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http://www.authorstream.com/Presentation/sourabh_mrsc-179216-spss-week7-education-ppt-powerpoint/ | 1,524,539,509,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946453.89/warc/CC-MAIN-20180424022317-20180424042317-00506.warc.gz | 358,722,329 | 27,750 | # SPSS
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Category: Education
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## Presentation Transcript
### SPSS Tutorial :
SPSS Tutorial AEB 37 / AE 802 Marketing Research Methods Week 7
### Cluster analysis :
Cluster analysis Lecture / Tutorial outline Cluster analysis Example of cluster analysis Work on the assignment
### Cluster Analysis :
Cluster Analysis It is a class of techniques used to classify cases into groups that are relatively homogeneous within themselves and heterogeneous between each other, on the basis of a defined set of variables. These groups are called clusters.
### Cluster Analysis and marketing research :
Cluster Analysis and marketing research Market segmentation. E.g. clustering of consumers according to their attribute preferences Understanding buyers behaviours. Consumers with similar behaviours/characteristics are clustered Identifying new product opportunities. Clusters of similar brands/products can help identifying competitors / market opportunities Reducing data. E.g. in preference mapping
### Steps to conduct a Cluster Analysis :
Steps to conduct a Cluster Analysis Select a distance measure Select a clustering algorithm Determine the number of clusters Validate the analysis
### Defining distance: the Euclidean distance :
Defining distance: the Euclidean distance Dij distance between cases i and j xki value of variable Xk for case j Problems: Different measures = different weights Correlation between variables (double counting) Solution: Principal component analysis
### Clustering procedures :
Clustering procedures Hierarchical procedures Agglomerative (start from n clusters, to get to 1 cluster) Divisive (start from 1 cluster, to get to n cluster) Non hierarchical procedures K-means clustering
### Agglomerative clustering :
Agglomerative clustering
### Agglomerative clustering :
Agglomerative clustering Linkage methods Single linkage (minimum distance) Complete linkage (maximum distance) Average linkage Ward’s method Compute sum of squared distances within clusters Aggregate clusters with the minimum increase in the overall sum of squares Centroid method The distance between two clusters is defined as the difference between the centroids (cluster averages)
### K-means clustering :
K-means clustering The number k of cluster is fixed An initial set of k “seeds” (aggregation centres) is provided First k elements Other seeds Given a certain treshold, all units are assigned to the nearest cluster seed New seeds are computed Go back to step 3 until no reclassification is necessary Units can be reassigned in successive steps (optimising partioning)
### Hierarchical vs Non hierarchical methods :
Hierarchical vs Non hierarchical methods Hierarchical clustering No decision about the number of clusters Problems when data contain a high level of error Can be very slow Initial decision are more influential (one-step only) Non hierarchical clustering Faster, more reliable Need to specify the number of clusters (arbitrary) Need to set the initial seeds (arbitrary)
### Suggested approach :
Suggested approach First perform a hierarchical method to define the number of clusters Then use the k-means procedure to actually form the clusters
### Defining the number of clusters: elbow rule (1) :
Defining the number of clusters: elbow rule (1) n
### Elbow rule (2): the scree diagram :
Elbow rule (2): the scree diagram
### Validating the analysis :
Validating the analysis Impact of initial seeds / order of cases Impact of the selected method Consider the relevance of the chosen set of variables
SPSS Example
### Slide 19:
Number of clusters: 10 – 6 = 4
### Open the dataset supermarkets.sav :
Open the dataset supermarkets.sav From your N: directory (if you saved it there last time Or download it from: http://www.rdg.ac.uk/~aes02mm/supermarket.sav Open it in SPSS
### The supermarkets.sav dataset :
The supermarkets.sav dataset
### Run Principal Components Analysis and save scores :
Run Principal Components Analysis and save scores Select the variables to perform the analysis Set the rule to extract principal components Give instruction to save the principal components as new variables
### Cluster analysis: basic steps :
Cluster analysis: basic steps Apply Ward’s methods on the principal components score Check the agglomeration schedule Decide the number of clusters Apply the k-means method
### Analyse / Classify :
Analyse / Classify
### Select the component scores :
Select the component scores Select from here Untick this
### Output: Agglomeration schedule :
Output: Agglomeration schedule
### Number of clusters :
Number of clusters Identify the step where the “distance coefficients” makes a bigger jump
### The scree diagram (Excel needed) :
The scree diagram (Excel needed)
### Number of clusters :
Number of clusters Number of cases 150 Step of ‘elbow’ 144 __________________________________ Number of clusters 6
### Now repeat the analysis :
Now repeat the analysis Choose the k-means technique Set 6 as the number of clusters Save cluster number for each case Run the analysis
K-means
### K-means dialog box :
K-means dialog box Specify number of clusters
Final output
### Cluster membership :
Cluster membership
### Component meaning(tutorial week 5) :
Component meaning(tutorial week 5) 1. “Old Rich Big Spender” 3. Vegetarian TV lover 4. Organic radio listener 2. Family shopper 5. Vegetarian TV and web hater
### Cluster interpretation through mean component values :
Cluster interpretation through mean component values Cluster 1 is very far from profile 1 (-1.34) and more similar to profile 2 (0.38) Cluster 2 is very far from profile 5 (-0.93) and not particularly similar to any profile Cluster 3 is extremely similar to profiles 3 and 5 and very far from profile 2 Cluster 4 is similar to profiles 2 and 4 Cluster 5 is very similar to profile 3 and very far from profile 4 Cluster 6 is very similar to profile 5 and very far from profile 3
### Which cluster to target? :
Which cluster to target? Objective: target the organic consumer Which is the cluster that looks more “organic”? Compute the descriptive statistics on the original variables for that cluster
### Representation of factors 1 and 4(and cluster membership) :
Representation of factors 1 and 4(and cluster membership) | 1,274 | 6,374 | {"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} | 3.1875 | 3 | CC-MAIN-2018-17 | latest | en | 0.769059 |
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# homework_05 - are statistically independent what is the...
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% % 1.010 Fall 2008 Homework Set #5 Due October 16, 2008 (in class) 1. Two continuous variables X and Y have joint probability density function: # 1 f X , Y ( x , y ) = \$ 8 ( x + y ) 0 " x " 2 and 0 " y " 2 & 0 elsewhere a) Find and plot the marginal probability density function (PDF) of X . b) Find and plot the marginal cumulative distribution function (CDF) of X . c) Find and plot the conditional PDF of ( Y | X =1). d) Are X and Y independent? Comment. 2. According to schedule, Train A arrives at station S at 10:55 am and Train B departs from the same station at 11:05 am. Due to delays, the arrival time of Train A is uniformly distributed between 10:55 and 11:10 and the departure time of Train B is uniformly distributed between 11:05 and 11:15. If the arrival and departure times of the two trains
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Unformatted text preview: are statistically independent, what is the probability that a passenger on Train A misses the connection with Train B?[Hint: Let T A and T B be the times when train A arrives and train B departs, respectively. Plot the joint range of ( T A , T B ) on the ( T A , T B )-plane and find the region that corresponds to missing the connection.] 3. Show that the function below is the PDF of R , the distance between the epicenter of an earthquake and the site of a dam, when the epicenter is equally likely to be at any location along a neighboring fault (see figure below). You may restrict your attention to a length of fault l that is within a distance r o of the site because earthquakes at greater distances will have negligible effect at the site. f R ( r ) = 2 r ( r 2-d 2 )-1/2 , d ≤ r ≤ r o l Sketch the function. Site r o r o d Fault A B l /2 l /2...
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Jill Tulane University ‘16, Course Hero Intern | 740 | 3,047 | {"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} | 3.6875 | 4 | CC-MAIN-2018-22 | latest | en | 0.894193 |
http://www.isibang.ac.in/~manish/teaching/418h/418hw.htm | 1,545,005,351,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828018.77/warc/CC-MAIN-20181216234902-20181217020902-00083.warc.gz | 413,062,127 | 1,240 | # Homework for 418H
0.2 : 3,7
0.3 : 3, 4, 7, 9
## Main Course
1.1 : 1abd, 7, 8 13, 14, 22, 25, 35.
1.2 : 1, 4, 7, 18
1.3 : 4, 5, 10, 13, 15, 19
1.4 : 3, 11
1.6 : 3, 4, 6, 9, 18, 25, 26
1.7 : 5, 6.
2.1 : 4, 6, 7, 10, 15
2.2 : 7, 11, 14
2.3 : 2, 3, 9, 11, 13, 16, 19, 25
3.1 : 3, 5, 12, 16, 22, 33, 36, 37, 39
3.2 : 4, 5, 9, 10, 13, 14, 16, 18
3.3 : 3, 4, 7
3.4 : 5, 8, 9, 11
3.5 : 3, 4, 6, 10, 12
4.1 : 1, 4, 8(a), 10
4.2 : 4, 8, 10, 14
4.3 : 5, 6, 8, 13, 19, 22, 30, 35
4.4 : 3, 5, 6, 12, 18
4.5 : 14, 16, 23, 25, 29(Use Prop 23), 32, 36, 39.
7.1 : 3, 7, 11, 14, 26, 27.
7.2 : 3, 5, 12
7.3 : 1, 4, 5, 7, 16, 17, 20, 23, 24, 33
7.4 : 7, 8, 9, 10, 11, 15, 16, 25
7.5 : 3, 4, 5.
Supplemental Exercises
1) Let R be a commutative ring with identity and S a mupltiplicative set such every element of S is a unit in R. Show that ring of fractions of R by S is isomorphic to R.
2) Let R=Z/6Z and S={1,2,4}. Show that S is a multiplicative set. Compute the ring of fractions of R by S. How many elements does it have?
7.6 : 1, 4, 7
8.1 : 2(c), 3, 4(a), 11
8.2 : 1, 4, 5, 6, 8
8.3 : 1, 2, 8(a)
9.1 : 3, 4, 6, 13
9.2 : 1, 3, 5
9.3 : 2, 3 | 736 | 1,130 | {"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} | 3.25 | 3 | CC-MAIN-2018-51 | latest | en | 0.719257 |
https://www.jigidi.com/jigsaw-puzzle/D0K3HCYP/Ljapunow-Fractal | 1,493,592,271,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917125881.93/warc/CC-MAIN-20170423031205-00183-ip-10-145-167-34.ec2.internal.warc.gz | 913,757,549 | 9,187 | # Ljapunow Fractal
35 pieces
151 solves
It's a 'Pearler', is another expression like 'Bobby Dazzler'... Hence Perla Dazzlero... Mangling words, I just can't help myself... Sorry I didn't get back to you sooner.... Hugs.... :) :)
Roberto Dazzlero... yes, I 've met Bobby myself. Back,then there was no Perla. .... where on earth,did;you come up with that name? .... Ahhhhh, ... Oz! .... :-)))
Lol! So they kidnapped poor Ljapunow! ......now, THAT'S a worry!
Right!!!!!!!!!!!!!!!! We will have some lovely discussions, Good Buddy... Oh dear, I thought the ink flow stability in signatures was a Dept of Treasury top secret item.... When I worked there, it was mum's the word.... The would be counterfeiters were totally flummoxed... Until they kidnapped Ljapunow and now everyone knows the secret.... It's a worry.... :) :)
We meet Roberto Dazzlero when we were passing through NM... Nice bloke... He had a charming wife, Perla Dazzlero... Fun couple.... :) :)
Ah, so you know Robert Dazzler personally as does Sally! Perhaps, I, too, should call him " Bobby ", lol! I'm glad you like the colors and think it is a fine puzzle. I think fractals have been fun for both of us! You are very welcome, dear Shirley!
Do what??....Ahem, I mean, of course this is within my mathematical capabilities. Did you know the computations of the Ljapunow fractal are used to determine the authenticity of a signature? It determines the stability of the flow of ink in the signature. One that is counterfeit does not have the continual flow but displays a measurable discrepancy (chaotic).Quite useful, the Ljapunow fractal! When we finally meet, we'll be able to sit and discuss this at length..........rightttttttt!
I am so glad you think this is a real Robert Dazzler ( I don't know him well enough to use his nickname. I don't think he lives in this Hemisphere). Most welcome, dear one!!
Lol, a Mickey ears fractal... I like that, Robyn! It pleases me that you that you think this one is beautiful. I'm really glad you enjoyed it! You are very welcome, my friend... :-)
Why, thank you, my enthusiastic friend! I'm glad you liked the color and texture! I can't take credit for putting the Mickey ears in the image. They evolved as the fractal was being made. However, I did think they gave a humorous touch to the final image! Thanks, Kirsten!
Jill, I think our Sally has said it all, yes I agree it's a Bobby Dazzler, love the colours too, a very fine puzzle, Thanks Jill.
You do have a degree in higher mathematics, right????? Figuring out how to construct a fractal by mapping the regions of stability and chaotic behavior (measured using the Ljapunov exponent \lambda) in the a?b plane for given periodic sequences is nothing but phenomenal.... And the way you've shifted the iteration sequence.... Wow!!... Good Buddy, you're a genius.... :) :)
And yes, indeedy do... The purple is pretty fantastic, a real Bobby Dazzler.... Thanks Jillia.... :) :)
I'm so glad someone else saw the Mickey ears too. But I think your Mickey ears fractal is very beautiful. I don't know how to turn Mickey ears into anything this gorgeous. Thanks for sharing.
Oh my Jill!! That is SO beautiful!! The colours and textures are AMAZING!! Thanks so much. :)))
And I love your sense of humour. Only YOU would put a set of Mickey ears on a fractal!! Hee, hee.
I'm glad you found it to be interesting, KARLS! Thankyou for letting me know
Thank you, loveydear. I appreciate your comment,
VERY INTERESTING!
I like your creation J M! | 861 | 3,512 | {"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-2017-17 | longest | en | 0.961437 |
https://www.physicsforums.com/threads/normal-plane-to-a-curve.349570/ | 1,508,421,235,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823284.50/warc/CC-MAIN-20171019122155-20171019142155-00147.warc.gz | 963,353,067 | 15,811 | Normal Plane to a Curve.
1. Oct 27, 2009
Septcanmat
1. The problem statement, all variables and given/known data
At what point on the curve ⃗r(t) = (t^3, 3t, t^4) is the normal plane parallel to the plane 3x + 3y − 4z = 9 (the normal plane is the plane through the point ⃗r(t) which is normal to ⃗r′(t))
2. Relevant equations
I'm not really sure.
3. The attempt at a solution
(6t)(x-t^3) + (0)(y-3t) + (8t)(z-t^4) = 0
But that got me nowhere.
2. Oct 28, 2009
lanedance
i'm not sure what you attempted there...
first find vector normal to the plane given, then find the tangent vector of the curve.. and have a think about how they will be related
3. Oct 28, 2009
Septcanmat
I read somewhere that that would be the equation of a normal plane to a curve. But it didn't work.
I did what you said, and they'll be related in that they'll be parallel vectors. I tried doing what you said and setting them equal to each other, but I just got equations for t that seem insolvable. ( for instance, 0= 8t^4 + 9 +16t^6)
4. Oct 28, 2009
lanedance
that's not what i get, it works out ok... i'm not sure how you get the higher powers of t in your equation either
what do you get for the normal to the plane & for the tangent vector?
5. Oct 28, 2009
Septcanmat
The normal is the gradient, so I got (3,3,-4). And I got (3t^2,3,4t^3)/sqrt(9t^4+9+16t^6) for the tangent vector.
6. Oct 28, 2009
lanedance
both look good, but I see you are normalising the tangent vector to length 1, that's not needed here, as you just need to know when its parallel
so if p is normal to the plane, t is the tangent, you just need to know when t = c.p for any constant c, which shows they are parallel. This should lead to a reasonably easy equation set if you don't normalise the vector
Last edited: Oct 28, 2009
7. Oct 28, 2009
Septcanmat
Well alright then, lol. Thanks guys, got it all figured out now. The answer is (-1,-3,1). Or rather, I'm assuming that that's the correct answer because it's what I got and it matches up with one of the multiple choice options :P
8. Oct 28, 2009
lanedance
yep thats what i get | 652 | 2,114 | {"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} | 3.890625 | 4 | CC-MAIN-2017-43 | longest | en | 0.952669 |
https://ru.scribd.com/document/152429096/Radical-Worksheet | 1,566,286,746,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315258.34/warc/CC-MAIN-20190820070415-20190820092415-00528.warc.gz | 623,571,434 | 64,630 | You are on page 1of 38
Fractional Exponents
What if an exponent is a fraction?
An exponent of 1/2 is actually a square root And an exponent of 1/3 is a cube root An exponent of 1/4 is a 4th root And so on!
Why?
Let's see why in an example. First, the Laws of Exponents tell us how to handle exponents when we multiply:
## Example: x2x2 = (xx)(xx) = xxxx = x4
Which shows that x2x2 = x(2+2) = x4 So let us try that with fractional exponents:
Example: What is 9 9 ?
9 9 = 9(+) = 9(1) = 9 So 9 times itself gives 9. What do we call a number that, when multiplied by itself gives another number? The square root! See: 9 9 = 9 So 9 is the same as 9
Radical Notation: If n is a positive integer that is greater than 1 and a is a real number then,
where n is called the index, a is called the radicand, and the symbol is called the radical. The left side of this equation is often called the radical form and the right side is often called the exponent form. .
Note as well that the index is required in these to make sure that we correctly evaluate the radical. There is one exception to this rule and that is square root. For square roots we have,
## Lets do a couple of examples to familiarize us with this new notation.
Example 1 Write each of the following radicals in exponent form. (a) (b)
(c) Solution
(a)
(b)
(c)
As seen in the last two parts of this example we need to be careful with parenthesis. When we convert to exponent form and the radicand consists of more than one term then we need to enclose the whole radicand in parenthesis as we did with these two parts. To see why this is consider the following,
Only the term immediately to the left of the exponent actually gets the exponent. Therefore, the radical form of this is,
So, we once again see that parenthesis are very important in this class. Be careful with them.
Since we know how to evaluate rational exponents we also know how to evaluate radicals as the following set of examples shows.
## (b) (c) (d)
(e)
Solution To evaluate these we will first convert them to exponent form and then evaluate that since we already know how to do that.
(a) These are together to make a point about the importance of the index in this notation. Lets take a look at both of these.
So, the index is important. Different indexes will give different evaluations so make sure that you dont drop the index unless it is a 2 (and hence were using square roots).
(b)
(c)
(d)
(e) As we saw in the integer exponent section this does not have a real answer and so we cant evaluate the radical of a negative number if the index is even. Note however that we can evaluate the radical of a negative number if the index is odd as the previous part shows.
Lets briefly discuss the answer to the first part in the above example. In this part we made the claim that because . However, 4 isnt the
only number that we can square to get 16. We also have . So, why didnt we use -4 instead? There is a general rule about evaluating square roots (or more generally radicals with even indexes). When evaluating square roots we ALWAYS take the positive answer. If we want the negative answer we will do the following.
This may not seem to be all that important, but in later topics this can be very important. Following this convention means that we will always get predictable values when evaluating roots.
Note that we dont have a similar rule for radicals with odd indexes such as the cube root in part (d) above. This is because there will never be more than one possible answer for a radical with an odd index.
We can also write the general rational exponent in terms of radicals as follows.
Properties If n is a positive integer greater than 1 and both a and b are positive real numbers then,
1. 2.
3.
Note that on occasion we can allow a or b to be negative and still have these properties work. When we run across those situations we will acknowledge them. However, for the remainder of this section we will assume that a and b must be positive.
Also note that while we can break up products and quotients under a radical we cant do the same thing for sums or differences. In other words,
If you arent sure that you believe this consider the following quick number example.
If we break up the root into the sum of the two pieces we clearly get different answers! So, be careful to not make this very common mistake!
We are going to be simplifying radicals shortly so we should next define simplified radical form. A radical is said to be in simplified radical form (or just simplified form) if each of the following are true.
1. 2. 3. 4.
All exponents in the radicand must be less than the index. Any exponents in the radicand can have no factors in common with the index. No fractions appear under a radical. No radicals appear in the denominator of a fraction.
In our first set of simplification examples we will only look at the first two. We will need to do a little more work before we can deal with the last two.
Example 3 Simplify each of the following. Assume that x, y, and z are positive.
(a) (b)
(c)
(d)
(e) (f)
Solution
(a) In this case the exponent (7) is larger than the index (2) and so the first rule for simplification is violated. To fix this we will use the first and second properties of radicals above. So, lets note that we can write the radicand as follows.
So, weve got the radicand written as a perfect square times a term whose exponent is smaller than the index. The radical then becomes,
Now use the second property of radicals to break up the radical and then use the first property of radicals on the first term.
This now satisfies the rules for simplification and so we are done.
Before moving on lets briefly discuss how we figured out how to break up the exponent as we did. To do this we noted that the index was 2. We then determined the largest multiple of 2 that is less than 7, the exponent on the radicand. This is 6. Next, we noticed that 7=6+1.
Finally, remembering several rules of exponents we can rewrite the radicand as,
In the remaining examples we will typically jump straight to the final form of this and leave the details to you to check.
(b) This radical violates the second simplification rule since both the index and the exponent have a common factor of 3. To fix this all we need to do is convert the radical to exponent form do some simplification and then convert back to radical form.
(c) Now that weve got a couple of basic problems out of the way lets work some harder ones. Although, with that said, this one is really nothing more than an extension of the first example.
There is more than one term here but everything works in exactly the same fashion. We will break the radicand up into perfect squares times terms whose exponents are less than 2 (i.e. 1).
## Dont forget to look for perfect squares in the number as well.
Now, go back to the radical and then use the second and first property of radicals as we did in the first example.
Note that we used the fact that the second property can be expanded out to as many terms as we have in the product under the radical. Also, dont get excited that there are no xs under the radical in the final answer. This will happen on occasion.
(d) This one is similar to the previous part except the index is now a 4. So, instead of get perfect
squares we want powers of 4. This time we will combine the work in the previous part into one step.
## (e) Again this one is similar to the previous two parts.
In this case dont get excited about the fact that all the ys stayed under the radical. That will happen on occasion.
(f) This last part seems a little tricky. Individually both of the radicals are in simplified form. However, there is often an unspoken rule for simplification. The unspoken rule is that we should have as few radicals in the problem as possible. In this case that means that we can use the second property of radicals to combine the two radicals into one radical and then well see if there is any simplification that needs to be done.
## Now that its in this form we can do some simplification.
Before moving into a set of examples illustrating the last two simplification rules we need to talk briefly about adding/subtracting/multiplying radicals. Performing these operations with radicals is much the same as performing these operations with polynomials. If you dont remember how to add/subtract/multiply polynomials we will give a quick reminder here and then give a more in depth set of examples the next section.
Recall that to add/subtract terms with x in them all we need to do is add/subtract the coefficients of the x. For example,
Weve already seen some multiplication of radicals in the last part of the previous example. If we are looking at the product of two radicals with the same index then all we need to do is use the second property of radicals to combine them then simplify. What we need to look at now are problems like the following set of examples.
