230604191111-00174.warc.gz | 111,571,403 | 18,971 | # Tychonoff theorem
To start off, I will be giving a proof of the Tychonoff theorem of topology, the theorem says that if $X_i$ are compact topological spaces indexed by $i\in I$, then $X=\Pi_{i\in I}X_i$ with the product topology is compact.
The following proof has the main idea illustrated in the following simpler theorem (the Königs lemma):
Take a poset $P$ and assume that every point has finite rank, that is, if $p\in P$ then the cardinality of chains in $\{q\in P | q are bounded by a natural number, the least bound is called the rank, so the minimal elements have rank zero, those after them have rank 1, etc. If we assume that the poset $P$ has infinite elements but finitely many of a given rank $n$, then there exists an infinite chain $p_0 < p_1 (with the rank of $p_n$ equal to $n$).
The proof is simple but clever, first choose $p_0$ to be minimal such that it has infinitely many elements above it, it exists as there are infinite elements in the poset but finitely many minimal elements; after that choose $p_1$ minimal above $p_0$ having infinitely many elements above it (exists by similar considerations), etc. (strictly speaking “$q$ is above $p$” here should be defined as having a chain $p=p_0 with the required rank property).
The Tychonoff theorem can now be proved using the same idea with the technical padding of the analogy of “compact space” to “finite set” and transfinite recursion, the sets of rank $i$ being the space $X_i$.
In full suppose by contradiction that there is an open cover $\mathcal{U}$ of $X$ which has no finite subcover (we may assume it to consist of elements of the usual basis). Well order $I$ by a cardinal $\kappa$, may assume w.l.o.g that $I=\kappa$. Define with a transfinite recursion process on $\alpha\leq \kappa$ points $x_{\beta}\in X_{\beta}$ with $\beta<\alpha$, and such that $\Pi_{\beta<\alpha}\{x_{\beta}\}\times\Pi_{\beta\geq \alpha}X_{\beta}$ is not covered by a finite subcover of $\mathcal{U}$. At stage $\alpha=\kappa$ we get a contradiction.
At stage $\alpha=\delta$ a limit ordinal, no new element needs to be chosen, and the process goes through by the finiteness in the definition of the product topology (i.e. it has a basis of the form $\Pi_{i\in I}U_i$ where $U_i\subset X_i$ is open and is equal to $X_i$ for all but finitely many indexes), said differently if $\Pi_{\beta<\delta}\{x_{\beta}\}\times\Pi_{\beta\geq \delta}X_{\beta}$ is covered by a finite subcover then $\Pi_{\beta<\gamma}\{x_{\beta}\}\times\Pi_{\beta\geq \gamma}X_{\beta}$ for a $\gamma<\delta$.
At sucesor stage $\alpha+1$ we need to choose $x_{\alpha}\in X_{\alpha}$ with the infiniteness property, for this assume by contradiction that no such element exists, choose for each point $x\in X_{\alpha}$ an open neighborhood $U_x$ such that $\Pi_{\beta<\alpha}\{x_{\beta}\}\times U_x \times \Pi_{\beta> \alpha}X_{\beta}$ is covered by an open subcover and apply the compacity of $X_{\alpha}$
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I will use this blog to post some things in mathematics which I find exceptionally neat, I expect the pace to be at first artificially fast while I go through the stuff I have found during the previous years. | 890 | 3,184 | {"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": 43, "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-2023-23 | latest | en | 0.909949 |
https://math.stackexchange.com/questions/4833721/on-the-derivative-of-dirac-delta | 1,716,243,124,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058313.71/warc/CC-MAIN-20240520204005-20240520234005-00290.warc.gz | 332,939,719 | 36,349 | On the derivative of Dirac delta
Let $$\delta(x)$$ be the Dirac's delta, i.e. the "strange object" characterized by the two properties $$\delta(x)=0$$ for all $$x\neq 0$$ and $$\int \delta(x) f(x)\text{ d}x= f(0)$$ ($$\int f$$ is the integral over $$\mathbb{R}$$). Now, let's define the derivative of $$\delta(x)$$ as
$$\begin{equation*} \dot{\delta}(x)\triangleq \lim_{\tau \to 0} \frac{\delta(x+\tau)-\delta(x)}{\tau} \end{equation*}$$ For simplicity, I'm considering the univariate case $$x\in\mathbb{R}$$. Now let's try to see what happens when we compute a weighted integral of $$f(x)$$ involving $$\dot{\delta}(x)$$. I would write naively \begin{equation*} \begin{aligned} \int \dot{\delta}(x) f(x)\text{ d}x&= \int \left(\lim_{\tau \to 0} \frac{\delta(x+\tau)-\delta(x)}{\tau}\right)f(x)\text{ d}x\\ &= \lim_{\tau \to 0} \frac{1}{\tau}\left[\int \delta(x+\tau)\,f(x)\text{ d}x - \int \delta(x)\,f(x)\text{ d}x\right]\\ &= \lim_{\tau \to 0} \frac{1}{\tau}\left[f(0-\tau) - f(0)\right]=-\dot{f}(0)\\ \end{aligned} \end{equation*} provided that we can push the limit sign outside the integral (if I'm not wrong, here Lebesgue help us to see when this can be done); the difference between the two integrals is not in the form $$\pm \infty \mp\infty$$; the derivative of $$f(x)$$ exists in $$x=0$$.
If I'm not missing anything else, we can say that, just like $$\int \delta(x) f(x) \text{d}x=f(0)$$, we have $$\int \dot{\delta}(x) f(x)\text{ d}x=-\dot{f}(0)$$.
Question
Since the $$\delta(x)$$ is a "strange object", I'm not 100% sure about my conclusions. So, my question is: am I right? If not, where I'm wrong?
• You are wrong by a sign: $\lim_{\tau \to 0} \frac{1}{\tau}\left[f(0-\tau) - f(0)\right]=-\dot{f}(0)$, hence $\int \dot{\delta}(x) f(x)\text{ d}x=-\dot{f}(0)$. Dec 26, 2023 at 3:04
• Yes, though in general manipulation of weird things like this is easy to mess up. See en.wikipedia.org/wiki/Dirac_delta_function#Derivatives
– Eric
Dec 26, 2023 at 3:39
• You have missed one important properrty of Dirac Delta: $\delta(a x)=\frac1{|a|}\delta(x)$. This thing exactly makes it the most strange Dec 26, 2023 at 8:52
• where do we use $\delta(ax)$? Dec 26, 2023 at 9:54
Your argument shows that the derivative of the delta distribution, let's denote it by $$\delta',$$ is given by a limit of difference quotients, i.e. your $$\dot{\delta}.$$
1. For a distribution $$\phi,$$ its derivative $$\phi'$$ is defined by the relation $$\int \phi' f:=-\int \phi f'$$ for any test function $$f.$$ The motivation behind this definition is that when $$\phi$$ is a (continuously differentiable) function, this identity is just integration by parts.
2. The limit $$\lim_{n\to \infty} \phi_n$$ of a sequence of distributions $$\phi_n$$ is characterized by $$\int(\lim_{n\to \infty} \phi_n) f=\lim_{n\to \infty}\int \phi_n f$$ for any test function $$f.$$ This can either be taken as a definition of convergence for distributions or proven as a theorem.
3. For a distribution $$\phi$$ and $$\tau\in \mathbb{R}$$ there is its shift $$\phi(\cdot+\tau),$$ which is again a distribution given by the identity $$\int \phi(\cdot+\tau) f=\int \phi f(\cdot-\tau)$$ for any test function $$f.$$ Similarly to 1., for functions this is just a change of variables.
Using these facts, your proof shows $$\delta'=\lim_{\tau\to 0} \frac{\delta(\cdot+\tau)-\delta}{\tau},$$ an equality of two distributions, which is exactly the result we would expect from the corresponding equality for functions.
• (+1) You might consider correcting the OP's assertions about the "properties" that characterized the Dirac Delta. Namely, $\delta(x)=0$ for all $x\ne0$ is not at all correct. And $\int_{\mathbb{R}}\delta(x)f(x)\,dx=f(0)$ is only a notation, and not an integral (unless the Dirac Delta is regarded as a measure, in which case its derivative is not). Dec 26, 2023 at 15:43 | 1,258 | 3,858 | {"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": 35, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-22 | latest | en | 0.727252 |
www.almondburycs.org.uk | 1,579,985,304,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251681412.74/warc/CC-MAIN-20200125191854-20200125221854-00284.warc.gz | 742,640,752 | 9,174 | # Mathematics (Y5 - 8)
Mathematics contributes to the school curriculum by developing students’ abilities to calculate; to reason logically and to solve problems. Mathematics is important for learners in many other areas of study, particularly Technology and Science. It is also important in everyday living and in many forms of employment.
Mathematics enables learners to build a secure framework of mathematical reasoning, which they can use and apply with confidence. The subject goes beyond cultural differences and its importance is universally recognised. Mathematics helps us to understand and change the world.
We aim to give students confidence in their Mathematical abilities and we encourage them to reach their full potential.
Key Stage 2
The Key Stage 2 Curriculum teaches:
• Place value
• Statistics
• Multiplication and Division
• Perimeter and Area
• Fractions
• Decimals and Percentages
• Properties of Shapes
• Position and Direction
• Converting Units
• Measures and Volume
• Algebra
• Ratio
• Problem Solving
• Investigations
Children in Key Stage 2 have maths lessons each day, either single or double periods. In Year 3 and 4 children are taught maths in their classes and are put into sets for Year 5 and 6.
For every aspect taught in maths, children are engaged in fluency questions to build confidence in an area and then focus on reasoning and problem solving opportunities in order to prepare them for their end of Key Stage 2 SAT’s.
Each class/set complete weekly arithmetic tests and also practise times tables daily using the Times Table Rockstars programme. Children also have their own username and password to access this site at home.
Homework is given out weekly.
For any enquiries regarding the teaching of mathematics at Key Stage 2 then please contact Assistant Achievement Lead for Maths, Mrs R Haley at rachael.haley@kirkleeseducation.uk
Years 7-8
The curriculum at Key Stage 3 is designed to:
• Build on calculation methods and Maths knowledge from KS2
• Develop mathematical reasoning, so students can apply their skills in real life
• Increase expertise so that students learn to solve problems
All students have four lessons of Maths each week. Students are placed in sets by the end of September in year 7. The setting is reviewed regularly. We are constantly striving to keep our curriculum up to date and topical as well as engaging for students. Homework is set each week and is generally consolidation questions based on the topics covered in lessons. It should last 20-30 minutes. All students receive a revision guide and workbook in year 7 and have access to mathswatch revision website.
Equipment
All students are expected to have a pen, pencil and ruler for each lesson. Investment in a protractor, a pair of compasses and scientific calculator is desirable.
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We encourage all students to ask their class teacher for help, however, students are welcome to attend the drop- sessions in the library Monday to Thursday at 3pm -4pm. | 619 | 3,025 | {"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-2020-05 | longest | en | 0.92272 |
https://gradedpapers.net/3827cd2151bf6293/law-homework-5545 | 1,696,343,024,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511106.1/warc/CC-MAIN-20231003124522-20231003154522-00628.warc.gz | 296,401,532 | 39,811 | law homework 5545
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https://www.teacherspayteachers.com/Product/Area-of-Parallelograms-Trapezoids-and-Triangles-Google-Activity-68B-68C-68D-3701363 | 1,542,671,481,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746171.27/warc/CC-MAIN-20181119233342-20181120015342-00011.warc.gz | 990,915,649 | 22,666 | # Area of Parallelograms, Trapezoids and Triangles Google Activity 6.8B 6.8C 6.8D
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Tired of the same old task cards? Are you trying a new blended classroom? Need help integrating technology into math? May be your school is moving to one-to-one devices... Here is a chance to integrate technology and still provide a rigorous problem solving activity!
This download comes with directions on how to take my interactive lesson (which is shared via a link in the directions) and share it with your students. This lesson can be shared on Google Classroom or you can share it as a link on classroom websites such as Edmodo.
The interactive lesson is on Google Slides. Students can improve computer skills while being fully engaged in the math lesson. No copies or paper needed!
I also include another link for an answer key. This would be perfect as a partner activity, independent activity, enrichment, or review before STAAR! This project fully encompasses TEKS 6.8B 6.8C and 6.8D (ONLY FOCUSES ON AREA, NOT VOLUME)
The Lesson contains 17 slides:
Slide 1- is a title page and asks for the student name
Slide 2- Match the Equation to the Shape
Slide 3- Solve for the Area of a Parallelogram
Slide 4- Solve for the Area of a Trapezoid
Slide 5- Solve for the Area of a Triangle
Slide 6-8- Isolate the Variable, students follow guided steps to solve for the unknown height. One trapezoid, one parallelogram and one triangle.
Slide 9- 12 Open Ended Problem Solving- equations are posted at the bottom. One question involves using a digital ruler.
Slide 13-16 Test Prep- Multiple Choice
Slide 17- Reminder to share with your teacher.
Looking for more 6th grade Math Lessons and Activities? Check out these!
Multiply and Divide Integers Google Classroom Activity TEKS 6.3C 6.3D
Stuck in the 80s Digital Breakout Adding Subtracting Multiplying Divide Integers
Adding and Subtracting Integers Musical Mix-Up TEKS 6.3C 6.3D
Entire Learning Cycle on Math Properties of Operations TEKS 6.7D Inc. Technology
Prime Factorization Predicament Classroom Escape Room Challenge TEKS 6.7A
Exponent Task Cards QR Codes & Augmented Reality Reinforces Order of Operations
Exploring Multiplying by a Fraction TEKS 6.3B Comparing Factors to the Products
Fraction Division TEKS 6.3E 6.3A 6.2E Dividing Fractions Google Classroom
Fraction Multiplication TEKS 6.3E Multiply Fractions Google Classroom
Decimal Multiplication TEKS 6.3E Multiplying Decimals Google Classroom
Decimal Division TEKS 6.3E Problem Solving Dividing Decimals Google Classroom
TEKS 6.3E Bundle Multiply and Divide Decimals and Fractions Google Classroom
Lucky Ratios- A Hands-On Introduction to Ratios TEKS 6.4B 6.4E RATIOS ONLY
Powerful Pancake Proportions TEKS 6.5A 6.6A
Would You Rather? - An Intro to Unit Rates TEKS 6.4D 6.5A
World Record Converting Measurements Using Proportions and Unit Rates TEKS 6.4H
Prime Factorization Bingo TWO GAMES! TEKS 6.7A
TEKS 6.2A Classifying Numbers Interactive Math Journal and Google Activity
Multiplying Decimals QR Code Scavenger Hunt TEKS 6.3E Multiply Decimals Only
Super Six! A Multiplying Fractions Partner Game
Ratios Rates and Proportions Google Activity TEKS 6.4B 6.4C 6.4D 6.4E 6.5A
Percent Fraction and Decimals Scavenger Hunt with and Without QR Codes 6.4E 6.4G
Converting Fractions, Decimals and Percents Google Activity TEKS 6.4E 6.4G 6.5C
Percent To Win It! Convert a Fraction into a Percent TEKS 6.4E 6.4G 6.5C Google
The Art of Percent - Convert Fractions, Decimals and Percent Google Math and Art
Problem Solving with the Percent Proportion TEKS 6.5B Missing Part Whole Percent
Problem Solving with the Percent Proportion Find Missing Part, Whole or Percent
Ordering and Comparing Rational Numbers TEKS 6.2D Google Interactive
Compare and Order Rational Numbers Whole Class Activity with Sentence Stems 6.2D
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http://docplayer.net/25144924-1-each-of-the-first-6-months-the-entry-is-rent-expense-1-800-deferred-rent-liability-1-800.html | 1,544,512,695,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823588.0/warc/CC-MAIN-20181211061718-20181211083218-00205.warc.gz | 83,530,857 | 25,790 | # 1. Each of the first 6 months, the entry is: Rent expense 1,800 Deferred rent liability 1,800
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1 Buad 273 Chapter 17 Example Solutions Page 1 of 13 Example 1 Operating Lease with Inducement The matching principle requires that we record rent expense as the benefit of using the building is used to earn revenue. We spread the total cost evenly over the life of the lease as follows: Total number of months: 5*12 = 60 Total number of payments: = 54 Total amount paid \$2,000*54 = \$108,000 Monthly expense: 108,000/60 = \$1, Each of the first 6 months, the entry is: Rent expense 1,800 Deferred rent liability 1,800 At the end of 6 months, the deferred rent liability balance is 6*1,800 = \$10,800. This is amortized over the remaining 54 months, at 10,800/54=\$200 per month 2. Remaining 54 entries will be: Rent expense 1,800 Deferred rent liability 200 Cash 2,000 At the end of 54 months, the balance in the deferred rent liability account will be zero.
2 Buad 273 Chapter 17 Example Solutions Page 2 of 13 Example 2: A17-5 Requirement 1 #1 #2 #3 a. Lease term 7 years (1) 5 years 1 year b. Bargain purchase option \$1 n/a n/a c. Unguaranteed residual n/a n/a? d. Guaranteed residual n/a \$75,000 n/a e. Bargain renewal terms n/a n/a n/a f. Minimum net lease payments \$150,001 (2) \$596,500 (3) \$9,200 g. Contingent lease payments n/a n/a \$7.40/hour h. Interest rate to be used to discount 8% 10% 10% (1) Renewal term preceeds BPO (2) ((\$28,600 \$2,600) x 5) + ((\$11,500 \$1,500) 2) + \$1 (3) (\$104,300 5) + \$75,000. The full amount of the guaranteed residual is included. Requirement 2 Lease 1: 1. Title passes (BPO). YES 2. PV of the MLP: (Optional: we could also calculate this in one step using the CF function (see the financial calculator tutorial): CF0 = 0 C01 26,000 F01 5 C02 10,000 F02 1 C03 10,001 F03 1 NPV I=8 CPT gives 115,948 (\$1 difference due to rounding in multi-step calculation) 115,947/116,000 = 99.95% > 90%, YES 3. Lease term is 7/10 = 70% of life, less than 75%. NO Conclusion: Capital Lease. Copyright 2008 by McGraw-Hill Ryerson Limited. All rights reserved.
3 Buad 273 Chapter 17 Example Solutions Page 3 of 13 Example 2: A17-5 Lease 2: 1. Title passes/bpo? NO 2. PV of minimum lease payments (include GRV): BGN, 104,300PMT, 5N, 10I, 75,000FV CPT PV gives \$481, ,486/550,000 = 87% <90%, NO 3. Lease term 5/6 = 83%>75%, YES Conclusion: Capital Lease. Lease 3: 1. Title passes/bpo? NO 2. PV of minimum lease payments: 10,600/55,000 = 19%<90% NO 3. Lease term 1/5 = 20%<75% NO Conclusion: Operating Lease. Requirement 3 #1 Asset under capital lease ,947 Lease liability ,947 #2 Asset under capital lease ,486 Lease liability ,486 Lease liability ,300 Cash ,300 #3 Maintenance and insurance expense... 1,400 Rental expense, machinery... 9,200 Cash... 10,600 [The expense could all be recorded in rental expense] Copyright 2008 by McGraw-Hill Ryerson Limited. All rights reserved.
4 Buad 273 Chapter 17 Example Solutions Page 4 of 13 Example 3: Assignment 17-1 Requirement 1 Minimum lease payments: [(\$104,000 \$9,600) 5] + \$26,500 = (\$94,400 5) + \$26,500 = \$498,500. Only the guaranteed residual is included despite higher expectations of resale value. Requirement 2 The asset will be recorded at the present value of the minimum lease payments (including the guaranteed residual value), discounted at 10%: BGN, 94,400PMT, 5N, 10I, 26,500FV, CPT PV gives \$410,090 Requirement 3 If the residual value is unguaranteed, it won t be included in the minimum lease payments. Therefore, the asset will be recorded at the NPV of the lease payments: BGN, 94,400PMT, 5N, 10I, CPT PV gives \$393,635 Requirement 4 A leased asset cannot be recorded at more than its fair value. The asset must be recorded at \$375,000. Subsequent accounting for the lease must be at the implicit interest rate that discounts the stream of \$498,500 lease payments to a present value of \$375,000. Assuming the GRV, BGN, -375,000PV, 94,400PMT, 5N, 26,500FV, CPT I gives 15.35%. Copyright 2008 by McGraw-Hill Ryerson Limited. All rights reserved.
5 Buad 273 Chapter 17 Example Solutions Page 5 of 13 Example 4: Assignment 17-2 Requirement 1 The lease term is eight years. Guaranteed residual value, none. Unguaranteed residual value exists as the value of the asset to the lessor at the end of the lease term. There is no way to calculate this amount. BPO, none. Minimum net lease payments, (\$24,000 \$4,000) 8 years = \$160,000. Incremental borrowing rate, 8%. Requirement 2 To be a capital lease, the lease would have to meet one of the following three criteria, applied judgmentally: 1. Transfer of title No 2. Economic life vs lease term No; 8/12 < 75% 3. PV of MLP vs fair value PV of MLP: BGN, (24,000 4,000)Pmt, 8n, 8i, Cpt PV gives \$124,127. \$124,127 \$133,000 = 93%>90% Yes Since criterion 3 is met, this is a capital lease. Copyright 2008 by McGraw-Hill Ryerson Limited. All rights reserved.
6 Buad 273 Chapter 17 Example Solutions Page 6 of 13 Example 4: Assignment 17-2 Requirement 3 Date Opening balance Interest Payment Closing balance 1-Jan-X1 124,127 20, ,127 1-Jan-X2 104,127 8,330 20,000 92,457 1-Jan-X3 92,457 7,397 20,000 79,854 1-Jan-X4 79,854 6,388 20,000 66,242 1-Jan-X5 66,242 5,299 20,000 51,541 1-Jan-X6 51,541 4,123 20,000 35,664 1-Jan-X7 35,664 2,853 20,000 18,517 1-Jan-X8 18,517 1,483 20,000 - Beginning of fiscal year and lease term: Asset under capital lease ,127 Lease liability ,127 Lease liability... 20,000 Prepaid maintenance expense... 4,000 Cash... 24,000 End of the fiscal year: Maintenance expense... 4,000 Prepaid maintenance expense... 4,000 Interest expense... 8,330 Lease liability... 8,330 Amortization expense (\$124,127 8)... 15,516 Accumulated amortization... 15,516 Requirement 4 If the lease were cancellable, the lease term would be (effectively) one year, and the minimum lease payments, \$20,000 net; \$24,000 gross. The lease would not qualify for capitalization, and would be recorded annually as an operating lease: Rent expense... 24,000 Cash... 24,000 The annual payment could be split between rental and maintenance if Burrill wished. Copyright 2008 by McGraw-Hill Ryerson Limited. All rights reserved.
7 Buad 273 Chapter 17 Example Solutions Page 7 of 13 Example 5: Capital Lease, with Bargain Purchase Option In calculating the present value of the lease payments when there is a Bargain Purchase Option, we include the bargain purchase amount in the calculation. In the amortization schedule, the last payment is the bargain purchase amount. Note that the last payment is on January 1 20X5, but the lease term ends on December 31, 20X5. 1 Jan. X3 Date Opening balanceinterestpaymentclosing balance 1-Jan-03 15,756-5,000 10,756 1-Jan-04 10,756 1,183 5,000 6,939 1-Jan-05 6, ,000 2, Dec-05 2, ,000 - Inception of the lease record the asset and liability, and the first payment: 31 Dec. X3 Asset under capital lease 15,756 Lease liability 15,756 Lease liability 5,000 Cash 5,000 At year-end we record the expenses for the year interest and amortization: Interest expense 1,183 Lease liability 1,183 Amortization expense 15,756/5 1 3,151 Accumulated amortization 3,151 1 We include the full PV including BPO, and divide by the life of the asset rather than the lease life. Because of the BPO the lessee is sure to buy the asset, and we assume they will use it for its full useful life. If there is no BPO, we divide by the lease term, even if it is less than the life of the asset.
8 Buad 273 Chapter 17 Example Solutions Page 8 of 13 Example 5: Capital Lease, with Bargain Purchase Option 1 Jan. X4 Record the payment: Lease liability 5,000 Cash 5,000 Note that the liability is now \$6,939, as per the schedule 31 Dec. X4 Year-end adjustments: 1 Jan. X5 Interest expense 763 Lease liability 763 Amortization expense 3,151 Accumulated amortization 3,151 Record the payment: Lease liability 5,000 Cash 5,000 Note that the liability is now \$2,702, as per the schedule 31 Dec. 05 Year-end adjustments: Interest expense 298 Lease liability 298 Amortization expense 3,151 Accumulated amortization 3,151 This is the end of the lease term, so the lessee would purchase the asset: Lease liability 3,000 Cash 3,000 Note that the liability is now \$0, as per the schedule The lessee would also reallocate the asset and its accumulated amortization: Capital assets 15,756 Asset under capital lease 15,756 Accumulated amortization, leased asset 3,151*3 9,453 Accumulated amortization, asset 9,453 (Amortization would continue until the end of year 5)
9 Buad 273 Chapter 17 Example Solutions Page 9 of 13 Example 6: Capital Lease, Different Year-end In Example 5 we calculated the following amortization schedule. Date Opening balanceinterestpaymentclosing balance 1-Jan-03 15,756-5,000 10,756 1-Jan-04 10,756 1,183 5,000 6,939 1-Jan-05 6, ,000 2, Dec-05 2, ,000 - This schedule still applies to the lease. Given a May 31 year-end, we simply have to calculate the portion of interest expense that applies to each fiscal year. 1 Jan. X3 Inception of the lease record the asset and liability and first payment (as before): 31May X3 Asset under capital lease 15,756 Lease liability 15,756 Lease liability 5,000 Cash 5,000 At year-end we record the expenses for the year interest and amortization. Only 5 months have passed since the inception of the lease. 1 Jan. X4 Interest expense 1,183*5/ Lease liability 493 Amortization expense (15,756/5)*5/12 1,313 Accumulated amortization 1,313 Record the payment: 31 May X4 Lease liability 5,000 Cash 5,000 Year-end adjustments: Interest expense 1,183*7/ *5/12 1,008 Lease liability 1,008 Amortization expense (full year) 3,151 Accumulated amortization 3,151
10 Buad 273 Chapter 17 Example Solutions Page 10 of 13 Example 6: Capital Lease, Different Year-end 1 Jan. X5 Record the payment: 31 May X5 Lease liability 5,000 Cash 5,000 Year-end adjustments: 31 Dec. X5 Interest expense 763*7/ *5/ Lease liability 569 Amortization expense (full year) 3,151 Accumulated amortization 3,151 This is the end of the lease term, so the lessee would purchase the asset: Lease liability 3,000 Cash 3,000 Record interest expense: Interest expense 298*7/ Lease liability 174 Record amortization expense to disposal date: Amortization expense 3,151*7/12 1,838 Accumulated amortization 1,838 And reallocate the asset and its accumulated amortization: Capital assets 15,756 Asset under capital lease 15,756 Acc. amortization, leased asset 1,313+3,151+3, ,838 9,453 Accumulated amortization, asset 9,453
11 Buad 273 Chapter 17 Example Solutions Page 11 of 13 Example 7: Capital Lease with Guaranteed Residual Value Some leases have a Guaranteed Residual Value, which means that if the asset is not worth that specified amount at the end of the lease term, the lessee has to pay the difference to the lessor. When there is a guaranteed residual value, we include it in the calculation of the present value, just like we do for a BPO. Note that if there is an unguaranteed residual value, the lessee ignores it. It is of interest only to the lessor. The calculation of the PV of the lease would be: BGN, 10,000Pmt, 4,000FV, 3N, 9i, Cpt PV, which gives 30,680. Note how the following amortization schedule ends with a liability balance of 4,000 (the guaranteed residual value) at the end of year 4. We will show what happens to that balance when we do the final journal entries. Date Opening balanceinterestpaymentclosing balance 1-Jan-03 30,680-10,000 20,680 1-Jan-04 20,680 1,861 10,000 12,541 1-Jan-05 12,541 1,129 10,000 3, Dec-05 3, ,000 Journal entries: 1 Jan 20X3 Asset under capital lease 30,680 Lease liability 30,680 Lease liability 10,000 Cash 10, Dec. 20X3 Interest expense (schedule) 1,861 Lease liability 1,861 Amortization expense (30,680-4,000)/3 1 8,893 Accumulated amortization 8,893 1 When there is a guaranteed residual value, we subtract the GRV from the PV in calculating the amortization base, and divide by the lease life rather than the life of the asset. The reason for this will become clear when we do the end-of-lease entries at the end of year 3. 1 Jan 20X4 Lease liability 10,000 Cash 10, Dec. 20X4 Interest expense (schedule) 1,129 Lease liability 1,129 Amortization expense 8,893 Accumulated amortization 8,893 1 Jan 20X5 Lease liability 10,000 Cash 10,000
12 Buad 273 Chapter 17 Example Solutions Page 12 of 13 Example 7: Capital Lease with Guaranteed Residual Value 31 Dec. 20X5 Interest expense (schedule) 330 Lease liability (see note below) 330 Amortization expense 8,894 Accumulated amortization 8,894 Amortization expense was increased by \$1 to correct a rounding error, which will soon become clear. So what happens now? The lease is over, and here s what is on the books of the lessee. Lease Liability Asset under C.L. Accum. Amort. 10,000 30,680 30,680 8,893 10,000 1,861 8,893 10,000 1,129 8, ,000 30,680 26,680 How do we record the fact that the lease is done? Note that the remaining balances all balance out with equal Dr and Cr, and can be removed with a single journal entry. That s why we removed the guaranteed residual value from the amortization base (and also why we rounded the last amortization amount!) Assumption 1 Lessee does not have to pay Lease liability 4,000 Accumulated amortization 26,680 Asset under capital lease 30,680 Assumption 2 Lessee pays \$1,000 Same entry as Assumption 1, plus: Loss on lease contract 1,000 Cash 1,000 Note that under both assumptions, the asset is fully amortized and does not appear on the books any more.
13 Buad 273 Chapter 17 Example Solutions Page 13 of 13 Example 8: Sale and Leaseback Case 1: Capital Lease: We calculate the present value of the lease payments as usual: In BGN mode press 20n, 50,000Pmt, 10i, Cpt PV which gives 468,246 Record disposal of the building: Cash 500,000 Accumulated amortization 550,000 Building 1,000,000 Deferred gain (plug) 50,000 Record leaseback under capital lease, and initial payment: Leased building 468,246 Lease obligation 468,246 Lease obligation 50,000 Cash 50,000 Next payment, at year-end: Lease obligation 50,000 Cash 5,0000 Interest expense (468,246 50,000)*10% 41,825 Lease obligation 41,825 Amortization expense 468,246/20 23,412 Accumulated amortization 23,412 We also amortize the deferred gain on the same basis as the amortization of the asset: Deferred gain 50,000/20 2,500 Amortization expense 2,500 Case 2: Operating Lease: Record disposal of the building: Cash 500,000 Accumulated amortization 550,000 Building 1,000,000 Deferred gain (plug) 50,000 Record initial payment: Prepaid lease expense 50,000 Cash 50,000 Next payment, at year-end: Lease expense 50,000 Prepaid lease expense 50,000 Prepaid lease expense 50,000 Cash 50,000 We also amortize the deferred gain on the same basis as the asset: Deferred gain 50,000/20 2,500 Lease expense 2,500
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### GOVERNMENT OF MALAYSIA
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### Leasing Decisions CHAPTER 3
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### Financial Accounting I (FAC4861/3) Rendani Muthelo 16 December 2015
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### IFRS 16 - Leases. CPD Activity Organized by: Institute of Chartered Accountants of Pakistan UAE Chapter
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### Statement of Financial Accounting Standards No. 13
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### Accounting for Leases
CHAPTER 21 O BJECTIVES After reading this chapter, you will be able to: 1 Explain the advantages of leasing. 2 Understand key terms related to leasing. 3 Explain how to classify leases of personal property. | 10,811 | 41,195 | {"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-2018-51 | latest | en | 0.794116 |
http://woodenbooks.com/browse/sacred-geometry/SGE.00013.php | 1,537,696,858,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267159193.49/warc/CC-MAIN-20180923095108-20180923115508-00055.warc.gz | 273,874,061 | 3,765 | # Sacred Geometry
19 18 Pyramid Pie a marriage of everything There is perhaps no more famous a geometric object on Earth than the Great Pyramid at Giza in Egypt with its strange passages and enigmatic chambers. The five diagrams opposite show 1 The 51.8o slope of the Great Pyramid means the square of its height produces an area equal to that of each face. 2 The golden section in the pyramid, 1.618 see page 24. 3 Pi in the pyramid. Pi, or , defines the ratio between a circle s circumference and its diameter 3.14159.... 4 The pyramid squaring the circle see page 14. 5 A pentagram defining a net for the pyramid cut and fold Geometry means Earth measure and the Pyramid functions as a ridiculously accurate sundial, observatory, surveyor s tool and repository for standard weights and measures. Its perimeter is exactly half a degree of equatorial latitude. A 345 triangle fits the shape of the King s chamber below and gives the angle of slope of the second pyramid at Giza. Halfway between the two slopes is 51.4o, one seventh of a circle. a map of the Earth square of height area of face the golden section in the pyramid cosine of slope 0.618 pi in the pyramid perimeter 2 p height circumference of circle on height perimeter of pyramid base constructing the pyramid from a pentagram drawn in a circle | 291 | 1,310 | {"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} | 2.90625 | 3 | CC-MAIN-2018-39 | latest | en | 0.869553 |
https://isabelle.in.tum.de/repos/isabelle/rev/7af6723ad741 | 1,660,711,481,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572833.95/warc/CC-MAIN-20220817032054-20220817062054-00511.warc.gz | 305,354,541 | 2,436 | author huffman Fri, 08 Jul 2005 02:38:05 +0200 changeset 16751 7af6723ad741 parent 16750 282d092b1dbd child 16752 270ec60cc9e8
src/HOLCF/Sprod.thy file | annotate | diff | comparison | revisions
```--- a/src/HOLCF/Sprod.thy Fri Jul 08 02:37:42 2005 +0200
+++ b/src/HOLCF/Sprod.thy Fri Jul 08 02:38:05 2005 +0200
@@ -144,11 +144,14 @@
lemma surjective_pairing_Sprod2: "(:sfst\<cdot>p, ssnd\<cdot>p:) = p"
by (rule_tac p=p in sprodE, simp_all)
-lemma less_sprod: "p1 \<sqsubseteq> p2 = (sfst\<cdot>p1 \<sqsubseteq> sfst\<cdot>p2 \<and> ssnd\<cdot>p1 \<sqsubseteq> ssnd\<cdot>p2)"
+lemma less_sprod: "x \<sqsubseteq> y = (sfst\<cdot>x \<sqsubseteq> sfst\<cdot>y \<and> ssnd\<cdot>x \<sqsubseteq> ssnd\<cdot>y)"
apply (simp add: less_Sprod_def sfst_def ssnd_def cont_Rep_Sprod)
apply (rule less_cprod)
done
+lemma eq_sprod: "(x = y) = (sfst\<cdot>x = sfst\<cdot>y \<and> ssnd\<cdot>x = ssnd\<cdot>y)"
+by (auto simp add: po_eq_conv less_sprod)
+
lemma spair_less:
"\<lbrakk>x \<noteq> \<bottom>; y \<noteq> \<bottom>\<rbrakk> \<Longrightarrow> (:x, y:) \<sqsubseteq> (:a, b:) = (x \<sqsubseteq> a \<and> y \<sqsubseteq> b)"
apply (case_tac "a = \<bottom>")``` | 437 | 1,158 | {"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-2022-33 | latest | en | 0.287115 |
https://www.jiskha.com/display.cgi?id=1251064677 | 1,502,998,805,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886103910.54/warc/CC-MAIN-20170817185948-20170817205948-00060.warc.gz | 922,265,201 | 3,831 | posted by .
How do you do a ball dichotomy?
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More Similar Questions | 425 | 1,638 | {"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} | 2.859375 | 3 | CC-MAIN-2017-34 | latest | en | 0.930192 |
http://www.independent.co.uk/life-style/science-technoquest-5402635.html | 1,493,379,395,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122933.39/warc/CC-MAIN-20170423031202-00068-ip-10-145-167-34.ec2.internal.warc.gz | 585,275,606 | 15,992 | Questions and answers provided by Science Line's Dial-a-Scientist on 0345 600444
Q How fast would a coin dropped from the top of the Eiffel Tower be travelling by the time it reached the bottom? Could it hurt someone if it hit them?
A falling object doesn't accelerate indefinitely. It speeds up until it reaches its "terminal velocity", after which it will continue to fall at the same speed. A coin dropped from the Eiffel Tower has a terminal velocity of about 45 m/s (about 100 miles an hour) - which would certainly cause some damage to anyone it hit.
Q How much heat does the human body give off?
Curiously, men typically give off more heat than women - between 158 and 167 joules per square metre of skin in men and 150 to 158 joules per square metre in women. Adult men have about two square metres of skin, and so give out about 326 joules from their bodies each hour - a rather faint 0.09 watts.
Q When will the new millennium start? Why can't anyone agree on it?
The new millennium will not start until the year 2001. A man called Denys the Little can be blamed for the confusion. In the 6th century AD, he was making a table giving the dates of Easter when he decided to re- jig how the number of years was counted. Instead of counting years since the beginning of the reign of the Roman emperor Diocletian, he decided to start with the birth of Christ. However, because the number zero had yet to be invented, he called the first year of his era year one. When the use of the terms AD and BC came into use in the 11th century, the years went from 1BC straight to 1AD.
The effect on the start of the new millennium is that if the year one is the first year of the century, then the year two is the second, and so on up to the year 99, which is the 99th year of the century and the year 100 which is the 100th year. The start of the next century isn't until the year 101. It's the same with the millennium. The year 2000 is the 2000th year of the millennium while the year 2001 is the first year of the next. But that probably won't put a lot of people off celebrating on 31 December 1999.
Q When were fire alarms invented?
In 1852, D L Price filed the first patent for a fire alarm. It had a bi-metallic strip (two pieces of metal that expand at different rates when heated) that bent as the temperature rose until it made a connection in a circuit containing a bell. In the 1880s, a system was developed using a network of mercury thermometers around a house. As the temperature rose, the mercury rose until it made contact with two platinum electrodes which completed the circuit and set off the alarm. Smoke detectors weren't invented until the 20th century. In 1941, W Jaeger and E Meili from Switzerland developed a system very similar to that used today. They had a radioactive source whose radiation passed through an ionisation chamber creating ions and generating a current. As smoke entered the chamber, the number of radioactive particles decreased so the current decreased. This set off the alarm.
You can also visit the technoquest World Wide Web site at http://www.campus.bt.com/CampusWorld/pub/ScienceNet
Questions for this column can be submitted by email to sci.net@campus.bt.com | 741 | 3,219 | {"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-2017-17 | latest | en | 0.967245 |
https://nl.mathworks.com/matlabcentral/cody/problems/43326-solve-expression-i/solutions/1008490 | 1,575,917,211,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540521378.25/warc/CC-MAIN-20191209173528-20191209201528-00061.warc.gz | 483,393,340 | 15,553 | Cody
# Problem 43326. Solve expression I
Solution 1008490
Submitted on 11 Oct 2016 by Andriy Kavetsky
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = [2 -3 4]; y_correct = (1.+sin(x))./cos(x)+cos(x)/(1.+sin(x)); assert(isequal(your_fcn_name(x),y_correct))
2 Pass
x = 1:3:100; y_correct = (1.+sin(x))./cos(x)+cos(x)/(1.+sin(x)); assert(isequal(your_fcn_name(x),y_correct)) | 167 | 501 | {"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-2019-51 | latest | en | 0.649151 |
cmbosma.rbind.io | 1,716,636,719,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058822.89/warc/CC-MAIN-20240525100447-20240525130447-00163.warc.gz | 6,700,804 | 6,237 | # Create New Column Based on Values in Another Column in R
Here is a short post on how to create a new column with the values from one column based on the values in separate column. There are a few situations where this might be useful. One is if you have data in long format with your surveys questions in one column and you want to separate the scores of one of the surveys into a separate column for time-series/longitudinal analyses. Another situation might be that you have several conditions and you want to create a column of scores retaining scores of a subset of the conditions.
I recently needed to do this to tidy up some data in long format and noticed that there are no obvious blog posts or stack overflow entries addressing this data tidying scenario. The solutions you will find come with a caveat: if you create a new vector based on values in another vector, you will likely generate a vector with a length that does not match that of the data frame, precluding you from adding it to the data frame.
I will provide an abstract example and then a real-life example.
Below we have a table with columns varA and varB. We want to make a new column, varC, with values corresponding to rows with ‘c’ in the varA column.
varAvarB
a12
b34
c23
a34
b23
c13
Let's first illustrate the problem. For example, you would think that df\$newVar <- df\$varB[(varA == "c")] would create a new column with values from varB that correspond with varA. The problem is that this approach, and many other approaches, will create a vector the length of the cases that fit the parameters you have for the new vector. In this case, since there are only two rows with “c”, the length of the new vector would be 2.
Okay, now let's work on the solution. We can use the mutate and ifelse functions from the tidyverse to accomplish our task:
# Load in the `tidyverse` package for the `dplyr` package tidying functions and piping.
library(tidyverse)
df <- df %>%
mutate(varC = ifelse(varA == "b", varB, NA)
What does this code do?
1. From the data frame, create varC using and ifelse statement and assign it to the original data frame.
2. In the ifelse statement, we indicate that if values in varA equal “b”, then keep the corresponding value from column varB, else NA.
Here is the table you would get from this R code.
varAvarBvarC
a12NA
b3434
c23NA
a34NA
b2323
c13NA
Since these abstract examples can be a little tricky to apply to our own stuff, here's an example resembling a real data set, such as an experiential momentary assessment data set. Below is a table with data in long format. We want to make separate a column for negative affect subscale of the Positive and Negative Affect Schedule (PANAS-X) to conduct analyses solely on those scores. Although this can be accomplished with the data in wide format, you may want to keep it in long format depending on the analyses or data visualization you are interested in conducting.
idtime_indexpanas_x_subscalescore
0012pans_x_pa11
0013panas_x_na6
0022panas_x_pa2
0023panas_x_na4
0032panas_x_pa10
0033panas_x_na8
df <- df %>%
mutate(panas_x_na = ifelse(panas_x_subscale == "panas_x_na", score, NA)
Below is a data table amended with a vector containing values copied from the score column corresponding with “panas_x_na” in the panas_x_subscale column.
idtime_indexpanas_x_subscalescorepanas_x_na | 822 | 3,356 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-22 | latest | en | 0.900107 |
https://www.scribd.com/document/376871497/Capacitive-Current | 1,566,394,152,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316021.66/warc/CC-MAIN-20190821131745-20190821153745-00202.warc.gz | 966,101,211 | 71,517 | You are on page 1of 30
# Power System Analysis
Prof. A. K. Sinha
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 5
Transmission Line Capacitance
## (Refer Slide Time: 00:55)
Welcome to lesson 5 on Power System Analysis course. In this course we will talk about
the Transmission Line Capacitance, before we get into the calculation of transmission
line capacitance. I would like to answer those questions that I asked in lesson 4. First
question was why bundled conductors are used in EHV lines? Well the answer to this
question is bundling of conductors that is instead of using one single conductor; use of a
number of conductor connected by conducting frames, reduces electric field strength on
conductor surface, which in effect reduces the corona losses, which result in power loss
as well as radio interference and audible noise in the system. Bundling also increases the
effective radius of the conductor. And there by reduces the inductance of the
transmission line. This in effect well improve the regulation of the transmission line.
As seen from here, the effective radius for a three conductor bundle which are spaced at
a distance d from the center or d from each other with a radius r is given by 9th root of r
dash into d into d whole cube, which is equal to cube root of r dash d square. This is
much larger than r dash, which is used when a single conductor is used. And therefore,
bundling helps in reducing the inductance as it increases the effective radius. Second
question was, what is transposition?
## (Refer Slide Time: 0:03:12)
Well transposition means, exchanging the position of the three phase conductors along
the length of the line. That is when we have three phase conductors, which are not
equilaterally spaced. That is equal spacing between them is not possible due to physical
constraints of that transmission line construction. There in such a case we will get
unequal flux linkages with the three conductors.
Most of the time if you have seen the power line transmission towers. You will find that
the three phase conductors are either placed in a horizontal configuration, if it is a single
circuit line. Or they may be placed in a vertical configuration if it is a double circuit line.
Therefore the equilateral spacing is not there and the distance between the conductors
phase conductors are not equal. This results in different difference in flux linkages with
the three conductors, Rresulting in difference in inductances; in order to have a balance
three phase system.
What we do is we make these conductors a phase conductors go through all the three
positions. That is what we do in transposition. We exchange the position of the three
conductors along the length of the line. So, that each conductor or each phase conductor
is in all the three positions for equal length of the line, which means that the average flux
linkage of all the conductors or for all the three phase conductors has same.
So, if the spacing between three conductors are in unequal. Unequal flux linkages will
occur when the three phase conductors are even when the three phase currents has
balanced.
## (Refer Slide Time: 05:20)
Here we show the transposition the phase conductor a is in position 1 for 1 3rd length of
the line. It occupies position 2 for the next 1 3rd length of the line. And position 3 for the
last 1 3rd portion of the line. Similarly, phase b conductor is in position 2 for 1 3rd of
length of the line next 1 3rd length of the line it is in position 3.
And the last 1 3rd length of the line it is in position 1. In this way each phase conductor
is occupying all the three positions, which results in average flux linkages to be equal.
So, we say that by transposition each phase conductor occupies the three positions for 1
3rd length of the line. Therefore the average flux linkage and inductance become equal.
(Refer Slide Time: 06:28)
The third question was how the effect of earth return current is taken into account in
inductance calculation? Well J R Carson in his paper a 1923 said that the earth return
currents can be taken into account by means of considering earth return conductance.
That is we assume imaginary conductors, which are placed below the ground. These
conductors are suppose to have the same radius are the GMR as the overhead conductors
and they are placed directly below them in the ground.
The distance between the overhead conductor and these imaginary conductors is given
by the distance D e. Where D e or is given by an empirical relation 658.5 square root of
rho by f, where rho is the earth resistivity normally if earth resistivity is not known. Then
we assume it to be around 100 ohm meter. And f is the system frequency. So, having
answer these questions, now we will going to the main part of the lesson 5.
(Refer Slide Time: 08:07)
Now, this lessons as I said earlier we will be talking about transmission line capacitance
and it is calculation. So, here we will first start with electric field and voltage calculation.
Next we will take up the transmission line capacitance calculation for a single phase line
with solid conductors. Then we will take up three phase line with equal spacing. And
finally, we will talk about three phase line with bundle conductors and unequal spacing.
Well as we have learnt in the lessons 3 and 4 about calculating the resistance and
inductance of the transmission line. The transmission line also has capacitance, this is
mainly because the line conductors have a voltage difference between the phases. That is
phase conductors have voltage difference and there is a difference of voltage between the
phase conductor and the ground.
And these conductors are separated by a dielectric medium for overhead line this is air.
And in case of cables, this can be some kind of a dielectric, which can be impregnated
paper or it can be XLPE that is cross linked polyethylene insulation or some other kind
of insulation. Therefore we have two conductors at different voltages and there is a
dielectric between them, which results into a capacitance between the conductors or
between the conductors and the ground.
(Refer Slide Time: 10:10)
Now, for calculating the transmission line, capacitance we have to go through first using
Gauss’s law to calculate the electric field strength. Once we are calculated the electric
field strength, we find out the voltage between the conductors and then we find out the
capacitance by relationship C equal q by V. So, will start with the Gauss’s law and
calculation of electric field strength.
## (Refer Slide Time: 10:43)
Well Gauss’s law states that, total electric flux leaving a close surface is equal to total
charge within the volume enclose by the close surface, which basically leads to that
statement that normal electric flux density integrated over the closed surface will be
equal to the charge enclosed by thus closed surface. Mathematically we can write this as
the surface integral over the closed surface of electric flux density D or the norm the
electric flux density which is normal to the surface into d s will be equal to or integrated
over the surface. Will be equal to the surface integral of epsilon E, which is again normal
to the surface into d s that is over the whole surface.
Now, this epsilon is the permittivity of the medium and E is the electric field strength at
the surface. So, that is what we have done is electric flux density D is replaced by epsilon
E, because D equal epsilon E, that we know from electric field theory. So, this is equal to
the total charge enclosed by the volume, enclose by this close surface, this we can see in
this figure very clearly.
## (Refer Slide Time: 12:36)
Let us say we have a conductor, which is of very long conductor of radius r and this is
charge with q Coulomb’s per meter. Here we have assume q to be a positive charge it
can be positive or negative a does not matter. So, here we have assume there is conductor
of radius r is having a charge of q Coulomb’s per meter and it is very long conductor.
Now, since this is perfect conductor electric field inside is going to be zero. Now, for
finding out the electric field, outside the conductor what we need to do is we take up
concentric cylindrical volume of 1 meter long length. So, this volume shown here this is
a concentric cylinder of 1 meter length at with a radius of X. Now since the electric field
lines will be radial to this surface, which is of this concentric cylinder.
Therefore, there will be no tangential component of the electric field. Now, we can find
out with this, what is the fields strength at the surface of the this concentric cylinder.
Now, using Gauss’s law, we will use the integration surface integration of epsilon E d s.
So, epsilon E x that is field strength at this surface of this concentric cylinder integrated
over this close surface. Closed surface area is to going be equal to twice pi X into 1
meter length of this cylinder. So, epsilon Ex into twice pi X into 1 is equal to the charge
enclosed which is q into 1 meter length of the line. Because q coulomb per meter is the
charge on the conductor from this we can calculate the electric field strength, at the
surface of the concentric cylinder of radius X.
## (Refer Slide Time: 15:15)
So, E x equal to q by twice pi epsilon X volt per meter. Now, using this Ex or the
relationship for the field strength at any point we can always find out the voltage
difference between two points. So, if you want to find out the voltage difference between
2 points P 1 and P 2, which are at a distance D 1 and D 2. From the center of the
conductor, what we need to do is integrate from D 1 to D 2 of Ex d x over D 1 to D 2.
## So, E x the integral Ex to E x d x to from D 1 to D 2 will give me a the voltage difference
V 1 2. So, V 1 2 equal to integral D 1 integral of q by twice pi epsilon X d x from D 1 to
D 2 this is equal to q by twice pi epsilon log n D 2 by D 1 volts. Here epsilon is as I said
earlier is the permittivity of the medium and this equal epsilon r into epsilon 0, where
epsilon 0 is the permittivity of the free space and for here also it is almost the same,
epsilon 0 equal to 8.854 into 10 to the power minus 12 Farad per meter. Now, real take
since we have now found out the electric field strength and the voltage between any 2
points for any charge conductor. We can extended to a multi conductor system.
Now, let us a that we have a system of n conductors with any conductor k having radius r
k and has a charge q k per meter length of the conductor. Then we can find out the
voltage between 2 conductors i and j due to charge on this conductor k. We write this as
V i j k. That is voltage between conductor i and j due to charge on conductor k. This will
be equal to q k by twice pi epsilon log n the distance d of k to j, k to j and divided by
distance of k to i.
As we can see using the old relationship, that we have found here; q by twice pi epsilon
log n D 2 by D 1. Now, if you want to find out the voltage difference due to charge from
all the conductors, which are in this system. Then what we need to do is use the
superposition theorem. That is add the voltages due to the charge from each of this
conductors. So, V i j in that case due to charges on all the conductor will be equal to
summation k equal to 1 2 n of q k by twice pi epsilon log n D j k by D i k volts. So, this
way we can find out the voltage between any two conductors or any two points due to
charge on various conductors in a multi conductor system.
## (Refer Slide Time: 19:49)
Now, let us try to apply this to find out the voltage and then the capacitance of a single
phase line. Let us say we have a single phase line with two conductors are with
conductor x and conductor y. Conductor x has a radius r x and has a charge q Coulomb
per meter. Conductor y has a radius r y with a charge of minus q Coulomb per meter
because conductor y will be the written conductor.
So, it will be carrying the written currents. So, the charge will be negative after charge on
the conductor x. So, now we using the same formula we can find out V x y will be given
by one twice pi epsilon q log n D y x by D x x. That is because of the charge on this
conductor the voltage difference between these two will be given by this relationship 1
by twice pi epsilon q log n D y x by D x x.
And the voltage between these two conductors that is V x to V y. So, V x y will be equal
to minus 1 by twice pi epsilon minus q log n D y y by D x y. So, this can be reduce to q
by twice pi epsilon log n D y x into D x y divided by D x x into D y y. That is this minus
is converted to plus by changing reversing these D x y and D y y; that is taking the
inverse of this. So, we get D y x into D x y divided by log n D x x by D y y. So, finally,
we get the voltage between to conductors by the relation q by twice pi epsilon log n D y
x into D x y divided by D x x into D y y.
## Which if we substitute the values we get V x y equal to q by pi epsilon log n D by square
root of r x, r y. Because, this would have become D square and this would be r x r y. So,
this square is taken out that becomes 2 into q and this 2 pi epsilon. So, this 2 well cancel
out and we get finally, q by pi epsilon log n D by square root of r x into r y.
Now, once we have found out the voltage, we can find out the capacitance very easily.
Capacitance C x y that is capacitance between the two conductors x and y is equal to q
by voltage between the two conductor. So, this will be equal to pi epsilon divided by log
n D by square root r x r y. That is what we have done is divided q by this term. And then
we get this C x y is pi epsilon divided by log n D by square root of r x r y.
## (Refer Slide Time: 0023:33)
So, this if we consider the radius of the two conductors to be equal then r x will be equal
to r y; and therefore, square root of r x into r y equal r. Therefore we can write C x y
equal to pi epsilon divided log n D by r Farad per meter this is a line capacitance
between line to line. Now, if we see the system, in the center of this, we will get 0
voltage because, this will be positive potential, this will be equal negative potential.
So, by symmetry at the center we will get a 0 potential or the potential of the ground or
the neutral. Therefore we can find out the voltage to the neutral as V x n equal to V y n
equal to V x y divided by 2. This will be half the voltage because the voltage of one
conductor is positive another conductor is negative. So, the zero potential line will be in
between at the half, so V x n equal to V y n equal to V x y divided 2. Therefore we can
find out the capacitance to the neutral or capacitance to ground as C n which will be
equal to C x n. And which will be all again equal to C y n that is capacitance from to
ground for conductor x will be same as capacitance to ground for conductor y.
And this will be equal to q by V x n, which is equal to 2 times C x y. As we can see here
between x and y we have a capacitance C x y. And if we have taken the neutral at the
center, then we have now capacitance C x n, between x and n. And capacitance C y n
between n and y. And these capacitance is will be equal to 2 times the capacitance
between x and y, because they are in series. So, 2 x y and 2 x, the 2 C x y and 2 C x y in
series will give me C x y.
## So, C y n or C x n which is equal to C n capacitance to ground for any conductor, after
for any of the 2 phase conductor will be equal 2 times C x y. This equal to twice pi
epsilon divided by log n D by r Farad per meter for line to neutral capacitors. Now, we
will take up the case of 3 phase system, because we know that post of the power system
that we have to the specially transmission systems are 3 phase systems. Therefore will
take the case of a three phase system.
## (Refer Slide Time: 27:15)
Will start with the three phase line with equilateral spacing, because as we have seen for
inductance calculation. We can always convert, if the line is transpose, we can always
convert any system into the equivalent 3 phase equivalent equilateral spacing of the
conductors by finding out the equivalent distance d e q.
So, here we have equilateral spacing conductor a b and c each with a distance d from
each other. We also assume that this system consist of only three conductors. So, the
some of the total charges will be equal to 0. That is q a plus q b plus q c equal 0. Again
as earlier we can find out the voltage between any two conductors.
## (Refer Slide Time: 28:16)
So, let us find out the voltage between a and b. So, V a b will be equal to again using the
same relationship that we are used earlier. It will be equal to 1 by twice pi epsilon into
the q to charge on conductor a q a log n D b a by D a a. Due to charge q b it will be plus
q b log n D b b by D a b that distance of conductor b to conductor b divided by distance
of the conductor b to conductor a. And due to charge q c log n D b c by D a c, that is
distance from conductor c to conductor b and distance from conductor c to conductor a.
Now, this if we substitute the values of D b a D a b or D b a and D a a D b b and so on.
This can be written as V a b equal to one twice pi epsilon into q a log n D by r D b a is
equal to D a b, which is equal to D. And D a a is its self distance that is r the radius of the
conductor plus q b into log n D b b again.
Since the conductor radius we have assume to be same that is 3 conductors are same
radius or diameter. Therefore, D b b is equal to r Dab we will be equal to D plus qc log n
D b c which is equal to D and D a c, which is equal to D. So, this is q c log and D by D
which will be q c log and 1; therefore, this term will go to 0. Therefore we have V a b is
equal to 1 by twice be epsilon into q a log n D by r plus q b log n r by D.
(Refer Slide Time: 30:31)
Similarly, we can find out the voltage between conductor a and conductor c. So, V a c
will be equal to 1 by twice by epsilon into due to charge q a, q a log n the distance
between a and c D c a the distance between a to a. So, D a plus q b log n distance of b to
c that is D c b divided by D distance of b to a. So, that is D a b plus q c log n distance of
c to c divide by distance of c to a, that is equal to D c c by D c a.
This will be equal to 1 by twice pi epsilon q a log n D by r that substituting the distances
D by r plus q b log n D by D plus qc log n r by D, again this term since it is D by D log n
one. So, this term goes to 0, so we have V a c is equal to 1 by twice pi epsilon into q a
log n D by r plus qc log n r by D.