## Example 4 Multiply each of the following. Assume that x is positive.
(a)
(b)
(c) Solution In all of these problems all we need to do is recall how to FOIL binomials. Recall,
With radicals we multiply in exactly the same manner. The main difference is that on occasion well need to do some simplification after doing the multiplication
(a)
As noted above we did need to do a little simplification on the first term after doing the multiplication.
(b) Dont get excited about the fact that there are two variables here. It works the same way!
Again, notice that we combined up the terms with two radicals in them.
## (c) Not much to do with this one.
Notice that, in this case, the answer has no radicals. That will happen on occasion so dont get excited about it when it happens.
The last part of the previous example really used the fact that
If you dont recall this formula we will look at it in a little more detail in the next section.
Okay, we are now ready to take a look at some simplification examples illustrating the final two rules. Note as well that the fourth rule says that we shouldnt have any radicals in the denominator. To get rid of them we will use some of the multiplication ideas that we looked at above and the process of getting rid of the radicals in the denominator is called rationalizing the denominator. In fact, that is really what this next set of examples is about. They are really more examples of rationalizing the denominator rather than simplification examples.
Example 5 Rationalize the denominator for each of the following. Assume that x is positive.
(a)
(b)
(c)
(d) Solution There are really two different types of problems that well be seeing here. The first two parts illustrate the first type of problem and the final two parts illustrate the second type of problem. Both types are worked differently.
(a) In this case we are going to make use of the fact that . We need to determine what to multiply the denominator by so that this will show up in the denominator. Once we figure this out we will multiply the numerator and denominator by this term.
## Here is the work for this part.
Remember that if we multiply the denominator by a term we must also multiply the numerator
by the same term. In this way we are really multiplying the term by 1 (since and so arent changing its value in any way.
(b) Well need to start this one off with first using the third property of radicals to eliminate the fraction from underneath the radical as is required for simplification.
Now, in order to get rid of the radical in the denominator we need the exponent on the x to be a 5. This means that we need to multiply by so lets do that.
(c) In this case we cant do the same thing that we did in the previous two parts. To do this one we will need to instead to make use of the fact that
When the denominator consists of two terms with at least one of the terms involving a radical we will do the following to get rid of the radical.
So, we took the original denominator and changed the sign on the second term and multiplied the numerator and denominator by this new term. By doing this we were able to eliminate the radical in the denominator when we then multiplied out.
(d) This one works exactly the same as the previous example. The only difference is that both terms in the denominator now have radicals. The process is the same however.
Rationalizing the denominator may seem to have no real uses and to be honest we wont see many uses in an Algebra class. However, if you are on a track that will take you into a Calculus class you will find that rationalizing is useful on occasion at that level.
We will close out this section with a more general version of the first property of radicals. Recall that when we first wrote down the properties of radicals we required that a be a positive number. This was done to make the work in this section a little easier. However, with the first property that doesnt necessarily need to be the case.
## Here is the property for a general a (i.e. positive or negative)
where is the absolute value of a. All that you need to do is know at this point is that absolute value always makes a a positive number.
## For square roots this is,
This will not be something we need to worry all that much about here, but again there are topics in courses after an Algebra course for which this is an important idea so we needed to at least acknowledge it.
Lets briefly discuss the answer to the first part in the above example. In this part we made the claim that because . However, 4 isnt the
only number that we can square to get 16. We also have . So, why didnt we use -4 instead? There is a general rule about evaluating square roots (or more generally radicals with even indexes). When evaluating square roots we ALWAYS take the positive answer. If we want the negative answer we will do the following.
This may not seem to be all that important, but in later topics this can be very important. Following this convention means that we will always get predictable values when evaluating roots.
Note that we dont have a similar rule for radicals with odd indexes such as the cube root in part (d) above. This is because there will never be more than one possible answer for a radical with an odd index.
We can also write the general rational exponent in terms of radicals as follows.
Properties If n is a positive integer greater than 1 and both a and b are positive real numbers then,
1. 2.
3.
Note that on occasion we can allow a or b to be negative and still have these properties work. When we run across those situations we will acknowledge them. However, for the remainder of this section we will assume that a and b must be positive.
Also note that while we can break up products and quotients under a radical we cant do the same thing for sums or differences. In other words,
If you arent sure that you believe this consider the following quick number example.
If we break up the root into the sum of the two pieces we clearly get different answers! So, be careful to not make this very common mistake!
We are going to be simplifying radicals shortly so we should next define simplified radical form. A radical is said to be in simplified radical form (or just simplified form) if each of the following are true.
1. 2. 3. 4.
All exponents in the radicand must be less than the index. Any exponents in the radicand can have no factors in common with the index. No fractions appear under a radical. No radicals appear in the denominator of a fraction.
In our first set of simplification examples we will only look at the first two. We will need to do a little more work before we can deal with the last two.
Example 3 Simplify each of the following. Assume that x, y, and z are positive.
(a) (b)
[Solution]
[Solution]
(c)
[Solution]
(d)
[Solution]
(e) (f)
[Solution]
[Solution]
Solution
(a) In this case the exponent (7) is larger than the index (2) and so the first rule for simplification is violated. To fix this we will use the first and second properties of radicals above. So, lets note that we can write the radicand as follows.
So, weve got the radicand written as a perfect square times a term whose exponent is smaller than the index. The radical then becomes,
Now use the second property of radicals to break up the radical and then use the first property of radicals on the first term.
This now satisfies the rules for simplification and so we are done.
Before moving on lets briefly discuss how we figured out how to break up the exponent as we did. To do this we noted that the index was 2. We then determined the largest multiple of 2 that is less than 7, the exponent on the radicand. This is 6. Next, we noticed that 7=6+1.
Finally, remembering several rules of exponents we can rewrite the radicand as,
In the remaining examples we will typically jump straight to the final form of this and leave the details to you to check.
(b) This radical violates the second simplification rule since both the index and the exponent have a common factor of 3. To fix this all we need to do is convert the radical to exponent form do some simplification and then convert back to radical form.
(c) Now that weve got a couple of basic problems out of the way lets work some harder ones. Although, with that said, this one is really nothing more than an extension of the first example.
There is more than one term here but everything works in exactly the same fashion. We will break the radicand up into perfect squares times terms whose exponents are less than 2 (i.e. 1).
## Dont forget to look for perfect squares in the number as well.
Now, go back to the radical and then use the second and first property of radicals as we did in the first example.
Note that we used the fact that the second property can be expanded out to as many terms as we have in the product under the radical. Also, dont get excited that there are no xs under the radical in the final answer. This will happen on occasion.
(d) This one is similar to the previous part except the index is now a 4. So, instead of get perfect squares we want powers of 4. This time we will combine the work in the previous part into one step.
## (e) Again this one is similar to the previous two parts.
In this case dont get excited about the fact that all the ys stayed under the radical. That will happen on occasion.
(f) This last part seems a little tricky. Individually both of the radicals are in simplified form. However, there is often an unspoken rule for simplification. The unspoken rule is that we should have as few radicals in the problem as possible. In this case that means that we can use the second property of radicals to combine the two radicals into one radical and then well see if there is any simplification that needs to be done.
## Now that its in this form we can do some simplification.
Before moving into a set of examples illustrating the last two simplification rules we need to talk briefly about adding/subtracting/multiplying radicals. Performing these operations with radicals is much the same as performing these operations with polynomials. If you dont remember how to add/subtract/multiply polynomials we will give a quick reminder here and then give a more in depth set of examples the next section.
Recall that to add/subtract terms with x in them all we need to do is add/subtract the coefficients of the x. For example,
Weve already seen some multiplication of radicals in the last part of the previous example. If we are looking at the product of two radicals with the same index then all we need to do is use the second property of radicals to combine them then simplify. What we need to look at now are problems like the following set of examples.
## Example 4 Multiply each of the following. Assume that x is positive.
(a)
[Solution]
(b)
[Solution]
(c) Solution
[Solution]
In all of these problems all we need to do is recall how to FOIL binomials. Recall,
With radicals we multiply in exactly the same manner. The main difference is that on occasion well need to do some simplification after doing the multiplication
(a)
As noted above we did need to do a little simplification on the first term after doing the multiplication.
(b) Dont get excited about the fact that there are two variables here. It works the same way!
Again, notice that we combined up the terms with two radicals in them.
## (c) Not much to do with this one.
Notice that, in this case, the answer has no radicals. That will happen on occasion so dont get excited about it when it happens.
The last part of the previous example really used the fact that
If you dont recall this formula we will look at it in a little more detail in the next section.
Okay, we are now ready to take a look at some simplification examples illustrating the final two rules. Note as well that the fourth rule says that we shouldnt have any radicals in the denominator. To get rid of them we will use some of the multiplication ideas that we looked at above and the process of getting rid of the radicals in the denominator is called rationalizing the denominator. In fact, that is really what this next set of examples is about. They are really more examples of rationalizing the denominator rather than simplification examples.
Example 5 Rationalize the denominator for each of the following. Assume that x is positive.
(a)
[Solution]
(b)
[Solution]
(c)
[Solution]
(d) Solution
[Solution]
There are really two different types of problems that well be seeing here. The first two parts illustrate the first type of problem and the final two parts illustrate the second type of problem. Both types are worked differently.
(a) In this case we are going to make use of the fact that . We need to determine what to multiply the denominator by so that this will show up in the denominator. Once we figure this out we will multiply the numerator and denominator by this term.
## Here is the work for this part.
Remember that if we multiply the denominator by a term we must also multiply the numerator by the same term. In this way we are really multiplying the term by 1 (since and so arent changing its value in any way. )
(b) Well need to start this one off with first using the third property of radicals to eliminate the fraction from underneath the radical as is required for simplification.
Now, in order to get rid of the radical in the denominator we need the exponent on the x to be a 5. This means that we need to multiply by so lets do that.
(c) In this case we cant do the same thing that we did in the previous two parts. To do this one we will need to instead to make use of the fact that
When the denominator consists of two terms with at least one of the terms involving a radical we will do the following to get rid of the radical.
So, we took the original denominator and changed the sign on the second term and multiplied the numerator and denominator by this new term. By doing this we were able to eliminate the radical in the denominator when we then multiplied out.
(d) This one works exactly the same as the previous example. The only difference is that both terms in the denominator now have radicals. The process is the same however.
Rationalizing the denominator may seem to have no real uses and to be honest we wont see many uses in an Algebra class. However, if you are on a track that will take you into a Calculus class you will find that rationalizing is useful on occasion at that level.
We will close out this section with a more general version of the first property of radicals. Recall that when we first wrote down the properties of radicals we required that a be a positive number. This was done to make the work in this section a little easier. However, with the first property that doesnt necessarily need to be the case.
## Here is the property for a general a (i.e. positive or negative)
where is the absolute value of a. If you dont recall absolute value we will cover that in detail in a section in the next chapter. All that you need to do is know at this point is that absolute value always makes a a positive number.
## For square roots this is,
This will not be something we need to worry all that much about here, but again there are topics in courses after an Algebra course for which this is an important idea so we needed to at least acknowledge it.
## Now For Something Light:
Here is another multiplication trick for you. This is called, Russian Peasant Algorithm and it is a LOT of fun.
The way most people learn to multiply large numbers looks something like this:
86 x 57 -----602 + 4300 -----4902
If you know your multiplication facts, this "long multiplication" is quick and relatively simple. However, there are many other ways to multiply. One of these methods is often called the Russian peasant algorithm. You don't need multiplication facts to use the Russian peasant algorithm; you only need to double numbers, cut them in half, and add them up. Here are the rules:
Write each number at the head of a column. Double the number in the first column, and halve the number in the second column. If the number in the second column is odd, divide it by two and drop the remainder. If the number in the second column is even, cross out that entire row. Keep doubling, halving, and crossing out until the number in the second column is 1. Add up the remaining numbers in the first column. The total is the product of your original numbers.
## Let's multiply 57 by 86 as an example: Write each number at the head of a column. 57 86
Double the number in the first column, and halve the number in the second column. 57 114 86 43
If the number in the second column is even, cross out that entire row. 57 114 86 43
Keep doubling, halving, and crossing out until the number in the second column is 1. 57 114 228 456 912 1824 3648 86 43 21 10 5 2 1
Add up the remaining numbers in the first column. 57 114 228 456 912 1824 + 3648 4902 Real Russian peasants may have tracked their doublings with bowls of pebbles, instead of columns of numbers. (They probably weren't interested in problems as large as our example, though; four thousand pebbles would be hard to work with!) Russian peasants weren't the only ones to use this method of multiplication. The ancient Egyptians invented a similar process thousands of years earlier, and computers are still using related methods today. 86 43 21 10 5 2 1
## Why does Russian peasant multiplication work? Let's calculate 9 * 8 as an example: 9 18 36 72 8 4 2 1
72 is the only remaining number in the left-hand column, so our answer is 72. Notice that we were multiplying by 2 on one side, and by 1/2 on the other side. 2 * 1/2 = 1, so the overall product did not change:
9*8
= 18 * 4
= 36 * 2 = 72 * 1.
We were grouping numbers in a different way, not changing the answer. If we multiply 8 * 9, we should get the same answer. Can we explain our answer the same way? 8 16 9 4
32 + 64 72
2 1
When we cut 9 in half, we dropped the remainder because 9 is an odd number. Because we have "lost" a one, the product of each row should be smaller from now on. Let's find the difference between the first row and the second row: 8*9 - 16*4 = 72 - 64 = 8. We can rewrite the subtraction as a sum:
8*9
= 16 * 4 + 8.
Because our product has decreased by 8, we have to add 8 back in again at the end. We can think of the addition as restoring 1 group of 8, for the remainder of 1 that we dropped earlier. In a different problem, we might have to restore several different groups of numbers. | 6,150 | 27,656 | {"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} | 4.8125 | 5 | CC-MAIN-2019-35 | latest | en | 0.942404 |
https://help.sumologic.com/docs/metrics/metrics-operators/eval/ | 1,723,323,294,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640822309.61/warc/CC-MAIN-20240810190707-20240810220707-00116.warc.gz | 223,911,061 | 11,314 | # eval Metrics Operator
The eval operator evaluates a time series based on a user-specified arithmetic or mathematical function.
## Syntax
``eval expr([REDUCER BOOLEAN EXPRESSION | _value] [_granularity])``
• `expr` is basic arithmetic or mathematical function: +, -, *, /, sin, cos, abs, log, round, ceil, floor, tan, exp, sqrt, min, max
• `_value` is the placeholder for each data point in the time series.
• `REDUCER BOOLEAN EXPRESSION` is an expression that takes all the values of a given time series, uses a function to reduce them to a single value, and evaluates that value. The supported functions are:
• `avg`. Returns the average of the time series.
• `min`. Returns the minimum value in the time series.
• `max`. Returns the maximum value in the time series.
• `sum`. Returns the sum of the values in the time series.
• `count`. Returns the count of data points in the time series.
• `pct(n)`. Returns the nth percentile of the values in the time series.
• `latest`. Returns the last data point in the time series.
• `stddev`. Returns standard deviation of the points in the time series.
• `_granularity`. Returns the length of the quantization bucket in milliseconds. You can use this placeholder in your query.
## Examples
Example 1
This query returns the value of the `CPU_Idle` metric, multiplied by 100.
``metric=CPU_Idle | eval _value * 100``
Example 2
This query sets the value of each point in a single time series to the average of all values in that time series.
``metric=CPU_Idle | eval avg``
For example, if you have this series, where the points are `(timestamp, value)`:
``m1: (0, 1) (1, 2) (2, 3)m2: (0, 3) (1, 6) (2, 9)``
then `eval avg` would produce:
``m1: (0, 2) (1, 2) (2, 2)m2: (0, 6) (1, 6) (2, 6)``
Example 3
This query returns the rate of change per second for the metric.
``metric=CPU_Idle | sum | eval 1000 * _value / _granularity``
Status
Legal
Privacy Statement | 528 | 1,920 | {"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} | 2.828125 | 3 | CC-MAIN-2024-33 | latest | en | 0.690267 |
https://www.slideshare.net/Rohan04/cryptography-17257580 | 1,492,957,938,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917118707.23/warc/CC-MAIN-20170423031158-00637-ip-10-145-167-34.ec2.internal.warc.gz | 965,266,137 | 36,832 | Upcoming SlideShare
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# Cryptography
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### Cryptography
1. 1. CryptographyPresented To:- Presented By:-Mr. Anurag Sharma Rohan Jain , CS B ,1000210077
2. 2. Overview• Cryptography………………..……….An introduction• Objectives………………………………….Brief Aspect• Terminology……………..…..To Make You Familiar• How………………….…………………….Demonstration• Methods………………………….......Public & Private• Implementation………………..Hashing Algorithm• Cryptography in Networking Security………....…• Applications……………………………Real Life Scope
3. 3. Introduction The Word Cryptology is made up of “Kryptos", which means hidden and “Logos" which means word. In Laymen Words, it is an Art and science of protecting data. Technically, It involves logical transformation of information. The Principles of Cryptography are today applied to the encryption of fax ,television, and computer network communications and many other fields. Since the secure exchange of computer data is of great importance to banking, government, and commercial communications as well as for individuals.
4. 4. Objectives1)Confidentiality :-The information cannot be understood byanyone for whom it was unintended.2) Integrity :-The information cannot be altered in storage ortransit between sender and intended receiver without thealteration being detected. (Data Is Not Corrupted).3)Non-repudiation :-The creator/sender of the informationcannot deny at a later stage his or her intentions in thecreation or transmission of the information.4) Authentication :-The sender and receiver can confirm eachother’s identity and the origin/destination of the information.(Source Of Data IS Genuine).
5. 5. TerminologyR O H A N J A I N Plain Text(Can Be Variable Length) Encryption Using Key Using Algorithms (MD4,MD5,SHA-1,RSA) Cipher Text 00 B8 3c Ef G0 Xh 99 3d 2f Using Algorithm (Same As Decryption Using Same Used To Encrypt the Text) KeysR O H A N J A I N
6. 6. HOW…??A Simple Demonstration:-Substitution CipherTo Encode:-> S E C R E TKey :-> Offset the 3rd letter so the alphabets begin with it. So starting with:- ABCDEFGHIJKLMNOPQRSTUVWXYZ and sliding everything by 3, we get:- DEFGHIJKLMNOPQRSTUVWXYZABC So D=A, E=B, F=C…..and so on.Encoded:-> V H F U H WTo Decode:-> Provide anyone the key i.e., =>Offset the 3rd letter so the alphabets begin with it.
7. 7. Cryptography Methods : Modern Cryptographyo Symmetric-Key Cryptography.o Asymmetric-Key Cryptography.o Cryptanalysis.
8. 8. Private(Symmetric) CryptographyIn symmetric-key encryption each end already has a secret key(code) that it can use to encrypt a packet of information before itis sent over the network to another computer.
9. 9. Private Cryptography Methods:-DES (Data Encryption Standard) AES(Advanced Encryption Standard)Older NewerBreakable UnbreakableSmaller Key (56-bit Encryption). Bigger Key(128/192bit /256 bit Encryption).7*10^16 Key Combinations. 3*10^35 Key Combinations.Smaller Block Size (64 bits). Larger Block Size (128bits).For DES with 64 bits, the maximum amount For AES with 128 bits, the maximumof data that can be transferred with a single amount of data that can be transferredencryption key is 32GB. with a single encryption key is 256 EB.
10. 10. Public/Asymmetric CryptographyAsymmetric/Public encryption uses two different keys at oncei.e., combination of a private key and a public key. The privatekey is known only to your computer while the public key isgiven by your computer to any computer that wants tocommunicate securely with it. To decode anencrypted Message acomputer must use thepublic key provided byoriginating computer,and its own private key
11. 11. Cryptanalysis The Study of methods to break Cryptosystems. Often targeted at obtaining a key. Cryptanalysis Attacks:-o Brute force o Trying all key values in the keyspace.o Frequency Analysis o Guess values based on frequency of occurrence.o Dictionary Attack o Find plaintext based on common words.
12. 12. Implementation of Encryption Keys :Hash FunctionA hash function is any algorithm or subroutine that maps largedata sets of variable length to smaller data sets of a fixedlength. For example, a persons name, having a variablelength, could be hashed to a single integer.Basic Idea:-Input Number 10,667Hashing Function Input# x 143Hash Value 1,525,381Public keys generally use more complex algorithms and verylarge hash values for encrypting, including 40-bit or even 128-bit numbers. A 128-bit number has a possible 2128.
13. 13. • The values returned by a hash function are called hash values, hash codes, digest ,hash sums, checksums or simply hashes.• A Cryptographic hash function (specifically, SHA-1) at work. Note that even small changes in the source input (here in the word "over") drastically change the resulting output. V U
14. 14. Cryptography in Networking Transport Layer Security (TLS) and its predecessor, Secure Sockets Layer (SSL), are cryptographic protocols that provide communication security over the Internet. Several versions of the protocols are in widespread use in applications such as web browsing, electronic mail, Internet faxing, instant messaging and voice-over-IP (VoIP). When youre accessing sensitive information, such as an online bank account or a payment transfer service like PayPal or Google Checkout.
15. 15. The client request the SSL connection by sending the request.Server provides it’s secure certificate to client to show it’sauthenticity.Client validates the certificate and request a one time sessionwith server.Server completes the SSL handshake and the session begins.
16. 16. Applications• ATM Cards• E-Commerce• Computer Passwords• Electronic Fund Transfer• Digital Signatures• Network Security• Storage Integrity
17. 17. References http://en.wikipedia.org/wiki/Cryptographyhttp://en.wikipedia.org/wiki/Cryptographic_hash functionhttp://computer.howstuffworks.com/encryption.ht mhttp://en.wikipedia.org/wiki/Secure_Sockets_Layer
18. 18. Queries ?? | 1,478 | 6,205 | {"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-2017-17 | latest | en | 0.809192 |
https://www.physicsforums.com/threads/significance-of-spring-mass-in-shm.275961/ | 1,610,868,018,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703509973.34/warc/CC-MAIN-20210117051021-20210117081021-00642.warc.gz | 944,672,450 | 13,439 | # Significance of spring mass in SHM.
## Homework Statement
So, we are considering a spring-mass system, in which the mass at the end of the spring, M, is comparable to the mass of the spring, m.
Using Newtons laws, I have to calculate, how significant the mass of the spring is.
## Homework Equations
Mass at the end of the spring, M.
Mass of the spring, m.
Spring constant, k.
Newtons second law, and Hooke's law.
## The Attempt at a Solution
Actually, I have tried quite lot different approaches, but they don't seem to give me anything useful.
My latest attempt was to take a differential piece of mass of the spring, and calculate it's acceleration, in hope of getting something which i could integrate, but it didn't seem to work out.
I was asked this question by a high school student, whom I have to help writing a larger assignment.
I solved this problem rather easily using Lagrangian mechanics, but this is not available to the student, so I have to do it with Newtonian mechanics, which doesn't seem too easy.
I really appreciate any help I can get.