(Refer Slide Time: 32:01)
Now, if we look at this equilaterally space conductors, here the voltage V a b is equal to
root 3 times V a n at an angle of plus 30 degrees from the x axis. So, this is equal to root
3 times V a n into root 3 by 2 plus j half. That is we have converted this from polar
coordinate to rectangular. Similarly V a c if you see is equal to minus of V c a. That is
negative of V c a, this is equal to again root 3 times V a n. And this if you see V a c’s at
minus 30 degree from the x axis. So, root 3 V a n with an angle of minus 30 degrees.
This is equal to route 3 V a n into root 3 by 2 minus j half.
## (Refer Slide Time: 33:31)
So, now if we add these two voltages V a b plus V a c we get this as equal to root 3 times
V a n. That is if we see if we add these two this j plus j half and minus j half will cancel
out and we have got two times root 3 by 2 into times V a n. So, that two will cancel out
and root 3 into root 3 will give us 3. So, this will be equal to 3 V a n. Therefore, we will
get V a n that is voltage of phase a to the neutral will be equal to 1 3rd of the V a b plus
V a c.
So, substituting the values from V a b for V a b and V a c as we have calculated earlier.
We will get this as 1 3rd of 1 by twice pi epsilon into 2 times q a log n D by r plus q b
plus q c log n r by D. That is if we see here V a c is q a log n D by r plus qc log n r by D
and V a b is 1 by twice by epsilon q a log n D by r plus q b log n r by D. So, if we add we
will get twice q a log n D by r and q b log n r by D plus q c log n r by D that is what we
have got here.
So, we have two times q a log n r by D plus q b plus qc log n r by D. So, from this
relationship, we can find out now V a n is equal to 1 by twice pi epsilon into; now q b
plus qc is equal to minus q a, because q a plus q b plus qc equal to 0. So, this becomes
twice q a log n r D by r minus q a log n r by D, which can be put as plus q a log n r D by
r. And therefore, we get 3 q a log n D by r.
And this three will cancel with this 1 by 3. So, we have got 1 by twice pi epsilon q a log
n D by r. So, once we have calculated the voltage between phase conductor a and the
neutral, which is at zero potential. Then we have got C a n or the voltage or the
capacitance between conductor a and the neutral or the ground will be equal to q a by V
a n, which comes out to be equal to twice pi epsilon divided by log n D by r Farad per
meter. This is line to neutral capacitance or line to ground capacitance.
(Refer Slide Time: 36:57)
## In general, we instead of having D as we have seen earlier, we if the conductor spacing
are not equal. That is we do not have a equilateral configuration of conductors, then other
configurations can always be converted into an equivalent equilateral configuration there
the distance D equal to D e q that is the equivalent equilateral configuration distance. So,
D e q is equal to cube root of D a b into D b c into D a c; that is what we had seen in
earlier lecture on 3 phase line that is in lessons 4.
## (Refer Slide Time: 37:52)
Now, let us take a more general case, where we have considered a 3 phase bundle
conductor line with unequal spacing. We have here a 3 phase line with 2 conductors in a
bundle for each phase. And this configuration for the three phase system is now
horizontal configuration. That is distance from a to b that is centre of conductor bundle a
to conductor bundle b is D a b.
The distance between centre of conductor bundle b and conductor bundle c is D b c. And
distance from the centre of the conductor bundle a to centre of conductor bundle c is D a
c. Whereas each conductor has a radius r, here we are assuming that all the conductors
have the same radius. And the bundle distance that is distance between the centre of the
two conductors is d.
As we had said earlier this d is generally much larger than r, it is normally 10 times or
more r. In general we have around 30 centimeter or 40 centimeter distance between the
conductors for bundle conductor lines. So, d can be of the order of 30 to 40 centimeters
where as the conductor radius will be of the order of 2 to 3 centimeter. So, now again we
need to find out the voltage difference between the two phase conductors.
## (Refer Slide Time: 39:38)
So, we find out V a b, now here what we have assumed ((Refer Time: 39:41)) is that
conductor a is having a total charge of q u. And since we have two conductor bundle. So,
each conductor or the sub conductor in the bundle will be having a charge q by 2. So, it
will be having q a by 2, this will be having q a by 2. This conductor will be having
charge q b by 2, this will be having charge q b by 2, because all of them are of the same
So, c will be having charge q c by 2 this sub conductor and this subsequently conductor
will be having charge q c by 2. So, now in order to find out V a b again what we need to
do is find out the voltage between a and b due to the charges on each of these conductors
or sub conductors.
## So, V a b will be equal to 1 by twice pi epsilon. Now q 2 charge on conductor sub
conductor one of phase a it will be q a by 2, that is the charge on that sub conductor into
log n. D b a by D a a that is the distance between a to b divided by distance of a to a that
will be again the radius of the conductor. Plus q a by 2 due to the sub conductor the
second sub conductor in phase a that is, this conductor this sub conductor it will be q a
by 2 into log n D b a dash by D a a dash.
Now, due to this sub conductor the voltage V a b will be given by q b by 2 log n D b b
divided by Dab again for the other sub conductor of phase b this will be plus q b by 2 log
n D b b divided Dab dash. Similarly, for the two sub conductors charge on the two sub
conductors of phase c. We will have components qc by 2 log n D b c by D a c plus qc by
two log n D b c dash by D a c dash.
## (Refer Slide Time: 42:16)
This after substituting the values will be given by q a by 2 log n Dab by r. That is that a a
plus log n D a b by D, which is the distance D a a dash. Now, here what we are saying
that D b a and D b a dash are almost same, because the distance between the two phase
conductors is much larger compared to the distance between two sub conductors of a
bundle.
So, this small approximation we have made, that is D a b is the distance between the
centre of the two conductors. So, D this is q a by 2 log n D a b by r plus log n D a b by
D. Similarly for conductor b or the charge due to conductor b will be q b by 2 into log n r
by D a b plus log n D by D a b plus due to charge on conductor c q c by 2 log n D b c by
D a c plus log n D b c by D a c.
Now, this when we add these terms finally, can be written as 1 by twice pi epsilon q a
log n D a b by r d square root of r d plus q b by log n square root of r d by D a b plus q c
log n D b c by D a c. Now, this if we remember the inductance relationship is very
similar to that relationship.
## (Refer Slide Time: 44:23)
And therefore this can be again written in terms of capacitance is equal to, capacitance
between neutral can be written as twice pi epsilon log n D e q by D s, where we have
already defined D e q, as cube root of Dab into D b c into D a c.
Therefore, capacitance to neutral is given by the same relationship except that we have
now replaced r by D s, Where D s is the effective radius of the bundle conductor and D e
q is the equivalent equilateral spacing for that conductor. So, for two conductor bundle as
we had see earlier or in this case the D s is equal to the self distance. D s is equal to
square root of r into D for a 3 conductor bundle the self D s is equal to cube root of r into
D square and for a 4 conductor bundle.
It is equal to 1.0491 4th root of rd cube this we had seen in the last lesson. Now, once we
have calculated the capacitance. Now, we can find out the charging current, because
there is voltage and the there is capacitance between the conductor and the ground. So,
there will be some current, which will be flowing in the system and this current, which is
flowing in the capacitance is called charging current for the system.
(Refer Slide Time: 46:24)
Now, this transmission line has a capacitance C a n for phase a C b n for phase b n c, c n
for phase c. Now, this capacitance is we can find out that admittance, admittance will be
equal to twice pi f into c. That is we will have y will be equal to minus j omega c.
So, we can find out the admittance between the phase conductor and neutral or the
ground. So, if we are finding between two conductors x and y. So, then we will find out
the admittance for C x y as Y x y and the charging current will be Y x y into V x y. This
is equal to j omega C x y into V x y Amperes. So, once we have calculated the charging
current, we can also find out how much is the volt Ampere reactive, which is supplied by
this charging capacitance of the transmission line.
So, the capacitance of the transmission line will. In fact, be supplying reactive power to
the system. And this will be given by Q c, which is equal to V x y square divided by X c
or Y x y into V square x y or V x y square. So, we can find out the charging volt Ampere
reactive as omega C x y into V x y square. So, this will provide as the volt Ampere
reactive which is generated by the transmission line.
And if we are looking for the charging current to the ground, then this will be equal to y
into V a n. This will be equal to j omega Can into the voltage between line and the
neutral. So, once we know this current we can find out by multiplying this by another V
that is j omega C a n into V l n square will give us the reactive power generated by this
line.
(Refer Slide Time: 49:22)
So, for a single phase system we have the charging reactive power supplied by the line
given by y into V a n square. This is equal to omega C a n into V l n square volt ampere
reactive of the 3 phases system this will be equal to 3 times the charging volt ampere
reactive generated by a single phase line or between each phase and neutral of the line.
So, this will be equal to three times omega C a n into V l n square, which is equal to
omega C a n into V l l square that is if you take line to line voltage or the system voltage
for a 3 phase system. Then charging reactive power generating by the line will be given
by omega C a n V l l square. Now, that we have seen have we can calculate the
capacitance and also find out the charging current and the reactive power generated by
the line. Let us take one example to see have we calculate these quantities for a
Transmission system.
(Refer Slide Time: 50:43)
So, let us this example we have a three phase 400 kV 50 Hertz 350 kilometer overhead
transmission line that has flat horizontal spacing with three identical conductors. That is
we have three identical conductors placed in a flat horizontal spacing. The conductors
have on outside diameter of 3.28 centimeter that is the diameter of the conductor is 3.28
centimeter and the distance between the adjacent conductors is 12 meter. Now, for this
system determine the capacitive reactance to neutral in Ohms per meter per phase and
the capacitive reactance for the line in Ohms per phase.
## (Refer Slide Time: 51:36)
So, for solving this, we will take case of a fully transpose 3 phase line. So, we have the
fully transpose 3 phase line. The voltage is given by the relationship V a b equal to 1 by
twice pi epsilon this is epsilon into q a into log n D e q by r plus q b into log n r by D e q,
where D e q is equal to D 1 2 into D 2 3 into D 3 1 into the power 1 by 3 that is cube root
of D 1 2 D 2 3 into D 3 1. Now, substituting this values we get this as equal to 12 meters
into 12 meters into 24 meters. That is the distance between the conductor a and c, the
cube root of this will give us equal to 15.119 meter. Similarly, we can write the
relationship for V a c and V a c will be equal to 1 by twice pi epsilon into q a into log n
D e q by r plus q c into log n r by D e q.
## So, substituting again the values for the system.
(Refer Slide Time: 53:07)
We have substituting the values we will get V a b and V a c and we know V a b plus V a
c equal to 3 V a n. And also we know that q b plus qc equal to minus q a therefore,
adding V a b and V a c we will get V a n equal to q a divided by twice pi epsilon log n D
e q by r.
## Therefore the capacitance to neutral will be equal to q a by V a n, which will be equal to
twice pi epsilon log n D e q by r. And this will after substituting value comes out to be
8.163 into 10 to the power minus 6 micro Farad per meter. And we can find out the
admittance the charging admittance of the line to neutral, which will be equal to twice pi
f into C n.
So, after substituting the value, we will get this as equal to 2.565 into 10 to the power
minus 9 Siemens per meter per phase. Now, we given the length of the line as 350
kilometers therefore, Y n equal to 8.978 into 10 to the power minus 4 Siemens per phase.
That is multiplying this by 350 kilometer which is 350 into 10 to the power 3 meters.
So, this gives as this much and the reactance will be equal to 1 by admittance, which
comes out be equal to 1.1138 into 10 to the power 3 Ohms per phase. That is 1.1138 kilo
Ohms per phase. That is all per this lecture, we will continue with capacitance
calculation in the next lessons. Where will talk about how we take into account the effect
of earth for capacitance calculation.
Thank you.
## Transmission Line Capacitance (Contd.)
Welcome to lessons 6 on Power System Analysis. In this lesson we are going to discuss
Transmission Line Capacitance, which we were discussing in lessons 5, we are
continuing with it.
(Refer Slide Time: 56:01)
Here what we will do is we will consider the effect of earth on transmission line
capacitance. Actually when we have this transmission lines, the phase conductors are in
above the ground in over transmission system and the distance between the phase
conductors. And the distance between phase conductor and ground are of the same
magnitude. And therefore, the earth which acts as equi-potential surface thus affect the
electric field lines and there by the capacitance.
So, we need to consider the effect of earth and calculating the capacitance for
transmission lines. So, in these lessons we will discuss how the earth or the ground
affects the capacitance of a single phase transmission line and how it affects the 3 phase
transmission line.
(Refer Slide Time: 57:18)
Then we will talk about a double circuit line and we will talk about, how we calculate the
series and shunt impedance of the double circuit line.
## (Refer Slide Time: 57:33)
Such a system we are showing here, here we have a three phase line. This is one circuit
and this another circuit which is placed here. So, this is a double circuit system, here we
are showing this double circuit line. This is one circuit, this is another circuit connecting
two bus bars or two substations.
Now, the current flowing in one circuit is I p 1, current flowing in the other circuit is I p
2 where I p is basically a three element vector, which consists of I a, I b and I c. So, we
have I a 1, I b 1, I c 1 after circuit 1 and I a 2, I b 2, I c 2 for the circuit 2. Now, this
current flowing here from one circuit will be I p 1 plus I p 2, which gets divide into I p 1
I p 2 and may be x getting connected on this side. This is the system that we are showing
we may we have learnt about we will do a lessons 7.
Thank you. | 8,790 | 34,156 | {"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} | 3.609375 | 4 | CC-MAIN-2019-35 | latest | en | 0.94947 |
https://nrich.maths.org/public/topic.php?code=-25&cl=4&cldcmpid=324 | 1,582,289,138,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145529.37/warc/CC-MAIN-20200221111140-20200221141140-00056.warc.gz | 474,204,337 | 6,633 | # Resources tagged with: Number theory
Filter by: Content type:
Age range:
Challenge level:
### There are 22 results
Broad Topics > Numbers and the Number System > Number theory
### The Public Key
##### Age 16 to 18 Challenge Level:
Find 180 to the power 59 (mod 391) to crack the code. To find the secret number with a calculator we work with small numbers like 59 and 391 but very big numbers are used in the real world for this.
### More Sums of Squares
##### Age 16 to 18
Tom writes about expressing numbers as the sums of three squares.
### Mod 7
##### Age 16 to 18 Challenge Level:
Find the remainder when 3^{2001} is divided by 7.
### Pythagorean Golden Means
##### Age 16 to 18 Challenge Level:
Show that the arithmetic mean, geometric mean and harmonic mean of a and b can be the lengths of the sides of a right-angles triangle if and only if a = bx^3, where x is the Golden Ratio.
### An Introduction to Number Theory
##### Age 16 to 18
An introduction to some beautiful results of Number Theory (a branch of pure mathematics devoted primarily to the study of the integers and integer-valued functions)
### Sums of Squares and Sums of Cubes
##### Age 16 to 18
An account of methods for finding whether or not a number can be written as the sum of two or more squares or as the sum of two or more cubes.
### Modulus Arithmetic and a Solution to Differences
##### Age 16 to 18
Peter Zimmerman, a Year 13 student at Mill Hill County High School in Barnet, London wrote this account of modulus arithmetic.
### Novemberish
##### Age 14 to 16 Challenge Level:
a) A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. (b) Prove that 11^{10}-1 is divisible by 100.
### 2^n -n Numbers
##### Age 16 to 18
Yatir from Israel wrote this article on numbers that can be written as $2^n-n$ where n is a positive integer.
### An Introduction to Modular Arithmetic
##### Age 14 to 16
An introduction to the notation and uses of modular arithmetic
### Number Rules - OK
##### Age 14 to 16 Challenge Level:
Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number...
### Binomial Coefficients
##### Age 14 to 18
An introduction to the binomial coefficient, and exploration of some of the formulae it satisfies.
### There's a Limit
##### Age 14 to 18 Challenge Level:
Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely?
### Really Mr. Bond
##### Age 14 to 16 Challenge Level:
115^2 = (110 x 120) + 25, that is 13225 895^2 = (890 x 900) + 25, that is 801025 Can you explain what is happening and generalise?
### Always Perfect
##### Age 14 to 16 Challenge Level:
Show that if you add 1 to the product of four consecutive numbers the answer is ALWAYS a perfect square.
### Ordered Sums
##### Age 14 to 16 Challenge Level:
Let a(n) be the number of ways of expressing the integer n as an ordered sum of 1's and 2's. Let b(n) be the number of ways of expressing n as an ordered sum of integers greater than 1. (i) Calculate. . . .
##### Age 16 to 18
What is the relationship between the arithmetic, geometric and harmonic means of two numbers, the sides of a right angled triangle and the Golden Ratio?
### Never Prime
##### Age 14 to 16 Challenge Level:
If a two digit number has its digits reversed and the smaller of the two numbers is subtracted from the larger, prove the difference can never be prime.
### Data Chunks
##### Age 14 to 16 Challenge Level:
Data is sent in chunks of two different sizes - a yellow chunk has 5 characters and a blue chunk has 9 characters. A data slot of size 31 cannot be exactly filled with a combination of yellow and. . . .
### Diophantine N-tuples
##### Age 14 to 16 Challenge Level:
Can you explain why a sequence of operations always gives you perfect squares?
### A Little Light Thinking
##### Age 14 to 16 Challenge Level:
Here is a machine with four coloured lights. Can you make two lights switch on at once? Three lights? All four lights?
### Euler's Squares
##### Age 14 to 16 Challenge Level:
Euler found four whole numbers such that the sum of any two of the numbers is a perfect square... | 1,061 | 4,346 | {"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} | 3.890625 | 4 | CC-MAIN-2020-10 | latest | en | 0.829654 |
https://zbmath.org/?q=an%3A0214.30301 | 1,627,553,194,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153854.42/warc/CC-MAIN-20210729074313-20210729104313-00066.warc.gz | 1,165,549,733 | 9,833 | # zbMATH — the first resource for mathematics
Large prime factors of binary forms. (English) Zbl 0214.30301
##### MSC:
11N32 Primes represented by polynomials; other multiplicative structures of polynomial values 11N36 Applications of sieve methods
Full Text:
##### References:
[1] {\scG. Greaves}, The divisor sum problem for binary cubic forms, Acta Arith.{\bf17} to appear. · Zbl 0198.37903 [2] Richert, H.-E, Selberg’s sieve with weights, Mathematika, 16, 1-23, (1969) · Zbl 0192.39703 [3] Hooley, C, On the greatest prime factor of a quadratic polynomial, Acta math., 117, 281-299, (1967) · Zbl 0146.05704 [4] Goldfeld, M, On the number of primes p for which p + a has a large prime factor, Mathematika, 16, 23-27, (1969) · Zbl 0201.05301 [5] Ramachandra, K, A note on numbers with a large prime factor, J. London math. soc., 1, 303-306, (1969) · Zbl 0179.07301 [6] Nagell, T; Nagell, T, () · Zbl 0997.01010 [7] Nagell, T, Généralisation d’un théorème de Tchebycheff, J. math. pures appl., 4, 8, 343-356, (1921) · JFM 48.1173.01 [8] Hooley, C, On binary cubic forms, J. reine angew. math., 226, 30-87, (1967) · Zbl 0163.04605 [9] Selberg, A, On an elementary method in the theory of primes, Norske vid. selsk. forh. (Trondheim), 19, 64-67, (1947) · Zbl 0041.01903 [10] Halberstam, H; Roth, K.F, Sequences, (1966), Oxford Univ. Press New York/London · Zbl 0141.04405 [11] Prachar, K, Primzahlverteilung, (1957), Springer-Verlag · Zbl 0080.25901
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 601 | 1,824 | {"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-2021-31 | latest | en | 0.694545 |
http://formulas.ultrafractal.com/reference/reb/REB_TrapShapeSerpentineCurve.html | 1,708,826,097,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474573.20/warc/CC-MAIN-20240225003942-20240225033942-00644.warc.gz | 13,370,005 | 3,603 | ## reb Class REB_TrapShapeSerpentineCurve
```Object
common:Generic
common:TrapShape
reb:REB_TrapShapeSerpentineCurve
```
`class TrapShape:REB_TrapShapeSerpentineCurve`
This shape uses the serpentine curve function.
This variant of the function has a signed and absolute value version of the distance. The function also has a complex conjugate variant. The transformed value is the complex return value of the function.
Ultra Fractal Source
``` class REB_TrapShapeSerpentineCurve(common.ulb:TrapShape) {
; This shape uses the serpentine curve function.<br>
; <p>
; This variant of the function has a signed and absolute value
; version of the distance. The function also has a complex conjugate variant.
; The transformed value is the complex return value of the function.
public:
import "common.ulb"
; Constructor
func REB_TrapShapeSerpentineCurve(Generic pparent)
TrapShape.TrapShape(pparent)
endfunc
; Call this for each iteration being trapped.
float func Iterate(complex pz)
TrapShape.Iterate(pz)
float theta = 0
float d = 0
theta = atan2(pz)
int sgn = 1
if @sgn
sgn = -1
endif
complex af1 = 0.2*@a*cotan(theta)
complex af2 = 0.2*@b*sin(theta)*cos(theta)
complex as = (af1 + sgn*flip(af2))
if @absval
d = abs(cabs(pz) - cabs(as))
else
d = cabs(pz - as)
endif
m_LastZ = pz - as
return d
endfunc
default:
title = "Serpentine Curve"
int param v_trapshapeserpentinecurve
caption = "Version (Trap Shape Serpentine Curve)"
default = 101
hint = "This version parameter is used to detect when a change has been made to the formula that is incompatible with the previous version. When that happens, this field will reflect the old version number to alert you to the fact that an alternate rendering is being used."
visible = @v_trapshapeserpentinecurve < 101
endparam
float param a
caption = "Polar parameter"
default = 0.2
hint = "Affects spread and scale of trap"
endparam
float param b
caption = "2nd Polar parameter"
default = 1.0
hint = "Affects spread and scale of trap"
endparam
bool param sgn
caption = "Conjugate transform"
default = false
endparam
param absval
caption = "Absolute Value"
default = false
endparam
}
```
Constructor Summary
`REB_TrapShapeSerpentineCurve()`
`REB_TrapShapeSerpentineCurve(Generic pparent)`
Constructor
Method Summary
` float` `Iterate(complex pz)`
Call this for each iteration being trapped.
Methods inherited from class common:TrapShape
`GetColorChannel, GetTextureValue, GetTransformedPoint, Init, IterateSilent, SetThreshold`
Methods inherited from class common:Generic
`GetParent`
Methods inherited from class Object
Constructor Detail
### REB_TrapShapeSerpentineCurve
`public REB_TrapShapeSerpentineCurve(Generic pparent)`
Constructor
### REB_TrapShapeSerpentineCurve
`public REB_TrapShapeSerpentineCurve()`
Method Detail
### Iterate
`public float Iterate(complex pz)`
Call this for each iteration being trapped.
Overrides:
`Iterate` in class `TrapShape` | 788 | 2,921 | {"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-2024-10 | latest | en | 0.683407 |
https://metanumbers.com/580526 | 1,623,791,769,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621627.41/warc/CC-MAIN-20210615211046-20210616001046-00051.warc.gz | 365,115,696 | 7,610 | ## 580526
580,526 (five hundred eighty thousand five hundred twenty-six) is an even six-digits composite number following 580525 and preceding 580527. In scientific notation, it is written as 5.80526 × 105. The sum of its digits is 26. It has a total of 3 prime factors and 8 positive divisors. There are 274,968 positive integers (up to 580526) that are relatively prime to 580526.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 26
• Digital Root 8
## Name
Short name 580 thousand 526 five hundred eighty thousand five hundred twenty-six
## Notation
Scientific notation 5.80526 × 105 580.526 × 103
## Prime Factorization of 580526
Prime Factorization 2 × 19 × 15277
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 580526 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 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 580,526 is 2 × 19 × 15277. Since it has a total of 3 prime factors, 580,526 is a composite number.
## Divisors of 580526
8 divisors
Even divisors 4 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 916680 Sum of all the positive divisors of n s(n) 336154 Sum of the proper positive divisors of n A(n) 114585 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 761.923 Returns the nth root of the product of n divisors H(n) 5.06634 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 580,526 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 580,526) is 916,680, the average is 114,585.
## Other Arithmetic Functions (n = 580526)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 274968 Total number of positive integers not greater than n that are coprime to n λ(n) 45828 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 47510 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 274,968 positive integers (less than 580,526) that are coprime with 580,526. And there are approximately 47,510 prime numbers less than or equal to 580,526.
## Divisibility of 580526
m n mod m 2 3 4 5 6 7 8 9 0 2 2 1 2 2 6 8
The number 580,526 is divisible by 2.
• Arithmetic
• Deficient
• Polite
• Square Free
• Sphenic
## Base conversion (580526)
Base System Value
2 Binary 10001101101110101110
3 Ternary 1002111022222
4 Quaternary 2031232232
5 Quinary 122034101
6 Senary 20235342
8 Octal 2155656
10 Decimal 580526
12 Duodecimal 23bb52
20 Vigesimal 3cb66
36 Base36 cfxq
## Basic calculations (n = 580526)
### Multiplication
n×i
n×2 1161052 1741578 2322104 2902630
### Division
ni
n⁄2 290263 193509 145132 116105
### Exponentiation
ni
n2 337010436676 195643320761771576 113576034428548205928976 65933840962667375795124721376
### Nth Root
i√n
2√n 761.923 83.4207 27.6029 14.2156
## 580526 as geometric shapes
### Circle
Diameter 1.16105e+06 3.64755e+06 1.05875e+12
### Sphere
Volume 8.19509e+17 4.235e+12 3.64755e+06
### Square
Length = n
Perimeter 2.3221e+06 3.3701e+11 820988
### Cube
Length = n
Surface area 2.02206e+12 1.95643e+17 1.0055e+06
### Equilateral Triangle
Length = n
Perimeter 1.74158e+06 1.4593e+11 502750
### Triangular Pyramid
Length = n
Surface area 5.83719e+11 2.30568e+16 473997
## Cryptographic Hash Functions
md5 d696c43d822a0ecd41f177fcb329dcac de760ff4722db15d6fcb20b3a2d476fc6f94358b edc5bc61404012986ee94188fe2977c306ff6e425d6dde17d4fe0d83c780636d 540d420e0d8b1f7d78977e0d377e942c0cfd7b7dec861195e754ca4f15252d3da6aafe4206340b5662d8db679c65a13d2f75a63f53169a5374a05bdbf73effd3 39837cc799e1d1d94ee3c66aa5136bdf7c8ad7f5 | 1,459 | 4,175 | {"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} | 3.375 | 3 | CC-MAIN-2021-25 | latest | en | 0.811022 |
2015.icmsao.org | 1,586,280,573,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371803248.90/warc/CC-MAIN-20200407152449-20200407182949-00360.warc.gz | 739,537 | 5,991 | # Keynote Speakers
Wang Chien Ming National University of Singapore Title: Analogies between central finite difference, Hencky bar chain and Eringen’s nonlocal beam models for buckling and vibration. Abstract: This keynote lecture points out the analogies between the central finite difference beam model, Hencky bar-chain model and Eringen’s nonlocal beam model. The finite difference beam model is obtained by discretizing the differential governing equation of the beam via finite difference equations for approximating the derivatives. The Hencky bar-chain model comprises finite rigid beam segments connected by frictionless hinges with elastic rotational springs. Eringen’s nonlocal theory allows for the effect of small length scale effect which becomes significant when dealing with nanobeams. Based on the mathematically similarity of the governing equations of these three beam models, analogies exist between them. The consequence is that one could readily obtain the buckling and vibration solutions of beams without solving the differential equation as well as it allows one to calibrate Eringen’s small length scale coefficient $e_0$. As an example, for an initially stressed vibrating beam with simply supported ends, it is found via this analogy that Eringen’s small length scale coefficient $\sqrt{\frac{1}{6}-\frac{1}{12}\frac{\sigma_0}{\bar{\sigma_m}}}$ where $\sigma_0$ is the initial stress and $\bar{\sigma_m}$ is the m-th mode buckling stress of the corresponding local Euler beam. It is shown that $e_0$ varies with respect to the initial axial stress, from $1/\sqrt{12}$ at the buckling compressive stress to $1/\sqrt{6}$ when the axial stress is zero and it monotonically increases with increasing initial tensile stress. The small length scale coefficient $e_0$ however, does not depend on the vibration/buckling mode considered. Keywords: buckling, finite difference beam model, Hencky bar chain model, nonlocal beam theory, repetitive cells, small length scale coefficient, vibration Juan I. Ramos University of Malaga Title: Local linearization methods in space and time for nonlinear wave propagation. Abstract: In this keynote lecture, linearization methods for nonlinear ordinary (odes) and partial differential equations (pdes) are presented. For initial-value problems of odes, these linearization methods provide locally exponential solutions, whereas, for boundary-value problems of odes, local spatial linearization subject to continuity of the dependent variables and their first derivative provides locally analytical solutions. For one-dimensional transport equations, upon first discretizing the time derivative and then linearizing the resulting equations with respect to the previous time level, two-point nonlinear boundary-value problems are obtained. These problems become locally linear and integrable upon local linearization in space. For multidimensional problems, time discretization followed by time linearization and approximate factorization of the resulting multidimensional nonlinear elliptic problem into a sequence of one-dimensional ones yields nonlinear two-point boundary-value problems which upon local spatial linearization result in linear ones that have local analytical solutions. The advantages and disadvantages of time linearization are discussed in detail, and the methods are applied to solve initial-value problems of nonlinear dynamical systems including nonlinearities that are not differentiable, differentialdifference equations, delay equations, jerk dynamics, nonlinear two-point boundary-value problems with regular singular points at a boundary, one- and two-dimensional advectionreaction diffusion problems arising in combustion theory, one-dimensional nonlinear acoustic phenomena, and the break-up and formation of solitary waves from both the equalwidth and the generalized regularized-long wave equations. Comparisons between the results obtained with these methods and standard second-order accurate finite-difference discretization and three-point, compact, fourth-order accurate discretization of the spatial derivatives are also presented. Keywords: time linearization, local spatial discretization, nonlinear evolution equations, nonlinear wave propagation, reaction-diffusion equations, nonlinear dynamics Mohamed Guma El-Tarhuni American University of Sharjah Title: The Road to 5G Wireless Communications Systems Abstract:There have been tremendous developments in mobile radio communications systems over the past few decades as a response to the wide acceptance of wireless communication services including basic voice applications, higher date rates applications. We have witnessed an accelerated trend on adopting new technologies to support the ever increasing number of users and From basic voice call services, support of data applications, seamless worldwide connectivity, and enhanced Quality of Services. In this talk, we will present an overview of the recent developments in mobile radio systems as we move towards the fifth generation (5G) by the year 2020. The main features expected from 5G and targeted applications are highlighted. Finally, the enabling technologies and challenges facing 5G systems are discussed. | 969 | 5,229 | {"found_math": true, "script_math_tex": 8, "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": 8, "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-2020-16 | latest | en | 0.862691 |
https://www.expertsmind.com/questions/cyclical-variation-in-time-series-analysis-30112634.aspx | 1,679,594,777,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945182.12/warc/CC-MAIN-20230323163125-20230323193125-00378.warc.gz | 813,609,938 | 14,071 | ## Cyclical variation in time series analysis, Financial Management
Assignment Help:
Cyclical Variation
By cyclical variations, we refer to the long-term movement of the variable about the trend line. Therefore, does the movement of the actual series about a trend line observed in figure (1) suggest a pattern of cyclical variation? Yes, it does. A more appropriate example of cyclical variation would be the pattern of business cycle. The whole cycle encompassing the phenomena of the business activity reaching a high, its gradual slow down, a depression and then the recovery can be observed. But one of the points to which we have to pay attention is that the business and the economic cycles may not be periodic in nature. That is, in two different business cycles, at equal intervals of time, the pattern observed may be different.
Generally, the behavior of the variable is considered to be cyclic, only if the movements recur after a period of more than one year. The cyclical variation is shown in the figure.
Figure
In the figure, we observe the component of the time series moving towards and away from the secular trend.
#### Importance of capital budgeting decision, Q. Importance of Capital Budgetin...
Q. Importance of Capital Budgeting Decision? 1. Such Decision affect the profitability of the Firm: - Capital Budgeting decision influences the long-term profitability of a fir
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Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report! | 780 | 3,777 | {"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-2023-14 | longest | en | 0.907362 |
https://www.calculationworld.com/mulch-calculator | 1,553,543,080,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204300.90/warc/CC-MAIN-20190325194225-20190325220225-00210.warc.gz | 722,967,628 | 8,066 | ### How to Use Mulch Calculator
Enter the depth in Inches. Enter width in feet or inches. Enter length in feet or inches
The cubic yards required will be calculated
# Mulch Calculator
T
W
W
L
L
### How To calculate how much Mulch is needed
This calculator can calculate the amount of mulch you need for a rectangle area, triangular area and a round/circular area.
Example 1: Rectangle Area
• Measure the length of the area, the length is 6.
• Then we measure the width, which is 4 feet and the depth is: 3 inches.
• We then get the area of the rectangle Length X Width: 6 X 4 = 24
• We divide the depth by 12 to convert inches to feet
• Now we have to get the volume of the area with = Depth X Area: (3/12) X 24 = 6
• Now we divide the volume by 27: 6/27 = 0.22 cubic yards are needed of pea gravel
Example 2: Triangle Area
• We measure the depth:4 inches
• After measuring the depth, measure the length of the triangle, each side of the triangle is equal at 3 feet.
• We calculate the area with 1/4 X Square root[(3+3+3) X (3+3-3)X(3+3-3)X(3+3-3)] = 3.89
• Now we multiply the area by depth: 3.89 X (4/12) = 1.296
• Once more divide the depth by 12 since its in inches
• We than divide the volume by 27: 1.296/27 = 0.048 cubic yards of pea gravel
Example 3: Circle Area
• Measure the depth at 3 inches
• Then the width is measured at 4 feet
• We calculate the area PI(3.14) X (width/2): 3.14 X (4/2)^2 = 12.56
• We then multiply the area by depth to get the volume: 12.56 X 3/12 = 3.14
• Once again divide the depth by 12: 3/12 in the above equation
• Now we get the cubic yards: 3.14/27 = 0.116
• 0.116 cubic yards are needed
What is mulch? Mulch can be any material spread on top of soil in planting beds as a covering. Generally the material used is organic; it can also be more permanent mulch such as plastic (polyethylene) sheeting.
There are many benefits to using mulch, similar to pea gravel. Technically any organic material can be used as mulch for ground cover.
Mulch is often said to improve the appearance of the garden/landscape. The material helps retain moisture levels in the soil to lessen irrigation/watering needed and acts as insulation to protect the roots from extreme heat or cold. It also effectively suppresses common garden weeds. Mulching also helps extend your growing season by allowing you to move your planting date earlier, wield earlier harvests and grow crops longer into the season.
A benefit of using organic mulch is that it decomposes over time which supplies the surrounding soil with fertilization and nutrients. Depending on your mulch material, the decomposition of certain mulches can either add acidic levels to your soil (Pine Straw Mulch) or make it neutral/sweet (Alkaline) using Hardwood Bark Mulch. While more expensive, it is best for amending your soil.
An inexpensive common option is using old hay; taking care to build up to a depth of 30cm approximately to prevent the grain’s own seeds from sprouting weeds. For general flower beds, simple bark mulch suffices, spreading them across the bed up to 10-15cm. The flip side is that the mulch will need to be replaced. Ideally gardens should be mulched twice a year – once in spring and once in the fall.
Pea gravel here has an advantage in that it does not need to be replaced as often. Calculationworld also has a Pea Gravel Calculator. Plastic sheeting or Plastic mulch can also be more durable and has been used in commercial vegetable production for over 50 years. With technology, many options become available.
Clear vinyl sheets aids in solarization, ridding the soil of unwelcomed pathogens, insects and diseases in a non-chemical, organic and affordable way. Whereas Black Plastic mulch is great for vegetable crops that love heat (tomatoes, strawberries, pumpkins, and melons).
White Plastic Mulch acts as a cooling agent to allow plantation of cooler weather corps in warmer climates (cauliflower, cabbage, peas, and broccoli).With innovation, Infrared Transmitting (IRT) mulch have combined the ability of weed control like Black plastic mulch, while still pigmented to reduce the transmission of light to discourage weeds which can be an issue with Clear plastic mulch. A wide variety of other colored mulches, including red, yellow, green, blue, and silver are available, each with a differing effect on radiation levels in the crop canopy and light reflection.
### Common materials used as mulch
• Grass clippings
• Wood chips
• Cocoa hulls
• Hay
• Crushed seashells
• Pine needles
• Leaves
• Stone
• Straw
• Bark
• Sawdust
• Shredded bark
• Cardboard | 1,157 | 4,588 | {"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} | 4.65625 | 5 | CC-MAIN-2019-13 | latest | en | 0.904597 |
https://everything2.com/user/GameTheory/writeups/strategic+form | 1,539,964,618,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512411.13/warc/CC-MAIN-20181019145850-20181019171350-00052.warc.gz | 683,254,054 | 7,300 | In game theory, strategic or normal form games are used to describe the effect of a single, simultaneous decision by each participant. Thus a strategic form game is more akin to a single turn in chess than the entire chess match; despite this simplicity many interesting (and famous) problems can be formulated in terms of strategic form.
## Strategic form for two player zero-sum games
Definition 1: A two player zero-sum game in strategic (or normal) form X,Y,A consists of two strategy sets X and Y, corresponding to the players, and a function A : X × Y → R representing the pay-off. The game is finite if both X and Y are finite sets.
Definition 2: A play of the game consists of Player 1 choosing a strategy x ∈ X and Player 2 simultaneously choosing a strategy y ∈ Y . Player 1 is then awarded A(x, y) in winnings, and Player 2 awarded -A(x, y) (i.e, Player 2 loses whatever Player 1 wins; this is the zero-sum condition).
Definition 3: The payoff matrix or game matrix for the game with X = {x1, . . . , xm}, Y = {y1, . . . , yn} and payoff function A is given by
```a11 . . . a1n
. . .
. . .
. . .
am1 . . . amn
```
where aij = A(xi, yj), that is, the (i, j)th entry of A determines the winnings for Player 1 and losses for Player 2 when Player 1 chooses strategy xi ∈ X and Player 2 chooses strategy yj ∈ Y . As a shorthand, we may describe Player 1 as choosing the row and Player 2 as choosing the column.
## Strategic form for two player general sum games
The zero-sum condition is not vital to the strategic form, but its relaxation necessitates an increase in complexity of the notation, since the payoff to Player 2 is no longer immediate from knowledge of the payoff to Player 1.
Definition 4: The 2 player general sum game where Player 1 has strategy set X = {x1, . . . , xm} and Player 2 has strategy set Y = {y1, . . . , yn} such that we can represent Player 1’s payoff A(x, y) by a matrix A and Player 2’s B(x, y) by a matrix B (both m × n ) is described as being a game in strategic form X, Y,A,B:
``` Player 1 Player 2
a11 . . . a1n b11 . . . b1n
. . . . . .
. . . . . .
. . . . . .
am1 . . . amn bm1 . . . bmn
```
Thus any zero-sum game in strategic form X,Y,A is a game of strategic form X,Y,A,B where B=-A i.e., (bij)=(-aij) . The more compact Bimatrix notation is often employed, and this class of games referred to as Bimatrix games.
## Strategic form for n player general sum games
With some further adjustment of notation, strategic form can accomodate any number of participants.
Definition 5: A finite n-player game in strategic form X1, . . . ,Xn consists of strategy sets X1, . . . ,Xn corresponding to the players; and real valued functions a1 . . . an : X1 × . . . × XnR such that the payoff to player i when the strategies chosen by each player j is xj ∈ Xj is given by ai(x1, . . . , xn).
Of course, bimatrix games then become the special case n=2 (identifying X with X1, Y with X2 and the functions A,B with a1, b1. There is no convenient notation for an n player game; further, for large numbers of players individual return may be less interesting than group dynamics, in which case the coalitional form may be a more suitable model.
Part of A survey of game theory- see project homenode for details and links to the print version. | 894 | 3,377 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2018-43 | latest | en | 0.93502 |
https://www.jobilize.com/chemistry/course/1-5-measurement-uncertainty-accuracy-and-precision-by-openstax?qcr=www.quizover.com&page=1 | 1,600,808,369,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400206763.24/warc/CC-MAIN-20200922192512-20200922222512-00133.warc.gz | 936,038,097 | 21,357 | # 1.5 Measurement uncertainty, accuracy, and precision (Page 2/11)
Page 2 / 11
Whenever you make a measurement properly, all the digits in the result are significant. But what if you were analyzing a reported value and trying to determine what is significant and what is not? Well, for starters, all nonzero digits are significant, and it is only zeros that require some thought. We will use the terms “leading,” “trailing,” and “captive” for the zeros and will consider how to deal with them.
Starting with the first nonzero digit on the left, count this digit and all remaining digits to the right. This is the number of significant figures in the measurement unless the last digit is a trailing zero lying to the left of the decimal point.
Captive zeros result from measurement and are therefore always significant. Leading zeros, however, are never significant—they merely tell us where the decimal point is located.
The leading zeros in this example are not significant. We could use exponential notation (as described in Appendix B) and express the number as 8.32407 $×$ 10 −3 ; then the number 8.32407 contains all of the significant figures, and 10 −3 locates the decimal point.
The number of significant figures is uncertain in a number that ends with a zero to the left of the decimal point location. The zeros in the measurement 1,300 grams could be significant or they could simply indicate where the decimal point is located. The ambiguity can be resolved with the use of exponential notation: 1.3 $×$ 10 3 (two significant figures), 1.30 $×$ 10 3 (three significant figures, if the tens place was measured), or 1.300 $×$ 10 3 (four significant figures, if the ones place was also measured). In cases where only the decimal-formatted number is available, it is prudent to assume that all trailing zeros are not significant.
When determining significant figures, be sure to pay attention to reported values and think about the measurement and significant figures in terms of what is reasonable or likely when evaluating whether the value makes sense. For example, the official January 2014 census reported the resident population of the US as 317,297,725. Do you think the US population was correctly determined to the reported nine significant figures, that is, to the exact number of people? People are constantly being born, dying, or moving into or out of the country, and assumptions are made to account for the large number of people who are not actually counted. Because of these uncertainties, it might be more reasonable to expect that we know the population to within perhaps a million or so, in which case the population should be reported as 3.17 $×$ 10 8 people.
## Significant figures in calculations
A second important principle of uncertainty is that results calculated from a measurement are at least as uncertain as the measurement itself. We must take the uncertainty in our measurements into account to avoid misrepresenting the uncertainty in calculated results. One way to do this is to report the result of a calculation with the correct number of significant figures, which is determined by the following three rules for rounding numbers:
give two properties of liquid
what is measurement
is the comparison of an unknown quantity with a fixed quantity of the same kind
How does an element differ from a compound? How are they similar?
an element is an indivisible particles that can take part in a reaction and consist of smaller or tiny particles i.e proton, neutrons and electron while a compound is when two or more element chemically combine together. They are similar when they are homogeneous compound. they take the same rxn.
Yusuf
an element is s chemically pure substance containing a particular type of atoms.. A COMPOUND is a substance containing atoms from different elements..
Inemesit
How to get the Lewis formula of SeCl+3
hi,I'm new here can I join the conversation
EZEA
what is the structural formula for starch
Starch is a mixture (of chemicals) of amylose and amylopectin. Both are macromolecules and polymers. You can search on wikipedia.
Abdelkarim
what is the roles of filter bed
Fathmat
what is the roles of Alu m
Fathmat
what is the roles of chlorine
Fathmat
Roles can be classified or correlate it to different areas: For example: Chlorine can be used in reactions (in industry) to manufacture HCl, which then can be used for other things. Or in swimming pools to kill bacteria. Or as a component in compounds with pharmaceutical roles (drugs). For Al:
Abdelkarim
Its dentisty value is suitable to be used in alloys (mixture of metals) in aircraft bodies. Also, Aluminium foils, Tin cans,.. Some of them are also in Al overhead cables in streets and long roads.
Abdelkarim
what is chemistry
Maxamed
what is the meaning of exceedingly
it is an adverb which means extremely
Rohini
what is atomic chemistry?
Lewis structure for no3
Lewis structure for no3
what is weak acid
It is an acid which partially ionises in water.
Abdelkarim
what is incandescence
Clifton
what makes it glow
Clifton
why is it red, irange and yellow in color
Clifton
hello am new here and I want to join you
Aliyu
hello
Clifton
hi
Aliyu
too
Gillian
hello i am new here please i want to join this group
Paul
Hi, I'm also new here
Salaudeen
Hi
Keeya
hello guys !!
Sourav
what is pressure?
The force applied to suction Area of the body
Ahmed
Matter composed of exceedingly small paticle called atom.
Yushao
questions related to metals
occurrence and preparation of the representatives metals
Regina
list the 20, periodic table and their symbols
hydrogen:h helium;he lithium:l beryllium:be Boron:b Carbon;C Nitrogen:n Oxygen:O FLUORINE:f Neon:n Sodium:s Magnesium:mg Aluminum:a Silicon:s Phosphorus:p Sulphur:s Chlorine:c Argon;a Potassium:p Calcium:c
Benita
Hydrogen, helium, lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine, neon, sodium, magnesium, aluminium, silicon, phosphorus, sulphur, chlorine, argon, potassium, calcium
Cudjoe
what is a solute
Any substance that is disolved in a liqid solvent to create a solution
Fifa
sorry liquid
Fifa
it's a liquid substance
Fathmat
hello group
Ayomide
is the substance that dissolves in the solvent
Amos
so is HCl ionic compound
No, covalent compound ➡️ molecule. As both H and Cl are non-metals and and form covalent bind by sharing valence e-. But can fully ionice in water forming H+ (a proton, a reason for acidity) and Cl- (anion =Chloride) Hydrogen Chloride is a gas at room; Hydrochloric acid = HCl (aq), dissolved in w
Abdelkarim
Form covalenr bond*
Abdelkarim
The question marks are an emoji in the first sentence is an unread emoji. HCl Covalent compund -> molecule
Abdelkarim
Hi.
Queen
Hi
Calvin
Yh
Cudjoe
yes
Amos
what is chemistry
is the study of composition of substances and the way they behave under different conditions
Amos
how do calculate n1 though n6 any help on understanding the concept
Clifton
is the study of properties of matter and it's component
Grace
where can I get the test bank or mcqs ? any idea ? | 1,672 | 7,032 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "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} | 4.25 | 4 | CC-MAIN-2020-40 | longest | en | 0.918031 |
http://www.numbersaplenty.com/947763551 | 1,591,164,962,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347432237.67/warc/CC-MAIN-20200603050448-20200603080448-00262.warc.gz | 193,712,288 | 3,489 | Search a number
947763551 = 7592294827
BaseRepresentation
bin111000011111011…
…011100101011111
32110001101101011202
4320133123211133
53420111413201
6234013510115
732325600020
oct7037334537
92401341152
10947763551
11446a95869
122254a233b
1312147ab47
148dc30c47
1558313c6b
hex387db95f
947763551 has 8 divisors (see below), whose sum is σ = 1101517440. Its totient is φ = 798599448.
The previous prime is 947763541. The next prime is 947763563. The reversal of 947763551 is 155367749.
It is a happy number.
It is a sphenic number, since it is the product of 3 distinct primes.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-947763551 is a prime.
It is a Duffinian number.
It is a junction number, because it is equal to n+sod(n) for n = 947763496 and 947763505.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (947763541) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 1147001 + ... + 1147826.
It is an arithmetic number, because the mean of its divisors is an integer number (137689680).
Almost surely, 2947763551 is an apocalyptic number.
947763551 is a deficient number, since it is larger than the sum of its proper divisors (153753889).
947763551 is a wasteful number, since it uses less digits than its factorization.
947763551 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 2294893.
The product of its digits is 793800, while the sum is 47.
The square root of 947763551 is about 30785.7686439692. The cubic root of 947763551 is about 982.2755502975.
The spelling of 947763551 in words is "nine hundred forty-seven million, seven hundred sixty-three thousand, five hundred fifty-one". | 541 | 1,829 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-24 | latest | en | 0.849379 |
http://crewtonramoneshouseofmath.blogspot.com/2011_04_01_archive.html | 1,527,204,341,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866894.26/warc/CC-MAIN-20180524224941-20180525004941-00476.warc.gz | 78,135,354 | 30,374 | Here you will see students as young as 4 and 5 years old doing algebra and "advanced" math, without ever knowing it's supposed to be hard.
You are invited to learn how to use this method...
## Thursday, April 28, 2011
### Patient Problem Solving.
Yeah, what he said:
I don't have patients I have students..but still really captures a few issues in a short period of time. I watch this every couple of months...
I need to put more math problem solving videos up.
Here are a few:
Problem Solving with Crewton Ramone.
Crewton Ramone Problem Solving w/ Fractions.
Crewton Ramone Problem Solving With Sarah
Crewton Ramone Simple Systems of Equations
Find us on Face Book...I'm the only Crewton Ramone.
## Monday, April 25, 2011
### Mortensen Math Squeeks Back To Life...
Here is a video from the early 90's...
People always wonder about the nature of the relationship between Matt Dillon and Miss Kitty, likewise the relationship between Crewton Ramone and Mortensen Math. Let's just say we go way, way back...
Due to a shall we say, a slight altercation between myself and the Marketing Company that produced the promotional video I was edited out of the vid with the little boy who does the Algebra, even though it was my idea to use him and I'm the one that taught his parents, (particularly his mother) how to use the kit they just bought.
I'm bringing up another generation of testimonials on Maui as we speak. For example:
Multiplication and algebra are the two things we focused on, when I met her she was still adding on her fingers...now: 100's. Before she got a 28 on one test...we haven't even done 10 lessons yet. It's not that Crewton Ramone is so awesome, it's that the methodology IS.
Here is a FAQ at the house of math.
## Wednesday, April 20, 2011
### Crewton Ramone I just wanna Play Math all day.
Play. Play Math.
I've said it over and over again. Play is important. It's crucial. Note how he talks about problem solvers having to use their hands. Mechanics. Engineers. Kids with blocks. Something happens when you get the tactile sense involved. Kinesthetic learner or not, adding the sense of touch and being able to get your hands on it makes such a big difference mere English words do not adequately convey the importance of it.
2011 and the world of mathematics has still not caught on to how important manipulatives are for increasing comprehension and learning of math concepts. It's still something for special education or minorities. "Real" math teachers don't use manipulatives and certainly not on adults. I note the "real" math teachers are failing miserably as a whole, with a very few bright spots here and there.
Here is yet another TED talk on the subject of play:
"Almost all creativity involves purposeful play." ~Abraham Maslow American psychologist 1908-1970
"Whoever wants to understand much must play much." ~Gottfried Benn German physician 1886-1956
"Play gives children a chance to practice what they are learning." ~Fred Rogers American television personality 1928-2003
"People tend to forget that play is serious." ~David Hockney Contemporary British painter
"Do not…keep children to their studies by compulsion but by play." ~Plato Greek philosopher 427-347 BCE
"Necessity may be the mother of invention, but play is certainly the father." ~Roger von Oech Contemporary American creativity guru
Not to hit you over the head with it, with these math quotes but perhaps maybe, you just might want this to become part of your teaching philosophy. I try to keep these quotes in mind when I play math with my students. Grander men than me have left behind these words so that I might also understand the crucial role of play.
"We didn't do anything dad, we just played with blocks." ~My favorite quote from a little girl who went from "F" to "A".
Here are a few blog posts as a reminder:
http://crewtonramoneshouseofmath.blogspot.com/2010/12/4-year-old-math-enrichment.html
http://crewtonramoneshouseofmath.blogspot.com/2010/09/make-it-fun.html
http://crewtonramoneshouseofmath.blogspot.com/2011/02/math-rich-environment-ii.html
### Crewton Ramone Quikie on Ordered Pairs.
Pre-calc concepts are not hard either if you can SEE the relationships.
Here is a short video that shows how an 11 year old can begin to MASTER the concepts related to constant rate problems and make ordered pairs into simple linear equations without being given formulas to memorize...in fact given time they will be able to come up with the formulas themselves.
We never once talked about dy/dx. He quickly began to see that the fastest way to do it was to take the largest number that would divide equally into each of them and is just beginning to find out that the result is dy/dx but he is doing it by multiplying and subtracting...which by the way is division. Do you see how math really is a language? If you are a layman that paragraph might be jibberish, but if you are a teacher I encourage you to pause and reflect. If you are a homeschooler with math issues of your own take a while and play with the concepts and you will see that it becomes easier the more you play with it.
This student did 15 or 20 minutes for a few sessions over a few days, later one 30 minute session of which this video is part and we haven't been back since but we will and each time we do more understanding is created.
More at my house of math.
### Crewton Ramone Playing with Radicals.
First off apologies to public school teachers who can't watch YouTube a school the screencasting software is buggin' out and won't let me copy it to a screencast without it being herky jerky video or bad audio...so for now all I have is this youtube video.
It shows me doing a few problems with radicals. Adding and subtracting (combining) them and doing other operations is actually pretty easy when you keep the 5 basic concepts in mind and can see what you are doing.
You might want to look at the square numbers page at the house of math, too.
You might also note there is a new tab on the navigation bar there for percentages. It's password protected though. Progress is being made. Albeit slowly. All it has is two Youtube vids showing how to do percentages with base ten blocks on it at the moment but there's more to come.
Meantime the completing the square page which certainly is related is growing too.
Square numbers are fun and easy.
## Wednesday, April 13, 2011
### Response to Radiation in Hawaii Milk.
Look, first things first. Drinking water limits and limits for milk are two very different things.
The limits for drinking water are 3 pCi/L that is 3 picocuries per liter, and assumes long term exposure. The average person doesn't drink as much milk as water, nor do you bathe in it, brush your teeth with it, cook (as much) with it etc. The amount is minute because it assumes constant exposure to potable, drinkable water.
The milk they sampled contains 18 pCi/L which would be 6 times the limit for drinking water but not 6 times for milk for I-131. (The limit for milk is 4,700 pCi/L or 261 times higher.) For Cesium-134 they found 24 pCi/L, and for Cesium-137 19 pCi/L, which would be 8 times and 6 1/3 times over the limit for WATER respectively.
This translates into 600%, 800% and 633% for each of the three isotopes respectively; however you CANNOT then ADD the percentages and come up with 2033% over the limit. What kind of math is that? It's not a cumulative total for radiation in general, and besides it's milk, NOT drinking water. No one isotope was 2033% over the limit for water or milk.
That said, how did the radioactive isotopes get into the cow to get into the milk? I would assume from drinking water and eating grass contaminated with those isotopes. We have no measurements for rainwater in Hawaii available to the public.
What is the ratio of isotopes in pCi/L that a cow has to consume to produce milk that has a pCi/L of the isotopes. I doubt it's one to one, because the "experts" have constantly repeated how poorly the uptake for those isotopes is in humans and animals. Now, I don't know what the ratio is and though I bet there is a study or two on the subject I have yet to find it.
If for example, if the ratio is 2 to 1 then the cow would have had to consume 36 pCi/L of water/grass I-131. For Cesium-134 48 pCi/L, and for Cesium-137 38 pCi/L...if it's 3 to 1 or 4 to 1 you can hopefully do the math yourself. Maybe it's 10 to 1: I don't know. Were the cows drinking straight undiluted rainwater? Doubtful. Was the grass soaked in rainwater that just contained 48 pCi/L of Cesium-134...again doubtful. I would think that the rainwater that fell on the cows had many multiples of 3 pCi/L of I-131 because it would be diluted in water troughs and on the grass and feed that they consumed. This is all conjecture on my part. I have no numbers with which to work, only a little logic, reasoning and critical thinking skills of which certain bureaucrats seem to be blissfully bereft.
I look at rainwater readings elsewhere and find them to be in the from 79 pCi/L in MA to 543 pCi/L in CA (San Fransisco) for I-131 but in all cases they also found Cesium and Te-132. On the day they measured 543 pCi/L of I-131 they found 13 pCi/L of Te-132 and Ce-137 in San Fransisco along with a few other isotopes. The point is, for it to show up in the milk it has to be in the rainwater in many multiples of what it is in the milk, but we have no measurements for rainwater in Hawaii that I have seen. NONE. ZIP. NADA. ZILTCH. ZED. ZERO. Not any. Squat.
All I can gather from the exceedingly limited data I can find is that it has been more than 18 pCi/L of I-131 in the rainwater. I doubt it was 500 pCi/L of I-131, on the other hand I doubt it was only 36 pCi/L of I-131. I am thinking the range is some where between 100 and 400 pCi/L of I-131 in our rainwater but that's a guess. Problem is rainwater IS drinking water for some of the state where catchment and surface water make up a good deal of the water supply.
Again the amounts are minuscule in the main water supply because even so the contaminated water mixes with the drinking water in ratios that would knock down the pCi/L count by factors: Waikamoi Reservoir is a 30 million gallon (MG) reservoir and Kahakapao Reservoir is 100 MG. One gallon is 3.7854 liters. So doing a little math even if 100,000 liters of contaminated rainwater got directly into the reservoir when it was only about half full it would be diluted far below the 3 pCi/L for the isotopes. It makes for some nice math problems: 100,000 liters of 200 pCi/L I-131 water mixes with 200,000,000 liters of uncontaminated water...assuming a somewhat even distribution what is the new contamination level?
I just made these numbers up to make the math easy. We have had a wet winter and the reservoirs are better than half full and 100,000 liters would be an extremely heavy rainfall so that would be a conservative (read unlikely) guesstimate using made up numbers erring on the side of contamination. Possible but not plausible and you'd still be 30 times under the limit for drinking water. A thought exercise for the sake of example.
After all that, people on catchment do not have the luxury of large numbers, and rainfall directly on you is not diluted...100 to 400 pCi/L of I-131 and even the smaller amounts of the other isotopes that are basically here permanently due to their much longer half lives are not amounts to be flippantly ignored. Cesium accumulates. It also bio-accumulates. Stay out of the rain for a while and don't drink rainwater. Especially if you are pregnant or a small child.