## Answers and Replies
LowlyPion
Homework Helper
Welcome to PF.
I would think if you know the length of the spring that you can calculate the kinetic energy along its length by integration on the basis of the velocities all along it's length. Then you can relate that to the kinetic energy of the attached mass at the end. | 314 | 1,395 | {"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-2021-04 | latest | en | 0.974776 |
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in the given figure, what value of x will make POQ a straight line
Asked by Prashanth Balasubramaniam | 13 Oct, 2012, 10:47: AM
For POQ to be a straight line , we have
?POR + ?QOR = 1800
Given that ?POR = 4x-36 and ?QOR = 3x-20
Therefore, we get
4x - 36 + 3x - 20 = 180
7x - 56 = 180
7x = 236
x= 33.70
Therefore, POQ will be a straight line if x = 33.70
Answered by | 13 Oct, 2012, 06:55: PM
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# Re: Partial Functions and Logics
Patrice Chalin writes:
[.. stuff about Larch deleted ...]
Consider a Z version of the first example used in [Jon95]. Actually, because
of the way types are defined I must slightly change the example (reasons
for doing so are given in Note below).
[Unit]
axdef
u: Unit
where
Unit = {u}
end
axdef
f: ZZ -+> Unit
where
f i = if i = 0 then u else f (i - 1)
end
This spec is ill formed, because there is no declaration of the
variable i. But let's suppose that you wrote (forall i : ZZ @ ...)
From this specification (let us call it SPEC1) we can prove that
forall i: ZZ @ f i = u
This property follows from the fact that the value of a term is
always an element of the carrier set of the type of the term
[Spi88, pp.61--62,].
This isn't so: the value of an expression is a member of the carrier
of its type only if the expression is defined. Expressions can also
be undefined.
In this example, as always, we have to ask what functions f satisfy
your specification. Certainly the (total) constant function from ZZ
to Unit satisfies it; but so may some other functions -- we just can't
tell, because Z does not tell us whether E1 = E2 is true or false in
the case where one or both of them is undefined.
[...]
In Z, functions are modeled as sets of ordered pairs.
Agreed.
SPEC1 does not uniquely determine f: there are many
sets satisfying the definition of f in SPEC1; viz. { },
{(0,u)}, {(1,u), (-1,u)}, etc.
Actually, SPEC1 may or may not determine f, depending on the answers
To consider a simpler example,
axdef
g: ZZ -+> ZZ
where
g 0 = 0
end
does not allow us to deduce that (0,0) \in g unless we know that
0 \in \dom g.
True
By the customary Z stylistic guidelines'', the SPEC1 definition of f
would be considered incomplete. Spivey writes, Partial functions may be
defined by giving their domain and their value for each argument in the
domain''; by doing so we completely determine the partial function being
defined [Spi89, p.43,]. Thus, in general, partial functions can be defined
using recursive equations only if their domains have been fixed (outside of
the recursive equation). Hence, a proper'' definition of f would be, e.g.
axdef
f: ZZ -+> Unit
where
dom f = {- 1, 0, 1}
f i = if i = 0 then u else f (i - 1)
end
Again a missing quantifier here -- it should be (forall i: {0, 1} @ ...)
I think.
In which case there is a unique solution f = {(-1,u),(0,u),(1,u)}.
OK.
This brings me to an interesting observation that I made several years ago.
In the semantics of programming languages one tends to require that, e.g., a
function definition, have a single denotation. For example, the denotation
of F in
function F(i:integer):integer
begin
if i = 0 then F := 0
else F := F(i-1)
end;
would be the appropriate least fixed point: {i: @ (i,u)} {i: @
(-i,u)}. In contrast to this approach, it would seem that in the
semantics of specifications languages, we tend to speak of the
structures (or models) that satisfy a given specification. For
example, there are many (nonisomorhpic) models that satisfy SPEC1:
in particular one model will assign { } to f, another will assign
{(0,u)} to f, etc.
Again, we do not know whether these are models or not
Hence we do not seek a single denotation for
components of a specification. With respect to a given
specification, what usually concerns us are those properties that
are logical consequences of that specification: i.e. properties
that hold in all models that satisfy the specification. Thus, the
specification of g does not allow us to deduce that (0,0) \in g
since this statement does not hold in all models that satisfy the
definition of g.
This is a good point, but one that is slightly undermined by Z's
treatment of partial functions.
... it would seem on closer investigation that Z' might have some
difficulty in fixing denotations of its recursive functions.'' [Jon95]
In the light of what I stated above, it would not seem necessary
that Z' fix denotations for its recursively defined functions
since this is not the kind of interpretation that is applied to Z
specifications.
It's true that Z does not necessarily fix on one interpretation of a
recursive definition, but that doesn't make the problem go away. We
want to develop progams that contain recursive functions, and we'll
need to prove that the recursion terminates. All that Z seems to say is
that this is a completely separate problem from defining functions in
mathematics!
Reference(s): | 1,151 | 4,511 | {"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} | 3.375 | 3 | CC-MAIN-2023-06 | latest | en | 0.90289 |
https://www.physicsforums.com/threads/coefficients-of-expansion-and-compressibility.369516/ | 1,508,720,328,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825497.18/warc/CC-MAIN-20171023001732-20171023021732-00686.warc.gz | 959,669,668 | 16,720 | # Coefficients of expansion and compressibility
1. Jan 14, 2010
### Phyisab****
1. The problem statement, all variables and given/known data
I need to derive the following relation:
$$\alpha_{S}=\alpha_{P}\frac{1}{1-\frac{\kappa_{T}}{\kappa_{S}}}$$
2. Relevant equations
Hopefully you can see that my notation |P means at constant pressure, I could not find a better way to do this, any ideas?
$$\alpha_{S}=\frac{1}{V}\frac{\partial V}{\partial T} |P$$
$$\alpha_{P}=\frac{1}{V}\frac{\partial V}{\partial T} |S$$
$$\kappa_{T}=\frac{1}{V}\frac{\partial V}{\partial P} |T$$
$$\kappa_{S}=\frac{1}{V}\frac{\partial V}{\partial P} |S$$
3. The attempt at a solution
I have no intuition about this problem, so i have just been trying everything I can think of and nothing works. I think the first step is pretty obvious:
$$\alpha_{s}=\frac{\alpha_{P}}{\frac{\partial V}{\partial T}|P}\frac{\partial V}{\partial T}|S$$
after this it just seems like a guessing game, trying to apply the proper identity to lead me to the answer. Note that I used one of the Maxwell relations somewhere in here. Here is my last ditch effort. I'm pretty sure the identity I made up is not true, but this seems to have gotten me closest to the answer, and it would take me days to type up all my false leads.
$$dV = \frac{\partial V}{\partial T}|P dT + \frac{\partial V}{\partial P}|T dP$$
From here, I made the almost certainly false conclusion that
$$\frac{\partial V}{\partial T}|S = \frac{\partial V}{\partial T}|P + \frac{\partial V}{\partial P}|T\frac{\partial P}{\partial T}|S$$
Plugging this back into my first step gives:
$$\alpha_{s}= \frac{\alpha_{P}}{1 + \frac{V\kappa_{T}\frac{\partial P}{\partial T}|S}{\frac{\partial V}{\partial T}|S}}$$
Which seems very close to me, but I still can't finish it off and I'm pretty sure I cheated to get there anyway. Any help would be so greatly appreciated, this has got to be a pretty simple problem and it is driving me insane!
2. Jan 14, 2010
### Mapes
Your "almost certainly false conclusion" is fine; you just differentiated both sides with respect to T at constant S. (Try working it through carefully with the chain rule to prove this to yourself, noting that lone differential terms like dT can be assumed to be negligible when compared to partial derivatives.) Try proceeding from there, using partial derivatives and Maxwell relations as necessary.
3. Jan 14, 2010
### Phyisab****
That means the following variations on the fundamental identity all apply:
$$\frac{\partial V}{\partial T}|S=\frac{\partial V}{\partial T}|P+\frac{\partial V}{\partial P}|T\frac{\partial P}{\partial T}|S$$
$$\frac{\partial V}{\partial P}|S=\frac{\partial V}{\partial P}|T+\frac{\partial V}{\partial T}|P\frac{\partial T}{\partial P}|S$$
$$0=\frac{\partial S}{\partial T}|P+\frac{\partial S}{\partial P}|T\frac{\partial P}{\partial T}|S$$
And also the Maxwell relations. Are there others I haven't thought of? I just can't see this problem as anything other than an exercise in pure trial and error.
4. Jan 14, 2010
### Mapes
Trial and error at the beginning, but increased savviness and intuition about which identities to use by the end, after working through many problems.
5. Jan 14, 2010
### Phyisab****
Here is where I am now:
$$\alpha_{S}=\frac{\alpha_{P}}{1-\frac{\frac{\kappa_{T}}{\kappa_{S}}\frac{\partial P}{\partial T}|S \frac{\partial V}{\partial P}|S}{\frac{\partial V}{\partial T}|S}}$$
Now all that is left is to show that
$$\frac{\frac{\partial P}{\partial T}|S \frac{\partial V}{\partial P}|S}{\frac{\partial V}{\partial T}|S}$$
is equal to one. No problem right? How deep can this rabbit hole possibly go.
$$\frac{\frac{\partial P}{\partial T}|S \frac{\partial V}{\partial P}|S}{\frac{\partial V}{\partial T}|S}$$
$$=\frac{\left(\frac{\partial V}{\partial P}|T+\frac{\partial V}{\partial T}|P\frac{T}{P}|S\right)\left( \frac{\frac{\partial V}{\partial T}|S-\frac{\partial V}{\partial T}|P}{\frac{\partial V}{\partial P}|T}\right)}{\frac{\partial V}{\partial T}|S}$$
(To Be Continued...)
6. Jan 14, 2010
### Mapes
Surely it is no problem to show that
$$\left(\frac{\partial P}{\partial T}\right)_S \left(\frac{\partial V}{\partial P}\right)_S \left(\frac{\partial T}{\partial V}\right)_S=1$$
!!!
7. Jan 14, 2010
### Phyisab****
Thanks a ton Mapes. But clearly I am not thinking about all this as well as I could. I knew
$$\left(\frac{\partial P}{\partial T}\right)|V \left(\frac{\partial V}{\partial P}\right) |T \left(\frac{\partial T}{\partial V}\right)|P=1$$,
which is clearly related to what you just said. From what should I try to derive this relation?
8. Jan 14, 2010
### Phyisab****
Nevermind I got it thanks again mapes. | 1,416 | 4,713 | {"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} | 3.390625 | 3 | CC-MAIN-2017-43 | longest | en | 0.895743 |
https://www.homeworklib.com/question/1781946/in-all-parts-of-this-problem-assume-the-planet-is | 1,620,568,149,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988986.98/warc/CC-MAIN-20210509122756-20210509152756-00188.warc.gz | 822,677,317 | 15,861 | # In all parts of this problem, assume the planet is located at its mean distance from...
In all parts of this problem, assume the planet is located at its mean distance from the Sun as shown in Chapter 5 of the textbook. (These same distances can be found at http://theanswermachine.tripod.com/id2.html.)
How long would it take a message sent as radio waves from Earth to reach the following?
(a) Mars when it is nearest to Earth
minutes
(b) Pluto when farthest from the Earth
minutes
When nearest - 33.9 million miles - 3.03 minutes.
at the farthest- 294.16 million miles - 26.35 minutes.
Both have elliptical orbits, hence the great difference in the distance.
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Free Homework App | 1,551 | 6,426 | {"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-2021-21 | latest | en | 0.926583 |
https://www.maplesoft.com/support/help/errors/view.aspx?path=GroupTheory%2FSearchPerfectGroups | 1,680,403,885,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950373.88/warc/CC-MAIN-20230402012805-20230402042805-00482.warc.gz | 972,196,056 | 41,650 | SearchPerfectGroups - Maple Help
Home : Support : Online Help : Mathematics : Group Theory : SearchPerfectGroups
GroupTheory
SearchPerfectGroups
search for perfect groups satisfying specified properties
Calling Sequence SearchPerfectGroups(spec, formopt)
Parameters
spec - expression sequence of search parameters formopt - (optional) an option of the form form = X, where X is one of "id" (the default), "permgroup", "fpgroup" or "count".
Description
• The SearchPerfectGroups(spec) command searches Maple's database of perfect groups for groups satisfying properties specified in a sequence spec of search parameters. The valid search parameters may be grouped into several classes, as follows.
• Use the form = X option to control the form of the output from this command. By default, an expression sequence of IDs for the PerfectGroups database is returned. This is the same as specifying form = "id". To have an expression sequence of groups, either permutation groups, or finitely presented groups, use either the form = "permgroup" or form = "fpgroup" options, respectively. Finally, the form = "count" option causes SearchPerfectGroups to return just the number of groups in the database satisfying the constraints implied by the search parameters.
• Note that the IDs returned in the default case are the IDs of the groups within the PerfectGroups database. These may differ from the IDs for the same group if it happens to be present in another database, such as the SmallGroups database, which has its own set of group IDs.
Boolean Search Parameters
• Boolean search parameters p, such as simple, can be specified in one of the forms p = true, p = false, or just p (which is equivalent to p = true). If the boolean search parameter p is true, then only groups satisfying the corresponding predicate are returned. If the boolean search parameter p is false, then only groups that do not satisfy the predicate are returned. Leaving a boolean search parameter unspecified causes the SearchPerfectGroups command to return groups that do, and do not, satisfy the corresponding predicate.
• The supported boolean search parameters are described in the following table.
simple describes the class of simple groups quasisimple describes the class of quasi-simple groups indecomposable describes the class of directly indecomposable groups frobenius describes the class of Frobenius groups perfectorderclasses describes the class of groups with perfect order classes
Numeric Search Parameters
• Maple supports search parameters that describe numeric invariants of finite groups. All have positive integral values. A numeric search parameter p may be given in the form p = n, for some specific value n, or by indicating a range, as in p = a .. b. In the former case, only groups for which the numeric parameter has the value n will be returned. In the case in which a range is specified, groups for which the numeric invariant lies within the indicate range (inclusive of its end-points) are returned. In addition, inequalities of the form p < n (p > n) or p <= n (p >= n) are supported.
• The supported numeric search parameters are listed in the following table.
order indicates the order (cardinality) of the group classnumber indicates the number of conjugacy classes of the group compositionlength indicates the number of composition factors of the group degree indicates the degree of the permutation representation stored in the perfect groups database elementordersum indicates the sum of the orders of the elements of the group maxelementorder indicates the largest among the orders of the elements of the group
Subgroup Group Search Parameters
• A subgroup of a perfect group is typically not perfect. Therefore, subgroups of perfect groups are indicated by using their ID from the database of small groups.
• Several subgroup search parameters are supported. These describe the isomorphism type of various subgroups of a group by specifying the Small Group ID (as returned by the IdentifySmallGroup command).
• For a subgroup or quotient search parameter p, passing an equation of the form p = [ord,id] causes the SearchSmallGroups command to return only groups whose subgroup, or quotient group, corresponding to p are isomorphic to the small group ord/id to be returned. Passing an equation of the form p = ord causes the SearchSmallGroups command to return only groups whose subgroup, or quotient group, corresponding to p have order ord.
• The following table describes the supported subgroup search parameters.
center specifies the SmallGroup ID (or order) of the center of a group socle specifies the SmallGroup ID (or order) of the socle of a group fittingsubgroup specifies the SmallGroup ID (or order) of the Fitting subgroup of a group
• It is important to understand that the option values for subgroups are the IDs within the small groups database, while the IDs returned by the SearchPerfectGroups command are the IDs of groups within the perfect groups database.
Quotient Group Search Parameters
• Quotient groups of perfect groups are perfect so, unlike subgroups, quotient groups are specified by their IDs in the perfect groups database.
centralquotient specifies the PerfectGroup ID (or order) of the central quotient of a group
Examples
> $\mathrm{with}\left(\mathrm{GroupTheory}\right):$
Count the total number of groups in the database.
> $\mathrm{SearchPerfectGroups}\left('\mathrm{form}'="count"\right)$
${1097}$ (1)
Find the quasisimple perfect groups up to order $500$ that are not simple.
> $\mathrm{SearchPerfectGroups}\left(\mathrm{order}=1..500,'\mathrm{simple}'=\mathrm{false},'\mathrm{quasisimple}'\right)$
$\left[{120}{,}{1}\right]{,}\left[{336}{,}{1}\right]$ (2)
Find the quasisimple perfect groups up to order $500$ that are not simple, returned as finitely presented groups.
> $\mathrm{SearchPerfectGroups}\left(\mathrm{order}=1..500,'\mathrm{simple}'=\mathrm{false},'\mathrm{quasisimple}','\mathrm{form}'="fpgroup"\right)$
$⟨{}{\mathrm{a1}}{,}{\mathrm{a2}}{,}{\mathrm{a3}}{}{\mid }{}{{\mathrm{a3}}}^{{2}}{,}{{\mathrm{a1}}}^{{2}}{}{{\mathrm{a3}}}^{{-1}}{,}{{\mathrm{a2}}}^{{3}}{,}{{\mathrm{a3}}}^{{-1}}{}{{\mathrm{a2}}}^{{-1}}{}{\mathrm{a3}}{}{\mathrm{a2}}{,}{\mathrm{a1}}{}{\mathrm{a2}}{}{\mathrm{a1}}{}{\mathrm{a2}}{}{\mathrm{a1}}{}{\mathrm{a2}}{}{\mathrm{a1}}{}{\mathrm{a2}}{}{\mathrm{a1}}{}{\mathrm{a2}}{}⟩{,}⟨{}{\mathrm{a1}}{,}{\mathrm{a2}}{,}{\mathrm{a3}}{}{\mid }{}{{\mathrm{a3}}}^{{2}}{,}{{\mathrm{a1}}}^{{2}}{}{{\mathrm{a3}}}^{{-1}}{,}{{\mathrm{a2}}}^{{3}}{,}{{\mathrm{a3}}}^{{-1}}{}{{\mathrm{a2}}}^{{-1}}{}{\mathrm{a3}}{}{\mathrm{a2}}{,}{\mathrm{a1}}{}{\mathrm{a2}}{}{\mathrm{a1}}{}{\mathrm{a2}}{}{\mathrm{a1}}{}{\mathrm{a2}}{}{\mathrm{a1}}{}{\mathrm{a2}}{}{\mathrm{a1}}{}{\mathrm{a2}}{}{\mathrm{a1}}{}{\mathrm{a2}}{}{\mathrm{a1}}{}{\mathrm{a2}}{,}{{\mathrm{a1}}}^{{-1}}{}{{\mathrm{a2}}}^{{-1}}{}{\mathrm{a1}}{}{\mathrm{a2}}{}{{\mathrm{a1}}}^{{-1}}{}{{\mathrm{a2}}}^{{-1}}{}{\mathrm{a1}}{}{\mathrm{a2}}{}{{\mathrm{a1}}}^{{-1}}{}{{\mathrm{a2}}}^{{-1}}{}{\mathrm{a1}}{}{\mathrm{a2}}{}{{\mathrm{a1}}}^{{-1}}{}{{\mathrm{a2}}}^{{-1}}{}{\mathrm{a1}}{}{\mathrm{a2}}{}{{\mathrm{a3}}}^{{-1}}{}⟩$ (3)
Find the quasisimple perfect groups up to order $500$ that are not simple, returned as permutation groups.
> $\mathrm{SearchPerfectGroups}\left(\mathrm{order}=1..500,'\mathrm{simple}'=\mathrm{false},'\mathrm{quasisimple}','\mathrm{form}'="permgroup"\right)$
$⟨\left({1}{,}{2}{,}{5}{,}{3}\right)\left({4}{,}{7}{,}{6}{,}{8}\right)\left({9}{,}{13}{,}{11}{,}{14}\right)\left({10}{,}{15}{,}{12}{,}{16}\right)\left({17}{,}{19}{,}{18}{,}{20}\right)\left({21}{,}{24}{,}{23}{,}{22}\right){,}\left({1}{,}{4}{,}{2}\right)\left({3}{,}{5}{,}{6}\right)\left({7}{,}{9}{,}{10}\right)\left({8}{,}{11}{,}{12}\right)\left({13}{,}{16}{,}{17}\right)\left({14}{,}{15}{,}{18}\right)\left({19}{,}{21}{,}{22}\right)\left({20}{,}{23}{,}{24}\right){,}\left({1}{,}{5}\right)\left({2}{,}{3}\right)\left({4}{,}{6}\right)\left({7}{,}{8}\right)\left({9}{,}{11}\right)\left({10}{,}{12}\right)\left({13}{,}{14}\right)\left({15}{,}{16}\right)\left({17}{,}{18}\right)\left({19}{,}{20}\right)\left({21}{,}{23}\right)\left({22}{,}{24}\right)⟩{,}⟨\left({1}{,}{2}{,}{5}{,}{3}\right)\left({4}{,}{7}{,}{6}{,}{8}\right)\left({9}{,}{13}{,}{11}{,}{14}\right)\left({10}{,}{15}{,}{12}{,}{16}\right){,}\left({1}{,}{4}{,}{2}\right)\left({3}{,}{5}{,}{6}\right)\left({7}{,}{9}{,}{10}\right)\left({8}{,}{11}{,}{12}\right){,}\left({1}{,}{5}\right)\left({2}{,}{3}\right)\left({4}{,}{6}\right)\left({7}{,}{8}\right)\left({9}{,}{11}\right)\left({10}{,}{12}\right)\left({13}{,}{14}\right)\left({15}{,}{16}\right)⟩$ (4)
Count the number of non-simple, quasisimple perfect groups in the database.
> $\mathrm{SearchPerfectGroups}\left('\mathrm{simple}'=\mathrm{false},'\mathrm{quasisimple}','\mathrm{form}'="count"\right)$
${54}$ (5)
Find the quasi-simple perfect groups in the databaswe whose centre has order equal to $4$.
> $\mathrm{SearchPerfectGroups}\left('\mathrm{quasisimple}','\mathrm{centre}'=4\right)$
$\left[{80640}{,}{2}\right]{,}\left[{80640}{,}{3}\right]{,}\left[{80640}{,}{4}\right]{,}\left[{116480}{,}{1}\right]$ (6)
Find the quasi-simple perfect groups in the databaswe whose centre is isomorphic to SmallGroup( 4,2 ).
> $\mathrm{SearchPerfectGroups}\left('\mathrm{quasisimple}','\mathrm{centre}'=\left[4,2\right]\right)$
$\left[{80640}{,}{2}\right]{,}\left[{116480}{,}{1}\right]$ (7)
> $\mathrm{IdentifySmallGroup}\left(\mathrm{Centre}\left(\mathrm{PerfectGroup}\left(80640,2\right)\right)\right)$
${4}{,}{2}$ (8)
Take care to recognise the difference between the ID of a group in the perfect groups database and the (normally different!) ID in the small groups database when a group is present in both databases.