This is FRANCE where the numbers are smaller and they are much farther away:
The people who have catchment tanks should have been warned, one good rainfall can take a tank from 3/4 full to full and then the dilution is only 1:4. I know quite a few people who use catchment
Up until March I collected rainwater on Maui for watering plants and occasionally for drinking...fresh vog free rainwater is delicious. This is no longer the case. Over time the longer lived isotopes could be cause for concern in our drinking water but right now this is not the case. It is indeed cause for caution when it comes to direct exposure to rainwater. The "nothing to worry about" mantra becomes annoying.
It would be nice if our government would do it's (expletive removed) job and inform the people of the measurements in the rainwater instead of just the lack thereof. Still, it's better than Canada were they aren't even testing the milk because they are so certain there is no cause for concern. The point is it's not the milk itself but the amount of radiation it took to get the cows to produce contaminated milk in the first place. Also to be noted is not all the rain has radiation in it all the time, but obviously it has had some in it recently or they wouldn't be detecting it in the milk and that's the problem: we don't know how much or when.
## Saturday, April 9, 2011
### Math Might Be fun
Here is a short vid with a student who is figuring out that math might be fun after all. At Crewton Ramones House Of Math it's hard not to have fun with it...
This has been a great week with two of my former "F" students (both of whom who happen to be teenage girls) getting perfect scores on homework and tests and quizzes. One parent quipped "I guess you're worth the money..."
Yeah.
This little girl is well spoken, therefore I know I can teach her math because English is a messed up language and math is a beautiful one.
Here you can see they have a set of base ten blocks but they are not suitable for teaching all the math concepts the combo kit can. Also note 1000 as a cube which makes third power mathematics difficult at best and fourth power mathematics impossible...
Here you see the algebra again being used to teach addends and multiplication. By know the method should be familiar to some. Each lesson is tailored to the individual student but the method is the same. The problems are presented the same way in the same order...
She has some friends to help her count the rectangles. x2 + 10x + 24 is "no problem..."
Here is the screencast covering the lesson:
The password protected pages keep growing with contect. The price is still a buck a month but now you have to buy 12 months worth to get it for a buck a month or you can buy one month (which ends up being 60 days or so...) for 3 bucks...just added a cute vid of an 11 year old and 6 year old playing algebra together. The footage is raw but it is charming...or so I have been told.
There is about 4 hours of password protected screencasts and many pages of pdf's for that price and like I said more being added all the time...if you buy a combo kit or tutoring the passwords are free.
## Monday, April 4, 2011
### Yet More Factoring With Positive 3rd Power Polynomials
Here is a session with an 11 year old, I tacked a short segment on the end of if with a 15 ear old high school student so there can be no question it's basically the same stuff...the textbook problem at the end uses M's instead of X as if that makes it more interesting...
Again if you want to see negatives you need a password...
More advanced algebra at Crewton Ramone's House Of Math...
### Even More Factoring with 3rd and 4th Power Polynomials
Factoring is fun and easy or hard and incomprehensible depending on how you learn it. I once had a high school student tell his mother he was math learning disabled due to his inability to "get" problems like these. He quit tutoring one lesson before we got to these...
This student on the other hand is 11 and he'll never know this stuff is supposed to be hard. Note how we stay positive until we really have a handle on how to make these go together. Textbooks always jumble the degree of difficulty and start of with negatives and positives. Then wonder why students don't "get" it...that and they haven't figured out how to show third and fourth power algebra in two dimensions...
There are a lot of posts on factoring polynomials on this blog and also on my website. If you want to see how to do ones that contain negative coefficients as well as how to do division and more you need a password and you will find it on the password protected pages (advanced algebra tab) at Crewton Ramone's House Of Math...
### Crewton Ramone Start In The Concrete!
Start in the concrete, there's no better way to cement the facts into their heads than to start with hands on experience of the math...
Some people find it amusing that kids who have trouble with addends and multiplication can factor third power polynomials...that's because they fail to understand proper application of the 5 basic concepts. Math is just counting. Multiplication and addition as well as 3rd power algebra are just counting.
The main idea here though is start in the concrete and watch your success rate go up...
Here is a short screencast covering this lesson. | 3,955 | 16,983 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-22 | latest | en | 0.966857 |
https://www.thestudentroom.co.uk/showthread.php?page=88&t=3321327 | 1,529,516,406,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863830.1/warc/CC-MAIN-20180620163310-20180620183310-00419.warc.gz | 948,054,313 | 45,453 | You are Here: Home >< Maths
Edexcel C2 20th May 2015 *Official Thread* watch
1. Anyone formula?
2. which proof do we need to know?
someone help
3. (Original post by Sam_the-man)
Someone please tells every formula I need to know
Volume of a Cylinder = Pie * r^2 * h
Volume of cuboid = LHW
Surface Area of a Cylinder = 2*h*pie*r +2*pie*r^2
(Optimization)
r*theta = Arc Length
theta*1/2 * r^2 = Area of sector
1/2*a*b*SinC = Area of Triangle
Tan Theta = Sin Theta/Cos Theta
(a+b)/2 * h = Area of Trapezium
4. (Original post by thiss1)
which proof do we need to know?
someone help
Sn = a(1-r^n) / 1-r
You need to know the proof for that.
Sn = a(1-r^n) / 1-r
You need to know the proof for that.
is it just that ?
6. Remainder theorem too!
7. (Original post by thiss1)
is it just that ?
For the C2 paper, yes that is it.
For the C2 paper, yes that is it.
9. (Original post by Hsakbo)
Remainder theorem too!
I thought that was more a method than proof. But if you are talking about things like that then yes remainder theorem and maximum/minimum (double derivative)
10. (Original post by thiss1)
What about them? You just need to learn the rules for that
Volume of a Cylinder = Pie * r^2 * h
Volume of cuboid = LHW
Surface Area of a Cylinder = 2*h*pie*r +2*pie*r^2
(Optimization)
r*theta = Arc Length
theta*1/2 * r^2 = Area of sector
1/2*a*b*SinC = Area of Triangle
Tan Theta = Sin Theta/Cos Theta
(a+b)/2 * h = Area of Trapezium
Trying to memorize them on the way to school xD
Volume of a Cylinder = Pie * r^2 * h
Volume of cuboid = LHW
Surface Area of a Cylinder = 2*h*pie*r +2*pie*r^2
(Optimization)
r*theta = Arc Length
theta*1/2 * r^2 = Area of sector
1/2*a*b*SinC = Area of Triangle
Tan Theta = Sin Theta/Cos Theta
(a+b)/2 * h = Area of Trapezium
Thanks man
13. Bro! did you just started C2 today? :??
14. I DONT HAVE A SPARE CALCULATOR
15. When you do area under a curve, even if the answer is negative do you give the absolute (positive) value?
16. Just so you know guys, the R papers are not the replacement papers, they are REGIONAL papers.
When edexcel withdrew the FP1 exam (and others) in June 2013, the replacement paper looked identical to the original paper. The R paper is for students outside the UK. IAL papers are for different timezones. (Itd be unfair if someone sat the same paper in India and finished 5 hours before we started it haha).
But the papers have NOT been leaked.
17. (Original post by dripper)
When you do area under a curve, even if the answer is negative do you give the absolute (positive) value?
you do, otherwise you may have a negative area or they could cancel out. even if the y values are negative, you have to assume they are positive
18. (Original post by dripper)
When you do area under a curve, even if the answer is negative do you give the absolute (positive) value?
If they say find the area yes give the abosolute answer. If they evaluate the integral blah then give whatever cones out of the integral. Remember peopke that the calc can check ur answers!!
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19. (Original post by bubblegumcat)
that's the differentiation stuff right? I might be able to help
Posted from TSR Mobile
Yep with those shapes
20. Can anybody clarify that area thing for me please?
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Edexcel C3 Maths Unofficial Markscheme
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Exam discussions | 1,007 | 3,643 | {"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} | 3.796875 | 4 | CC-MAIN-2018-26 | latest | en | 0.918377 |
https://docs.duckietown.com/daffy/course-intro-to-drones/pid-controllers/project/pid-tuning.html | 1,695,408,830,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506421.14/warc/CC-MAIN-20230922170343-20230922200343-00726.warc.gz | 235,408,665 | 6,761 | # Part 2: Tuning#
Write brief answers to all exercises in answers_pid.md.
## Problem 1: Ziegler-Nichols Method#
Imagine you are flying your drone and observing its flight for tuning it. Ideally, you would tune the $$K_p$$ by slowly increasing its value between flights until you can see the drone moving up and down with uniform oscillations. The final $$K_p$$ value that causes uniform oscillations is termed as $$K_u$$, the ultimate gain. While, the time difference between these two peaks during oscillations is termed as $$T_u$$, the ultimate period.
Exercises
1. Given $$K_u$$ = 500, $$T_u$$ = 10. Use your $$K_u$$ and $$T_u$$ values to compute $$K_p$$, $$K_i$$, and $$K_d$$ using Ziegler-Nichols Method.
## Problem 2: Flying with Velocity Control#
In velocity control, we use planar velocity measure from the camera as the process variable. The keyboard keys are used to set the setpoints.
Exercises
1. Now suppose you are flying in velocity mode over a blank white poster board. How do you expect the drone to behave, and why will it behave this way? hint: Think about why we fly over a highly textured planar surface. | 278 | 1,134 | {"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": 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} | 3.296875 | 3 | CC-MAIN-2023-40 | longest | en | 0.9186 |
https://www.math-edu-guide.com/CLASS-8-Fundamental-Concept-Of-Algebraic-Expression.html | 1,716,987,275,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059239.72/warc/CC-MAIN-20240529103929-20240529133929-00284.warc.gz | 757,086,037 | 6,379 | CLASS-8FUNDAMENTAL CONCEPT OF ALGEBRAIC EXPRESSION
FUNDAMENTAL CONCEPTS OF ALGEBRAIC EXPRESSION –
As we all aware of the basic terms and concepts of algebra, now we will learn about some advanced form of Algebra.
Literals & Constants
A symbol, such as a, b, c, d, x, y, and z are representing an unspecified number or member of a class of objects is called a literal or a variable.
A symbol that has a fixed value is known as a constant. Numbers such as 3, 52, 0, 5/9, 4.7, 28.231, - 12, and – 2.89 have a specific value, so they are constant.
Operation on Literals
Since literals represent numbers, we can add, subtract, multiply, & divide literals and numbers.
Addition-1) a + b is to be considered as the sum where a & b are literals.
2) a + a = 2a, a + a + a = 3a, a + a + a + a = 4a, etc.
3) b + 0 = b, 0 + b = b
Subtraction– 1) a – b is to be considered as the subtract where a & b are literals.
2) b – b = 0 or a – a = 0
3) 4 subtracted from a or b are a – 4 and b – 4.
Multiplication– 1) ab means a X b or ‘a’ multiplied by ‘b’
2) 4b means 4 X b or 4 multiplied with ‘b’
3) a.1 = 1.a = a
a
Division- 1) a ÷ b or --------- means ‘a’ divided by ‘b’
b
5
2) 5 ÷ b or --------- means 5 divided by ‘b’
b
a
3) a ÷ 8 or --------- means ‘a’ divided by 8
8
a b
4) -------- = 1 or --------- = 1
a b
a b
5) --------- = a or --------- = b
1 1 | 508 | 1,486 | {"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} | 4.71875 | 5 | CC-MAIN-2024-22 | latest | en | 0.764776 |
https://splendidwritings.com/application-of-the-pearson-correlation-and-chi-square-test/ | 1,652,775,395,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517018.29/warc/CC-MAIN-20220517063528-20220517093528-00639.warc.gz | 612,336,553 | 22,982 | Application of the Pearson Correlation and Chi-Square Test
The chi-square test of independence is used to determine whether two or more samples of cases differ on a nominal level variable. A Pearson correlation is used to determine the relationship between two continuous variables. Both the chi-square test of independence and correlation are widely used in the analysis of public health data.
The purpose of this assignment is to practice calculating and interpreting the Pearson correlation coefficient and a chi-square test of independence. After analyzing the data, communicate the results in a PowerPoint presentation. Refer to the Using and Interpreting Statistics: A Practical Text for the Behavioral, Social, and Health Sciences textbook and instructional videos for assistance completing this assignment.
Part 1
Use SPSS and the Topic 2 “Health Behavior Data Set” and complete the following:
1. Conduct a Pearson correlation to determine the relationship between age and annual income.
2. Conduct a chi-square test to determine the relationship between sex and smoking status.
3. Export the SPSS output for the Pearson correlation and chi-square tests.
Part 2
Create an 8-10 slide PowerPoint to discuss the findings for either the chi-square or correlation findings. For the presentation of your PowerPoint, use Loom to create a voice-over or a video. Refer to the topic resources for additional guidance on recording your presentation with Loom. Include an additional slide for the Loom link at the beginning, and an additional slide for references at the end.
Include the following:
1. Explain why the statistical test is most appropriate for analyzing the data and whether the assumptions were met.
2. What are the null and alternative hypotheses?
3. What is the decision rule?
4. What is the test statistic and p-value?
5. How do you interpret the results? (What was done? What was found? What does it mean? What suggestions are there for the creation of a health promotion intervention?)
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http://galileo.phys.virginia.edu/outreach/8thGradeSOL/NeonChargeST.htm | 1,537,512,273,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156857.16/warc/CC-MAIN-20180921053049-20180921073449-00364.warc.gz | 105,695,235 | 2,413 | University of Virginia
Physics Department
## Neon Bulbs and Motion of Charge
A Physical Science Activity
### Student Activity
Materials
• Small neon light bulb
• Several drinking straws
• Scissors
• Glass or plastic rod
• Wool, silk or fur
• Styrofoam cup
Procedure to make light shield:
Make a light shield for the neon bulb. It will be easier to see what is happening inside the bulb with a white backing made out of straw. Note: The bulb can also be taped to a white or a black piece of paper.
1. Cut about a four cm length of straw. Cut away half of it for about one cm.
2. Cut slits about a half cm up the straw.
3. Bend the leads of the neon bulb out at right angles and slide the neon bulb into the cut away part of the straw, leads first, so that you can see both electrodes of the bulb and the leads stick out from the slits in the straw.
Procedure:
Investigating charge transfer with the neon bulb.
To use the neon bulb, hold one lead in your hand and touch the other lead to a charged object while watching the electrodes. Try the following tests: (NEVER STICK A WIRE INTO AN 110-V ELECTRICAL SOCKET!):
• Charge a Styrofoam cup by charging the glass rod and then discharging it on the cup. The rod should be charged by rubbing with fur or silk and then discharged by touching the cup. Don't let anything else touch the cup besides the table it rests on. Touching something will discharge the stored charge on the cup. Repeat this process four or five times. Then touch one lead to the cup while holding the other lead. Note which electrode flashes and write it down.
• Rub part of a balloon with wool. Touch one lead to the charged portion of the balloon while holding the other lead. Note: This will only work on extremely dry days when enough charge can be built up on the balloon. Note which electrode flashes and write it down.
• Try rubbing your shoes across carpet and touching one lead of the bulb on a neutral object while holding the other lead. Note: Again this will only work on extremely dry days allowing sufficient charge to build up. Note which electrode flashes and write it down.
Data Sheet
Questions:
1. Why does the bulb light in each of those tests?
2. For each test, tell which object, your hand or the charged object, was losing electrons. You should be able to determine this by observing which electrode flashed.
3. When you rub your feet on the floor what is happening? | 537 | 2,422 | {"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-2018-39 | latest | en | 0.891975 |
https://mathematica.stackexchange.com/questions/245856/ndsolve-with-embedded-piecewise-function-for-state-variable | 1,716,028,444,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057379.11/warc/CC-MAIN-20240518085641-20240518115641-00806.warc.gz | 357,887,216 | 40,087 | # NDSolve with embedded piecewise function for state-variable
Given ODE:
$$\dot{x}=-2 e^{-x^2} x + U$$
where:
$$U=-u_{as}$$
$$u_{as} = \begin{cases} \frac{e}{\psi_+-e}, & \mbox{if } e\mbox{>=0} \\ \frac{e}{e-\psi_-}, & \mbox{if } e\mbox{<0} \end{cases}$$
$$e=x$$
$$\psi_+=\frac{1-\delta}{T \cdot t+1}+\delta$$
$$\psi_-=-\frac{0.5+\delta}{3T \cdot t+1}-\delta$$
There is my code:
(***)
Clear["Derivative"]
ClearAll["Global*"]
pars = {xs = -1/4, xe = 1, T = 5, \[Delta] = 0.1}
e = x[t]
\[Psi]plus = (1 - \[Delta])/(T t + 1) + \[Delta]
\[Psi]minus = -(1/2 + \[Delta])/(3 T t + 1) - \[Delta]
uas = Piecewise[{{e/(\[Psi]plus - e), e >= 0}, {e/(e - \[Psi]minus),
e < 0}}]
U = -uas
sys = NDSolve[{x'[t] == -2 E^-x[t]^2 x[t] + U, x[0] == xs}, {x}, {t,
0, 3}]
Plot[{Evaluate[x[t] /. sys], \[Psi]plus, \[Psi]minus}, {t, 0, 3},
PlotRange -> Full, PlotPoints -> 100]
Plot[{Evaluate[U /. sys]}, {t, 0, 1}, PlotRange -> Full,
PlotPoints -> 100]
When I run NDSolve, I get a error on the 3rd second of the calculation.
NDSolve::smpf: Failure to project onto the discontinuity surface when computing Filippov continuation at time 2.359060463990643.
I want to understand what the reason for this error is and how to avoid it in the numerical calculations.
I would be grateful for the help.
• It looks like a numerical problem at t=2.35. Decrease the range {t, 0, 2.35} and NDSolve evaluates without error message May 11, 2021 at 8:33
• @UlrichNeumann Yes, if we reduce the time interval of the calculation. But I need calculation at large intervals {t, 0,100} and above. It is necessary to correct the error and avoid it in the future.
– dtn
May 11, 2021 at 8:36
• @UlrichNeumann The problem disappeared when I added to the code Method -> {"DiscontinuityProcessing" -> False}. But I still did not understand what was the reason for the error.
– dtn
May 11, 2021 at 8:42
• Try AccuracyGoal -> 32 or even AccuracyGoal -> Infinity. May 11, 2021 at 14:47
• +1. Consider reporting it to WRI support. I feel there is room for improvement here. There's no real discontinuity in the vector field, just a suspected one arising from Piecewise. But the vector field changes continuously as crosses the equilibrium $x=0$. I suspect it's this singularity in the vector field at the boundary between the pieces that leads to the failed projection. (Raising AccuracyGoal keeps the numerical solution from crossing the boundary, and the true solution does not cross it either.) May 11, 2021 at 20:58
Perhaps StreamPlot shows you the reason:
StreamPlot[{1, -2 E^-x ^2 x -
Which[x >= 0, x /(0.1 + (0.9/(1 + 5 t)) - x ),x < 0, x /(0.1 + (0.6/(1 + 15 t)) + x ), True, 0]}, {t, 0,5}, {x, -.5, .5}]
The trajectory x->0 seems to be unstable!
• The problem disappeared when I added to the code Method -> {"DiscontinuityProcessing" -> False}.
– dtn
May 11, 2021 at 8:56
• Yes that's already clear. Trajectory x~0 isn't attractive, because of the discontinuity of the ode near x~0 May 11, 2021 at 9:04
• Ok, so the reason is internal, not algorithmic.
– dtn
May 11, 2021 at 9:12
• The reason is your discontinuous ode near x==0. Method -> {"DiscontinuityProcessing" -> False}seems to ignore this fact! May 11, 2021 at 9:16
• That's what I'm talking about.
– dtn
May 11, 2021 at 9:18 | 1,119 | 3,272 | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-22 | latest | en | 0.760744 |
https://documen.tv/question/a-2-80-kg-mass-is-dropped-from-a-height-of-4-50-m-find-its-potential-energy-pe-when-it-is-3-00-m-24029307-77/ | 1,628,081,816,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154805.72/warc/CC-MAIN-20210804111738-20210804141738-00140.warc.gz | 204,380,775 | 18,074 | ## A 2.80 kg mass is dropped from a height of 4.50 m. Find its potential energy (PE) when it is 3.00 m above the ground.
Question
A 2.80 kg mass is dropped from a
height of 4.50 m. Find its potential
energy (PE) when it is 3.00 m
above the ground.
in progress 0
6 days 2021-07-22T16:16:27+00:00 2 Answers 1 views 0
Explanation
The energy that the ball had instantaneously prior to being dropped was Potential Energy (EP) only and gravitational potential energy is calculated as Ep = mgh = (2.80)(9.81)(4.50) = 123.606 Joules.
Because no outside energy was added and no work was done, the energy instantaneously prior to being dropped is equal to the energy when the mass is 3 m above the ground only at that point, the energy consists of two parts, gravitational Potential Energy (EP) and Kinetic Energy (EK). So, considering the point just prior to the drop as point 1 and the point 3 m above the ground as point 2, we can say:
EP1 = EP2 + EK2
so EK2 = EP1 – EP2
and EP2 = (2.80)(9.81)(3.00) = 82.404
So EK2 = 123.606 – 82.404 = 41.2 Joules | 324 | 1,047 | {"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-2021-31 | latest | en | 0.945253 |
http://math.stackexchange.com/questions/58277/decompose-rigid-motion-affine-transform-into-parts | 1,469,322,709,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823805.20/warc/CC-MAIN-20160723071023-00066-ip-10-185-27-174.ec2.internal.warc.gz | 161,234,204 | 17,846 | # Decompose rigid motion affine transform into parts
I have an affine transform from $R^3$ to $R^3.$ It is described as Rotation about Z axis, rotation about X axis, a translation, rotation about Z axis, and lastly a scaling (same in all 3 dimensions).
It is therefore a similarity. Now, I would like to represent this affine transform as the following composition instead: Translation, rotation about X, rotation about Y, rotation about Z, and lastly scale.
Thus, given a similarity, how do I find the 3 angles, the translation, and the scale? I know this representation is not unique, but any one should do.
If it helps, I have access to all parameters in the first representation.
-
The given similarity is of the form $$S:\quad x\mapsto Ax+b\ ,$$ where the matrix $A$ and the vector $b$ are known to us. The scaling factor $\rho>0$ (if not given in advance) and the rotational part $T$ of $S$ can be read off from the matrix $A$: If $a_{\cdot1}$ is the first column of $A$ then $\rho=|a_{\cdot1}|$. The matrix $$T:={1\over\rho} A$$ is orthogonal and describes the rotational part of $S$. Now we should represent $T$ as a product of three matrices in the following way: $$T=\left[\matrix{c_3&-s_3&\cr s_3&c_3&\cr&&1\cr}\right]\left[\matrix{c_2&&s_2\cr &1&\cr -s_2&&c_2\cr}\right]\left[\matrix{1&&\cr &c_1&-s_1\cr&s_1&c_1\cr}\right]=:P\ .$$ Here $c_i=\cos\phi_i$, $s_i=\sin\phi_i$. If you compute the product $P$ of the three rotation matrices you will see that it is not a hopeless task to find the $c_i$ and $s_i$; e.g., one has $p_{31}=-s_2$.
I assume now that you have found three admissible angles $\phi_i$. It remains to find a translation vector $c$ such that $$Ax+b\ \equiv \rho\bigl(T(x+c)\bigr)\ .$$ As $A=\rho T\$ it is enough to take care of the point $x=0$. This leads to the equation $b=\rho\ Tc$ which can easily be solved for $c\>$: Since $T$ is an orthogonal matrix one gets $$c\ = {1\over\rho}\ T'b\ .$$ | 591 | 1,929 | {"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} | 3.796875 | 4 | CC-MAIN-2016-30 | latest | en | 0.859962 |
https://finance.yahoo.com/news/17-41-time-buy-energy-212709049.html | 1,544,423,894,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823318.33/warc/CC-MAIN-20181210055518-20181210081018-00244.warc.gz | 629,201,908 | 102,654 | U.S. Markets open in 7 hrs 52 mins
# At \$17.41, Is It Time To Buy Energy Transfer Equity LP (ETE)?
Energy Transfer Equity LP (NYSE:ETE) is trading with a trailing P/E of 16x, which is lower than the industry average of 20.6x. While this makes ETE appear like a great stock to buy, you might change your mind after I explain the assumptions behind the P/E ratio. In this article, I will explain what the P/E ratio is as well as what you should look out for when using it. Check out our latest analysis for Energy Transfer Equity
### What you need to know about the P/E ratio
P/E is often used for relative valuation since earnings power is a chief driver of investment value. It compares a stock’s price per share to the stock’s earnings per share. A more intuitive way of understanding the P/E ratio is to think of it as how much investors are paying for each dollar of the company’s earnings.
P/E Calculation for ETE
Price-Earnings Ratio = Price per share ÷ Earnings per share
ETE Price-Earnings Ratio = 17.41 ÷ 1.085 = 16x
The P/E ratio itself doesn’t tell you a lot; however, it becomes very insightful when you compare it with other similar companies. We want to compare the stock’s P/E ratio to the average of companies that have similar characteristics as ETE, such as size and country of operation. One way of gathering a peer group is to use firms in the same industry, which is what I’ll do. ETE’s P/E of 16x is lower than its industry peers (20.6x), which implies that each dollar of ETE’s earnings is being undervalued by investors. Therefore, according to this analysis, ETE is an under-priced stock.
### Assumptions to watch out for
Before you jump to the conclusion that ETE is the perfect buying opportunity, it is important to realise that our conclusion rests on two assertions. Firstly, our peer group contains companies that are similar to ETE. If this isn’t the case, the difference in P/E could be due to other factors. For example, if you compared higher growth firms with ETE, then its P/E would naturally be lower since investors would reward its peers’ higher growth with a higher price. The second assumption that must hold true is that the stocks we are comparing ETE to are fairly valued by the market. If this does not hold, there is a possibility that ETE’s P/E is lower because our peer group is overvalued by the market.
### What this means for you:
Are you a shareholder? You may have already conducted fundamental analysis on the stock as a shareholder, so its current undervaluation could signal a good buying opportunity to increase your exposure to ETE. Now that you understand the ins and outs of the PE metric, you should know to bear in mind its limitations before you make an investment decision.
Are you a potential investor? If you are considering investing in ETE, looking at the PE ratio on its own is not enough to make a well-informed decision. You will benefit from looking at additional analysis and considering its intrinsic valuation along with other relative valuation metrics like PEG and EV/Sales.
PE is one aspect of your portfolio construction to consider when holding or entering into a stock. But it is certainly not the only factor. Take a look at our most recent infographic report on Energy Transfer Equity for a more in-depth analysis of the stock to help you make a well-informed investment decision. Since we know a limitation of PE is it doesn't properly account for growth, you can use our free platform to see my list of stocks with a high growth potential and see if their PE is still reasonable.
To help readers see pass the short term volatility of the financial market, we aim to bring you a long-term focused research analysis purely driven by fundamental data. Note that our analysis does not factor in the latest price sensitive company announcements.
The author is an independent contributor and at the time of publication had no position in the stocks mentioned. | 875 | 3,979 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-51 | latest | en | 0.959072 |
http://nrich.maths.org/public/leg.php?code=31&cl=1&cldcmpid=229 | 1,371,719,361,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368711005985/warc/CC-MAIN-20130516133005-00001-ip-10-60-113-184.ec2.internal.warc.gz | 183,930,866 | 10,152 | # Search by Topic
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https://www.convertunits.com/from/kilogram+calorie/to/zeptojoule | 1,627,564,936,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153857.70/warc/CC-MAIN-20210729105515-20210729135515-00372.warc.gz | 689,601,745 | 13,262 | ## ››Convert kilogram calorie to zeptojoule
kilogram calorie zeptojoule
## ››More information from the unit converter
How many kilogram calorie in 1 zeptojoule? The answer is 2.3890295761862E-25.
We assume you are converting between kilogram calorie and zeptojoule.
You can view more details on each measurement unit:
kilogram calorie or zeptojoule
The SI derived unit for energy is the joule.
1 joule is equal to 0.00023890295761862 kilogram calorie, or 1.0E+21 zeptojoule.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between kilogram calories and zeptojoules.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of kilogram calorie to zeptojoule
1 kilogram calorie to zeptojoule = 4.1858E+24 zeptojoule
2 kilogram calorie to zeptojoule = 8.3716E+24 zeptojoule
3 kilogram calorie to zeptojoule = 1.25574E+25 zeptojoule
4 kilogram calorie to zeptojoule = 1.67432E+25 zeptojoule
5 kilogram calorie to zeptojoule = 2.0929E+25 zeptojoule
6 kilogram calorie to zeptojoule = 2.51148E+25 zeptojoule
7 kilogram calorie to zeptojoule = 2.93006E+25 zeptojoule
8 kilogram calorie to zeptojoule = 3.34864E+25 zeptojoule
9 kilogram calorie to zeptojoule = 3.76722E+25 zeptojoule
10 kilogram calorie to zeptojoule = 4.1858E+25 zeptojoule
## ››Want other units?
You can do the reverse unit conversion from zeptojoule to kilogram calorie, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Kilogram calorie
The large calorie or kilogram calorie approximates the energy needed to increase the temperature of 1 kg of water by 1 °C. This is about 4.185 kJ, and exactly 1000 small calories.