For example, the alternating group of degree $5$ has ID [ 60, 5 ] in the database of small groups (it is the fifth group of order $60$) , but ID [60, 1] in the perfect groups database (it is the first perfect group of order $60$).
> $\mathrm{IdentifySmallGroup}\left(\mathrm{Alt}\left(5\right)\right)$
${60}{,}{5}$ (9)
> $\mathrm{SearchPerfectGroups}\left(\mathrm{socle}=\left[60,5\right]\right)$
$\left[{60}{,}{1}\right]$ (10)
> $\mathrm{IdentifySmallGroup}\left(\mathrm{PerfectGroup}\left(60,1\right)\right)$
${60}{,}{5}$ (11)
Compatibility
• The GroupTheory[SearchPerfectGroups] command was introduced in Maple 2019. | 3,080 | 10,431 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 30, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-14 | latest | en | 0.71029 |
https://brilliant.org/problems/circles-that-touch-the-axis/ | 1,537,870,437,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267161350.69/warc/CC-MAIN-20180925083639-20180925104039-00129.warc.gz | 471,427,773 | 11,730 | # Circles that touch the axis
Level 2
A circle passes through the point $$(2,-4)$$ and touches both the x-axis and the y-axis. If the centers of the two circles which satisfy these conditions can be represented as $$(a,b)$$ and $$(c,d)$$,what is the value of $$a-b+c-d$$?
× | 79 | 276 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-39 | latest | en | 0.889164 |
https://www.physicsforums.com/threads/speed-of-catching-a-falling-ruler.395600/ | 1,544,997,777,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827998.66/warc/CC-MAIN-20181216213120-20181216235120-00376.warc.gz | 995,690,027 | 16,073 | # Homework Help: Speed of catching a falling ruler
1. Apr 15, 2010
### bobbajobb
1. The problem statement, all variables and given/known data
Okay, I've been set a piece of coursework to compare the reaction times of catching a ruler, with, or without a certain variable. (I've chosen either eye.)
I'm having trouble working out the exact amount of time it takes the person to catch the ruler. (Yeah, I have attempted it. :) )
2. Relevant equations
(Explained in section 3.)
3. The attempt at a solution
The ruler weighs 0.025kg.
I hold the ruler at 1.04m. (An average of 10 times holding the ruler at a neutral position)
To calculate the gravitational potential energy I do:
0.025 x 1.04 x 10
This gives me = 0.26 (Joules)
The ruler falls 12 cm, (the person caught it at 18cm)
So 0.025 x 0.12 x 10 = the kinetic energy = 0.03
The formula for speed is :
Speed squared = kinetic energy/(0.5 times the weight)
0.03 divided by 0.0125 = 2.4
The square root of 2.4 = 1.549193338
To two decimal places = 1.55m/s
And if the ruler travelled 12 cm at 1.55m/s
0.12/1.55 = 0.0774 seconds……
I must have gone wrong somewhere...because I swear its 'physically' (pun intended) impossible for them to catch it that quickly?
2. Apr 15, 2010
### Stonebridge
If the ruler falls 12cm you find the time taken using
S=ut + ½gt²
s=12cm, that is 0.12m
g=9.8m/s²
and u=0 if it started from rest.
The reaction time will be between 1 and 2 tenths of a second, which seems about right.
3. Apr 15, 2010
### bobbajobb
Sorry, would you mind using words to write it out? This is only GCSE so i'm not too good with all the proper letters and stuff :(
4. Apr 15, 2010
### snshusat161
so, you are using work- kinetic energy theorem. I'm trying to understand your problem, if anywhere I go wrong then let me know by posting here.
The ruler weighs 0.025kg.
so m = 0.025 kg
I hold the ruler at 1.04m.
so h = 1.04 m
Correct!
I'm unable to understand this sentence and your solution from here. And most probably you are wrong from here. Also remember that potential energy is calculated from the bottom. I mean height is calculated from the ground level. so if the body falls 12 cm then potential energy will be mg (h -12). Secondly, loss in potential energy is equal to gain in kinetic energy.
5. Apr 15, 2010
### bobbajobb
Ahhh sorry. The way the test works is that I hold the '30cm' mark directly above their fingers, so its almost touching. So the person's fingers are on '18cm' on the ruler.
6. Apr 15, 2010
### bobbajobb
And as you might have seen. I don't actually use the potential energy at all, I'm just showing that I know how to, to gain extra marks.
7. Apr 15, 2010
### Stonebridge
Have you done the equations of uniform motion at GCSE yet? See below.
If you are trying to use kinetic and potential energy considerations you will not get the time directly.
There is a set of 4 equations you can use to solve problems of objects accelerating when you know some of these:
s=distance travelled
t=time taken
u=initial velocity
v=final velocity
a=acceleration
I mentioned one of them in my post. This was the one to use in this case. g is acceleration due to gravity, =9.8m/s/s
8. Apr 15, 2010
### snshusat161
okay, then potential energy at the point where you stop the scale is
= 0.025 * 10 * 0.92
= 0.23 Joule
Now loss in energy = 0.26 - 0.23
=0.03 Joule
This much energy is converted into kinetic energy
0.03 = (1/2) 0.025. v^2
v = sqrt of (2 * 0.03)/0.023
Now try to calculate. But remember don't trust blindly on me cause I'm noob like you in physics.
9. Apr 15, 2010
### snshusat161
on second line 0.92 is because the height of the scale from the ground is (1.04 - 0.12) m
10. Apr 15, 2010
### Staff: Mentor
The weight or mass and energy don't enter into this problem at all. Basically all you're doing is measuring how far the ruler falls before the other person catches it. From this distance you can calculate the person's reaction time. The governing equation is s = (1/2)gt2, with g ~ 9.8 m/sec2.
11. Apr 15, 2010
### bobbajobb
Guys I'm seriously confused now :(
So is what I'm doing in my original post....correct? Or at least somewhat? I don't need to go into detail about drag or anything for GCSE do i...?
12. Apr 15, 2010
### bobbajobb
What I'm trying to say is....have I just gone about it in a different way? Could someone use the g ~ 9.8m/sec^2 thing for my example in my first post and see if it gets the same outcome?
13. Apr 15, 2010
### snshusat161
so I should continue with my last post.
after calculation, I've got v = 1.61 m/s
Now v = u + a.t
1.6 = 0 + 10 t
t = 1.6/10
t = 0.16 second.
Or you can also get this result by using equation
v^2 = u^2 + 2.a.s
v^2 = 2*10*0.12
then apply,
v = u + a.t
Or simply you can use as told by most of here,
s = u*t + (1/2) a* t^2
=> s = (1/2)a * t^2
=> 0.12 = (1/2) 10 * t^2
=> t = 0.154 second
Last edited: Apr 15, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | 1,520 | 5,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2018-51 | latest | en | 0.915666 |
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# See figure 411 find x y 0 1 figure 411 unit interval
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Unformatted text preview: ned for uo such that fX (uo ) > 0 by fY |X (v |uo ) = fX,Y (uo , v ) fX (uo ) − ∞ < v < +∞. (4.6) Graphically, the connection between fX,Y and fY |X (v |uo ) for uo fixed, is quite simple. For uo fixed, the right hand side of (4.6) depends on v in only one place; the denominator, fX (uo ), is just a constant. So fY |X (v |uo ) as a function of v is proportional to fX,Y (uo , v ), and the graph of fX,Y (uo , v ) with respect to v is obtained by slicing through the graph of the joint pdf along the line u ≡ uo in the u − v plane. The choice of the constant fX (uo ) in the denominator in (4.6) makes fY |X (v |uo ), as a function of v for uo fixed, itself a pdf. Indeed, it is nonnegative, and ∞ ∞ fY |X (v |uo )dv = −∞ −∞ fX,Y (uo , v ) 1 dv = fX (uo ) fX (uo ) ∞ fX,Y (uo , v )dv = −∞ fX (uo ) = 1. fX (uo ) In practice, given a value uo for X , we think of fY |X (v |uo ) as a new pdf for Y , based on our change of view due to observing the event X = uo . There is a bit of difficulty in this interp...
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## This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Tech.
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https://www.doorsteptutor.com/Exams/NCO/Class-6/Questions/Topic-Mental-Ability-0/Subtopic-Basic-Arithmetic-0/Part-38.html | 1,524,469,773,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945855.61/warc/CC-MAIN-20180423070455-20180423090455-00480.warc.gz | 769,233,114 | 9,758 | # Mental Ability-Basic Arithmetic (NCO- Cyber Olympiad (SOF) Class 6): Questions 216 - 219 of 268
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## Question number: 216
» Mental Ability » Basic Arithmetic
MCQ▾
### Question
The greatest common factor of 9, 18 and 27 is
### Choices
Choice (4) Response
a.
9
b.
3
c.
27
d.
36
## Question number: 217
» Mental Ability » Basic Arithmetic
Appeared in Year: 2011
MCQ▾
### Question
Ms. Gupta’s class measured the rainfall each day of the school week. The given table shows the results. Which day had the most rain?
Rainfall for the week October 21 - 27 Day Rainfall (inches) Monday 1.05 Tuesday 0.0 Wednesday 0.76 Thursday 0.9 Friday 1.1
### Choices
Choice (4) Response
a.
Friday
b.
Wednesday
c.
Thursday
d.
Monday
## Question number: 218
» Mental Ability » Basic Arithmetic
Appeared in Year: 2011
MCQ▾
### Question
Atul put 4 pyramids on the right side of a balance scale, as shown in the figure. Each pyramid weighs 3 kg. Atul has cubes that weigh 2 kg each. How many cubes should he put on the left side of the scale so that both sides weigh the same?
### Choices
Choice (4) Response
a.
6
b.
12
c.
5
d.
8
## Question number: 219
» Mental Ability » Basic Arithmetic
Appeared in Year: 2011
MCQ▾
### Question
Beena bought sofa for her drawing room that covered two third of the floor of drawing room with tiles. The drawing room floor has an area of 126 square meters. How many square meters area has sofa covered with tiles?
### Choices
Choice (4) Response
a.
85
b.
84
c.
82
d.
81
f Page | 495 | 1,797 | {"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-2018-17 | latest | en | 0.825473 |
https://rdrr.io/cran/MultOrdRS/src/R/MultOrdRS-package.R | 1,685,410,969,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644915.48/warc/CC-MAIN-20230530000715-20230530030715-00532.warc.gz | 553,340,668 | 9,276 | # R/MultOrdRS-package.R In MultOrdRS: Model Multivariate Ordinal Responses Including Response Styles
#' Model Multivariate Ordinal Responses Including Response Styles
#'
#' A model for multivariate ordinal responses. The response is modelled
#' using a mixed model approach that is also capable of the inclusion
#' of response style effects of the respondents.
#'
#'
#' @name MultOrdRS-package
#' @docType package
#' @author Gunther Schauberger\cr \email{gunther.schauberger@@tum.de}\cr
#' \url{https://orcid.org/0000-0002-0392-1580}
#' @keywords multivariate ordinal response style adjacent categories cumulative
#' @references Schauberger, Gunther and Tutz, Gerhard (2021): Multivariate Ordinal Random Effects Models Including Subject and Group Specific Response Style Effects,
#' \emph{Statistical Modelling}, \url{https://journals.sagepub.com/doi/10.1177/1471082X20978034}
#' @examples
#' data(tenseness)
#'
#' ## create a small subset of the data to speed up calculations
#' set.seed(1860)
#' tenseness <- tenseness[sample(1:nrow(tenseness), 300),]
#'
#' ## scale all metric variables to get comparable parameter estimates
#' tenseness$Age <- scale(tenseness$Age)
#' tenseness$Income <- scale(tenseness$Income)
#'
#' ## two formulas, one without and one with explanatory variables (gender and age)
#' f.tense0 <- as.formula(paste("cbind(",paste(names(tenseness)[1:4],collapse=","),") ~ 1"))
#' f.tense1 <- as.formula(paste("cbind(",paste(names(tenseness)[1:4],collapse=","),") ~ Gender + Age"))
#'
#'
#'
#' ####
#' ####
#'
#' ## Multivariate adjacent categories model, without response style, without explanatory variables
#' m.tense0 <- multordRS(f.tense0, data = tenseness, control = ctrl.multordRS(RS = FALSE, cores = 2))
#' m.tense0
#'
#' \donttest{
#' ## Multivariate adjacent categories model, with response style as a random effect,
#' ## without explanatory variables
#' m.tense1 <- multordRS(f.tense0, data = tenseness)
#' m.tense1
#'
#' ## Multivariate adjacent categories model, with response style as a random effect,
#' ## without explanatory variables for response style BUT for location
#' m.tense2 <- multordRS(f.tense1, data = tenseness, control = ctrl.multordRS(XforRS = FALSE))
#' m.tense2
#'
#' ## Multivariate adjacent categories model, with response style as a random effect, with
#' ## explanatory variables for location AND response style
#' m.tense3 <- multordRS(f.tense1, data = tenseness)
#' m.tense3
#'
#' plot(m.tense3)
#'
#'
#'
#' ####
#' ## Cumulative Models
#' ####
#'
#' ## Multivariate cumulative model, without response style, without explanatory variables
#' m.tense0.cumul <- multordRS(f.tense0, data = tenseness, control = ctrl.multordRS(RS = FALSE),
#' model = "cumulative")
#' m.tense0.cumul
#'
#' ## Multivariate cumulative model, with response style as a random effect,
#' ## without explanatory variables
#' m.tense1.cumul <- multordRS(f.tense0, data = tenseness, model = "cumulative")
#' m.tense1.cumul
#'
#' ## Multivariate cumulative model, with response style as a random effect,
#' ## without explanatory variables for response style BUT for location
#' m.tense2.cumul <- multordRS(f.tense1, data = tenseness, control = ctrl.multordRS(XforRS = FALSE),
#' model = "cumulative")
#' m.tense2.cumul
#'
#' ## Multivariate cumulative model, with response style as a random effect, with
#' ## explanatory variables for location AND response style
#' m.tense3.cumul <- multordRS(f.tense1, data = tenseness, model = "cumulative")
#' m.tense3.cumul
#'
#' plot(m.tense3.cumul)
#'
#' ################################################################
#' ## Examples from Schauberger and Tutz (2020)
#' ## Data from the German Longitudinal Election Study (GLES) 2017
#' ################################################################
#'
#' ####
#' ## Source: German Longitudinal Election Study 2017
#' ## Rossteutscher et al. 2017, https://doi.org/10.4232/1.12927
#' ####
#'
#' data(GLES17)
#'
#' ## scale data
#' GLES17[,7:11] <- scale(GLES17[,7:11])
#'
#' ## define formula
#' f.GLES <- as.formula(cbind(RefugeeCrisis, ClimateChange, Terrorism,
#' Globalization, Turkey, NuclearEnergy) ~
#' Age + Gender + Unemployment + EastWest + Abitur)
#'
#' ## fit adjacent categories model without and with response style parameters
#' m.GLES0 <- multordRS(f.GLES, data = GLES17, control = ctrl.multordRS(RS = FALSE, cores = 6))
#' m.GLES <- multordRS(f.GLES, data = GLES17, control = ctrl.multordRS(cores = 6))
#'
#' m.GLES0
#' m.GLES
#'
#' plot(m.GLES, main = "Adjacent categories model")
#'
#'
#' ## fit cumulative model without and with response style parameters (takes pretty long!!!)
#' m.GLES20 <- multordRS(f.GLES, data = GLES17, model="cumul",
#' control = ctrl.multordRS(opt.method = "nlminb", cores = 6, RS = FALSE))
#'
#' m.GLES2 <- multordRS(f.GLES, data = GLES17, model="cumul",
#' control = ctrl.multordRS(opt.method = "nlminb", cores = 6))
#'
#' m.GLES20
#' m.GLES2
#'
#' plot(m.GLES2, main = "Cumulative model")
#'
#'}
NULL
#' Tenseness data from the Freiburg Complaint Checklist (tenseness)
#'
#' Data from the Freiburg Complaint Checklist. The data contain all 8 items corresponding to the scale
#' \emph{Tenseness} for 1847 participants of the standardization sample of the Freiburg Complaint Checklist.
#' Additionally, several person characteristics are available.
#'
#' @name tenseness
#' @docType data
#' @format A data frame containing data from the Freiburg Complaint Checklist with 1847 observations.
#' All items refer to the scale \emph{Tenseness} and are measured on a 5-point Likert scale where low numbers
#' correspond to low frequencies or low intensitites of the respective complaint and vice versa.
#' \describe{
#' \item{Clammy_hands}{Do you have clammy hands?}
#' \item{Sweat_attacks}{Do you have sudden attacks of sweating?}
#' \item{Clumsiness}{Do you notice that you behave clumsy?}
#' \item{Wavering_hands}{Are your hands wavering frequently, e.g. when lightning a cigarette or when holding a cup?}
#' \item{Restless_hands}{Do you notice that your hands are restless?}
#' \item{Restless_feet}{Do you notice that your feet are restless?}
#' \item{Twitching_eyes}{Do you notice unvoluntary twitching of your eyes?}
#' \item{Twitching_mouth}{Do you notice unvoluntary twitching of your mouth?}
#' \item{Gender}{Gender of the participant}
#' \item{Household}{Does participant live alone in a houshold or together with others?}
#' \item{WestEast}{is the participant from East Germany (former GDR) or West Germany?}
#' \item{Age}{Age in 15 categories, treated as continuous variable}
#' \item{Abitur}{Does the participant have Abitur (a-levels)?}
#' \item{Income}{Income in 11 categories, treated as continuous variable}
#' }
#' @source
#' ZPID (2013). PsychData of the Leibniz Institute for Psychology Information ZPID. Trier: Center for Research Data in Psychology.
#'
#' Fahrenberg, J. (2010). Freiburg Complaint Checklist [Freiburger Beschwerdenliste (FBL)]. Goettingen, Hogrefe.
#' @keywords datasets
#' @examples
#' \donttest{
#' data(tenseness)
#'
#' ## create a small subset of the data to speed up calculations
#' set.seed(1860)
#' tenseness <- tenseness[sample(1:nrow(tenseness), 300),]
#'
#' ## scale all metric variables to get comparable parameter estimates
#' tenseness$Age <- scale(tenseness$Age)
#' tenseness$Income <- scale(tenseness$Income)
#'
#' ## two formulas, one without and one with explanatory variables (gender and age)
#' f.tense0 <- as.formula(paste("cbind(",paste(names(tenseness)[1:4],collapse=","),") ~ 1"))
#' f.tense1 <- as.formula(paste("cbind(",paste(names(tenseness)[1:4],collapse=","),") ~ Gender + Age"))
#'
#'
#'
#' ####
#' ####
#'
#' ## Multivariate adjacent categories model, without response style, without explanatory variables
#' m.tense0 <- multordRS(f.tense0, data = tenseness, control = ctrl.multordRS(RS = FALSE))
#' m.tense0
#'
#'
#' ## Multivariate adjacent categories model, with response style as a random effect,
#' ## without explanatory variables
#' m.tense1 <- multordRS(f.tense0, data = tenseness)
#' m.tense1
#'
#' ## Multivariate adjacent categories model, with response style as a random effect,
#' ## without explanatory variables for response style BUT for location
#' m.tense2 <- multordRS(f.tense1, data = tenseness, control = ctrl.multordRS(XforRS = FALSE))
#' m.tense2
#'
#' ## Multivariate adjacent categories model, with response style as a random effect, with
#' ## explanatory variables for location AND response style
#' m.tense3 <- multordRS(f.tense1, data = tenseness)
#' m.tense3
#'
#' plot(m.tense3)
#'
#'
#'
#' ####
#' ## Cumulative Models
#' ####
#'
#' ## Multivariate cumulative model, without response style, without explanatory variables
#' m.tense0.cumul <- multordRS(f.tense0, data = tenseness, control =
#' ctrl.multordRS(RS = FALSE), model = "cumulative")
#'
#' m.tense0.cumul
#'
#' ## Multivariate cumulative model, with response style as a random effect,
#' ## without explanatory variables
#' m.tense1.cumul <- multordRS(f.tense0, data = tenseness, model = "cumulative")
#' m.tense1.cumul
#'
#' ## Multivariate cumulative model, with response style as a random effect,
#' ## without explanatory variables for response style BUT for location
#' m.tense2.cumul <- multordRS(f.tense1, data = tenseness,
#' control = ctrl.multordRS(XforRS = FALSE), model = "cumulative")
#'
#' m.tense2.cumul
#'
#' ## Multivariate cumulative model, with response style as a random effect,
#' ## with explanatory variables
#' ## for location AND response style
#' m.tense3.cumul <- multordRS(f.tense1, data = tenseness, model = "cumulative")
#' m.tense3.cumul
#'
#' plot(m.tense3.cumul)
#'}
NULL
#' German Longitudinal Election Study 2017 (GLES17)
#'
#' Data from the German Longitudinal Election Study (GLES) from 2017 (Rossteutscher et
#' al., 2017, https://doi.org/10.4232/1.12927). The GLES is a long-term study of the German electoral process.
#' It collects pre- and post-election data for several federal elections, the
#' data used here originate from the pre-election study for 2017.
#'
#' @name GLES17
#' @docType data
#' @format A data frame containing data from the German Longitudinal Election Study with 2036 observations.
#' The data contain socio-demographic information about the participants as well as their responses to items about specific political fears.
#' \describe{
#' \item{RefugeeCrisis}{How afraid are you due to the refugee crisis? (Likert scale from 1 (not afraid at all) to 7 (very afraid))}
#' \item{ClimateChange}{How afraid are you due to the global climate change? (Likert scale from 1 (not afraid at all) to 7 (very afraid))}
#' \item{Terrorism}{How afraid are you due to the international terrorism? (Likert scale from 1 (not afraid at all) to 7 (very afraid))}
#' \item{Globalization}{How afraid are you due to the globalization? (Likert scale from 1 (not afraid at all) to 7 (very afraid))}
#' \item{Turkey}{How afraid are you due to the political developments in Turkey? (Likert scale from 1 (not afraid at all) to 7 (very afraid))}
#' \item{NuclearEnergy}{How afraid are you due to the use of nuclear energy? (Likert scale from 1 (not afraid at all) to 7 (very afraid))}
#' \item{Age}{Age in years}
#' \item{Gender}{0: male, 1: female}
#' \item{EastWest}{0: West Germany, 1: East Germany}
#' \item{Abitur}{High School Diploma, 1: Abitur/A levels, 0: else}
#' \item{Unemployment}{1: currently unemployed, 0: else}
#' }
#' @references Rossteutscher, S., Schmitt-Beck, R., Schoen, H., Wessels, B., Wolf, C., Bieber, I., Stovsand, L.-C., Dietz, M., and Scherer, P. (2017).
#' Pre-election cross section (GLES 2017). \emph{GESIS Data Archive, Cologne, ZA6800 Data file Version 2.0.0.}, \doi{10.4232/1.12927}.