## ››Definition: Zeptojoule
The SI prefix "zepto" represents a factor of 10-21, or in exponential notation, 1E-21.
So 1 zeptojoule = 10-21 joules.
The definition of a joule is as follows:
The joule (symbol J, also called newton meter, watt second, or coulomb volt) is the SI unit of energy and work. The unit is pronounced to rhyme with "tool", and is named in honor of the physicist James Prescott Joule (1818-1889).
## ››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! | 792 | 2,632 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2021-31 | latest | en | 0.434064 |
https://mathoverflow.net/questions/241542/cyclic-modules-over-local-rings | 1,611,850,934,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704847953.98/warc/CC-MAIN-20210128134124-20210128164124-00191.warc.gz | 412,565,050 | 29,443 | # Cyclic modules over local rings
Let $R$ be a local ring. If two cyclic right $R$-modules are epimorphic images of each other, are these modules necessarily isomorphic?
• If $R$ is implicitly assumed commutative, the tag ac.commutative-algebra should be used. If not, perhaps it could be specified too. – YCor Jun 6 '16 at 13:35
• @YCor: The wording "right R-module" usually indicates that one is in the non-commutative setting. – Todd Leason Jun 6 '16 at 19:07
• I think this is an interesting and nontrivial question. It would probably get a better reception if you added some context/motivation (why local?, result essentially clear when $R$ is commutative, for what rings do you have a positive answer, ...) – Frieder Ladisch Jun 7 '16 at 13:43
I assume that $R$ is commutative. Let $M=R/I$ and $N=R/J$ be cyclic and let $\phi:M\to N$ and $\psi:N\to M$ be surjective. Then $\tau:=\psi\circ\phi$ is a surjective endomorphism of $R/I$. In particular, $a:=\tau(1)$ is a generator of $R/I$, i.e., $Ra+I=R$. Thus, there is $b\in R$ with $ba\in 1+I$. Since $\tau$ is multiplication by $a$, multiplication by $b$ is a (left) inverse of $\tau$. Thus, $\tau$ and therefore $\phi$ is injective.
Edit: Here is a proof for non-commutative rings which works under the additional hypothesis that $R$ is left Noetherian. So let $I,J,\phi,\psi,\tau,a,b$ as above. Then $I,J$ are only left ideals. Since $ba\in 1+I$ and $R$ is local, $ba$ is invertible. Replace $b$ by $(ba)^{-1}b$. Then $ba=1$. Since $ab$ is a non-zero idempotent and $R$ is local, we have $ab=1$. Thus $a$ is a unit of $R$. Since $x\mapsto xa$ induces an endomorphism of $R/I$ we have $Ia\subseteq I$. I claim $Ia=I$. Otherwise $$I\subsetneq Ia^{-1}\subsetneq Ia^{-2}\subseteq\ldots$$ would violate left Noetherianity. The claim implies that also $Ia^{-1}\subseteq I$. Thus, $x\mapsto xa^{-1}$ yields an inverse of $\tau$.
Edit 2: I think, I now have a counterexample. From the arguments above it suffices to construct a local ring $R$ and two left ideals $I,J$ such that there are units $a,b$ with $Ia\subseteq J$ and $Jb\subseteq I$ but there is no unit $c$ with $Ic=J$. Then $R/I$ and $R/J$ are epimorphic images of each other without being isomorphic.
For this, let $k$ be any commutative field, $t,u,v$ transcendental elements and $R:=k(t)\oplus k(u,v)$ with multiplication $$(f_1(t)+g_1(u,v))(f_2(t)+g_2(u,v))=f_1(t)f_2(t)+[f_1(u)g_2(u,v)+g_1(u,v)f_2(v^2)].$$ In other words, $R$ is the ring of matrices $$\left(\matrix{f(u)&g(u,v)\\0&f(v^2)}\right)$$ It is clearly local with maximal ideal $\{f=0\}$. Now put $$I:=k(u)[v]\text{ and }J:=k(u)[v]\cdot v.$$ Then $I\cdot t=k(u)[v]\cdot v^2\subseteq J$ and $J\subseteq I$ but there is no $f(t)$ with $I\cdot f(t)=J$.
• Thank you. But under commutative condition over the ring $R$, it will be clear and then we do not need local condition. But my question is in non commutative case.Thanks – Najmeh Dehghani Jun 7 '16 at 7:58
• Do you know of examples of a local ring $R$ with a right (or left) ideal $I \leq R$ and a unit $u\in R^*$ such that $uI < I$ ($Iu < I$) (strict containment)? I would guess that this is impossible, but I can't show it. – Frieder Ladisch Jun 7 '16 at 13:39
• Your proof in the Edit gives more precise results that could be helpful in the local case. But for the record let me note: Every surjective endomorphism of a left noetherian module over any ring (I think not even an identity is needed) is an isomorphism. – Todd Leason Jun 8 '16 at 10:33
• @Todd Leason: Thanks, I was not aware of that. After searching I found that modules with that property are called Hopfian and your statement is a very popular homework assignment. I also constructed a counterexample to the original problem. See edit 2. – Friedrich Knop Jun 8 '16 at 18:04
• Dear all, Thank you so much for your comments on my question. In fact under right Noetherian condition over the ring R, the result will be clear also. To see this, let R be a right Noetherian ring. Then every finitely generated module is Noetherian. Note that every Noetherian module is Hopfian (An R-module M is called Hopfian if every epimorphism on M is an isomorphism). So this imply that over a right Noetherian ring R, two finitely generated R-modules which are epi-morphic images of each others are isomorphic. – Najmeh Dehghani Jun 9 '16 at 10:10 | 1,358 | 4,335 | {"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} | 3.359375 | 3 | CC-MAIN-2021-04 | latest | en | 0.850352 |
https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_16&direction=prev&oldid=54045 | 1,638,715,348,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363189.92/warc/CC-MAIN-20211205130619-20211205160619-00503.warc.gz | 183,048,391 | 10,577 | 2006 AMC 12B Problems/Problem 16
Problem
Regular hexagon $ABCDEF$ has vertices $A$ and $C$ at $(0,0)$ and $(7,1)$, respectively. What is its area?
$\mathrm{(A)}\ 20\sqrt {3} \qquad \mathrm{(B)}\ 22\sqrt {3} \qquad \mathrm{(C)}\ 25\sqrt {3} \qquad \mathrm{(D)}\ 27\sqrt {3} \qquad \mathrm{(E)}\ 50$
Solution
To find the area of the regular hexagon, we only need to calculate the side length.
Drawing in points $A$, $B$, and $C$, and connecting $A$ and $C$ with an auxiliary line, we see two 30-60-90 triangles are formed.
Points $A$ and $C$ are a distance of $\sqrt{7^2+1^2} = \sqrt{50} = 5\sqrt{2}$ apart. Half of this distance is the length of the longer leg of the right triangles. Therefore, the side length of the hexagon is $\frac{5\sqrt{2}}{2}\cdot\frac{1}{\sqrt{3}}\cdot2 = \frac{5\sqrt{6}}{3}$.
The apothem is thus $\frac{1}{2}\cdot\frac{5\sqrt{6}}{3}\cdot\sqrt{3} = \frac{5\sqrt{2}}{2}$, yielding an area of $\frac{1}{2}\cdot10\sqrt{6}\cdot\frac{5\sqrt{2}}{2}=25\sqrt{3}$.
2006 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions | 486 | 1,232 | {"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": 17, "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} | 4.5625 | 5 | CC-MAIN-2021-49 | latest | en | 0.591856 |
http://www.java-forums.org/new-java/10321-need-help-using-algorithm-statements-print.html | 1,430,929,175,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1430458866468.44/warc/CC-MAIN-20150501054106-00031-ip-10-235-10-82.ec2.internal.warc.gz | 463,896,293 | 3,184 | # Need help - using algorithm statements
• 07-22-2008, 03:54 PM
javanewbie
Need help - using algorithm statements
Good day everyone.
I saw a Java program, a PalindromeTest program, with the following codes below:
Code:
```Public Class Palindrometest { Public Static String Testpalindrome(string Palindrome) { If (Ispalindrome(palindrome) == True) { Return "\"" + Palindrome + "\" Is A Palindrome"; } Else { Return "\"" + Palindrome + "\" Is Not A Palindrome"; } } [COLOR="Blue"] Public Static Boolean Ispalindrome(string Palindrome) { Palindrome = Palindrome.tolowercase(); If (Palindrome.length() <= 1) Return True; Char Firstchar = Palindrome.charat(0); Char Lastchar = Palindrome.charat(palindrome.length()-1); If (Character.isletter(firstchar) && Character.isletter(lastchar)) { If (Firstchar != Lastchar) { Return False; } Else { String Substring = Palindrome.substring(1,palindrome.length()-1); Return Ispalindrome(substring); } } Else If (!Character.isletter(firstchar)) { String Substring = Palindrome.substring(1); Return Ispalindrome(substring); } Else { String Substring = Palindrome.substring(0,palindrome.length()-1); Return Ispalindrome(substring); } }[/COLOR] Public Static Void Main(string[] Args) { System.out.println(testpalindrome("i Enjoy Tacos.")); System.out.println(testpalindrome("damn! I, Agassi, Miss Again! Mad!")); System.out.println(testpalindrome("yo, Banana Boy!")); System.out.println(testpalindrome("a Dog! A Panic In A Pagoda!")); Food.")); } }```
i'm just curious what would possible the algorithm behind this codes, especially on how the programmer uses the Character.isLetter, Substring, etc.
i just want to be familiarize with the following codes given. sorry for being a newbie here in java. that's why im asking if you could help me find out the algorithm used in the program.
thanks and have a nice day.
• 07-22-2008, 04:11 PM
Norm
A first attempt to understand what a program does would be to read the API doc for each method that you're not familiar with.
Then use a paper and pencil to play computer and "execute" the code line by line to see what happens. That's pretty much how an experienced programmer would do. Look at each line and do what the computer would do for each method. We can probably do most of it in our heads, but as a beginner, writing it down, drawing pointers, etc would help you visualize what the code is doing.
• 07-23-2008, 11:20 AM
Eranga
In advanced level of a programmer they used of dry run on the application. That's the word they do. Before run any application they simply execute the line-by-line of the code by his/her head and see the result. Then run the application to second it. :) | 701 | 3,052 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2015-18 | longest | en | 0.156723 |
https://socratic.org/questions/how-do-you-simplify-3x-2-3x-6-x-2-4 | 1,585,408,982,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370491998.11/warc/CC-MAIN-20200328134227-20200328164227-00212.warc.gz | 729,634,763 | 6,196 | # How do you simplify (3x^2-3x-6)/(x^2-4)?
May 12, 2018
$\frac{3 \left(x + 1\right)}{x + 2}$
#### Explanation:
$\text{factorise the numerator/denominator and }$
$\text{cancel any common factors}$
$\textcolor{m a \ge n t a}{\text{factor numerator}}$
$\text{take out a "color(blue)"common factor } 3$
$\Rightarrow 3 \left({x}^{2} - x - 2\right)$
$\text{the factors of - 2 which sum to - 1 are - 2 and + 1}$
$= 3 \left(x - 2\right) \left(x + 1\right)$
$\textcolor{m a \ge n t a}{\text{factor denominator}}$
${x}^{2} - 4 \text{ is a "color(blue)"difference of squares}$
•color(white)(x)a^2-b^2=(a-b)(a+b)
$\Rightarrow {x}^{2} - 4 = {x}^{2} - {2}^{2} = \left(x - 2\right) \left(x + 2\right)$
$\Rightarrow \frac{3 {x}^{2} - 3 x - 6}{{x}^{2} - 4}$
$= \frac{3 \cancel{\left(x - 2\right)} \left(x + 1\right)}{\cancel{\left(x - 2\right)} \left(x + 2\right)}$
$= \frac{3 \left(x + 1\right)}{x + 2}$
$\text{with restriction } x \ne - 2$ | 401 | 941 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 16, "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} | 4.71875 | 5 | CC-MAIN-2020-16 | latest | en | 0.471054 |
https://discusstest.codechef.com/t/nplq18c-wrong-answer/20620 | 1,627,732,442,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154089.6/warc/CC-MAIN-20210731105716-20210731135716-00213.warc.gz | 228,772,021 | 4,351 | Can anyone Help me with my solution for this question ? I went some correct submissions and checked output of correct submission and mine and both were same. So can anyone provide any failing test case for my solution ?
My solution : https://www.codechef.com/viewsolution/20331295
By going through your code, what I understood that is that you didn’t get the problem actually or maybe you failed to implement your idea correctly. Just for clarity, the problem asks for the next number greater than or equal to n, which is prime and lies in the range l to r. But what you have actually done in your code is that, you first found out the next prime number to n and then checked for something. No, you don’t need to find just the next prime to n, you need to find the next prime to n which lies in the range of l to r.
Since the given range of l to r is <= 1e5, you can use array to store all the prime numbers from l to r. Then you can use a dictionary to store the next prime numbers of the numbers in the range l to r, because the number between l and r are our concern. If n < l, then you simply need to print the next prime to l if it exists or print -1. If n > r or there is no prime number in after n which is < r, then simply print -1, otherwise you can find the prime number next to n within the given range of l to r from the dictionary in O(1) time, which is necessary for the large volume of queries.
You can find my solution if it help.
Hope this clears your doubt about the solution. Feel free to ask if you have any more doubts.
Actually what did is to find next prime greater than or equal to n and then store it in variable temp. Then I checked if temp lies in range l to r. If it does then I print temp else -1. This is just clarification. I found out what is wrong with my logic. Thanks for help.
// | 422 | 1,822 | {"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-2021-31 | latest | en | 0.951188 |
https://www.jiskha.com/search?query=Counting+and+probability | 1,542,365,034,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743007.0/warc/CC-MAIN-20181116091028-20181116113028-00352.warc.gz | 896,654,988 | 16,615 | # Counting and probability
10,602 results
1. ## Elementary Mathematics 2
Counting by ones, counting by tens, and counting by groups and singles: I was rather curious about how these methods of counting can be used to coordinate concepts and oral and written names for numbers.
asked by Tee on March 2, 2008
2. ## math
What are the similarities between organized counting and permutation? Explain using examples. I can only think of one similarity: they both use probability problems using counting P(event)= Number of favourable outcomes/Total number of possible outcomes.
asked by Jenna on October 15, 2016
3. ## statistics
12. The probability that the Dow Jones stock index will close above 18000 at the end of the year is an example of (a) classical probability. (b) subjective probability. (c) independent probability. (d) priory probability. 13. The probability calculated
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4. ## statistics
15. The probability that a recently offered stock by a technology company will double in value within the next three months is 90%. The conclusion regarding this probability estimate was reached based on the opinion of the experts in the technology field
asked by meri on June 30, 2015
5. ## computer programing
okay I need a little help here on next program this what it say for first one which I have done. Create a counting program that counts backward from 100 to 1 in increments of 10. #include < stdio.h > int main () { Int i; while ( I > = 1 ) { printf ( “ %
asked by Poohboi on February 5, 2016
6. ## Math
What is the difference between counting the spaces between whole numbers and counting the tick marks?
asked by Mary on October 17, 2012
7. ## Statistics
An electronics company is about to launch a new product. If the serial number for each piece produced has the following format: LLNNN where L stands for any letter in the English alphabet and N is a number from 0 to 9, please answer the following: a)What
asked by Ewuraadjua on May 16, 2013
8. ## algebra 1
When studying probability, a method used for counting the number of possible outcomes is a ______. a.tree diagram b.dependent event c.relative frequency d.simulation A?
asked by Lily on September 21, 2010
9. ## Pre Algebra
Please help! This is due tomorrow... I don't understand how "Counting Outcomes and Theoretical Probability" works! There's a question that says: You toss two coins. Find P (one head and one tail). Please help!
asked by Lily on May 20, 2008
10. ## government
why are voting machines used? a. to eliminate the election process b. to increase the number of persons needed to administer elections c. to minimize vote-counting errors d. to encourave manual vote counting i think its c.
asked by jere on December 26, 2007
what does it mean readability formulas? I don't understand the concept can someone explain it to me please. As I understand it, this is how they come up with these readability numbers: ~~By counting the number of paragraphs in a piece of writing. ~~By
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12. ## fundamental concepts
Sarah counts her handful of marbles one at a time into a bowl. Each time she puts a marble into the bowl she says the next number name in sequence. This is an example of A. perceptual subitizing. B. classification. C. rote counting. D. rational counting.
asked by Daniela on February 4, 2015
13. ## algebra
The sum of the swuares of the first n counting numbers: 1² + 2² + 3² + 4² + 5² + 6² + ... +n² is given by the formula S = n(n+1)(2n+1) ------------ 6 Find the sum of the squares of all the counting numbers between 15 and 35. - answer is 12445; how
asked by jadacess on February 9, 2008
14. ## Reiny/ MATH Help
THANK YOU VERY MUCH!!! I forgot to say thank you for this answer the other night. There are 900 of those counting numbers with no restriction. Let's find all those that DON'T have a 2 in them there are 8 possible in the left spot , (no 2 or 0) but 9
asked by TO: Reiny on February 2, 2012
15. ## Probability :(
Brenda’s Bike Booth, at the mall, allows people to order a bike in their choice of 4 colors (Atomic Tangerine, Banana, Café Noir, or Dollar Bill) and the options of with or without a basket; with or without a horn. Given all of these options, what’s
asked by Katie on June 30, 2016
16. ## Probability
Brenda’s Bike Booth, at the mall, allows people to order a bike in their choice of 4 colors (Atomic Tangerine, Banana, Café Noir, or Dollar Bill) and the options of with or without a basket; with or without a horn. Given all of these options, what’s
asked by Katie on June 30, 2016
17. ## Grammar (Writeacher)
Revise the sentences below in which "who" or "whom" is used incorrectly. Then, indicate how "who" or "whom" is used in each sentence. 1. To play Kick the Can, the group must make a decision about whom will be counting. 2. Everyone else must hide from the
asked by Anonymous on May 15, 2013
18. ## finite math
A probability experiment was conducted in which the sample space of the experiment is S={1,2,3,4,5,6,7,8,9,10,11,12}, event F= {6,7,}, and event G={9,10,11,12}. Assume that each outcome is equally likely. List the outcomes of F and G?. find P(F or G) by
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19. ## finite math
A probability experiment was conducted in which the sample space of the experiment is S={1,2,3,4,5,6,7,8,9,10,11,12}, event F= {3,6,7,}, and event G={9,10,12}. Assume that each outcome is equally likely. List the outcomes of F and G. find P(F or G) by
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20. ## finite math
A probability experiment was conducted in which the sample space of the experiment is S={1,2,3,4,5,6,7,8,9,10,11,12}, event F= {6,7,}, and event G={9,10,11,12}. Assume that each outcome is equally likely. List the outcomes of F and G?. find P(F or G) by
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21. ## maths
In a recent election for class president, Monika received 7 of the 10 votes and Alfred received 3 of the 10 votes that were cast by the class. When the machine was counting the votes, it malfunctioned and instead of giving the vote to the correct person,
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22. ## math
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23. ## statistics
Basic Concepts of Probability and Counting 1. License plates are made using 3 letters followed by 2 digits. How many different plates can be made if repetition of letters and digits is allowed? (26^3)(10^2) = (17576)(100) = 1,757,600
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24. ## Statistics
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25. ## Math
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26. ## com155
• Use at least five adverbs and five adjectives to write a brief review of a movie, sporting event, musical performance, or television show. • Bold each adverb (even the ones you are not counting as your original five, so that all adverbs are bolded in
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27. ## science(chem)
I have a question about distillation. Or heating in general. When something says heat for 1hr or 1/2 hr do you start counting when you turn on the hotplate? Or do you start counting when the mixture is boiling? I have a feeling that this issue is affecting
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28. ## Probability
Roll two fair dice. what is the probability that you get a 2 and a 5 without regard to which is on which die? What is the probability of at least one 2 or one 5? What is the probability of a sum of 7? "what is the probability that you get a 2 and a 5
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29. ## Counting and Probability
What is the difference between the sum of the first 400 even counting numbers and the sum of the first 400 odd counting numbers?
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30. ## algebra
true or false 1. fraction cant be written as decimal. 2. natural numbers are referred to as counting numbers.whole numbers consists of counting numbers.whole numbers consist of counting numbers as well as the number 0 3. intergers do not include negative
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31. ## statistics
The below table shows the probabilities generated by rolling one die 50 times and noting the up face. What is the probability of getting an odd up face? roll 1 Probability 0.22 roll2 probability 0.10 roll 3 probability 0.18 roll 4 probability 0.12 roll 5
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32. ## MS.SUE DAMON REED PSYDAG PLZ HELP ON MATH PLZZZ!!!
1. The table shows the results of spinning a four-colored spinner 50 times. Find the experimental probability and express it as a decimal. P(not red) = ? (1 point) a 0.6 b 0.4 c 0.2 d 0.3 3. The table below shows the results of flipping two coins. How does
asked by Alexia Marsel on May 20, 2016
33. ## Managerial accounting
[Figure 5-1]. - The receiving department of Owen has three activities: unloading, counting goods, and inspecting. Unloading requires a forklift that is leased for \$15,000 per year. The forklift is used only for unloading. The fuel for the forklift is
asked by Anonymous on January 8, 2013
34. ## Algebra 1
Can you please check my work and help me with the parts that I don't understand? For each event, choose the most appropriate term and solve the problem. 1. In a sweepstakes with nine hundred entries, the first winner selected receives the grand prize, the
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35. ## math, help
I need help or hint on the setup: what is the probability that at least 2 students in a class of 36 have the same birthday? This is 1 minus the probability that all students have different birthdays. Suppose that the birthday of a student is completely
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36. ## Probability
How do I do this probability math problem,--- In 2000, the population of a certain country was about 195 million. The overall birth rate was 18.8 births per 100. The overall death rate was 11.8 deaths per 1000. Based on births & deaths alone(not counting
asked by AJC on July 23, 2010
37. ## Math - Statistics
I roll 12 4-sided dice. What is the probability that I roll at least one 2? What is the probability that I roll exactly three 2s? So i take it, on any give roll, the probability of rolling a 2 is .25 (25%); the probability of rolling something else must be
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38. ## math
a liscence plate for a car must consist of a letter followed by two numbers. Each letter mus be an R, U or N. Each number must be a 2 or 6. Repetition of digits is permitted. Use the counting principle to determine the number of points in the sample.
asked by Joyce on June 12, 2011
39. ## Cards Probability
As shown above, a classic deck of cards is made up of 52 cards, 26 are black, 26 are red. Each color is split into two suits of 13 cards each (clubs and spades are black and hearts and diamonds are red). Each suit is split into 13 individual cards (Ace,
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40. ## math probability
this is a probability question... Suppose you are asked to choose a whole number between 1 and 13 inclusive. (a) what is the probability that it is odd?...7/13 (b) What is the probability that it is even?...6/13 (c) what is the probability that it is a
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41. ## Algerbra
The athletic department just received a shipment of footballs and basketballs. You don’t know how many there are, and don’t feel like counting. Here’s what you do know: each football costs 12 dollars, each basketball costs 15 dollars, and the credit
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42. ## Math
A number cube is rolled 100 times. The results are shown in the table below. Outcome: 1, 2, 3, 4, 5, 6. Number of times rolled: 22, 18, 9, 11, 19, 21. Find the experimental probability and express it as a percent. P(even) = ? A. 50% B. 40% C. 29%* How does
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43. ## Math
A number cube is rolled 100 times. The results are shown in the table below. Outcome: 1, 2, 3, 4, 5, 6. Number of times rolled: 22, 18, 9, 11, 19, 21. Find the experimental probability and express it as a percent. P(even) = ? A. 50% B. 40% C. 29%* How does
asked by Xx on May 17, 2017
44. ## Probability-Cards
As shown above, a classic deck of cards is made up of 52 cards, 26 are black, 26 are red. Each color is split into two suits of 13 cards each (clubs and spades are black and hearts and diamonds are red). Each suit is split into 13 individual cards (Ace,
asked by peter on October 8, 2016
45. ## Cards
As shown above, a classic deck of cards is made up of 52 cards, 26 are black, 26 are red. Each color is split into two suits of 13 cards each (clubs and spades are black and hearts and diamonds are red). Each suit is split into 13 individual cards (Ace,
asked by Anonymous on October 8, 2016
46. ## Economics/Math
1. Assume that q and z are two random variables that are perfectly positively correlated. q takes the value of 20 with probability 0.5 and the value of zero with probability 0.5, while z takes the value of 10 with probability 0.5 and the value of zero with
asked by Susan on December 5, 2009
47. ## statistics
26. The law of permutations, combinations, and filling slots are the counting rules used to (a) determine the total number of possible outcomes (b) determine all possible combinations of “n” objects from a total of N objects (c) determine the odds of
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48. ## math
In a game, a player tosses a coin 4 times. If the player gets 3 or 4 heads, he/she wins. What is the theoretical probability of winning this game? I just need to know the outcomes. I don't know how to get them. Please and Thank you. Coin 1: 50% Heads Coin
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49. ## Finite Math
An urn contains sixteen red balls and seventeen white balls. A sample of three balls is selected at random and the number of red balls observed. Determine the probability distribution for this experiment: Number of Red Balls/Probability 0...Probability=?
asked by Amanda on April 5, 2012
50. ## Statistics
A total of 16 mice are sent down a maze, one by one. From previous experience, it is believed that the probability a mouse turns right is .38 a) What is the probability that exactly 8 of these 16 mice turn right? b) What is the probability that 8 or fewer
51. ## counting up method
how do you do the counting up method
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52. ## math
Even counting numbers less than 18
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53. ## algebra
what are the even counting numbers less than 18
asked by Anonymous on October 9, 2013
54. ## math
what are the even counting numbers less than 10
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55. ## math
A= (x/7
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56. ## math
What is counting up method
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57. ## Math
In an experiment, the probability of the event E is known to be .4. Also the probability of the event F is .8,and the probability of E∪F is 1. i) Compute the probability of these events: P(E∩F)= P(E-F)= ii) Suppose the experiment is run twice in
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58. ## statistics
24. Suppose there are five traffic lights that you need to pass while driving from your work to school. The probabilities that you will stop for these red lights are: 0 red light with probability 0.05, 1 red light with probability 0.45, 2 red lights with
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59. ## PreCalculus
How do you find the probability of three events happening at the same time? Please walk me through the steps. If you are in ship and get hit by lightning the probability is 1/100000. And the probability of falling down into the ocean and getting eaten by a
asked by Emma on July 30, 2014
60. ## Math
A mini license plate for a toy car must consist of two numbers followed by a letter. Each number must be a 5, 7 or 9. Each letter must be a C, A or N. Repetition of digits is NOT permitted. Use the counting principle to determine the number of points in
asked by Jen on July 29, 2011
61. ## Statistics
There is a 0.0416 probability that a best-of-seven contest will last four games, a 0.0804 probability that it will last five games, a 0.2132 probability that it will last six games, and a 0.6648 probability that it will last seven games. Find
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62. ## math
what is the difference between doubles plus one and counting on?