#'
#' Schauberger, Gunther and Tutz, Gerhard (2021): Multivariate Ordinal Random Effects Models Including Subject and Group Specific Response Style Effects,
#' \emph{Statistical Modelling}, \url{https://journals.sagepub.com/doi/10.1177/1471082X20978034}
#' @source
#' \url{https://gles-en.eu/} and
#' \doi{10.4232/1.12927}
#' @keywords datasets
#' @examples
#' \donttest{
#' ###############################################################
#' ## Examples from Schauberger and Tutz (2020)
#' ## Data from the German Longitudinal Election Study (GLES) 2017
#' ###############################################################
#'
#' ####
#' ## Source: German Longitudinal Election Study 2017
#' ## Rossteutscher et al. 2017, https://doi.org/10.4232/1.12927
#' ####
#'
#' data(GLES17)
#'
#' ## scale data
#' GLES17[,7:11] <- scale(GLES17[,7:11])
#'
#' ## define formula
#' f.GLES <- as.formula(cbind(RefugeeCrisis, ClimateChange, Terrorism,
#' Globalization, Turkey, NuclearEnergy) ~
#' Age + Gender + Unemployment + EastWest + Abitur)
#'
#' ## fit adjacent categories model without and with response style parameters
#' m.GLES0 <- multordRS(f.GLES, data = GLES17, control = ctrl.multordRS(RS = FALSE, cores = 6))
#' m.GLES <- multordRS(f.GLES, data = GLES17, control = ctrl.multordRS(cores = 6))
#'
#' m.GLES0
#' m.GLES
#'
#' plot(m.GLES, main = "Adjacent categories model")
#'
#'
#' ## fit cumulative model without and with response style parameters (takes pretty long!!!)
#' m.GLES20 <- multordRS(f.GLES, data = GLES17, model="cumul",
#' control = ctrl.multordRS(opt.method = "nlminb", cores = 6, RS = FALSE))
#'
#' m.GLES2 <- multordRS(f.GLES, data = GLES17, model="cumul",
#' control = ctrl.multordRS(opt.method = "nlminb", cores = 6))
#'
#' m.GLES20
#' m.GLES2
#'
#' plot(m.GLES2, main = "Cumulative model")
#'
#'}
NULL
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https://cupdf.com/document/synthetic-geometry-and-number-resmath133nonmetric-modelspdf-synthetic-geometry.html | 1,713,593,883,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817491.77/warc/CC-MAIN-20240420060257-20240420090257-00646.warc.gz | 170,657,788 | 55,930 | SYNTHETIC GEOMETRY AND NUMBER SYSTEMS When the foundations of Euclidean and non-Euclidean geometry were reformulated in the late 19 th and early 20 th centuries, the axiomatic settings did not use the primitive concepts of distance and angle measurement which are central to the exposition in the course notes (an idea which goes back to some writings of G. D. Birkhoff in the nineteen thirties). For example, the treatment in D. Hilbert’s definitive Foundations of Geometry involved primitive concepts of betweenness , and congruence of segments , congruence of angles (in addition to the usual primitive concepts of lines and planes). Of course, Hilbert’s approach states its axioms in terms of these concepts, and ultimately one can prove that the approach in these notes is equivalent to Hilbert’s (and all other approaches for that matter). The Hilbert approach provides the setting for Greenberg’s book, and Appendix B of Greenberg discusses several issues related in this approach as they apply to hyperbolic geometry. The purpose of this document is to relate Greenberg’s perspective with that of the course notes. In a very lengthy Appendix we shall consider one additional aspect of nonmetric approaches to geometry; namely, finite geometrical systems. Comparing the metric and nonmetric approaches In Mo¨ ıse, both the Hilbert and Birkhoff approaches are discussed at length, with the latter as the primary setting. As noted in comments on page 138 that book, the underlying motivation for the Birkhoff approach is that the concepts of linear and angular measurement have been central to geometry on a theoretical level since the development of algebra; as Mo¨ ıse suggests, this approach did not appear in the Elements because the Greek mathematics at the time because the latter’s grasp of algebra was extremely limited, so that even very simple algebraic issues were studied geometrically. To quote Mo¨ ıse, the adoption of linear and angular measurement as undefined concepts “describe[s] the methods that in fact everybody uses.” A second advantage is that the Birkhoff approach leads to a fairly rapid development of classical geometry, which minimizes the amount of time and effort needed at the beginning to analyze the statements on betweenness and separation which may seem self-evident and possibly too simple to worry about (compare the comments in the second paragraph on page iii and the third paragraph on page 60 of Mo¨ ıse). In a classical approach along the lines of Hilbert’s development, many of the justifications for such results are not at all transparent and require long, delicate arguments which are often not helpful for understanding the big picture. On the other hand, the modern definitions of the real number system, due to R. Dedekind (1831–1916) and G. Cantor (1845–1918) in the second half of the 19 th century, require some fairly sophisticated concepts which are completely outside the scope of Greek mathematics and closely related to the notion of continuity, and by a general principle of scientific thought called Ockham’s razor (don’t introduce complicated auxiliary material to explain something unless this is simply unavoidable or saves a great deal of time and effort) it is also highly desirable to look for alternative formulations which do not use the full force of the real number system’s continuity properties and (in a quotation cited on page 571 of Greenberg) demonstrate that “the true essence of geometry can develop most naturally and economically.” The key to passing back and forth between the two approaches is summarized very well in the following sentence on page 573 of Greenberg: 1
21
# SYNTHETIC GEOMETRY AND NUMBER SYSTEMSmath.ucr.edu/~res/math133/nonmetric-models.pdf · SYNTHETIC GEOMETRY AND NUMBER SYSTEMS When the foundations of Euclidean and non-Euclidean geometry
Aug 18, 2018
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SYNTHETIC GEOMETRY AND NUMBER SYSTEMS
When the foundations of Euclidean and non-Euclidean geometry were reformulated in the late19th and early 20th centuries, the axiomatic settings did not use the primitive concepts of distanceand angle measurement which are central to the exposition in the course notes (an idea whichgoes back to some writings of G. D. Birkhoff in the nineteen thirties). For example, the treatmentin D. Hilbert’s definitive Foundations of Geometry involved primitive concepts of betweenness,and congruence of segments, congruence of angles (in addition to the usual primitive concepts oflines and planes). Of course, Hilbert’s approach states its axioms in terms of these concepts, andultimately one can prove that the approach in these notes is equivalent to Hilbert’s (and all otherapproaches for that matter). The Hilbert approach provides the setting for Greenberg’s book,and Appendix B of Greenberg discusses several issues related in this approach as they apply tohyperbolic geometry. The purpose of this document is to relate Greenberg’s perspective with thatof the course notes. In a very lengthy Appendix we shall consider one additional aspect of nonmetricapproaches to geometry; namely, finite geometrical systems.
Comparing the metric and nonmetric approaches
In Moıse, both the Hilbert and Birkhoff approaches are discussed at length, with the latter asthe primary setting. As noted in comments on page 138 that book, the underlying motivation forthe Birkhoff approach is that the concepts of linear and angular measurement have been central togeometry on a theoretical level since the development of algebra; as Moıse suggests, this approachdid not appear in the Elements because the Greek mathematics at the time because the latter’sgrasp of algebra was extremely limited, so that even very simple algebraic issues were studiedgeometrically. To quote Moıse, the adoption of linear and angular measurement as undefinedconcepts “describe[s] the methods that in fact everybody uses.”
A second advantage is that the Birkhoff approach leads to a fairly rapid development of classicalgeometry, which minimizes the amount of time and effort needed at the beginning to analyze thestatements on betweenness and separation which may seem self-evident and possibly too simple toworry about (compare the comments in the second paragraph on page iii and the third paragraphon page 60 of Moıse). In a classical approach along the lines of Hilbert’s development, many of thejustifications for such results are not at all transparent and require long, delicate arguments whichare often not helpful for understanding the big picture.
On the other hand, the modern definitions of the real number system, due to R. Dedekind(1831–1916) and G. Cantor (1845–1918) in the second half of the 19th century, require some fairlysophisticated concepts which are completely outside the scope of Greek mathematics and closelyrelated to the notion of continuity, and by a general principle of scientific thought called Ockham’srazor (don’t introduce complicated auxiliary material to explain something unless this is simplyunavoidable or saves a great deal of time and effort) it is also highly desirable to look for alternativeformulations which do not use the full force of the real number system’s continuity properties and(in a quotation cited on page 571 of Greenberg) demonstrate that “the true essence of geometrycan develop most naturally and economically.”
The key to passing back and forth between the two approaches is summarized very well in thefollowing sentence on page 573 of Greenberg:
1
Every Hilbert plane [a system satisfying all the axioms except perhaps eitherthe Dedekind Continuity Axiom or the Euclidean parallel postulate, or possiblyboth] has a field hidden in its geometry.
At the end of Section II.5 we mentioned that a similar statement is true for many abstract planeswhich satisfy the standard axioms of incidence and the Euclidean Parallel Postulate, and in par-ticular this is true if the plane lies inside a 3-space satisfying the corresponding assumptions. Theprinciple in the quotation leads directly to a four step approach to the systems he calls Hilbertplanes (systems which satisfy all the axioms except perhaps either the Dedekind Continuity Axiomor the Euclidean parallel postulate, or possibly both); this approach is outlined on page 588.
REFERENCES. The approach taken in Greenberg’s book is designed to be very closely connectedthat of the following more advanced textbook:
R. Hartshorne. Geometry: Euclid and Beyond. Springer-Verlag, New York, 2000.
Additional background references for this material are the books by Moıse and Forder, the bookby Birkhoff and Beatley, and the paper by Birkhoff; these are given in Unit II of the course notes.The latter also contain links references to many other relevant sources. Some further referencesfor more specialized topics in Addenda A and B will be listed at the end of the latter. This willbe the starting point for our discussion, and we shall begin by describing the additional featuresof this “hidden algebra” if the plane also satisfies Hilbert’s axioms of betweenness and congruence.Much of this material appears in Greenberg, but it is dispersed throughout different sections, andit seems worthwhile to gather everything together in one place. Similarly, our discussion of non-Euclidean systems will include a chart summarizing the various places in Greenberg which dealwith non-Euclidean Hilbert planes.
The level of the discussion in this document is (unavoidably) somewhat higher than that ofthe course notes; in many places it is probably close to the level of an introductory graduate levelalgebra course.
Euclidean geometry without the real number system
At the end of Section II.5 we noted that one can introduce useful algebraic coordinate systemsinto systems which satisfy the 3-dimensional Incidence Axioms (in Section II.1) and the ParallelPostulate (in Section II.5); since the planes of interest to us are all equivalent to planes whichlie inside 3-spaces, the coordinatization result also applies to the planes that we shall considerhere. If the Parallel Postulate is true, this yields algebraic coordinate structures on all systemsatisfying all the Hilbert axioms for Euclidean geometry except perhaps the Dedekind ContinuityAxiom (see pages 134–135 and 598–599 of Greenberg). The general results of Section II.5 in theclass notes state that the coordinates take values in an algebraic system called a division ring or askew-field; informally, in such a system one can perform addition, subtraction, multiplication anddivision by nonzero coordinates, but the multiplication does not necessarily satisfy the commutativemultiplication property xy = yx.
If one also assumes that the plane or 3-space satisfies Hilbert’s axioms for betweenness andcongruence (see pages 597–598 of Greenberg), then the coordinate system has additional structurecalled an ordering (special cases are described in Section 1.5 of Moıse, and the ties to geometry arediscussed on pages 117–118 of Greenberg). This means that there is a subset of positive elementswhich is closed under addition and multiplication, and also has the property that for every nonzero“number” x, either x or −x is positive. A standard algebraic argument shows the square of every
2
nonzero element in an ordered division ring is positive. The coordinate system also turns out tohave the Pythagorean property described in the following definition:
Definition. An ordered division ring K is Pythagorean if for each element x in the system thereis a positive element y such that y2 = 1 + x2.
It is a straightforward exercise to prove the following result:
PROPOSITION. Suppose that K is a Pythagorean field and that a1, · · · , an are nonzeroelements of K. Then there is unique y ∈ K such that y is positive and
y2 =√
a21
+ · · · + a2n
.
We shall be interested mainly in ordered division rings which are ordered fields (so that xy
and yx are always equal); for the sake of completeness, we note that the books by Forder andHartshorne describe examples of ordered division rings which are not ordered fields. Standard re-sults in projective geometry show that the algebraic commutative law of multiplication is equivalentto a condition known as Pappus’ Hexagon Theorem; further information on this result appears inUnit IV of the class notes and the following online document:
http://math.ucr.edu/∼res/progeom/pgnotes05.pdf
The Pappus Hexagon Theorem is a bit complicated to state (cf. also Greenberg, Advanced Project3, pp. 99–100); however, there is a more easily stated, and extremely useful, hypothesis whichimplies commutativity of multiplication and also considerably more:
THEOREM. Suppose that we are given a plane or 3-space E which satisfies all the Hilbertaxioms except (perhaps) the Dedekind Continuity Axiom and has coordinates given by the orderedPythagorean division ring K. Then the following are equivalent:
(i) The plane or space E satisfies the Archimedean Continuity Axiom (see Greenberg, page599).
(ii) The ordered division ring K satisfies the commutative law of multiplication and the Archi-medean Property (see Greenberg, page 601).
It turns out that the conditions in (ii) hold if and only if K is order-preservingly isomorphicto a subfield of the real numbers.
The Line-Circle and Two Circle Properties
As noted in Section III.6 of the course notes, the classical results in Euclidean geometry(including straightedge-and-compass constructions) require the two results on intersections of circlewith a line or another circle named in the heading above; both are implicit in the Elements but notstated explicitly. The proofs of these results in the course notes rely on the fact that every positivereal number has a positive square root (which is also a real number). In fact, the arguments inSection III.6 go through if we have a system with coordinates in a Pythagorean ordered field whichalso satisfies the commutative law of multiplication and the condition in the following definition:
Definition. A Pythagorean ordered field K is said to be surd-complete if every positive elementof K has a positive square root in K.
In fact, the validity of the Line-Circle and Two Circle Properties turns out to be equivalent tothe surd-completeness of K.
3
Of course, the real number system is surd-complete. Also, Sections 19.6–19.7 and 31.2 of Moısedescribe a countable subfield Surd of the real numbers which is surd-complete and is in fact theunique minimal surd-complete subfield of the real numbers. This field also has the following basicproperty (not stated or used explicitly in Moıse, but closely related to the results on the “impossible”classical construction problems); it is an elementary exercise to derive this from material on fieldextensions in introductory graduate level algebra courses.
PROPOSITION. Let α be a nonzero element of the surd field Surd. Then there is a uniquenonzero monic polynomial p(x) with rational coefficients such that the following hold:
(i) The surd α is a root of p.
(ii) If q is a nonzero polynomial with rational coefficients such that q(α) = 0, then q is amultiple of p (hence the degree of p is minimal among all polynomials for which α is a root).
(iii) The degree of p is a power of 2.
The results in Chapter 19 of Moıse show that a classical construction problem can be done bymeans of straightedge and compass construction if and only if the following holds:
If we begin with points, lines, and circles whose defining equations only involve elementsof Surd, then the defining numerical data for the constructed objects also lie in Surd. —More generally, if the original data lie in an ordered field F which is surd-complete, thenthe defining numerical data for the constructed objects also lie in F.
We have already noted that an ordered field must be surd-complete in order to carry out theclassical geometrical discussion of circles and constructions. By definition, a surd-complete field isautomatically Pythagorean, and further consideration yields the following:
THEOREM. Suppose that we are given a plane or 3-space E which satisfies all the Hilbertaxioms except (perhaps) the Dedekind Continuity Axiom and has coordinates given by the orderedPythagorean division ring K. Then the following are equivalent:
(i) The Line-Circle and Two Circle Properties are valid in the plane or space E.
(ii) The ordered division ring K is surd-complete.
As noted in the final paragraph on page 131 of Greenberg, it is possible to construct anArchimedean ordered field K which is Pythagorean but not surd-complete, and from this onecan conclude that it is impossible to prove the Two Circle Property from Hilbert’s axioms ofincidence, betweenness and congruence, even if one also assumes the coordinate field K satisfiesthe Archimedean Property .
As suggested by the discussion on pages 129–131 of Greenberg, this result implies that thevery first proposition in Euclid’s Elements (the existence of an equilateral triangle with a given linesegment as one of its edges) cannot be proved without making some additional assumption likethe Two Circle Property.
Numberless non-Euclidean geometry
A central theme in Greenberg’s book is to do as much of neutral and non-Euclidean geometryas possible without using the full force of the Dedekind Continuity Axiom, and one objective ofAppendix B in Greenberg is to describe coordinates in systems which satisfy all of Hilbert’s axiomsexcept perhaps the Dedekind Continuity Axiom or the Parallel Postulate (or both). Greenbergthen discusses ways in which such coordinate systems can shed light on some basic questions about
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these geometric systems; the latter involves several ideas well beyond the advanced undergraduatelevel, and because of this many parts of the exposition reflect the need to be sketchy and vagueabout various points.
In connection with this discussion of non-Euclidean geometry without the real numbers, itseems appropriate to summarize the other locations throughout the book which discuss the conse-quences of assuming everything but the Parallel Postulate or Dedekind Continuity.
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Page(s) Topic(s)
129 Definition of a Hilbert plane (i.e., satisfying Hilbert’s axioms exceptpossibly either the Parallel Postulate or Dedekind Continuity.
161–162 Statement that a Hilbert plane is the default setting for Chapter 4.
173 Statement of the converse to the Triangle Inequality for Hilbert planessatisfying the Two Circle Property.
175 Theorems on the role of the Parallel Postulate in a Hilbert plane.
176–191 Proofs of analogs to the results in Section V.3 of the course notes inHilbert planes for which the Parallel Postulate does not necessarilyhold; there are comments on the role of the Archimedean Property atthe end, including a reference for an example of a Hilbert plane inwhich the angle sum of a triangle always exceeds 180◦.
200 Exercise 33 discusses some issues about Hilbert planes for which theArchimedean property does not hold.
213–214 The logical equivalence (in Hilbert planes) of the Parallel Postulate andthe axiom of C. Clavius (1538–1612) — namely, that parallel lines areeverywhere equidistant — is discussed.
220–221 A version of an observation due to Proclus Diadochus (410–485) forHilbert planes is stated and proved.
249–254 Proofs of analogs to the results at the beginning Section V.4 of the notes(through AAA congruence) are given for non-Euclidean Hilbert planes.
254–259 The behavior of parallel lines in non-Euclidean Hilbert planes (eitherasymptotic or having a common perpendicular) is discussed; this isclosely related to the final parts of Section V.4 in the course notes.
408–409 The geometric symmetries (automorphisms in Greenberg) of a Hilbertplane are defined and discussed, and the significance of the Archimedeanproperty is mentioned.
471 Exercise 69 fills in some details for the discussion on pages 408–409.
571–596 This is Appendix B.
599–601 This summarizes the axioms for geometric and algebraic systems whichare central to Greenberg’s book.
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Final remarks
The construction of Hilbert’s Field of Ends (in Part I of Appendix B) reflects the relationshipbetween hyperbolic geometry and projective geometry that is apparent in the Beltrami-Klein modelfor the hyperbolic plane (see Greenberg, pages 333–346). One way of describing this relationship isdescribed below (this requires concepts from projective geometry and can be skipped if the readerwishes to do so):
If we view the Beltrami-Klein model Hyp as the open unit disk in R2 and take the usual
extension of R2 to the projective plane RP
2, then the points of the latter which do not lie in Hyp
may be viewed as points at infinity where various pencils of parallel lines in Hyp meet. The idealpoints for asymptotically parallel lines are the points on the circle which is the boundary of Hyp
in R2. Using this, it is possible to interpret some crucial properties of the real number system
(arithmetic operations and order) in terms of the geometry of Hyp. This process can be imitatedin an arbitrary Hilbert plane as follows: A general result of A. N. Whitehead shows that every 3-dimensional system satisfying the axioms of incidence and betweenness has a reasonable embeddinginside a projective 3-space over an ordered division ring; for the sake of completeness we shall givethe reference:
A. N. Whitehead. The Axioms of Descriptive Geometry . Cambridge Univ. Press,New York, 1905. [The cited results appear in Chapter III. — This book is freely availableon the Internet via a Google Book Search; the online address is much too long to fit ona single line, but one can get the link by doing a Google search for whitehead axioms
descriptive geometry.]
Not surprisingly, the coordinates obtained in this manner turn out to be equivalent to the coordi-nates given by Hilbert’s construction.
It is also possible to find reasonable embeddings of certain incidence geometries inside projectivespaces even if one does not have a concept of betweenness. The following paper are basic references:
S. Gorn. On incidence geometry . Bulletin of the American Mathematical Society 46
(1940), 158–167.
O. Wyler. Incidence geometry . Duke Mathematical Journal 20 (1953), 601–601.
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Addendum A: Finite affine and hyperbolic planes
At the end of Section II.1 in the course notes there is a discussion about abstract geometricalsystems which are finite and satisfy the standard Incidence Axioms. Clearly one can also considerfinite incidence geometries in which either the Euclidean Parallel Postulate or some negation of itholds. For example, one might assume the strongest possible negation (the Strong or Universal
Hyperbolic Parallel Postulate:
Given a point x and a line L not containing x, then there are at least two lines M and N
through x which do not meet L.
As in the main discussion above, we shall first discuss finite planes for which the Euclidean ParallelPostulate holds, and afterwards we shall discuss finite planes in which various negations of theEuclidean Parallel Postulate hold.
Finite affine planes
If we define a finite affine plane to be a finite (incidence) plane in which the Euclidean ParallelPostulate holds, the following two questions arise immediately:
1. Do the defining conditions yield interesting consequences? In particular, one would like tohave results which are fairly simple to state but not immediately obvious from the originalassumptions.
2. Do such systems arise in contexts of independent interest? (Compare the remark by J.L. Coolidge, quoted on page 35 of the course notes, in document geometrynotes2a.pdf:The unproved postulates ... must be consistent, but they had better lead to somethinginteresting .)
The following simple result suggests an affirmative answer to the first question:
THEOREM. Let (P,L) be a finite incidence plane. Then the following hold:
(i) The number of points in P is a perfect square (which must be at least 4 since P has atleast three points).
(ii) If the positive integer n ≥ 2 is such that P has n2 points, then every line in P containsexactly n points, and every point in P lies on exactly (n + 1) lines.
A reference for this result is Exercise 7 on page 33 of the following document:
http://math.ucr.edu/∼res/progeom/pgnotes04.pdf
If GF(n) is a finite field with n elements (for example, we can take GF(p) to be Zp if p is a prime),then the coordinate plane GF(n)2 with the usual lines (namely, all subsets of the form x+V wherex is an arbitrary vector and V is an arbitrary 1-dimensional vector subspace) is an affine planewith n2 elements; standard results from (graduate level) abstract algebra courses imply that suchfields exist if and only if n is a prime power.