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63. ## factorial help!!!
i just need help with factorials I assume you know what a factorial is: 1! = 1 2! = 1x2 = 2 3! = 1x2x3 = 6 N! = 1x2x3x...x(N-1)xN They occur often in probability theory. Please ask a more specific question about factorials. I have no idea what else to tell
asked by tiffanie robinson on August 31, 2005
64. ## Probaility
The rates of on time flights for commercial jets are continuously tracked by the US Department of Transportation. Recently, Southwest Air had the best rate of 80% of its flight time on time. A test is conducted by randomly selecting 15 southwest flights
asked by Mary on December 5, 2006
65. ## MATH-SO LOST :(
A mini license plate for a toy car must consist of a number followed by two letters. Each letter must be a C, A or R. Each number must be a 3 or 7. Repetition of letters is permitted. Use the counting principle to determine the number of points in the
asked by Jen on July 27, 2011
66. ## math
how do you explain this questionA mini license plate for a toy car must consist of a letter followed by two numbers. Each letter must be a T, O or Y. Each number must be a 4 or 6. Repetition of digits is permitted. Use the counting principle to determine
asked by kozy on June 6, 2011
67. ## early childhood
1.) the best example of an embedded learning experience is a.) counting from one to 20 just to practice b.) saying the alphabet while a teacher points to the letters c.) counting out napkins for each child at snack time. d.) tracing letters on a worksheet.
asked by sahar mostafa on March 19, 2013
68. ## math
list the set of: counting numbers from 20 to 30
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69. ## math
i've got a problem the question is what is another name for counting numbers
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70. ## math
in the question 4x-(3-x) =7(x-3) =10 why r u not counting 4x3 as -12? israel
asked by israel on June 10, 2011
71. ## Maths
How can you use skip counting to multiply 5 x 6? Explain.
asked by Parimala on August 2, 2013
72. ## Maths
list the set of: counting numbers from 20 to 30
asked by centa on October 23, 2016
73. ## chemistry
How to solve this problem in counting atoms 2 pb(CN)2
asked by bob on December 7, 2015
74. ## maths
Show the jumps of five in counting from one to hundreds.
asked by Anonymous on June 6, 2014
75. ## Statistics
Event A occurs with probability 0.1 and event B with probability 0.5 A) What is the maximum probability that the intersection of A and B can have? B) What is the minimum probability that the intersection of A and B can have? C) If it is know that P(A
asked by Rebekah on September 6, 2014
76. ## statistics
Suppose that each unit of a system is up with probability 2/3 and down with probability 1/3. Different units are independent. For each one of the systems shown below, calculate the probability that the whole system is up (that is, that there exists a path
asked by sam on February 18, 2014
77. ## biology
One rule of probability can be expressed as the following: The probability of two independent events occurring simultaneously is the product of the probability of their occurring separately. If, for example, you had a pair of dice and rolled each die one
asked by liam on November 11, 2009
78. ## biology
One rule of probability can be expressed as the following: The probability of two independent events occurring simultaneously is the product of the probability of their occurring separately. If, for example, you had a pair of dice and rolled each die one
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79. ## Statistics
Achusband and wife with brown eyes who have 0.75 probability of having children with brown eyes, 0.125 probability of having children with blue eyes, and 0.125 probability of having children with green eyes. (a) What is the probability that their first
asked by Neveah on February 6, 2016
80. ## Economics/Statistics
1. Assume that q and z are two random variables that are perfectly positively correlated. q takes the value of 20 with probability 0.5 and the value of zero with probability 0.5, while z takes the value of 10 with probability 0.5 and the value of zero with
asked by Susan on December 6, 2009
81. ## Genetics
Phenylketonuria, a Metabolic disease in humans, is caused by a recessive allele, k. If two heterozygotes marry and plan a family of six children: a)what is the probability that all will be unaffected? b)what is the probability that three will be unaffected
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82. ## math
specify the sets b roster {odd counting numbers less than 11}
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83. ## 3rd grade math II
I need to write a sentence for the following Reasoning: How can counting by 10's help you multiply by 10's?
asked by Alexis on December 15, 2010
84. ## 3rd grade math
The question reads: Reasoning: How can counting by 10's help you multiply by 10's?
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85. ## math
How many counting numbers between 100 and 1000 have at least one 2 as a digit?
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86. ## log graphs
when you're graphing a log graph what are is the counting pattern?
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87. ## math
Able starts off counting at 13, and counts by 7. What is the 11th number that Able will say?
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88. ## Probability -- Help?
Sampling one at a time with replacement from a bag with 8 blue, 7 red, and 5 green. What is the probability of getting at least two red in 3 draws? The book says the probability is 0.2818 but doesn't explain why. Any help???? Unfortunately, the probability
asked by Cheri on December 1, 2006
suppose that al knows that the probability of an investigation is 0.70 how low must the probability of water's completing its research on time be so that the probability of litres being awarded the contract is at least 0.65?
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90. ## Math
A spinner has numbers 1-16 on it. Probability of 6? 1/16 Probability not 6 15/16 Probability even number 8/16 Probability multiple of 3 5/16? Probability not a multiple of 3 11/16?
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91. ## Math
You flip a coin 20 times and get tails 15 times you flip the coin 80 more times.what do you expect to happen to the experimental probability of getting tails as you increase the number of trials a. The experimental probability will get closer to 50% b. The
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92. ## 12th grade
one type of probability used everyday is: a)experienced probability b)scientific probability c)frequency of occurence d)none of the above
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93. ## math
Draw a number line counting by 10s for 240
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94. ## help me in math!
is -5 a real #? rational #? irrational #? whole #? integer? counting (natural) #?
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95. ## Math
Use the counting principle to determine the number of different arrangements of bicycles that are possible.
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96. ## math
Allison says that every interger is a counting number. is she correct and explain Please help, thank u
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97. ## 3rd grade math
what is the next number in the sequence 1,3,9,27,______ I'm confused on the skip counting they use.
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98. ## s.s
What was the debate between North and South over counting slave populations?
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99. ## Math
What is the smallest counting number n so that 375n is a perfect square?
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100. ## data managment
what are the similarities and differences between organized counting and permutations, with examples?
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# Rationalise the denominator of (1)/((8+5sqrt(2)).
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Text Solution
((8-5sqrt(2)))/(12) ((8-5sqrt(2)))/(14) ((8+5sqrt(2)))/(14) ((8-5sqrt(2)))/(34)
B
Solution :
We have <br> (1)/((8+sqrt(2)))= (1)/((8+5sqrt(2))) xx ((8-5sqrt(2)))/((8-5sqrt(2))) =((8-5sqrt(2)))/((8)^(2) - (5sqrt(2))^(2)) <br> = ((8-2sqrt(2)))/(64-50) = ((8-5sqrt(2)))/(14)
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2:02 | 532 | 1,189 | {"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} | 3.8125 | 4 | CC-MAIN-2021-39 | latest | en | 0.451573 |
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# (Solved): CoordinatePlaneTrig: Problem 2 (1 point) Find the value of the six trigonometric functions of , w ...
CoordinatePlaneTrig: Problem 2 point) Find the value of the six trigonometric functions of , where is the angle formed by the positive -axis and the line segment fiom to Find the value of the six trigonometric functions of , where is the angle formed by the positive -axis and the line segment fiom to
We have an Answer from Expert | 141 | 584 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-23 | latest | en | 0.795805 |
http://mathoverflow.net/questions/55162/how-can-i-write-down-polynomial-relations-that-define-when-a-polynomial-is-a-squ/55175 | 1,398,073,691,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609539705.42/warc/CC-MAIN-20140416005219-00583-ip-10-147-4-33.ec2.internal.warc.gz | 142,669,661 | 19,148 | # How can I write down polynomial relations that define when a polynomial is a square?
It's easy to tell when a polynomial is squarefree (or not): that's just the question of the vanishing of the discriminant, which can be dealt with as the resultant of $f$ and $f'$. However, given a polynomial of degree $2n$ $f$, when is it of the form $g^2$ for $g$ a polynomial of degree $n$?
I've been trying to work out the relations on the coefficients that will guarantee this for a specific degree ($n=6$ is my case) but whenever I take the obvious equations in the coefficients of $g$ and of $f$ and try to use Groebner bases to eliminate the coefficients of $g$, I run out of memory and my software crashes. Is there a way to understand the locus of polynomials which are squares concretely without having to do a (seemingly unrealistically) big computation? Or perhaps a clever trick that will give these polynomial identities in a more computable way?
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Probably not the answer you are looking for, but for a square polynomial $gcd(f,f')$ has all roots of odd multiplicity, and if $2k_i-1$ is the multiplicity of the $x_i$th root then $\sum_i k_i=n$. The converse is also true, the condition $\sum_i k_i=n$ guarantees that $f$ doesn't have any simple roots. – Nick S Feb 11 '11 at 20:12
Not directly related to your question: what did you use as a Groebner solver? Last I checked (admittedly a while ago), Mathematica or Maple were vastly outperformed by more specialized software. This might be an interesting avenue to explore, regardless of a more theoretical solution. – Thierry Zell Feb 11 '11 at 22:54
Trivial observation: You must require the field to be algebraically closed, or the polynomial to be monic or something stronger. Else, $aX^2$ will reduce the undecidable is-$a$-a-square to your is-my-polynomial-a-square. – darij grinberg Feb 11 '11 at 23:06
Also, the answers require the field to have characteristic $0$. This seems to have a good reason: If the field has characteristic $2$, then you can reduce the question is-a-field-element-$a$-a-square to your is-my-polynomial-a-square (apply to the polynomial $X^2-a$). Of course, there is no polynomial equation that tells you when an element of a field of characteristic $2$ is a square. – darij grinberg Feb 11 '11 at 23:11
Note on darij's remark: Let $k$ have characteristic other than $2$. Let $f$ be a polynomial in $k[x]$. Then $f$ is square in $k[x]$ if and only if $f$ is square in $k^{\mathrm{alg}}[x]$ and the leading term of $f$ is square in $k$. So the issues about algebraic closure are comparatively mild. (This is evident from looking at Greg's explicit solution.) – David Speyer Feb 12 '11 at 17:34
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Say, for simplicity, you are working over $\mathbb{C}$ or in characteristic zero in general. Then you can guess one of the two values of $g(0)$ (say) and then compute the Taylor series of $\sqrt{f}$. The approach is similar to Hensel lifting: The equation for the first coefficient is non-linear; the equations for the others are all locally linear (so that you get explicit formulas for the coefficients of $g$ in terms of existing data).
I first misread Charles' question, but now that I have it right (I think), here is why I think that the above is still a solution. If you read the coefficients of a polynomial of degree $n$ as projective coordinates, then over $\mathbb{C}$ the set of squares of degree $2n$ is some projective variety $S$ in $\mathbb{C}P^{2n}$. Charles is interested in projective equations for this variety $S$.
For simplicity let's rescale the polynomial $f(x)$ so that $f(0) = 1$. (And I guess we're working the affine chart in which $f(0) \ne 0$ before the rescaling. It shouldn't change things much or at all.) Then you can assume that $g = \sqrt{f}$ also satisfies $g(0) = 1$, and you can make explicit expressions for its Taylor series. Then $g$ is a polynomial of degree $n$ if and only if its Taylor series vanishes in degree $n < k \le 2n$. I think that this gives you the desired equations.
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Maybe I'm not understanding properly, but wouldn't the conditions I want then be that for all $k>n$, the $k$th coefficient of $sqrt(f)$'s Taylor series vanishes? Doesn't that give me infinitely many conditions (a priori) and no obvious way to know which finite set of them suffice? – Charles Siegel Feb 11 '11 at 20:23
Of course once you reach $k=n$, you can square the result and see if it works. – Greg Kuperberg Feb 11 '11 at 20:26
Ok...perhaps you could be more explicit about what you're doing, because when I Taylor expand $\sqrt{f}$, I get that the degree of the coefficient of $x^i$ is $i$ in the coefficients of $f$ (well, with a $\sqrt{a_0}$ to some power in the denominator). I'm not sure what you're doing, or what the output should be, but I'm looking for a finite set of polynomials in the coefficients of $f$ whose vanishing determines that $f$ is a square. – Charles Siegel Feb 11 '11 at 20:38
It's true that I misinterpreted the problem, but I think that, at the expense of a $\sqrt{a_0}$, the solution still works. The coefficients of the Taylor series of $g$ vanish for $n < k \le 2n$ if and only if $g$ is a polynomial of degree $n$. – Greg Kuperberg Feb 11 '11 at 21:07
So unless I am misunderstanding the question, temporarily normalize so that the coefficient of $x^6$ in $f$ is 1. One is left with three degrees of freedom, coming from the quadradic, linear and constant terms of the degree 3 polynomial square root.
For concreteness, let $f(x) = x^6 + c_5x^5 + c_4x^4 + c_3x^3 + c_2x^2 + c_1x + c_0$
Then I work out that necessary relations on the coefficients are:
$c_2 = 2(\frac{1}{2}c_5)(\frac{1}{2}c_3-\frac{1}{4}c_4c_5+\frac{1}{16}c_5^3)+(\frac{1}{2}c_4-\frac{1}{8}c_5^2)^2$
$c_1 = 2(\frac{1}{2}c_4-\frac{1}{8}c_5^2)(\frac{1}{2}c_3 - \frac{1}{4}c_4c_5 + \frac{1}{16}c_5^3)$
$c_0 = (\frac{1}{2}c_3 - \frac{1}{4}c_4c_5 + \frac{1}{16}c_5^3)^2$
These are also sufficient since if they hold, then $f$ is the square of $x^3+(\frac{1}{2}c_5)x^2+(\frac{1}{2}c_4-\frac{1}{8}c_5^2)x + (\frac{1}{2}c_3-\frac{1}{4}c_4c_5+\frac{1}{16}c_5^3)$
Unless I did something wrong, it doesn't seem like these computations should be crashing the system. What are you using to run the Groebner calculations?
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This is equivalent to the solution that I describe, except of course working from the other end. The explicit calculation is nice; it shows quite clearly that it works. – Greg Kuperberg Feb 12 '11 at 2:33
@Greg: It is indeed essentially the same as what your answer says; I was interested in carrying through the solution concretely. – ARupinski Feb 12 '11 at 3:22
When I said "$n=6$", I meant the case where $\deg f=12$, which makes this much, much worse for a CAS to work out. – Charles Siegel Feb 12 '11 at 19:44
http://en.wikipedia.org/wiki/Square-free_polynomial gives a method for finding a square-free factorization of a polynomial (over characteristic zero field), ie $f=a_1\cdot a_2^2\cdots a_n^n$ where each $a_i$ is a square-free polynomial. Then $f$ is a perfect square iff $a_{2i+1}=1$ for each $i$.
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It seems to me that the OP almost contains the answer: the gcd of $f$ and $f^\prime$ (let's assume characteristic zero) contains all the irreducible factors of $f$ which appear with exponent greater than $1.$ This is surely enough to figure out if the polynomial is a square.
EDIT to answer the revised version of the question:
Write down $$\sum_{i=0}^n a_i x^i = (\sum_{j=0}^{n/2} b_j x^j)^2.$$ This gives a collection of $n+1$ quadratic equations in $3n/2 + 2$ variables. You now eliminate the $b_j$ to get the variety of perfect squares. Needless to say, this is not algorithmically very pleasant (the degree is going to be exponential in $n$), but you can use successive resultants or Grobner bases to do it for small degrees, and you might see a pattern.
Another Edit
If you have Mathematica, you can perform the above-mentioned experiments with the program below:
genpoly[deg_, name_, var_] := Sum[name[i] var^i, {i, 0, deg}]
quadraticeq[deg_, name1_, name2_, var_]:= Eliminate[MapThread[Equal, {CoefficientList[genpoly[2deg, name1, var], var],CoefficientList[Expand[genpoly[deg, name2, var]^2], var]}], Table[name2[i], {i, 0, deg}]]
(for example, to see what the variety is describing quadratic polynomials which are squares, you do: quadraticeq[1, a, b, x]
a and b are dummy variables, a[0], ..., a[2 deg] are the variables you care about. For quadratic polynomials you get (no surprise):
4 a[0] a[2]==a[1]^2
While for quartic polynomials you get:
a[0] a[3]^2==a[1]^2 a[4]&&-4 a[0] a[1] a[2]+8 a[0]^2 a[3]==-a[1]^3&&8 a[0] a[3] a[4]==a[1] (-a[3]^2+4 a[2] a[4])&&16 a[0] a[4]^2==-a[2] a[3]^2+4 a[2]^2 a[4]-2 a[1] a[3] a[4]&&8 a[1] a[4]^2==a[3] (-a[3]^2+4 a[2] a[4])&&a[0] (-4 a[2]^2+2 a[1] a[3])+16 a[0]^2 a[4]==-a[1]^2 a[2]&&a[0] (-4 a[2] a[3]+8 a[1] a[4])==-a[1]^2 a[3]
which is a little more painful.
-
That would be fine for determining if a given polynomial is a square, but I'm looking for the equations that define the locus of square polynomials, and I don't see how to do that in this way. – Charles Siegel Feb 11 '11 at 20:47
@Charles: everyone who has answered the question has misread it so far, so I've taken the liberty of making the title more precise. Hope I have captured your meaning. – Qiaochu Yuan Feb 11 '11 at 22:37
@Igor: This is essentially what I was doing in Macaulay2, but the elimination step was too bad by the time I got to 12th degree polynomials – Charles Siegel Feb 12 '11 at 19:46
@Charles: yes, mathematica seems to choke on this as well (not surprisingly -- macaulay 2 should be more efficient). But what are you trying to do with these equations? – Igor Rivin Feb 12 '11 at 22:33 | 2,936 | 9,726 | {"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} | 3.5 | 4 | CC-MAIN-2014-15 | latest | en | 0.939489 |
http://poncelet.math.nthu.edu.tw/disk3/cal03/html/jacobian/jacobian.html | 1,495,736,505,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608120.92/warc/CC-MAIN-20170525180025-20170525200025-00054.warc.gz | 288,926,187 | 8,199 | # JACOBIANS; CHANGING VARIABLES IN MULTIPLE INTEGRATION
During the course of the last few sections we have met several formulas for changing variables in multiple integration: to polar coordinates, to cylindrical coordinates, to spherical coordinates. The purpose of this section is to bring some unity into that material and provide a general prescription for other changes of variable.
We begin with a consideration of area. Let be a basic region in a plane that we are calling the -plane. (In this plane we denote the abscissa of a point by and the ordinate by .) Suppose that
are functions continuously differentiable on . As ranges over , the point generates a region in the -plane. If the mapping
is one-to-one on the interior of and the Jacobian
is never zero on the interior of , then
It is very difficult to prove this assertion without making additional assumptions. At this point we will simply assume this area formula and go on from there. Suppose now that we want to integrate some continuous function over If this proves difficult to do directly, then we can change variables to and try to integrate over instead:
The derivation of this formula from the area formula follows the usual lines. Break up into little basic regions These induce a decomposition of into little basic regions . We can then write
This last expression is a Riemann sum for
and tends to that integral as the maximum diameter of the tends to zero. This we can ensure by letting the maximum diameter of the tend to zero.
Problem 1. Evaluate
where is the parallelogram bounded by the lines
Solution. The boundaries suggest that we set
We want and in terms of and . Since
and
we have
The Jacobian is given by
Therefore
Problem 2. Evaluate
where is the first-quadrant region bounded by the curves
Solution. The boundaries suggest that we set
We want and in terms of and . Since
and
we have
The transformation has Jacobian
Therefore
Earlier we saw the formula for changing variables from rectangular coordinates to polar coordinates The formula reads
The factor in the double integral over is the Jacobian of the transformation ,:
When changing variables in a triple integral we make three coordinate changes:
If these functions carry a basic solid onto a solid , then, under conditions analogous to the two-dimensional case,
where now the Jacobian is a three-by-three determinant:
In this case the change of variables formula reads
# The Work Done by a Varying Force over a Curved Path
The work done by a constant force on an object that moves along a straight line is, by definition, the component of in the direction of the displacement multiplied by the length of the displacement vector :
comp ) .
We can write this more briefly as a dot product:
This elementary notion of work is useful, but it is not sufficient. Consider, for example, an object that moves through a magnetic field or a gravitational field. The path of the motion is then usually not a straight line but a curve, and the force, rather than remain constant, tends to vary from point to point. What we want now is a notion of work that applies to this more general situation. Let's suppose that an object moves along a curve
subject to a continuous force . (The vector field may vary from point to point, not only in magnitude but also in direction.) We will suppose that the curve is smooth; namely, we will suppose that the tangent vector is continuous and never zero. What we want to do here is define the total work done by along the curve .
To decide how to do this, we begin by focusing our attention on what happens over a short parameter interval . As an estimate for the work done over this interval we can use the dot product
In making this estimate we are using the force vector at and we are replacing the curved path from to by the line segment from to . If we set
total work done by from to and
total work done by from to
then the work done by from to must be the difference
Bringing our estimate into play, we are led to the approximate equation
which, upon division by , becomes
The quotients here are average rates of change, and the equation is only an approximate one. The notion of work is made precise by requiring that both sides have exactly the same limit as tends to zero; in other words, by requiring that
The rest is now determined. Since
and total work done on ,
we have
total work done on
In short, we have arrived at the following definition of work:
# Line Integrals
The integral on the right of the above equation can be calculated not only for a force function but for any vector field continuous on
DEFINITION LINE INTEGRAL
Let be a vector field continuous on a smooth curve
The line integral of over is the number
Note that, although we speak of integrating over , we actually carry out the calculations over the parameter set . If our definition of line integral is to make sense, the line integral as defined must be independent of the particular parametrization chosen for Within the limitations spelled out below this is indeed the case:
the line integral
is left invariant by every sense-preserving change of parameter.
Proof. Suppose that maps onto and that is positive and continuous on . We must show that the line integral over as parametrized by
equals the line integral over as parametrized by . The argument is straightforward:
Problem 1. Calculate
given that
and
Solution. Here
and
It follows that
Problem 2. Integrate the vector field over the twisted cubic
from to
Solution. The path of integration begins at and ends at In this case
Therefore
and
If a curve is not smooth but is made up of a finite number of adjoining smooth pieces , then we define the integral over as the sum of the integrals over the :
A curve of this type is said to be piecewise smooth.
All polygons are piecewise-smooth curves. In the next problem we integrate over a triangle. We do this by integrating over each of the sides and then adding up the results. Observe that the directed line segment that begins at and ends at can be parametrized by setting
Problem 3. Evaluate the line integral
if and is the triangle with vertices traversed counterclockwise.
Solution. The path is made up of three line segments:
We verify,
The integral over the entire triangle is the sum of these integrals:
When we integrate over a parametrized curve, we integrate in the direction determined by the parametrization. If we integrate in the opposite direction, our answer is altered by a factor of To be precise, let be a piecewise-smooth curve and let denote the same path traversed in the opposite direction. If is parametrized by a vector function defined on , then can be parametrized by setting
Our assertion is that
or more briefly that
We were led to the definition of line integral by the notion of work. It is clear that, if a force is continually applied to an object that moves over a piecewise-smooth curve then the work done by is the line integral of over :
Problem 4. An object, acted upon by various forces, moves along the parabola from the origin to the point One of the forces acting on the object is Calculate the work done by
Solution. We can parametrize the path by setting
Here
and
It follows that
If an object of mass moves so that at time it has position , then, from Newton's second law, the total force acting on the object at time must be
Problem 5. An object of mass moves from time to so that its position at time is given by the vector function
Find the total force acting on the object at time and calculate the total work done by this force.
Solution. Differentiation gives
The total force on the object at time is therefore
We can calculate the total work done by this force by integrating the force over the curve
Thus
# THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS
In general, if we integrate a vector field from one point to another, the value of the line integral depends upon the path chosen. There is, however, an important exception. If the vector field is a gradient field,
then the value of the line integral depends only on the endpoints of the path and not on the path itself. The details are spelled out in the following theorem.
THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS
Let be a piecewise-smooth curve that begins at and ends at . If the scalar field is continuously differentiable on an open set that contains the curve , then
Proof. If is smooth,
If is not smooth but only piecewise smooth, then we break up into smooth pieces
With obvious notation,
The theorem we just proved has an important corollary:
if the curve is closed (that is, if ), then and
Problem 1. Integrate the vector field over the circular arc
Solution. First we try to determine whether is a gradient. Note that has the form with
and . Since and are continuously differentiable everywhere and
we can conclude that is a gradient. Since the integral depends then only on the endpoints of and not on itself, we can simplify the computations by integrating over the line segment that joins these same endpoints.
We parametrize by setting
We then have
Alternative Solution. Once we have recognized that is a gradient we can determine by the methods as follows. Since
we have
The two expressions for can be reconciled only if
This means that
Since the curve begins at and ends at , we see that
Problem 2. Evaluate the line integral
where is the unit circle
and
Solution. Although is not a gradient, part of it,
is a gradient. Since we are integrating over a closed curve, the contribution of the gradient part is . The contribution of the remaining part is
This last integral is easy to evaluate:
This document created by Scientific WorkPlace 4.0. | 2,064 | 9,747 | {"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-2017-22 | longest | en | 0.928595 |
https://www.programmingalgorithms.com/category/sort/page/2 | 1,553,646,664,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912207146.96/warc/CC-MAIN-20190327000624-20190327022624-00248.warc.gz | 854,457,470 | 5,033 | # Sort
## Intro Sort
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Intro sort (a.k.a Introspective Sort) is hybrid sorting algorithm that provides both fast average performance and optimal worst-case performance. It ...
## Merge Sort Iterative
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Merge sort is a Divide and Conquer algorithm. It divides input array in two halves, calls itself for the two halves and then merges the two sorted ...
## Merge Sort Recursive
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Merge sort is a Divide and Conquer algorithm. It divides input array in two halves, calls itself for the two halves and then merges the two sorted ...
## Quick Sort Iterative
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Quick sort is an efficient sorting algorithm, serving as a systematic method for placing the elements of an array in order.
## Quick Sort Recursive
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Quick sort is an efficient sorting algorithm, serving as a systematic method for placing the elements of an array in order.
## Radix Sort
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Radix sort is a non-comparative integer sorting algorithm that sorts data with integer keys by grouping keys by the individual digits which share the ...
## Selection Sort
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Selection sort is a sorting algorithm. It works by selecting the smallest element of the array and placing it at the head of the array. Then the ...
## Shell Sort
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Shell sort (a.k.a Shell's Method and Diminishing Increment Sort) is an in-place comparison sort algorithm. The method starts by sorting pairs of ...
## Smooth Sort
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Smooth sort is a comparison-based sorting algorithm. A variant of heap sort, it was invented and published by Edsger Dijkstra in 1981. Like heap ... | 489 | 1,922 | {"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-2019-13 | latest | en | 0.791654 |
https://forums.sketchup.com/t/how-to-cover-frame/60658 | 1,652,965,860,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662527626.15/warc/CC-MAIN-20220519105247-20220519135247-00758.warc.gz | 331,240,475 | 7,712 | # How to cover frame
I want to cover this frame, seen here in the center of this image, with a plain, uniform wall, ie a surface. But somehow that proves impossible. I’ve tried glueing together a new wall by pulling lines across the frame, but it keeps only allowing me to cover parts, and what parts are covered and what parts aren’t seems completely arbitrary. How can I fix this issue?
As you can see the lines are primarily diagonal. That’s simply because that was the most effective way to increase the cover surface. Usually, with straight lines, no surface would appear at all.
It appears the opening isn’t planar.
Drawing diagonals only makes a mess but you can move the vertices on the right side of the hole onto the plane of the face on the left side. To do this, activate the move tool with nothing selected. Hold the mouse over one of the edges in the right side of the image, that appears to be perpendicular to the face to the left. Once you have a On Line inference (red square) you hold down shift; you have now locked the move tool in the direction of this edge. Click the vertex (endpoint), while still holding shift, and click the face to the left. The vertex should now have been moved to where the line of the edge intersect the plane of the face. Repeat for all the vertices around the hole (except those of the left face itself because they are already on the plane). Now you should be able to just draw one single edge and have the faces created.
I’m pretty sure the frame is planar, or at least was planar for a long time while I’ve been struggling with this problem. Often something could be wrong without you noticing, so I mean it’s fully possible that it’s become unplanar, but I’m very certain it’s been planar while I’ve been having this problem.
When you say vertices/edge, do you mean the corners? Or the general outer lines of the frame? The move tool will not give me a red square on the corners, but it will give me a read square on some other places. When I hold down shift and then move these points, they move backwards and forwards. See image.
You’re right. It wasn’t planar. I decided the best solution was to delete the shape and redo it. I couldn’t figure out the move tool solution.
There is something really strange going on here. When I create new lines, instead of going straight, they bend upwards for some reason. See image.
Sharing the .SKP model file (or a stripped-down version of it to shrink its size and remove confidential bits) may help others to understand what is happening. You can upload a file using the seventh icon from the right (an up-pointing arrow) when creating a post or reply.
Hello @retroryder,
An easy way to ensure that your initial face is always planar is by:
1. selecting the rectangle tool
2. place a rectangle of a size larger than needed in the field
(this will serve as the background or base of the initial face of the model.)
3. place another rectangle inside the first.
4. DO NOT DESELECT THE RECENT SHAPE
5. type the desired dimensions (length x width)
(this last rectangle will be planar)
6. push/pull the rectangle to the desired height.
7. delete the original background rectangle.
All faces of the newly formed rectangular solid will be coplanar and your modeling from that point forward should remain coplanar as long as you refer to the various inference tool tips that appear as you move through your work. Creating all of the diagonals only makes it more difficult to complete your tasks successfully.
1 Like
How large is the item you are modeling? There are some capture effects that can occur at very small sizes and some granularity effects that can happen at very large sizes or very far from the model origin.
As mentioned, if you can share the file or a sample where the issue occurs, the experts will be able to give you an exact answer.
This topic was automatically closed 91 days after the last reply. New replies are no longer allowed. | 855 | 3,956 | {"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-2022-21 | latest | en | 0.937736 |
https://engineering.stackexchange.com/tags/composite/hot | 1,632,302,096,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057337.81/warc/CC-MAIN-20210922072047-20210922102047-00427.warc.gz | 279,534,343 | 32,150 | # Tag Info
7
Cure Procedure Apply vacuum of 22 inches (74 kPa) Hg minimum. Apply 85 + 15 – 0 psig (586 + 103 – 0 kPa) pressure for laminates. Apply 45 + 15 – 0 psig (310 + 103 – 0 kPa) pressure for sandwich. I believe Steps #2 and #3 are providing you a tolerance for the applied pressure. So, for Step #2, it's telling you to apply 85 psig. You can go +15 psig higher (...
5
Since you used some specific words in your question, I will respond a certain way. Cracks are generated by mechanical action, and cannot be machined in. They may, in certain brittle materials, be a byproduct of machining. Using a method like EDM will alter the properties of a polymer nearby, as it will heat the polymer. If you are interested in ...
5
There are quite a few ways to find out the stiffness and/or strength of a composite material, with a good understanding of the materials that make up the composite material. 1. Pre-engineered composite materials Several materials, prevalent mainly in the aerospace industry, are pre-engineered and pre-designed. GY70/339 is a common carbon fiber epoxy used ...
4
Try this one. (Note: I don't know how to do tables in this site but please tabulate the data on your own.) Areas are: $0.75(4)(7)$ $\frac{\pi(2^2)}{2}(0.75)$ $-\pi(1.25^2)(0.75)$ $1(4)(2)$ $-\frac{\pi(1.25^2)}{2}(1)$ Note: Negative areas as holes that need to be subtracted from the system. X-Centroids are: $2$ $2$ $2$ $0.5$ $0.5$ Y-Centroids are: $-0.... 4 Yes it can be post cured - however, without support there is a risk of sagging or warping, particularly if it is a large and heavy part. Thermal curing is simply a way of speeding up the chemical hardening process, which in the case of typical epoxies may normally take place over several days, even weeks. The second benefit of heating the material is the ... 4 Some other answer have touched upon this, but I think it needs to be made explicit: Your mistake is in thinking that civil engineering is about making the lightest structure possible. It's not. Instead, it's about making the most cost-effective structure possible. Give me a material that weighs a billion tons per cubic centimeter but is cheap enough to let ... 3 The most important advantage of reinforced concrete over pure steel structure is availability. Many countries/regions on earth rely on imports for general goods production needs, the cost of steel can run prohibitively high compared to concrete. The next advantage is weight. In certain types of structure weight counts, such as underwater powerhouse. However, ... 3 I will make a slight add to the answers given here already, as follows. Take the case of a bridge. When loaded, some parts of it will be in compression, other parts will be in tension. The designer's job is to manage the stress levels in the most cost-effective manner. If a certain amount of concrete is capable of handling the compressive stresses less ... 3 No. You could use other polymers. Polyesters, vinylesters, and epoxies are all common in fibre-polymer composites and could be used. You can also use other special stuff if you need, eg, fire performance. Epoxies are common because if you're using carbon fibres (rather than a variety of glass) you are evidently looking for a high performance component. ... 3 An additional complication is the behaviour of the bullet itself on impact, which is not only difficult to model but can vary a lot according to the construction of the bullet, it's geometry and mass, and the angle and energy of impact. For example a low-velocity, unjacketed lead bullet might squash against the impact surface with little or no penetration, ... 3 Expanding on the comment by @grfrazee, it isn't as simple as with a static load. Several non-linear parameters change the response of the composite component. Extremely high strain rate increases the observed brittleness and decreases the strain-at-failure and strain-at-rupture of the impacted materials. There is a non-obvious limitation on penetration ... 3 To determine the properties of a composite laminate, the properties of each component and their relative volume fractions must be known. The properties of glass fiber laminate may be determined using the rule of mixtures. Consider a cross-section of the fiberglass laminate as shown below. White regions are the fiberglass, black regions are the epoxy matrix.... 2 I am not a chemist or material scientist, but it is my experience that reliable adhesion to a smooth Polyethylene surface is nearly impossible even with "plastic" rated epoxy. However where it is a fiber sheet, the epoxy, even with zero surface adhesion, will mechanically hold each fiber it encapsulates. I can not speak to what mechanical properties will ... 2 It isn't clear from your sketch whether the end supports are fixed or pinned, so I'm going to assume they were pinned. However, this is going to be a qualitative answer, so that doesn't actually matter too much. Assuming pinned supports, the original slab was therefore basically a simply-supported beam, with no moment at the ends and positive longitudinal ... 2 Curing conditions depend on the resin sytem. Some systems require a minimum temperature for a certain duration to cure, others cure completely at room temperature. If in doubt, you will need to ask the manufacturer of the system. It also depends on the requirements you have for the final part. If the final properties of the part are somehow safety relevant, ... 2 One problem is that you may not have good contact between adjacent metal fibres. In general resin/fibre composites aim to completely coat the fibres so it is entirely possible that a metal/plastic composite could have very low conductivity, depending on how the fibres are orientated. Equally fine metal fibres will have a layer of oxidation on the surface ... 2 Rather than thinking about a Young's modulus directly, the problem becomes easier to grasp if you think in terms of a relationship between the load and displacement, and subsequently between stress and the strain. If you consider the case where deformations are small, and just consider the stiff direction (tension/compression vertically and not horizontal ... 2 The bending stiffness will be determined by the second moment of area ($I$). The formula you provide$\int\int r^2 da$is for the Polar Moment of area ($J_p$), and is valid for torsional problems. Apart from little issue you are on the right track. Assuming that: x is the horizontal axis y is the vertical axis then you are after$I_{xx}$. Additionall, I'm ... 2 A partial listing ( in addition to the previous answer) ; for reinforced concrete only simple generic shapes are needed ( bars) while steel only needs "I" , "H" beams, gussets, fasteners, welding . Careful inventory control of the previous steel items. Coating of these shapes to reduce rust during construction. Inspection to verify the ... 2 Cool that you're considering CF, it's a fun material. So first, to answer your questions. If done right it can be fine, but even then it's the least preferable (in my opinion) way to bond FRPs (Fiber Reinforced Plastics). It's hard to do well without special tools, especially in curved and/or hollow profiles, since you need to support the material around ... 1 This is Work-in-progress How does drilling holes into carbon fibre affect its structural integrity? Drilling holes into carbon fibre has very adverse affects. The main reason is that usually the matrix is some sort of thermosetting resin, and as a result it is relatively brittle. The thing is that the drilling process, will separate fibres from the matrix (... 1 From Vacuum Bagging Wings Instruction Manual Purdue University Inner layer is foam with Kevlar at wing tip. Next fiberglass resin. Then Mylar with tape at tip. Then Perf Ply (perforated release film) and Breather. Both of these allow air to be removed from Mylar covered wing. Finally Vacuum Bag. In assembly procedure: Lay the fiberglass on the Mylar. ... 1 Have you heard of double shot moulding? In this process, the white is a separate molding that is actually embedded behind the black key. This would not show wear and might narrow down your material choices. Nowadays this is done with ABS or PBT. I don't know about your application though. 1 A unidirectional composite material cannot have the same modulus in all directions. Unidirectional means that all fibers run in the same direction. Obviously, this direction defines the axis for the first modulus of elasticity because here this parameter will have a maximum. T700 is a standard modulus carbon fiber with a tensile modulus of 230 GPa, now the ... 1 Let's approximately calculate the S of composite beam, knowing that aluminum section I is 146in^4.$ I_{COMP}= I_{alum}+ 2*A_{steel} *4.25^2 = 146+(8*0.25*2)4.25^2=146+72.25=218.25in^4 $The S of aluminum section is 35.5in^3 and the S of steel 72.25/4.5= 16.05in^3 At yield point the steel plates can take a moment of$ M_s= \sigma*S=18*16.05=288.9 k.in \$ ...
1
The difference in stiffness is important. First let's consider the aluminum W8x40 section without any steel plates: It can be loaded until the aluminum reaches a stress of 12 ksi. Then let's consider what happens if we bolt some very thin steel plates unto the flanges of the aluminum section. Then we can increase the load until the steel plates reach a ...
1
assuming you cannot compress more than the true density, is there a formula to determine how much force corresponds to the density achieved? I scanned through some literature and found this book to lead to useful information. Chapter 2 "Bulk Solid Characterization", section 4 "Compressibility", references several sets of equations, each set from a ...
1
Its not about the tension in the final product. [They are claiming] tension during fabrication buys them a much higher fiber loading. See 67 vol%.. a typical "wet layup" is maybe half that. Having the fibers very very straight probably contributes somewhat as well. Note also the 3+GPa claimed strength is for some "micro" composite. If you look elsewhere ...
1
Formulas from "The Behavior of Structures Composed of Composite Materials". Consider this source material for a much deeper understanding of composites. Does the answer depend on how the composite layer was prepared? For example, does the answer change if the composite layer is a pultruded carbon-fiber strip? Looks like the fiber ratio is the same as ...
1
Absolutely. Paper mache at it's core is an early predecessor of carbon fiber. The paper is the fiber structure and the paste (not sure of the technical term) is the equivalent to the epoxy. If you increase the 'strength' of the fibers you'll eventually reach a point that the paste is the weak point, beyond which increasing the strength of the paper won't ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2,453 | 10,881 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-39 | longest | en | 0.924418 |
https://www.accounks.com/c5-budgeting-and-control-sales-mix-and-quantity-variances-acca-f5/ | 1,628,173,986,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155925.8/warc/CC-MAIN-20210805130514-20210805160514-00685.warc.gz | 599,247,119 | 17,610 | # C5: Budgeting and Control | Sales Mix and Quantity Variances (ACCA F5)
Sales mix and quantity variances explains how to calculate, identify and explain sales mix and quantity variances and explains the relationship between the sales volume variances and the sales mix and quantity variances.
## Sales Mix and Quantity Variances
The sales volume variance consists of two variances – the sales mix variance and the sales quantity variance. They are useful “where management is in a position to control the sales mix.”
### a) Calculate, identify the cause of, and explain sales mix and quantity variances.
The sales mix and sales quantity variances “are only meaningful when management can control the proportions of the products being sold.”
Note: The sales mix variance occurs when “the proportions of the various products sold are different from those in the budget.”
To calculate the sales mix variance:
• Identify the total actual quantity of units sold
• Divide the total quantity of sales into the standard mix
• For each product, the difference between the actual quantity and the standard quantity is the mix variance
• Apply the standard profit per unit for the material to convert the mix variance into a money value
• The total materials mix variance is the total of the sales mix variance
Note: The sales quantity variance shows the “difference in contribution/profit because of a change in sales volume from the budgeted volume of sales.”
To calculate the sales quantity variance:
Calculate the total quantity of materials that should have been used for the actual number of units manufactured
• Calculate the weighted average standard profit per unit
• Calculate the difference between the actual total sales and the budgeted total sales
• Convert this into a money value by applying the weighted average standard profit per unit of sale
### b) Identify and explain the relationship of the sales volume variances with the sales mix and quantity variances.
The sales volume variance consists of two variances – the sales mix variance and the sales quantity variance. They are useful “where management is in a position to control the sales mix.” | 415 | 2,170 | {"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-2021-31 | latest | en | 0.852328 |
https://oeis.org/A333056 | 1,656,316,081,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103329963.19/warc/CC-MAIN-20220627073417-20220627103417-00662.warc.gz | 467,915,840 | 4,320 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A333056 Numbers k such that k, k+1 and k+2 have different prime signatures and d(k) = d(k+1) = d(k+2), where d(k) is the number of divisors of k (A000005). 2
59318, 72063, 72224, 184190, 185192, 215648, 300320, 355454, 362624, 384128, 548936, 550016, 640790, 682624, 707966, 723896, 758888, 828872, 828873, 858494, 860030, 888704, 901503, 963486, 963710, 993375, 1039742, 1039743, 1081214, 1248776, 1261897, 1340630 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Apparently most of the numbers k such that d(k) = d(k+1) = d(k+2) (A005238) are terms of A052214, i.e., k, k+1 and k+2 have the same prime signature. Of the first 10000 terms of A005238, 6406 are also in A052214, 3578 have a pair (k and k+1, k and k+2, or k+1 and k+2) with the same prime signature, and only 16 are in this sequence. LINKS Amiram Eldar, Table of n, a(n) for n = 1..10000 EXAMPLE 59318 is a term since d(59318) = d(59319) = d(59320) = 16, and the prime signatures of these 3 numbers are different: 59318 = 2 * 7 * 19 * 223, 59319 = 3^3 * 13^3, and 59320 = 2^3 * 5 * 1483 have 3 different ordered prime signatures (A124010): [1, 1, 1, 1], [3, 3], and [1, 1, 3]. MATHEMATICA psig[n_] := Sort @ FactorInteger[n][[;; , 2]]; d[sig_] := Times @@ (sig + 1); vsig = psig /@ Range[2, 4]; seqQ[v_] := Length@Union[v] == 3 && Length @ Union[d /@ v] == 1; seq = {}; Do[If[seqQ[vsig], AppendTo[seq, n - 3]]; vsig = Join[Rest[vsig], {psig[n]}], {n, 5, 10^6}]; seq CROSSREFS Subsequence of A005238. Cf. A000005, A052214, A124010, A333055. Sequence in context: A321814 A288886 A210172 * A224972 A104919 A135325 Adjacent sequences: A333053 A333054 A333055 * A333057 A333058 A333059 KEYWORD nonn AUTHOR Amiram Eldar, Mar 06 2020 STATUS approved
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Last modified June 27 03:29 EDT 2022. Contains 354888 sequences. (Running on oeis4.) | 819 | 2,219 | {"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} | 3.734375 | 4 | CC-MAIN-2022-27 | latest | en | 0.679992 |
https://www.airmilescalculator.com/distance/sbt-to-ura/ | 1,606,624,577,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141196324.38/warc/CC-MAIN-20201129034021-20201129064021-00282.warc.gz | 538,212,651 | 65,848 | # Distance between Sabetta (SBT) and Uralsk (URA)
Flight distance from Sabetta to Uralsk (Sabetta International Airport – Oral Ak Zhol Airport) is 1531 miles / 2464 kilometers / 1330 nautical miles. Estimated flight time is 3 hours 23 minutes.
Driving distance from Sabetta (SBT) to Uralsk (URA) is 2106 miles / 3390 kilometers and travel time by car is about 62 hours 56 minutes.
## Map of flight path and driving directions from Sabetta to Uralsk.
Shortest flight path between Sabetta International Airport (SBT) and Oral Ak Zhol Airport (URA).
## How far is Uralsk from Sabetta?
There are several ways to calculate distances between Sabetta and Uralsk. Here are two common methods:
Vincenty's formula (applied above)
• 1531.030 miles
• 2463.954 kilometers
• 1330.429 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 1527.439 miles
• 2458.175 kilometers
• 1327.308 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Sabetta International Airport
City: Sabetta
Country: Russia
IATA Code: SBT
ICAO Code: USDA
Coordinates: 71°13′9″N, 72°3′7″E
B Oral Ak Zhol Airport
City: Uralsk
Country: Kazakhstan
IATA Code: URA
ICAO Code: UARR
Coordinates: 51°9′2″N, 51°32′35″E
## Time difference and current local times
There is no time difference between Sabetta and Uralsk.
+05
+05
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 181 kg (400 pounds).
## Frequent Flyer Miles Calculator
Sabetta (SBT) → Uralsk (URA).
Distance:
1531
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
1531
Round trip? | 512 | 1,858 | {"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-2020-50 | latest | en | 0.814859 |
https://brainly.com/question/301485 | 1,484,816,537,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280504.74/warc/CC-MAIN-20170116095120-00565-ip-10-171-10-70.ec2.internal.warc.gz | 774,016,002 | 9,110 | 2015-02-11T18:16:14-05:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
X=20
4/15=x/75
1. cross multiply so.... ---------> 15 times x and 4X75
15x=300
2. divide by 15...... 300/15 ---------> x=20
and plug it back in to check
4/15=20/75 ------> 20/75 simplified is 4/15
so 4/15 = 4/15 is right
2015-02-11T18:16:39-05:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Solve it like a proportion. For example, if you had a/b = x/y, it wold be a(y) = b(x)
For this question you asked, it would be the same as: 4(75) = 15(x). You would simplify that and get 300 = 15x.
That would be solving x now. Divide 15 on both sides and you get the final answer:
20 = x
Hope this helps! :) Follow if you want. | 345 | 1,223 | {"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} | 3.84375 | 4 | CC-MAIN-2017-04 | latest | en | 0.914338 |
http://math.stackexchange.com/questions/58768/existence-of-a-subset-of-infinite-model-thats-not-definable | 1,469,661,737,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827080.38/warc/CC-MAIN-20160723071027-00183-ip-10-185-27-174.ec2.internal.warc.gz | 156,287,810 | 18,987 | # Existence of a subset of infinite model that's not definable
I'm working on the following problem: Show that any infinite model $\mathcal{M}$ has a subset of its domain that's not definable in $\mathcal{M}$ using parameters.
I tried to follow a contradiction proof but didn't get far. Because the language is unknown, I ruled out Tarski Undefinability of Truth and Fixed Point Lemma. It doesn't seem to have anything to do with the $\Sigma_n$ hierarchy either. Any hint on where to start?
Edit: $\mathcal{M}$ is model of a countable language
-
Depends on the language. But if language is "small" there are too many subsets. – André Nicolas Aug 21 '11 at 7:06
Sorry, I should have mentioned that it's a countable language. What do you mean by "small"? – nullgraph Aug 21 '11 at 7:18
Countable qualifies as small. If $\mathscr{L}$ is countable, how many formulas are there over $\mathscr{L}$, even allowing parameters? And $M$ has at least how many subsets? – Brian M. Scott Aug 21 '11 at 7:22
Ah I get it! There are only countably many formulas over $\mathcal{L}$, even allowing parameters and $M$ has uncountably many subsets, so some of them must not be definable. – nullgraph Aug 21 '11 at 7:29
There are more than countably many, if the model is large. But still not enough. – André Nicolas Aug 21 '11 at 7:49
The result (Ed: as first stated) is not true in general. But first of all, we give a result along the lines of the question which is true. (Ed: Conveniently, it answers the revised question.)
Let $L$ be a language which has a finite or denumerably infinite number of non-logical symbols, and let $T$ be a theory over $L$.
Let $M$ be a model of $T$ of cardinality $\kappa$, where $\kappa$ is infinite.
Then $M$ has $2^\kappa>\kappa$ subsets.
But there are only $\kappa$ expressions of the form $F(a_1,a_2,\dots,a_n,y)$, where $F(x_1,x_2,\dots,x_n,y)$ is a formula of the language, and the $a_i$ are (names of) elements of $M$.
Thus there are no more than $\kappa$ subsets of $M$ that are definable using parameters. There are in fact exactly $\kappa$. We use the formula $y=x$. If $a$ is (the name of) an element of $M$, then $\{a\}$ is definable by parameters.
The analysis can be extended to uncountable languages, by putting a suitable lower bound on the cardinality of the model $M$.
Counterexample (Ed: for the original question): Look at the language $L$ that has a predicate symbol for every subset of $\mathbb{N}$, and let $T$ be the theory over this language whose axioms are all sentences true in $\mathbb{N}$, with the natural interpretation for each predicate symbol. Then every subset of $\mathbb{N}$ is definable, even without parameters. For if $P_A$ is the predicate symbol corresponding to $A \subseteq \mathbb{N}$, then the formula $P_A(y)$ defines $A$.
Admittedly, $L$ is a rather large language. But such languages can be useful in Model Theory, for example in constructing non-standard models of analysis in which every "normal" function and relation on $\mathbb{R}$ can be automatically extended to the non-standard model.
-
To sum up Andre's great answer. This is a standard cardinality play, using two facts:
1. For every infinite cardinal $\kappa<2^\kappa$,
2. if $A$ is an infinite set then $\{B\subseteq A\big||B|<\aleph_0\}=|A|$.
Suppose $L$ is a countable language, therefore the set of formulae is countable (since every formula corresponds to a finite string over $\aleph_0$)
Now suppose the model has universe of cardinality $\kappa\ge\aleph_0$, then we have $2^\kappa>\kappa$ many subsets of the universe, in contrast every definable set corresponds to a formula and a finite set of parameters.
Therefore we have $\aleph_0\cdot\kappa=\kappa$ many definable sets in the model.
- | 1,016 | 3,744 | {"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} | 3.25 | 3 | CC-MAIN-2016-30 | latest | en | 0.920819 |
https://it.mathworks.com/matlabcentral/answers/525948-how-to-perform-mathematical-operation-in-columns-on-certain-elements-and-make-the-other-elements-zer | 1,623,810,820,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621699.22/warc/CC-MAIN-20210616001810-20210616031810-00122.warc.gz | 294,223,280 | 29,052 | # How to perform mathematical operation in columns on certain elements and make the other elements zero??
2 views (last 30 days)
Rabia Zulfiqar on 16 May 2020
Commented: dpb on 20 May 2020
For example I have this matrix:
[0 0
0 0
1 0
2 7
5 8
3 11
4 12
2 9]
The min and max value of column one are (1,5) and column 2 are (7,12). I want to perform a specific operation a*b when the min value occurs in each column and perform a*(b-2) to its next value. similarly when max value occurs I want to perform a/b at max value and a/(b-2) to its next value. All the other values must remain zero.
Here
a=10; a*b=10; a*(b-2)=-10
b=1 a/b=10; a/(b-2)=-10
The new matrix should be like:
Ans=[ 0 0
0 0
10 0
-10 10
10 -10
-10 0
0 10
0 -10]
dpb on 16 May 2020
Edited: dpb on 17 May 2020
Don't try to get too fancy...sometimes just deadahead is the simplest and best (and probably fastest, besides)...
a=10; % constants and compute insertion vectors
b=1;
vMIN=[a*b;a*(b-2)];
vMAX=[a/b;a/(b-2)];
A(A==0)=nan; % prelim fixup to exclude 0 from min/max
for i=1:size(A,2) % engine for each column...
[~,imn]=min(A(:,i)); % location min, max
[~,imx]=max(A(:,i));
A(imn:imn+1,i)=vMIN; % replace element and next
A(imx:imx+1,i)=vMAX;
end
A(isnan(A)=0; % restore the zeros...
NB1: Must do search for locations and save before either substitution...
NB2: Presumes min/max is not in last row of A or will get out-of-bounds addressing error.
dpb on 20 May 2020
No problem...glad to help. :)
### More Answers (1)
Stephen Cobeldick on 19 May 2020
Edited: Stephen Cobeldick on 20 May 2020
Something fancy...
>> M = [0,0;0,0;1,0;2,7;5,8;3,11;4,12;2,9]
M =
0 0
0 0
1 0
2 7
5 8
3 11
4 12
2 9
>> a = 10;
>> b = 1;
>> F = @(f,v) conv2(+(M==f(M./~~M,[],1)),[0;v],'same'); % requires >=R2016b
>> Q = F(@max,[a/b;a/(b-2)]) + F(@min,[a*b;a*(b-2)])
Q =
0 0
0 0
10 0
-10 10
10 -10
-10 0
0 10
0 -10
For earlier versions replace == with bsxfun.
dpb on 20 May 2020
" fastest approach I found was to split the vector indexing in your solution into two scalars"
The [;] operation has always been expensive. I get warnings about avoiding brackets all the time that I mostly ignore since not writing production code and I find them more legible...but if such code were buried in the bowels of a loop or iterative solution it then could be worth the less expressive form.
Interesting result... | 834 | 2,346 | {"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 | 0.780279 |
https://functions.wolfram.com/Bessel-TypeFunctions/KelvinKer2/06/01/02/0006/ | 1,721,921,024,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763860413.86/warc/CC-MAIN-20240725145050-20240725175050-00415.warc.gz | 225,829,098 | 10,997 | html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; }
KelvinKer
http://functions.wolfram.com/03.20.06.0008.01
Input Form
KelvinKer[\[Nu], z] == Sum[(1/(Subscript[z, 0]^k k!)) Sum[(-1)^(m + k) Binomial[k, m] Pochhammer[-\[Nu], k - m] Sum[(((-1)^i 2^(2 i - m) Pochhammer[-m, 2 (m - i)] Pochhammer[\[Nu], i])/(m - i)!) (KelvinKer[\[Nu], z] Sum[(((-1)^j (i - 2 j)!)/((2 j)! (i - 4 j)! Pochhammer[1 - i - \[Nu], 2 j] Pochhammer[\[Nu], 2 j])) (z/2)^(4 j), {j, 0, Floor[i/2]}] + (z/(2 Sqrt[2])) (KelvinKei[-1 + \[Nu], z] + KelvinKer[-1 + \[Nu], z]) Sum[(((-1)^j (i - 2 j - 1)!)/ ((2 j)! (i - 4 j - 1)! Pochhammer[1 - i - \[Nu], 2 j] Pochhammer[\[Nu], 2 j + 1])) (z/2)^(4 j), {j, 0, Floor[(i - 1)/2]}] + (z^2/4) KelvinKei[\[Nu], z] Sum[(((-1)^j (i - 2 j - 1)!)/((2 j + 1)! (i - 4 j - 2)! Pochhammer[1 - i - \[Nu], 2 j + 1] Pochhammer[\[Nu], 2 j + 1])) (z/2)^(4 j), {j, 0, Floor[(i - 1)/2]}] + (z^3/(8 Sqrt[2])) (KelvinKei[-1 + \[Nu], z] - KelvinKer[-1 + \[Nu], z]) Sum[(((-1)^j (i - 2 j - 2)!)/((2 j + 1)! (i - 4 j - 3)! Pochhammer[1 - i - \[Nu], 2 j + 1] Pochhammer[\[Nu], 2 j + 2])) (z/2)^(4 j), {j, 0, Floor[(i - 2)/2]}]) (z - Subscript[z, 0])^k, {i, 0, m}], {m, 0, k}], {k, 0, Infinity}] /; Abs[Arg[Subscript[z, 0]]] < Pi
Standard Form
Cell[BoxData[RowBox[List[RowBox[List[RowBox[List["KelvinKer", "[", RowBox[List["\[Nu]", ",", "z"]], "]"]], "\[Equal]", RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["k", "=", "0"]], "\[Infinity]"], RowBox[List[FractionBox[SubsuperscriptBox["z", "0", RowBox[List["-", "k"]]], RowBox[List["k", "!"]]], RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["m", "=", "0"]], "k"], RowBox[List[SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], RowBox[List["m", "+", "k"]]], RowBox[List["Binomial", "[", RowBox[List["k", ",", "m"]], "]"]], " ", RowBox[List["Pochhammer", "[", RowBox[List[RowBox[List["-", "\[Nu]"]], ",", RowBox[List["k", "-", "m"]]]], "]"]], " ", RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["i", "=", "0"]], "m"], RowBox[List[FractionBox[RowBox[List[SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], "i"], " ", SuperscriptBox["2", RowBox[List[RowBox[List["2", " ", "i"]], "-", "m"]]], " ", RowBox[List["Pochhammer", "[", RowBox[List[RowBox[List["-", "m"]], ",", RowBox[List["2", " ", RowBox[List["(", RowBox[List["m", "-", "i"]], ")"]]]]]], "]"]], " ", RowBox[List["Pochhammer", "[", RowBox[List["\[Nu]", ",", "i"]], "]"]]]], RowBox[List[RowBox[List["(", RowBox[List["m", "-", "i"]], ")"]], "!"]]], RowBox[List["(", " ", RowBox[List[RowBox[List[RowBox[List["KelvinKer", "[", RowBox[List["\[Nu]", ",", "z"]], "]"]], RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["j", "=", "0"]], RowBox[List["Floor", "[", FractionBox["i", "2"], "]"]]], RowBox[List[FractionBox[RowBox[List[SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], "j"], RowBox[List[RowBox[List["(", RowBox[List["i", "-", RowBox[List["2", "j"]]]], ")"]], "!"]], " "]], RowBox[List[RowBox[List[RowBox[List["(", RowBox[List["2", "j"]], ")"]], "!"]], " ", RowBox[List[RowBox[List["(", RowBox[List["i", "-", RowBox[List["4", " ", "j"]]]], ")"]], "!"]], " ", RowBox[List["Pochhammer", "[", RowBox[List[RowBox[List["1", "-", "i", "-", "\[Nu]"]], ",", 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MathML Form
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Rule Form
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Date Added to functions.wolfram.com (modification date)
2007-05-02 | 6,612 | 17,663 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-30 | latest | en | 0.182644 |
https://socratic.org/questions/how-do-you-find-the-slope-of-the-graph-of-3x-5y-10 | 1,722,658,882,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640356078.1/warc/CC-MAIN-20240803025153-20240803055153-00179.warc.gz | 428,627,642 | 6,068 | # How do you find the slope of the graph of 3x+5y=10?
May 5, 2018
$- \frac{3}{5}$ is the slope of the line.
#### Explanation:
Rearrange the equation into slope-intercept form to find the slope:
$y = m x + b$
$\text{m = slope}$
$\text{b = y-intercept}$
So rearrange the problem so that the $y$ is by itself on the left of the equal sign:
$3 x + 5 y = 10$
$5 y = - 3 x + 10$
$y = - \frac{3}{5} x + \frac{10}{5}$
$y = - \frac{3}{5} x + 2$
The slope is next to the $x$, so it is called $m$ in the slope-intercept form:
$y = \textcolor{b l u e}{m} x + b$
$y = \textcolor{b l u e}{- \frac{3}{5}} x + 2$
$- \frac{3}{5}$ is the slope of the line. | 253 | 653 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 14, "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} | 4.53125 | 5 | CC-MAIN-2024-33 | latest | en | 0.831364 |
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