In many important respects, affine geometry — and particularly finite affine geometry — isbest viewed as part os projective geometry (see Unit IV of the notes), and in particular the usualapproach to finite affine planes is to construct associated finite projective planes using the methodsof pages 47–48 and 62–65 of the following document:
http://math.ucr.edu/∼res/progeom/pgnotes03.pdf
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Finite projective planes have been studied extensively and effectively during the past 100 yearsor so, and they turn out to have important practical uses in mathematical statistics, especially inthe theory of experimental design. Further discussion and references are given on pages 37–39 ofthe course notes and on pages 79–82 and 84–86 of the previously cited online document ... pg-
notes04.pdf; the book by Bose in the bibliography is an important reference for the applicationsof finite projective planes and related structures.
Finite hyperbolic planes
Since there is a fairly extensive theory of finite affine and projective planes, it is natural tospeculate about finite analogs of hyperbolic planes. This topic has beens studied sporadically andonly to a limited extent, and the literature is somewhat scattered. Therefore the following summaryalmost surely overlooks some work on this question.
The most naıve and obvious approach to defining a finite non-Euclidean plane is to say it is afinite plane which does not satisfy the Euclidean Parallel Postulate. However, as in the precedingdiscussion of finite affine planes, there are immediate questions regarding the logical consequences ofsuch a definition or the existence of models which are relevant to questions of independent interest.It is possible to go even further and ask whether the given definition is enough by itself to yieldstructures which are worth studying in some degree of detail, but we shall not try to address thisquestion because it gets into subjective (but nevertheless important!) considerations.
BASIC CONSEQUENCES AND EXAMPLES. We shall begin by deriving a simple but noteworthyproperty of non-Euclidean planes:
PROPOSITION. Let (P,L) be a finite incidence plane in which there is a line L and a pointC 6∈ L such that there are at least two parallels to L through C. Then P contains (at least) fourpoints, no three of which are collinear.
Proof. Let A and B be two points of L, and let C be as above. Choose D and E such that CD
and CE are distinct lines which are parallel to L = AB (such lines exist by the hypothesis). Weclaim that the points A,B,C,D,E are distinct; certainly the first three are because C 6∈ AB, whilethe conditions on D and E also imply that neither of these points can be A, B or C, and D 6= E
because CD 6= CE. It will suffice to prove that {A,B,C,D,E} contains four points, no three ofwhich are collinear.
There are exactly 10 subsets of {A,B,C,D,E} which contain exactly 3 elements, and theymay be listed as follows:
{A,B,C}{A,B,D}{A,B,E}{A,C,D}{A,C,E}{A,D,E}{B,C,D}{B,C,E}{B,D,E}{C,D,E}
Since C 6∈ L = AB, we know that {A,B,C} is noncollinear. Also, both {A,B,D} and {A,B,E}are noncollinear because AB ∩ CD = AB ∩ CE = ∅ by assumption. Similarly, the sets {A,C,D}and {A,C,E} are not collinear for the same reason, and likewise for {B,C,D} and {B,C,E}. In a
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different direction, CD 6= CE implies that {C,D,E} must be noncollinear. Therefore, by processof elimination we see that the only three point subsets of {A,B,C,D,E} which might be collinearare {A,D,E} and {B,D,E}.
Since neither of these subsets is contained in {A,B,C,D} or {A,B,C,E}, it follows that eachof the latter is a subset of P containing four points, no three of which are collinear.
Note. It is easy to construct examples of (finite) planes which do not contain a subset offour or more noncollinear points such that every subset of three points is collinear (if every subsetof three points in X ⊂ P is collinear, it is a straightforward exercise to prove that X is collinear).The most obvious example of this sort is a plane with three points such that the lines are all subsetscontaining exactly two points, but we can also construct examples with any finite number of pointsas follows: Given an integer n ≥ 3, let P = {0, 1, · · · , n} and take L to be the family of all subsets{0, k} where k > 0 together with {1, · · · , n}. It is then a routine exercise to verify that (P,L) is anincidence plane, and since every subset with four or more points must contain at least three pointsin {1, · · · , n}, it follows that if X is a (noncollinear) subset of P containing 4 or more points, thenthere is a collinear subset of X which contains 3 points.
We have already raised questions whether the negation of the Euclidean Parallel Postulate is astrong enough assumption to yield a significant body of noteworthy results, and we have suggestedthe option of assuming the Universal Hyperbolic Parallel Postulate. Before doing so, we shall giveexamples of finite non-Euclidean planes which do not satisfy the Universal Hyperbolic ParallelPostulate. The basic idea is simple; namely, we take an affine plane (P,L) in which all lines haveat least three points, and we remove a line L0 from P.
Formally, if (P,L) is an affine incidence plane as above, let L0 be a line in P, and set Q
equal to P−L0. Now let M0 be a second line in P which meets L0 ar some point z0. If z1
and z2 are two distinct points of M0 other than |bfz0, then z1 and z2 lie on distinct lines
of L1 and L2 in P such that L1||L0 and L2||L0, and we claim that L2 ∩ Q is the uniqueline in Q which contains bfz2 and is disjoint from L1 ∩ Q. It follows immediately thatz2 ∈ L2 ∩ Q (if not, then z2 ∈ L0, so that z2 ∈ L0 ∩ M = {z0}, contradicting z2 6= z0),and clearly L1 ∩Q||L2 ∩Q because L1||L2. To prove uniqueness, we must show that if K
is a line in Q such that z2 ∈ K and K||L1 ∩ Q, then K = L2 ∩ Q.
Suppose that K = K ′capQ is such that z2 ∈ K and K||L1 ∩ Q. If K ′ ∩ L1 = 0, thenK ′ = L2 because P is affine, so that K = L2 ∩ Q. On the other hand, if K ′ ∩ L1 6= ∅,then the intersection must lie in L0. But this implies that L1 ∩ L0 is also nonempty,contradicting the choice of L1 as a line parallel to L0. This proves that K ′ = L2 andK = L2 ∩ Q.
Now let w0 be a second point of L0, let M1 be the unique parallel to M0 through w0, andlet w1 be a second point on M1, so that M1 ∩ L0 = {w0}. Then Q ∩ M1 and Q ∩ z0w1
are two lines in Q which pass through w1 and do not meet Q ∩ M0.
THE MINIMALITY PROPERTY. Even if we assume the Universal Hyperbolic Parallel Postulate,there are some “bloated” examples that fit the formal criteria but are really “too big” to be thoughtof as planes. For example, if (S,L,P) is an affine 3-space, then at least formally we can make S
into an incidence plane by simply decreeing that S is a plane and ignoring the family P. It followsimmediately that (S,L) is an incidence plane.
CLAIM. The system (S,L) satisfies the Universal Hyperbolic Parallel Postulate.
Proof of Claim. Let L be a line and let x be a point of S not on L. Then there is a uniqueparallel M to L through x in the affine 3-space (S,L,P). Let Q ∈ P be such that L ⊂ Q and
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x ∈ Q. Since Q is a proper subset of S, we can find some point y in S which does not lie in Q.It will suffice to prove that xy and L have no points in common. If there was some point z onboth, then x, z ∈ Q would imply that the line joining them — which is xy — would also lie in Q,contradicting our choice of y. Therefore xy and M are two lines through x which are disjoint fromL.
Still more examples of this type can be constructed by letting P = Fn, where F is a finite
field and n ≥ 4. As in the preceding examples, these systems have higher dimensional incidencestructures that are described on pages 31–36 of the following document:
http://math.ucr.edu/∼res/progeom/pgnotes02.pdf
Clearly we have turned higher dimensional incidence structures into planes by the formal trick ofsimply ignoring all higher dimensional structure. One way to avoid such questionable constructionsis to assume an additional property. It will be convenient to formulate this in terms of an auxiliaryconcept:
Definition. Let (P,L) be an incidence plane. A subset Q ⊂ P is flat if for each pair of distinctpoints x 6= y in Q, the line joining them is contained in Q. There is an obvious close relationshipbetween this condition and one of the 3-dimensional incidence axioms.
Definition. An incidence plane (P,L) is said to be minimal or irreducible provided the onlynoncollinear flat subset of P is P itself. We shall say that (P,L) is reducible if this condition doesnot hold.
Clearly all of the “bloated” examples are reducible; in fact, if we are given three noncollinearpoints in one of them, then the “plane” containing them (in the sense of the full incidence structure)is a proper, noncollinear, flat subset.
Before proceeding, we should note that, with one exception, all finite affine planes are irre-ducible and, with no exceptions, all finite projective planes are irreducible.
THEOREM. Let (P,L) be a finite incidence plane which is either a projective plane or an affineplane with more than 4 points. Then (P,L) is irreducible.
It is fairly straightforward to show that two affine planes with exactly 4 points have isomorphicincidence structures (the lines are the two point subsets in this case), and if we combine this withthe conclusion of the theorem we see that, up to incidence isomorphism, there is exactly one affineincidence plane which is reducible (the 4 point model turns out to be reducible, for every linecontains exactly two points, and therefore every subset of this model is flat in the sense of thedefinition).
Proof. Clearly there are two cases, depending upon whether the plane is projective or affine.We recall that a projective plane is one such that every pair of lines has a point in common, andevery line contains at least three points. Some basic properties of such objects are established inthe previously cited document ... pgnotes04.pdf.
Suppose first that (P,L) is projective, and let Q be a flat, noncollinear subset of P. Let A,B,C
be noncollinear points in Q, and let X ∈ P. If X ∈ AB or X ∈ BC then by flatness we know thatX ∈ Q, so suppose that X lies on neither line. It follows that the line XC is distinct from the lineAB, and hence it meets the latter in some point D; the points C and D are distinct, for otherwisethe two distinct points B and C = D would lie on the two distinct lines AB and AC. We knowthat D ∈ Q because D ∈ AB, and therefore we can conclude that the line CD = XC is containedin Q. But this means that X ∈ Q; therefore we have shown that Q contains every point of P , andhence we have Q = P.
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The preceding argument also works in the affine case provided the lines AB and CX have apoint in common, and hence in affine case we can say that a point X ∈ P lies in Q except perhapsif X lies on the unique line L through C such that AB||L. If we switch the roles of A and C in thisargument, we also see that a point X ∈ P lies in Q except perhaps if X lies on the unique line M
through A such that BC||L. Combining these, we see that a point X ∈ P lies in Q except perhapsif X ∈ L∩M . The lines L and M are distinct because one is parallel to AB and the other containsa point of AB, and therefore we have shown that Q contains all but at most one point of L ∩ M .We should note that these two lines do have a point in common, for if they did not then both lineswould be parallel to the nonparallel, nonidentical lines AB and BC, and this is impossible in anaffine plane. We shall denote this point by E.
At this step of the argument we shall finally use the assumption that each line in P containsmore than two points. The point E is the only point of P which might not be in Q. We know thatE ∈ L, so it will suffice to show that at least two points of L are contained in Q. By construction,we have C ∈ L∩Q, and the assumption on the order of P implies that there is a point Y ∈ L suchthat Y 6= X,C. Our reasoning thus far implies that Y ∈ Q, and therefore the flatness assumptionimplies that the entire line L is contained in Q; therefore E ∈ Q, so we have shown that every pointof P lies in Q.
Drawings to accompany this proof are posted in the following file:
http://math.ucr.edu/∼res/math133/irreducibleplanes1.pdf
To motivate the concept of irreducibility further, we shall also sketch a proof that
every Hilbert plane is also irreducible.
In fact, all that one needs to prove irreducibility are the Axioms of Incidence and Order (Between-ness and Plane Separation). An illustrated proof is given in the following online document:
http://math.ucr.edu/∼res/math133/irreducibleplanes2.pdf
In view of the preceding observations, we shall define a finite (synthetic) hyperbolic plane tobe an incidence plane which is irreducible and satisfies the Universal Hyperbolic Parallel Postulate.One example (in fact, the smallest possible such system) satisfying these conditions is given in theextremely readable paper by L. M. Graves which is cited in the Bibliography. This model contains13 points and 26 lines.
Additional examples, often satisfying stronger versions of the Universal Hyperbolic ParallelPostulate (specifically, how many parallels exist through a given point) and other desirably geo-metric conditions (for example, symmetry properties), appear in numerous other articles, includingthe 1963 paper by R. Sandler, the 1964 and 1965 papers by D. W. Crowe, the 1965 paper by M.Henderson, and the 1970 paper by S. H. Heath. Still further results along these lines appear in the1971 paper by R. Bumcrot, which also establishes some restrictions on the numerical data of finitehyperbolic planes.
Examples like the preceding ones are informative in several respects, but ultimately one wouldlike examples which are relevant to other geometrical topics of independent interest. In many cases,the outside interest arises from properties of the Beltrami-Klein model and the role of hyperbolicgeometry in a setting of F. Klein (the Erlangen Program), which was designed to provide a unifiedframework for the various types of geometry that existed when it was formulated in 1870–1872.Here is an online reference describing Klein’s influential views and their impact, followed by a linkto an English translation of Klein’s original paper:
http://en.wikipedia.org/wiki/Erlangen program
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http://www.ucr.edu/home/baez/erlangen/erlangen tex.pdf
The place of hyperbolic geometry in this organizational scheme is easy to describe. Namely, theBeltrami-Klein model for hyperbolic geometry provides the basis for integrating hyperbolic geom-etry into Klein’s framework.
Some very brief papers around 1940 suggested that there might not be any finite analogs ofthe classical hyperbolic plane aside from some trivial ones. The subsequent 1946 paper by R. Baerwas another early (and discouraging) step in the search for “extrinsically motivated” examples offinite hyperbolic planes. On the other hand, the 1962 paper by T. G. Ostrom produced examplessimilar to Graves’ which in many respects reflected the role of classical hyperbolic geometry inKlein’s Erlangen Program; a crucial link between Ostrom’s paper and Klein’s viewpoint is studiedin the 1955 paper by B. Segre. Other articles in this direction include the 1965 and 1966 papers byCrowe, the 1966 paper by R. Artzy, and the 1969 paper by G. I. Podol’nyı; the highly symmetricexamples of finite hyperbolic planes in previously cited articles are also closely related to Klein’sErlangen Program.
We could go into greater detail about results from the individual articles listed below, but tokeep the discussion relatively brief we shall merely conclude by mentioning the 1977 survey by J.Di Paola, which discusses finite hyperbolic planes in the more general context of finite geometries.
Bibliography
(Arranged by year of publication; probably incomplete)
R. C. Bose. On the construction of balanced incomplete block designs. Annals of Eugenics 9
(1939), 353–399.
F. P. Jenks. A new set of postulates for Bolyai-Lobachevsky geometry. I . Proceedings of the U.S. A. National Academy of Sciences 26 (1940), 277–279.
F. P. Jenks. A new set of postulates for Bolyai-Lobachevsky geometry. II . Reports of aMathematical Colloquium (2nd Series) 2 (1940), 10–14.
F. P. Jenks. A new set of postulates for Bolyai-Lobachevsky geometry. III . Reports of aMathematical Colloquium (2nd Series) 3 (1941), 3–12.
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J. Hjelmslev. Einleitung in die allgemeine Kongruenzlehre. III [Introduction to general congru-ence theory. III]. Danske Videnskabelige Selskab Mathematik-Fysik Meddelser 19 (1942), no. 12,50 pp.
B. J. Topel. Bolyai-Lobachevsky planes with finite lines. Reports of a Mathematical Colloquium(2nd Series) 5–6 (1944), 40–42.
R. Baer. Polarities in finite projective planes. Bulletin of the American Mathematical Society52 (1946), 77–93.
R. Baer. The infinity of generalized hyperbolic planes (Studies and Essays Presented to R.Courant, pp. 21–27). Interscience, New York, 1948.
W. Klingenberg. Projektive und affine Ebenen mit Nachbarelementen [Projective and affineplanes with neighboring objects]. Mathematische Zeitschrift 60 (1954), 384–406.
B. Segre. Ovals in a finite projective plane. Canadian Journal of Mathematics 7 (1955), 414–416.
T. G. Ostrom. Ovals, dualities, and Desargues’s Theorem. Canadian Journal of Mathematics7 (1955), 417–431.
E. Kleinfeld. Finite Hjelmslev planes. Illinois Journal of Mathematics 3 (1959), 403–407.
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D. W. Crowe. Regular polygons over GF(32). American Mathematical Monthly 68 (1961),762–765.
L. M. Graves. A finite Bolyai-Lobachevsky plane. American Mathematical Monthly 69 (1962),130–132.
T. G. Ostrom. Ovals and finite Bolyai-Lobachevsky planes. American Mathematical Monthly69 (1962), 899–901.
L. Szamko lowicz. On the problem of existence of finite regular planes. Colloquium Mathe-maticum 9 (1962), 245–250.
R. Sandler. Finite homogeneous Bolyai-Lobachevsky planes. American Mathematical Monthly70 (1963), 853–854.
H. J. Ryser. Combinatorial Mathematics (Mathematical Association of America, Carus Math-ematical Monographs No. 14). Wiley, New York, 1963.
D. W. Crowe. The trigonometry of GF(22n). Mathematika 11 (1964), 83–88.
D. W. Crowe. The construction of finite regular hyperbolic planes from inversive planes of evenorder. Colloquium Mathematicum 13 (1965), 247–250.
P. Dembowski and D. R. Hughes. On finite inversive planes. Journal of the LondonMathematical Society 40 (1965), 171–182.
M. Henderson. Certain finite nonprojective geometries without the axiom of parallels. Pro-ceedings of the American Mathematical Society 16 (1965), 115–119.
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R. Artzy. Non-Euclidean incidence planes. Israel Journal of Mathematics 4 (1966), 43–53.
D. W. Crowe. Projective and inversive models for finite hyperbolic planes. Michigan Mathe-matical Journal 13 (1966), 251–255.
M. Henderson. Finite Bolyai-Lobachevsky k-spaces. Colloquium Mathematicum 50 (1966),205–210.
E. Seiden. On a method of construction of partial geometries and partial Bolyai-Lobachevskyplanes. American Mathematical Monthly 73 (1966), 158–161.
M. Hall. Combinatorial Theory (2nd Ed.). Wiley, New York, 1967.
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H. Crapo and G.-C. Rota. On the Foundations of Combinatorial Theory. CombinatorialGeometries. MIT Press, Camnbridge, MA, 1970.
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(1970), 244–249.
S. H. Heath and C. R. Wylie. A geometric proof of the nonexistence of PG7. MathematicsMagazine 43 (1970), 192–197.
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S. H. Heath and C. R. Wylie. Some observations on BL(3, 3). Univ. Nac. Tucuman Ser. A20 (1970), 117–123.
J. W. Di Paola. Configurations in small hyperbolic planes. Annals of the New York Academyof Sciences 175 (1970), 93–103.
R. Bumcrot. Finite hyperbolic spaces. Atti del Convegno di Geometria Combinatoria e sueApplicazioni (Perugia, 1970), pp. 113–124. Universita degli Studia di Perugia, Perugia, 1971.
G. Sproar. The connection of block designs with finite Bolyai-Lobachevsky planes. MathematicsMagazine 46 (1973), 101–102.
H. Zeitler. Ovoide in endlichen projektiven Raumen der Dimension 3 [Ovoids in 3-dimensionalfinite projective spaces]. Mathematische-Physikalische Semesterberichte 22 (1975), 109–134.
F. Karteszi. Introduction to Finite Geometries [Translation by L. Vekerdi of the Hungarianoriginal, Bevezetes a veges geometriakba, Akademiai Kiado, Budapest, 1972], NOrth Holland Textsin Advanced Mathematics, Vol. 2. Elsevier/North Holland, New York, 1976.
J. W. Di Paola. Some finite point geometries. Mathematics Magazine 50 (1977), 79–83.
C. W. L. Garner. Conics in finite projective planes. Journal of Geometry 12 (1979), 132–138.
C. W. L. Garner. A finite analogue of the classical hyperbolic plane and Hjelmslev groups.Geometriæ Dedicata 7 (1978), 315–331.
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M. Barnabei and F. Bonetti. Two examples of finite Bolyai-Lobachevsky planes. Rendicontidi Matematica e delle sue Applicazioni (6) 12 (1979), 291–296.
S. R. Bruno. On certain geometric loci in finite hyperbolic spaces [Spanish]. MathematicaeNotae 27 (1979/80), 49–59.
C. W. L. Garner. Motions in a finite hyperbolic plane. The Geometric Vein [The CoxeterFestschrift],pp. 485–493. Springer-Verlag, New York−etc., 1981.
H. Zeitler. Finite non-Euclidean planes, Combinatorics ’81 (Rome, 1981), North-Holland Math-ematical Studies Vol. 78, pp. 805–817. North-Holland Publishing, Amsterdam, 1983.
R. Kaya and E. Ozcan. On the construction of Bolyai-Lobachevsky planes from projectiveplanes. Rendiconti del Seminario Matematico de Brescia 7 (1982), 427–434.
A. Delandtsheer. A classification of finite 2-fold Bolyai-Lobachevsky spaces. GeometriæDedicata 14 (1983), 375–394.
F. Karteszi and T. Horvath. Einige Bemerkungen bezuglich der Struktur von endlichen Bolyai-Lobatschefsky Ebenen [Some comments concerning the structure of finite Bolyai-Lobachevskyplanes]. Annales Universitatis Scientiarum Budapestinensis de R. Eotvos Nominatae Sectio Math-ematica 85 (1985), 263–270.
K. Gruning. Projective planes of odd order admitting orthogonal polarities. Results in Mathe-matics 7 (1986), 33–51.
18
S. Olgun. On the line classes in some finite Bolyai-Lobachevsky planes [Turkish]. Doga, TurkishJournal of Mathematics 10 (1986), 282–286.
H. Struve and R. Struve. Endliche Cayley-Kleinsche Geometrien [Finite Cayley-Klein geome-tries]. Archiv der Mathematik 48 (1987), 178–184.
W. Chernowitzo. Closed arcs in finite projective planes (Seventeenth Manitoba Conference onNumerical Mathematics and Computing, Winnipeg, 1987). Congressus Numerantium 62 (1988),69–77.
J. W. Di Paola. The structure of the hyperbolic planes S(2, k, k2 +(k−1)2). Ars Combinatoria25A (1988), 77–87.
C. W. T. Garner. Midpoints and midlines in a finite hyperbolic plane Combinatories ’86. Proceed-ings of the international conference on incidence geometries and combinatorial structures, Trento,Italy, 1986, Annals of Discrete Mathematics Vol. 37, pp. 181–187. North-Holland Publishing,Amsterdam, 1988.
C. W. L. Garner. Circles, horocycles and hypercycles in a finite hyperbolic plane. ActaMathematica Hungarica 56 (1990), 65–70.
S. Olgun. On some combinatorics of a class of finite hyperbolic planes. Doga, Turkish Journalof Mathematics 16 (1992), 134–147.
H. L. Skala. Projective-type axioms for the hyperbolic plane. Geometriæ Dedicata 44 (1992),255–272.
S. Olgun and I. Ozgur. On some finite hyperbolic 3-spaces. Turkish Journal of Mathematics18 (1994), 263–271.
19
F. Buekenhout (ed.). Handbook of Incidence Geometry. Elsevier Science Publishing, NewYork-(etc.), 1995.
C. C. Lindner and C. A. Rodger. Design theory. CRC Press Series on Discrete Mathematicsand its Applications. CRC Press, Boca Raton, FL, 1997.
S. Olgun, I. Ozgur, and I. Gunaltılı. A note on hyperbolic planes obtained from finiteprojective planes. Turkish Journal of Mathematics 21 (1997), 77–81.
B. Celik. On some hyperbolic planes from finite projective planes. International Journal ofMathematics and Mathematical Sciences 25:12 (2001), 757–762.
G. Korchimaros and A. Sonnino. Hyperbolic ovals in finite planes. Designs, Codes andCryptography 32 (2004), 239–249.
J. Malkevitch. Finite geometries. Available online at the following address:http://www.ams.org/features/archive/finitegeometries.html (Posted September, 2006.)
S. Olgun and I. Gunaltılı. On finite homogeneous Bolyai-Lobachevsky (B-L) n-spaces, n ≥ 2.International Mathematical Forum 2 (2007), 69–73.
V. K. Afanas’ev. Finite geometries. Journal of Mathematical Sciences 153 (2008), 856–868.
B. Celik. A hyperbolic characterization of projective Klingenberg planes. International Journalof Computational and Mathematical Sciences 2 (2008), 10–14.
20
Addendum B: Classical geometry and topological spaces
Another alternate approach to the foundations of geometry is to characterize them using thetheory of topological spaces, which is fundamental to much of modern mathematics. There havebeen several studies in this direction, but we shall only mention one pair of papers in which thenecessary mathematical background does not go beyond topics covered in standard undergraduatecourses for mathematics majors.
M. C. Gemignani. Topological geometries and a new characterization of Rn. Notre Dame
Journal of Formal Logic 7 (1966), 57–100.
M. C. Gemignani. On removing an unwanted axiom in the characterization of Rm using
topological geometries. Notre Dame Journal of Formal Logic 7 (1966), 365–366.
21 | 12,411 | 50,862 | {"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-2024-18 | latest | en | 0.949622 |
https://converterin.com/volume/board-foot-to-liter-l.html | 1,660,921,720,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573699.52/warc/CC-MAIN-20220819131019-20220819161019-00606.warc.gz | 186,137,427 | 8,469 | # BOARD FOOT TO LITER CONVERTER
FROM
TO
The result of your conversion between board foot and liter appears here
## BOARD FOOT TO LITER (board foot TO L) FORMULA
To convert between Board Foot and Liter you have to do the following:
First divide ((0.003785411784/231)*1728)/12 / 0.001 = 2.35973722
Then multiply the amount of Board Foot you want to convert to Liter, use the chart below to guide you.
## BOARD FOOT TO LITER (board foot TO L) CHART
• 1 board foot in liter = 2.35973722 board foot
• 10 board foot in liter = 23.59737216 board foot
• 50 board foot in liter = 117.9868608 board foot
• 100 board foot in liter = 235.9737216 board foot
• 250 board foot in liter = 589.934304 board foot
• 500 board foot in liter = 1179.868608 board foot
• 1,000 board foot in liter = 2359.737216 board foot
• 10,000 board foot in liter = 23597.37216 board foot
Symbol: board foot
No description
Symbol: L
No description
## More Conversions
Convertir Pie-tabla a LitroKonvertieren Board-foot bis Liter | 296 | 1,006 | {"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-2022-33 | latest | en | 0.727156 |
https://www.doubtnut.com/qna/647965866 | 1,720,875,169,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514494.35/warc/CC-MAIN-20240713114822-20240713144822-00818.warc.gz | 607,131,612 | 34,419 | Find (dy)/(dx) if y=cotx
Text Solution
Verified by Experts
y=cotx=cot(πx)180∴dydx=ddx(cotπx180)=−π180(cosec2πx180)−1.cosec2x
|
Updated on:21/07/2023
Knowledge Check
• Question 1 - Select One
The general solution of the differential equation dydx=cotxcoty is
Acos x =c cosec y
Bsin x = c sec y
Csin x = c cos y
Dcos x = c sin y
• Question 2 - Select One
The solution of the DE dydx+ylogycotx=0 is
Acosxlogy=C
Bsinxlogy=C
Clogy=Csinx
Dnone of these
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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 | 431 | 1,444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-30 | latest | en | 0.809471 |
https://www.geeksforgeeks.org/aptitude-algebra-question-7/?type=article&id=143015 | 1,708,842,302,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474581.68/warc/CC-MAIN-20240225035809-20240225065809-00564.warc.gz | 815,874,007 | 52,924 | Aptitude | Algebra | Question 7
If x1/3 + y1/3 – z1/3 = 0 then value of (x + y – z)3 + 27xyz is (A) 8 (B) 2 (C) 0 (D) 6
Explanation:
```x1/3 + y1/3 = z1/3
Now cubing both sides we get
x + 3x1/3y1/3(x1/3 + y1/3) + y = z
or, (x + y - z) = -3x1/3y1/3z1/3
Cubing again both sides, (x + y - z) = -27xyz. | 163 | 300 | {"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 | 4 | CC-MAIN-2024-10 | latest | en | 0.692163 |
https://www.coursehero.com/file/5985307/Assignment-8/ | 1,493,406,313,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123046.75/warc/CC-MAIN-20170423031203-00433-ip-10-145-167-34.ec2.internal.warc.gz | 854,598,142 | 50,208 | Assignment 8
Assignment 8 - CS 136 Fall 2009 Kate Larson Assignment 8...
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CS 136 Fall 2009 Kate Larson Assignment 8 Due Tuesday, November 24, 2009, 10pm Language level : C To submit : sieve.c , sieve-driver.c , pattern.c , pattern-driver.c , matrix.c and matrix-driver.c 1. The Sieve of Eratosthenes is a simple algorithm for finding all prime numbers between 2 and some number n . The algorithm works in the following way. First, cross out all numbers that are greater than 2 which are divisible by 2 (every second number). Then find the smallest number greater than 2 which has not been crossed out. In this case, it will be 3. Cross out all numbers that are greater than 3 which are divisible by 3. Then find the smallest number greater than 3 that has not been crossed out (in this case 5) and cross out all remaining numbers that are greater than 5 and divisible by 5. Continue this procedure until you have crossed out all numbers divisible by b n c . The numbers remaining are all prime. Write a C function void sieve( int A[], int n) where integer array A is of size at least n + 1 and which mutates A so that for 2 i n , A [ i ] has value 1 if i is prime and 0 otherwise. To submit: sieve.c and sieve-driver.c 2. Finding a pattern is a problem that arises in many situations, ranging from text editing pro- grams to analyzing DNA sequences. In this question you will implement a simple function that determines the number of times a particular pattern appears. Let
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This note was uploaded on 10/21/2010 for the course CS 136 taught by Professor Becker during the Fall '08 term at Waterloo.
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Assignment 8 - CS 136 Fall 2009 Kate Larson Assignment 8...
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Ask a homework question - tutors are online | 501 | 2,054 | {"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-2017-17 | longest | en | 0.914985 |
https://phet.colorado.edu/ro/simulation/area-model-algebra | 1,604,026,937,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107906872.85/warc/CC-MAIN-20201030003928-20201030033928-00113.warc.gz | 439,555,697 | 24,163 | Area Model Algebra
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Descriere
Build rectangles of various sizes and relate multiplication to area. Discover new strategies for multiplying algebraic expressions. Use the game screen to test your multiplication and factoring skills!
Studii de caz
• Develop and justify a method to use the area model to determine the product of a monomial and a binomial or the product of two binomials.
• Factor an expression, including expressions containing a variable.
• Recognize that area represents the product of two numbers and is additive.
• Represent a multiplication problem as the area of a rectangle, proportionally or using generic area.
• Develop and justify a strategy to determine the product of two multi-digit numbers by representing the product as an area or the sum of areas.
Standards Alignment
Common Core - Math
3.MD.C.7
Relate area to the operations of multiplication and addition.
3.MD.C.7c
Use tiling to show in a concrete case that the area of a rectangle with whole-number side lengths a and b + c is the sum of a × b and a × c. Use area models to represent the distributive property in mathematical reasoning.
3.MD.C.7d
Recognize area as additive. Find areas of rectilinear figures by decomposing them into non-overlapping rectangles and adding the areas of the non-overlapping parts, applying this technique to solve real world problems.
4.NBT.B.5
Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers, using strategies based on place value and the properties of operations. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
6.EE.A.3
Apply the properties of operations to generate equivalent expressions. For example, apply the distributive property to the expression 3 (2 + x) to produce the equivalent expression 6 + 3x; apply the distributive property to the expression 24x + 18y to produce the equivalent expression 6 (4x + 3y); apply properties of operations to y + y + y to produce the equivalent expression 3y.
6.EE.A.4
Identify when two expressions are equivalent (i.e., when the two expressions name the same number regardless of which value is substituted into them). For example, the expressions y + y + y and 3y are equivalent because they name the same number regardless of which number y stands for..
6.NS.B.4
Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Use the distributive property to express a sum of two whole numbers 1-100 with a common factor as a multiple of a sum of two whole numbers with no common factor. For example, express 36 + 8 as 4 (9 + 2)..
7.EE.A.1
Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients.
7.NS.A.2c
Apply properties of operations as strategies to multiply and divide rational numbers.
8.EE.C.7b
Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.
Versiune 1.2.1
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## Arrow
Code for labelling a parallel pair:
\overset{f}{\underset{g}{\rightrightarrows}}
$$\overset{f}{\underset{g}{\rightrightarrows}}$$
Code for a square 2-cell:
\require{AMScd} \begin{CD} A @>>> B\\ @VVV \Rightarrow @VVV\\ C @>>> D \end{CD}
$$\require{AMScd} \begin{CD} A @>>> B\\ @VVV \Rightarrow @VVV\\ C @>>> D \end{CD}$$
Code for a naturality square:
\require{AMScd} \begin{CD}
FA @>{\eta_A}>> GA\\ @V{Ff}VV @VV{Gf}V\\
FB @>>{\eta_B}> GB
\end{CD}
$$\require{AMScd} \begin{CD} FA @>{\eta_A}>> GA\\ @V{Ff}VV @VV{Gf}V\\ FB @>>{\eta_B}> GB \end{CD}$$
Code for a small chain map:
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
0 & \ra{f_1} & A & \ra{f_2} & B & \ra{f_3} & C & \ra{f_4} & D & \ra{f_5} & 0 \\
\da{g_1} & & \da{g_2} & & \da{g_3} & & \da{g_4} & & \da{g_5} & & \da{g_6} \\
0 & \ras{h_1} & 0 & \ras{h_2} & E & \ras{h_3} & F & \ras{h_4} & 0 & \ras{h_5} & 0 \\
\end{array}
$$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} 0 & \ra{f_1} & A & \ra{f_2} & B & \ra{f_3} & C & \ra{f_4} & D & \ra{f_5} & 0 \\ \da{g_1} & & \da{g_2} & & \da{g_3} & & \da{g_4} & & \da{g_5} & & \da{g_6} \\ 0 & \ras{h_1} & 0 & \ras{h_2} & E & \ras{h_3} & F & \ras{h_4} & 0 & \ras{h_5} & 0 \\ \end{array}$$
Code for a table:
\begin{array}{|c|c|c|c|}
\hline
t & a & b & l & e \\
\hline
a & b & c & d & e \\
a & b & c & d & e \\
a & b & c & d & e \\
a & b & c & d & e \\
\hline
\end{array}
$$\begin{array}{|c|c|c|c|} \hline t & a & b & l & e \\ \hline a & b & c & d & e \\ a & b & c & d & e \\ a & b & c & d & e \\ a & b & c & d & e \\ \hline \end{array}$$
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Posts 1 | 1,066 | 2,169 | {"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} | 3.203125 | 3 | CC-MAIN-2020-45 | latest | en | 0.21826 |
https://cookadvice.com/dont-have-23-cup-what-can-i-use/ | 1,709,124,842,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474715.58/warc/CC-MAIN-20240228112121-20240228142121-00336.warc.gz | 185,692,872 | 16,613 | # Don't Have 2/3 Cup What Can I Use
Staring down a recipe that calls for an elusive 2/3 cup measure can feel like solving a kitchen conundrum without the right clues. Fear not, culinary friends, for the secret to substituting this particular cup lies in the palms of your hands and the tools in your drawers.
From clever combinations of tablespoons and teaspoons to the scientific charm of water displacement, the path to precise portions without a 2/3 cup is simpler than it seems.
Let's unveil the kitchen hacks that keep your cooking on course, ensuring not a pinch nor a drop goes astray.
## Key Takeaways
• Use a 1/3 cup scoop and fill it twice to get 2/3 cup.
• Use eight tablespoons to measure 2/3 cup, leveling off each tablespoon for consistent measurements.
• Use the displacement method by filling a larger container with a known amount of water and using a smaller item to displace the water and measure 2/3 cup.
• Use visual estimation techniques by aiming for slightly less than two-thirds up a 1-cup measure or using marked pitchers for visual estimation.
## Use Smaller Measurements
Got no 2/3 cup measure? No sweat! Grab your 1/3 cup scoop. Fill and level it off once, then repeat. Boom, you've nailed 2/3 cup.
Missing a 1/3 cup? Take a 1/4 cup measure. Scoop three times, and toss in an extra two teaspoons to top it off. You're spot on with 2/3 cup.
This isn't just kitchen savvy; it's a lifesaver when you're in the thick of cooking and need to measure on the fly. Keep these hacks in your culinary toolkit, and you'll whip up recipes with the precision of a pro.
Happy cooking!
## The Tablespoon Trick
Hey there, kitchen adventurers! Ever find yourself ready to whip up something delicious but your measuring cup has gone AWOL? Fear not, because the tablespoon trick has got your back! Eight of these handy spoons will get you that 2/3 cup you need, and I'm going to tell you how to nail it every time.
Count Carefully
Grab your standard tablespoon – yes, that's the one that scoops up about 14.8 milliliters – and start counting. Eight is your magic number; no more, no less!
Level Off
If you're measuring dry goods, grab a knife and scrape off the excess for a perfect measure. It's like giving your ingredients a haircut for that just-right fit!
Consistency is Key
Repeat the same move with each tablespoon to keep your measures uniform. Think of it like a dance routine where every step is spot on!
Using this trick isn't only clever, it also cuts down on kitchen clutter. Say goodbye to a jumble of measuring cups!
Now, let's talk about another cool hack – the displacement method – for those times when you need to measure but a cup seems like a distant dream.
Keep cooking, keep smiling, and remember, where there's a whisk, there's a way!
## Displacement Method
Got a knack for the tablespoon twirl? Let's level up with the displacement method! It's a clever tactic to measure liquids when you're fresh out of measuring cups.
Picture this: you're in the middle of whipping up your grandma's secret sauce, and your measuring cup has pulled a disappearing act. No sweat!
Here's what you do. Grab a bigger container and fill it with a known amount of water—1 cup should do the trick. Now, find a smaller item with a volume you know, like a 1/3 cup scoop. Dip it into the water, and keep an eye on that water line. As it climbs back up to the 1-cup mark, you've just replaced 1/3 cup with the scoop. Fish out the scoop, and there you have it: 2/3 cup of water, ready for action.
This isn't just a party trick—it's precision in action. It's essential when your recipe needs to be spot-on, like in baking or when crafting that perfect sauce. Plus, it's a fun way to mix a little science into your kitchen hustle.
Now, ready for a quick pivot? Let's talk visual estimation—a handy skill for eyeballing measurements when you're in a pinch.
## Visual Estimation Techniques
Visual Estimation Techniques
Got no 2/3 cup measure? No stress! Visual estimation is your kitchen hack. It's using the things you know to figure out how much you need. Just remember, when you don't need to be super precise, these tricks are golden.
• Fill a 1 Cup Measure Just Right: Grab your 1-cup and eye it out. You're aiming for a smidge less than two-thirds up to the top. Use that rim as a ballpark.
• Master Marked Pitchers: Got a pitcher with lines? Spot the 2/3 line and you're all set. These guys are a roadmap to the right amount.
• Divide and Conquer: Picture your 1 cup split into thirds in your mind, then just scoop up two parts. Easy as pie.
These savvy moves keep you rolling, ensuring your dish comes out tops, even when you're short on gadgets.
## Scale and Weight Options
Lost your 2/3 cup measure? No sweat! Your trusty kitchen scale is here to save the day. Each ingredient is its own unique character with a different density, so they're not all going to weigh the same. Scooping up a 2/3 cup of sugar is going to tip the scales differently than a 2/3 cup of flour. To nail that precision, peek at a conversion chart or just use the handy one right here:
• Flour: 85 grams
• Sugar: 133 grams
• Butter: 151 grams
• Cocoa Powder: 67 grams
Keep this chart within arm's reach when you're prepping your ingredients. Aiming for that spot-on accuracy? Weighing is the way to go. Get that down pat, and you'll be whipping up dishes that are the talk of the town, every single time.
Happy cooking!
## Alternative Measurement Tools
Got a recipe calling for 2/3 cup and your measuring cup has gone AWOL? No sweat! Let's turn this into a chance to shine with some kitchen math magic. Knowing how to whip up the right volume with tools on hand is a game-changer. Here's how to nail that precise 2/3 cup measurement using what you've got.
• Double Up on 1/3 Cup: Got a 1/3 cup measure? Two scoops equal your 2/3 cup. Easy-peasy!
• 1/4 Cup + 1/8 Cup Combo: If you've got a 1/4 cup, add two 1/8 cup scoops to it. Remember, a 1/8 cup is the same as 2 tablespoons.
• 1/2 Cup Plus a Little More: Start with a 1/2 cup. Then, just toss in an extra 2 tablespoons (that's your 1/8 cup) to top it off.
Conquer these simple swaps and you'll be breezing through recipes, even when your kitchen's missing a few gadgets. Keep cooking stress-free and fun, because that's what it's all about! | 1,543 | 6,320 | {"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 | 0.928107 |
http://www.controleng.com/single-article/how-to-choose-industrial-wireless-components/0b8bac98f2db1800b8f86189d633fe12.html?OCVALIDATE= | 1,501,047,149,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425766.58/warc/CC-MAIN-20170726042247-20170726062247-00327.warc.gz | 376,594,774 | 20,856 | # How to choose industrial wireless components
## For industrial wireless networks, choosing the best accessories helps ensure a high-quality wireless network. Most problems in newly installed wireless networks result from poorly chosen or poor-quality accessories. Don’t make selection of wireless accessories an afterthought.
04/25/2013
Over the past decade, wireless technology has become one of the most efficient ways for industrial professionals to reduce operational expenses. The labor and material expenses of a cabled system can cost up to 500% more than a wireless system. Thanks largely to these potential savings, the industrial market has witnessed a rapid increase in the market acceptance of wireless.
While the radio device itself is obviously a key piece of equipment, installation practices and auxiliary equipment also affect the reliability of a wireless network. Poor installation practices, coupled with low-quality accessories, can lead to failure in a wireless network.
Power measurements
Wireless systems measure all radios’ output powers and components’ gains and losses with a common unit factor. This common factor is known as a decibel. A decibel (dB) is an abbreviation for the power ratio calculated using the formula below. Note that both P1 and P2 refer to the same power unit.
dB=10log(P1/P2)
In wireless communication, power is referred to as dBm. A dBm is calculated as a dB, but P2 always equals 1mW.
dBm=10log(P/(1 mW))
Table 1 shows some power levels commonly found in wireless communications.
Power levels common for wireless communications
Courtesy: Phoenix Contact
For every 3 dBm gain, the effect of power doubles. An increase of 6 dBm will double the effective line-of-sight range for a wireless link; dBi refers to the gain of isotropic antennas. An isotropic antenna radiates equally in all directions. dBi and dBm can be added, and the resulting sum is expressed in dBm.
The sum of all antenna gains, cable losses, radio transmission power, receiver sensitivity, and path loss is the “link budget.” The remaining single level is called the fade margin. This is the cushion between the received signal and the minimum receive threshold of the radio device.
Environmental changes can attenuate the transmitted signal, so when designing a wireless network, it is important to calculate a 10-20 dB fade margin into the link budget of the path to compensate for any such changes. Using higher gain antennas and lower loss coaxial cable, or increasing the transmission power (when possible), are two ways to increase the fade margin. Figure 1 shows an overview of the calculations.
Antenna selection
Proper antenna installation is critical and will affect the entire system’s performance. Choosing an antenna designed for use at the proper frequency and with matching impedance are the first steps. Select an antenna with an appropriate gain for the path it will be used for.
Polarization of antennas:
Polarization refers to the direction in which the radio emits energy through space. Antennas can be polarized in three ways: vertically, horizontally, and radially.
Cross-polarizing antennas mean that the receiver will accept only a fraction of the transmission power. The antenna’s angle determines what that fraction of emitted power is. For example, no power will be received by an antenna 90 degrees out of phase. An antenna 45 or 130 degrees out of phase will receive only half of the power.
However, cross-polarization is not necessarily a bad thing. Two neighboring or overlapping networks that are operating in the same frequency can result in interference. In this case, changing the polarization of one of the networks can overcome the intrusive signals between the two networks.
Universal antenna characteristics
Omni-directional antennas cover a 360-degree plane with nearly uniform characteristics across all directions (Figure 2). Some common applications for omni-directional antennas include cases where the position between the transmitter and receiver can change, moving applications or a multipoint network. They are also ideal when line of sight is obstructed, because the reflections can be used to send the signal from the transmitter to the receiver.
The best place to install an omni-directional antenna is on top of a mast or on a control cabinet. This allows it as much free space in all directions as possible. Unfortunately, however, it is not always possible. If an omni-directional antenna must be mounted on the side of a mast, the installer must observe specific measurements and distances to mount the antenna away from the mast for best possible signal.
If the omni-directional antenna is mounted to conductive material, such as a master control cabinet, its directional characteristics will be affected. The diameter and distance between the antenna and conductive material can alter the antenna’s coverage area significantly. Wall mounting should be avoided, as the wall has a great impact on the antenna’s transmission properties. If wall mounting is the only option, the installation should have a minimum of half a wavelength of the respective operating frequency of distance between the wall and antenna.
Yagi antennas
Yagi antennas (Figure 3) radiate power in a specific direction. This increases the range and reduces the chance of interference.
As the gain of a Yagi antenna increases, the energy becomes more focused. This causes the beam width to decrease, so proper alignment becomes even more critical. Aiming these antennas in the desired direction of communication (such as at the master station) is very important. Remote, fixed stations with line of sight that cover large distances should use directional antennas. The end of the antenna (farthest from the support mast) should face the associated station.
A master location with multiple slave radios must always have an omni-directional antenna, and the slave radios can have Yagi antennas to increase distance possibilities. During final alignment of the antenna heading, it should be oriented for maximum signal strength.
Directional antennas must be mounted securely. Strong winds might sway or wobble an unstable antenna, which could lead to serious misalignment.
Antenna cables
Many installers neglect the antenna coaxial cable. Using a low-quality cable can reduce efficiency or even damage the radio.
Every 3 dB of coaxial cable loss cuts the transmission distance in half. Choosing the correct coaxial cable depends on:
• The length of cable required to span the distance between transmitter and the antenna
• The amount of signal loss that can be tolerated
• Cost considerations.
Long-range transmission paths are likely to be weaker. A low-loss cable type will work best, especially if the cable must be more than 50 ft long. If you have a short-range system or only require a short coaxial cable, a less efficient (and costlier) cable might suffice.
The cable’s loss depends on the frequency. As the radio operating frequency increases, the cable loss also increases. Consult the coaxial cable datasheets to determine if it will work with your intended radio frequency.
The cable’s bending radius is another important consideration during the installation process. The bending radius is the minimum ratio to the inside curvature of the cable without kinking, damaging, or shortening its life span (see Figure 4). The cable’s datasheet typically includes the bending radius.
Power supplies
A radio will not operate properly if it does not have a sufficient power source. This can result in intermittent radio links, dropped data, or the need to reset devices. Yet many designers overlook power supplies when designing a wireless network.
Many radios will have operating modes or times that will demand a high amount of power. The power supply must provide enough current to all of the connected loads. It should also have at least 30% overhead to compensate for expansion and loads that draw variable power.
Installing a UPS as a backup power supply is also a good practice, especially for high-integrity sites. The UPS will act as a backup power supply if power at the site fails.
Choose the right wireless accessories
A high-quality wireless network needs to provide reliable communication, usually for many years of service. Most problems in newly installed wireless networks result from poorly chosen or poor-quality accessories, yet for many people, accessories are often an afterthought when planning their network. From planning to installation, choosing the right components for your wireless system is equally as important as choosing the main system.
- David Burrell is wireless product specialist, Phoenix Contact. Edited by Mark T. Hoske, content manager, CFE Media, Control Engineering, Plant Engineering, and Consulting-Specifying Engineer, mhoske@cfemedia.com.
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This course explains how maintaining power and communication systems through emergency power-generation systems is critical. | 2,172 | 11,339 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-30 | longest | en | 0.924365 |
https://quizstone.com/en/q17265/1-4-x-1-4-1-4-=/ | 1,653,093,311,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662534693.28/warc/CC-MAIN-20220520223029-20220521013029-00433.warc.gz | 545,106,792 | 5,863 | # 1/4 x 1/4 - 1/4 = ?
Find the answer below
### Explanation
In mathematics a fraction represents any number of equal parts. The numerator represents the number of equal parts, and the denominator, which cannot be zero, indicates how many of those parts make up a unit or a whole. For example, in the fraction 3/4, the numerator, 3, tells us that the fraction represents 3 equal parts, and the denominator, 4, tells us that 4 parts make up a whole. The word comes from the Latin "fractus", which means "broken".
### Statistics
Answer time 0s (0s). 0% have previously answered correct on this question. The question was created 2014-02-04. | 165 | 642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2022-21 | latest | en | 0.904774 |
https://study.com/academy/answer/the-diameter-of-the-moon-is-3480-km-a-what-is-the-surface-area-of-the-moon-b-how-many-times-larger-is-the-surface-area-of-the-earth.html | 1,580,297,208,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251796127.92/warc/CC-MAIN-20200129102701-20200129132701-00470.warc.gz | 661,417,337 | 23,378 | # The diameter of the Moon is 3480 km. (a) What is the surface area of the Moon? (b) How many...
## Question:
The diameter of the Moon is 3480 km.
(a) What is the surface area of the Moon?
(b) How many times larger is the surface area of the Earth?
## Solar system
The Solar System is formed by celestial bodies held in place by the gravitational attraction of the Sun. The principal components of the Solar System are the eight planets forming it, although there are several other celestial bodies (moons, asteroids, etc) also moving in the orbit of the Sun.
Ordering the planets attending to their size:
{eq}\bullet {/eq} Jupiter ({eq}69,911 \;\rm km {/eq}), over ten times the size of Earth.
{eq}\bullet {/eq} Saturn ({eq}58,232 \;\rm km) {/eq}, 9.5 times the size of Earth
{eq}\bullet {/eq} Uranus ({eq}25,362 \;\rm km {/eq}), 4 times the size of Earth
{eq}\bullet {/eq} Neptune ({eq}24,622 \;\rm km {/eq}), 3.9 times the size of Earth
{eq}\bullet {/eq} Earth {eq}(6,371 \;\rm km) {/eq}.
{eq}\bullet {/eq} Venus {eq}(6,052 \;\rm km) {/eq}, 0.95 times the size of Earth.
{eq}\bullet {/eq} Mars {eq}(3,390 \;\rm km) {/eq}, 0.53 times the size of Earth.
{eq}\bullet {/eq} Mercury {eq}(2,440 \;\rm km) {/eq}, 0.38 times the size of Earth.
{eq}\bullet {/eq} Moon ({eq}1740\;\rm km {/eq}), 0.27 times the size of Earth.
## Answer and Explanation:
The Moon is modeled as a sphere with radius,
{eq}R_M=\dfrac{D_M}{2}=\dfrac{3480\;\rm km}{2}=1740\;\rm km {/eq}.
a) The surface of the Moon is computed from the formula for the surface of a sphere,
{eq}S_{Moon}=4\pi R_M^2 {/eq},
evaluating the expression,
{eq}S_{Moon}=4\cdot 3.1416 \cdot 1740^2\;\rm km^2=\boxed{3.80\times 10^7 \;\rm km^2} {/eq}.
b) Let us compare the surface of the Moon with that of the Earth. The Earth is modeled as a sphere with radius,
{eq}R_E=6371\;\rm km {/eq}.
The Earth's surface is,
{eq}S_E=4\pi R_E^2=4\cdot 3.1416\cdot 6371^2\;\rm km^2=5.10\times 10^8 \;\rm km^2 {/eq}.
Dividing the surface of the Moon by the surface of the Earth,
{eq}\eta=\dfrac{S_M}{S_E}=\dfrac{3.80\times 10^7 \;\rm km^2}{5.10\times 10^8 \;\rm km^2}=\boxed{0.075} {/eq}.
The ratio is {eq}\eta=0.075 {/eq}, consequently, the surface of the Earth is {eq}\boxed{13.33} {/eq} times larger than the surface of the Moon. | 802 | 2,290 | {"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-2020-05 | longest | en | 0.765148 |
https://happylibus.com/doc/955/applications | 1,670,555,771,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711376.47/warc/CC-MAIN-20221209011720-20221209041720-00697.warc.gz | 312,976,493 | 10,802 | Applications
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Transcript
Applications
```2.5. APPLICATIONS
2.5
91
Applications
1. We follow the Requirements for Word Problem Solutions.
1. Set up a variable dictionary. Let θ represent the measure of the first
angle.
2. Set up an equation. A sketch will help summarize the information given
in the problem. First, we sketch two angles whose sum is 90 degrees.
The second angle is 6 degrees larger than 2 times the first angle, so the
second angle has measure 2θ + 6.
2θ + 6
θ
The angles are complementary, so their sum is 90 degrees. Thus the
equation is:
θ + (2θ + 6) = 90
3. Solve the equation. Simplify the left-hand side by combining like terms.
θ + (2θ + 6) = 90
3θ + 6 = 90
Subtract 6 from both sides of the equation, then divide both sides of the
resulting equation by 3.
3θ + 6 − 6 = 90 − 6
3θ = 84
3θ
84
=
3
3
θ = 28
4. Answer the question. To find the second angle, substitute 28 for θ in
2θ + 6 to get:
2θ + 6 = 2(28) + 6
= 62
Hence, the two angles are 28 and 62 degrees.
Second Edition: 2012-2013
CHAPTER 2. SOLVING LINEAR EQUATIONS
92
5. Look back. Let’s label the angles with their numerical values.
62◦
28◦
Clearly, their sum is 90◦ , so we have the correct answer.
3. We follow the Requirements for Word Problem Solutions.
1. Set up a variable dictionary. Let θ represent the measure of the first
angle.
2. Set up an equation. A sketch will help summarize the information given
in the problem. First, we sketch two angles whose sum is 180 degrees.
The second angle is 10 degrees larger than 4 times the first angle, so the
second angle has measure 4θ + 10.
4θ + 10
θ
The angles are supplementary, so their sum is 180 degrees. Thus the
equation is:
θ + (4θ + 10) = 180
3. Solve the equation. Simplify the left-hand side by combining like terms.
θ + (4θ + 10) = 180
5θ + 10 = 180
Second Edition: 2012-2013
2.5. APPLICATIONS
93
Subtract 10 from both sides of the equation, then divide both sides of
the resulting equation by 5.
5θ + 10 − 10 = 180 − 10
5θ = 170
5θ
170
=
5
5
θ = 34
4. Answer the question. To find the second angle, substitute 34 for θ in
4θ + 10 to get:
4θ + 10 = 4(34) + 10
= 146
Hence, the two angles are 34 and 146 degrees.
5. Look back. Let’s label the angles with their numerical values.
146◦
34◦
Clearly, their sum is 180◦ , so we have the correct answer.
5. In the solution, we address each step of the Requirements for Word Problem
Solutions.
1. Set up a Variable Dictionary. An example of three consecutive integers
is 19, 20, and 21. These are not the integers we seek, but they do give us
some sense of the meaning of three consecutive integers. Note that each
consecutive integer is one larger than the preceding integer. Thus, if k
is the length of the first side of the triangle, then the next two sides are
k + 1 and k + 2. In this example, our variable dictionary will take the
form of a well-labeled figure.
Second Edition: 2012-2013
CHAPTER 2. SOLVING LINEAR EQUATIONS
94
k+2
k
k+1
2. Set up an Equation. The perimeter of the triangle is the sum of the three
sides. If the perimeter is 483 meters, then:
k + (k + 1) + (k + 2) = 483
3. Solve the Equation. To solve for k, first simplify the left-hand side of the
equation by combining like terms.
k + (k + 1) + (k + 2) = 483
3k + 3 = 483
3k + 3 − 3 = 483 − 3
3k = 480
480
3k
=
3
3
k = 160
Original equation.
Combine like terms.
Subtract 3 from both sides.
Simplify.
Divide both sides by 3.
Simplify.
4. Answer the Question. Thus, the first side has length 160 meters. Because
the next two consecutive integers are k + 1 = 161 and k + 2 = 162, the
three sides of the triangle measure 160, 161, and 162 meters, respectively.
5. Look Back. An image helps our understanding. The three sides are
consecutive integers.
162 meters
160 meters
161 meters
Note that the perimeter (sum of the three sides) is:
160 meters + 161 meters + 162 meters = 483 meters
(2.1)
Thus, the perimeter is 483 meters, as it should be. Our solution is correct.
Second Edition: 2012-2013
2.5. APPLICATIONS
95
7. We follow the Requirements for Word Problem Solutions.
1. Set up a variable dictionary. Let x represent the unknown number.
2. Set up an equation. The statement “four less than eight times a certain
number is −660” becomes the equation:
8x − 4 = −660
3. Solve the equation. Add 4 to both sides, then divide the resulting equation
by 8.
8x − 4 = −660
8x − 4 + 4 = −660 + 4
8x = −656
8x
−656
=
8
8
x = −82
4. Answer the question. The unknown number is −82.
5. Look back. “Four less than eight times −82” translates as 8(−82) − 4,
which equals −660. The solution make sense.
9. In the solution, we address each step of the Requirements for Word Problem
Solutions.
1. Set up a Variable Dictionary. Let d represent the distance left for Alan
to hike. Because Alan is four times further from the beginning of the
trail than the end, the distance Alan has already completed is 4d. Let’s
construct a little table to help summarize the information provided in
this problem.
Section of Trail
Distance (mi)
Distance to finish
Distance from start
d
4d
Total distance
70
2. Set up an Equation. As you can see in the table above, the second column
shows that the sum of the two distances is 70 miles. In symbols:
d + 4d = 70
Second Edition: 2012-2013
CHAPTER 2. SOLVING LINEAR EQUATIONS
96
3. Solve the Equation. To solve for d, first simplify the left-hand side of the
equation by combining like terms.
d + 4d = 70
5d = 70
5d
70
=
5
5
d = 14
Original equation.
Combine like terms.
Divide both sides by 5.
Simplify.
4. Answer the Question. Alan still has 14 miles to hike.
5. Look Back. Because the amount left to hike is d = 14 miles, Alan’s
distance from the start of the trail is 4d = 4(14), or 56 miles. If we
arrange these results in tabular form, it is evident that not only is the
distance from the start of the trail four times that of the distance left to
the finish, but also the sum of their lengths is equal to the total length
of the trail.
Section of Trail
Distance (mi)
Distance (mi)
Distance to finish
Distance from start
d
4d
14
56
Total distance
70
70
Thus, we have the correct solution.
11. We follow the Requirements for Word Problem Solutions.
1. Set up a variable dictionary. Let p represent the percentage of Martha ’s
sixth grade class that is absent.
2. Set up the equation. The question is “what percent of the class size
equals the number of students absent?” The phrase “p percent of 36 is
2” becomes the equation:
p × 36 = 2
Or equivalently:
36p = 2
3. Solve the equation. Use a calculator to help divide both sides of the
equation by 36.
36p = 2
2
36p
=
36
36
p = 0.0555555556
Second Edition: 2012-2013
2.5. APPLICATIONS
97
4. Answer the question. We need to change our answer to a percent. First,
round the percentage answer p to the nearest hundredth.
Test digit
0.0 5 5 5555556
Rounding digit
Because the test digit is greater than or equal to 5, add 1 to the rounding
digit, then truncate. Hence, to the nearest hundredth, 0.0555555556 is
approximately 0.06. To change this answer to a percent, multiply by 100,
or equivalently, move the decimal two places to the right. Hence, 6% of
Martha’s sixth grade class is absent.
5. Look back. If we take 6% of Martha’s class size, we get:
6% × 36 = 0.06 × 36
= 2.16
Rounded to the nearest student, this means there are 2 students absent,
indicating we’ve done the problem correctly.
13. We follow the Requirements for Word Problem Solutions.
1. Set up a variable dictionary. Let x represent the length of the first piece.
2. Set up an equation. The second piece is 3 times as long as the first piece,
so the second piece has length 3x. The third piece is 6 centimeters longer
than the first piece, so the second piece has length x + 6. Let’s construct
a table to summarize the information provided in this problem.
Piece
Length (centimeters)
First
Second
Third
Total length
x
3x
x+6
211
As you can see in the table above, the second column shows that the sum
of the three pieces is 211 centimeters. Hence, the equation is:
x + 3x + (x + 6) = 211
Second Edition: 2012-2013
CHAPTER 2. SOLVING LINEAR EQUATIONS
98
3. Solve the equation. First, simplify the left-hand side of the equation by
combining like terms.
x + 3x + (x + 6) = 211
5x + 6 = 211
Subtract 6 from both sides of the equation, then divide both sides of the
resulting equation by 5.
5x + 6 − 6 = 211 − 6
5x = 205
5x
205
=
5
5
x = 41
4. Answer the question. Let’s add a column to our table to list the length
of the three pieces. The lengths of the second and third pieces are found
by substituting 41 for x in 3x and x + 6.
Piece
First
Second
Third
Total length
Length (centimeters)
Length (centimeters)
x
3x
x+6
41
123
47
211
211
5. Look back. The third column of the table above shows that the lengths
sum to 211 centimeters, so we have the correct solution.
15. In the solution, we address each step of the Requirements for Word Problem
Solutions.
1. Set up a Variable Dictionary. An example of three consecutive even
integers is 18, 20, and 22. These are not the integers we seek, but they
do give us some sense of the meaning of three consecutive even integers.
Note that each consecutive even integer is two larger than the preceding
integer. Thus, if k is the length of the first side of the triangle, then the
next two sides are k+2 and k+4. In this example, our variable dictionary
will take the form of a well-labeled figure.
Second Edition: 2012-2013
2.5. APPLICATIONS
99
k+4
k
k+2
2. Set up an Equation. The perimeter of the triangle is the sum of the three
sides. If the perimeter is 450 yards, then:
k + (k + 2) + (k + 4) = 450
3. Solve the Equation. To solve for k, first simplify the left-hand side of the
equation by combining like terms.
k + (k + 2) + (k + 4) = 450
3k + 6 = 450
3k + 6 − 6 = 450 − 6
3k = 444
444
3k
=
3
3
k = 148
Original equation.
Combine like terms.
Subtract 6 from both sides.
Simplify.
Divide both sides by 3.
Simplify.
4. Answer the Question. Thus, the first side has length 148 yards. Because
the next two consecutive even integers are k+2 = 150 and k+4 = 152, the
three sides of the triangle measure 148, 150, and 152 yards, respectively.
5. Look Back. An image helps our understanding. The three sides are
consecutive even integers.
152 yards
148 yards
150 yards
Note that the perimeter (sum of the three sides) is:
148 yards + 150 yards + 152 yards = 450 yards
(2.2)
Thus, the perimeter is 450 yards, as it should be. Our solution is correct.
Second Edition: 2012-2013
CHAPTER 2. SOLVING LINEAR EQUATIONS
100
17. We follow the Requirements for Word Problem Solutions.
1. Set up a variable dictionary. Let x represent the length of the first side
of the triangle.
2. Set up an equation. The second side is 7 times as long as the first side,
so the second side has length 7x. The third side is 9 yards longer than
the first side, so the second side has length x + 9. Let’s sketch a diagram
to summarize the information provided in this problem (the sketch is not
drawn to scale).
x+9
x
7x
The sum of the three sides of the triangle equals the perimeter. Hence,
the equation is:
x + 7x + (x + 9) = 414
3. Solve the equation. First, simplify the left-hand side of the equation by
combining like terms.
x + 7x + (x + 9) = 414
9x + 9 = 414
Subtract 9 from both sides of the equation, then divide both sides of the
resulting equation by 9.
9x + 9 − 9 = 414 − 9
9x = 405
9x
405
=
9
9
x = 45
4. Answer the question. Because the first side is x = 45 yards, the second
side is 7x = 315 yards, and the third side is x + 9 = 54 yards.
5. Look back. Let’s add the lengths of the three sides to our sketch.
54 yards
315 yards
Second Edition: 2012-2013
45 yards
2.5. APPLICATIONS
101
Our sketch clearly indicates that the perimeter of the triangle is Perimeter =
45 + 315 + 54, or 414 yards. Hence, our solution is correct.
19. We follow the Requirements for Word Problem Solutions.
1. Set up a variable dictionary. Let k represent the smallest of three consecutive odd integers.
2. Set up an equation. Because k is the smallest of three consecutive odd
integers, the next two consecutive odd integers are k + 2 and k + 4.
Therefore, the statement “the sum of three consecutive odd integers is
−543” becomes the equation:
k + (k + 2) + (k + 4) = −543
3. Solve the equation. First, combine like terms on the left-hand side of the
equaton.
k + (k + 2) + (k + 4) = −543
3k + 6 = −543
Subtract 6 from both sides, then divide both sides of the resulting equation by 3.
3k + 6 − 6 = −543 − 6
3k = −549
3k
−549
=
3
3
k = −183
4. Answer the question. The smallest of three consecutive odd integers is
−183.
5. Look back. Because the smallest of three consecutive odd integers is −183,
the next two consecutive odd integers are −181, and −179. If we sum
these integers, we get
−183 + (−181) + (−179) = −543,
so our solution is correct.
Second Edition: 2012-2013
CHAPTER 2. SOLVING LINEAR EQUATIONS
102
21. We follow the Requirements for Word Problem Solutions.
1. Set up a variable dictionary. Let θ represent the measure of angle A.
2. Set up an equation. A sketch will help summarize the information given
in the problem. Because angle B is 4 times the size of angle A, the degree
measure of angle B is represented by 4θ. Because angle C is 30 degrees
larger than the degree measure of angle A, the degree measure of angle
C is represented by θ + 30.
C
θ + 30
θ
4θ
A
B
Because the sum of the three angles is 180◦, we have the following equation:
θ + 4θ + (θ + 30) = 180
3. Solve the equation. Start by combining like terms on the left-hand side
of the equation.
θ + 4θ + (θ + 30) = 180
6θ + 30 = 180
Subtract 30 from both sides of the equation and simplify.
6θ + 30 − 30 = 180 − 30
6θ = 150
Divide both sides by 6.
150
6θ
=
6
6
θ = 25
4. Answer the question. The degree measure of angle A is θ = 25◦ . The
degree measure of angle B is 4θ = 100◦. The degree measure of angle C
is θ + 30 = 55◦ .
5. Look back. Our figure now looks like the following.
Second Edition: 2012-2013
2.5. APPLICATIONS
103
C
55◦
25◦
100◦
A
B
Note that
25 + 100 + 55 = 180,
so our solution is correct.
23. We follow the Requirements for Word Problem Solutions.
1. Set up a variable dictionary. Let k represent the smallest of three consecutive integers.
2. Set up an equation. Because k is the smallest of three consecutive integers, the next two consecutive integers are k + 1 and k + 2. Therefore,
the statement “the sum of three consecutive integers is −384” becomes
the equation:
k + (k + 1) + (k + 2) = −384
3. Solve the equation. First, combine like terms on the left-hand side of the
equaton.
k + (k + 1) + (k + 2) = −384
3k + 3 = −384
Subtract 3 from both sides, then divide both sides of the resulting equation by 3.
3k + 3 − 3 = −384 − 3
3k = −387
−387
3k
=
3
3
k = −129
4. Answer the question. The smallest of three consecutive integers is k =
−129, so the next two consecutive integers are −128 and −127. Therefore,
the largest of the three consecutive integers is −127.
5. Look back. If we sum the integers −129, −128, and −127, we get
−129 + (−128) + (−127) = −384,
so our solution is correct.
Second Edition: 2012-2013
104
CHAPTER 2. SOLVING LINEAR EQUATIONS
25. We follow the Requirements for Word Problem Solutions.
1. Set up a variable dictionary. Let x represent the unknown number.
2. Set up an equation. The statement “seven more than two times a certain
number is 181” becomes the equation:
7 + 2x = 181
3. Solve the equation. Subtract 7 from both sides, then divide the resulting
equation by 2.
7 + 2x = 181
7 + 2x − 7 = 181 − 7
2x = 174
2x
174
=
2
2
x = 87
4. Answer the question. The unknown number is 87.
5. Look back. “Seven more than two times 87” translates as 7 + 2(87), which
equals 181. The solution make sense.
27. In the solution, we address each step of the Requirements for Word Problem
Solutions.
1. Set up a Variable Dictionary. An example of three consecutive odd integers is 19, 21, and 23. These are not the integers we seek, but they
do give us some sense of the meaning of three consecutive odd integers.
Note that each consecutive odd integer is two larger than the preceding
integer. Thus, if k is the length of the first side of the triangle, then the
next two sides are k+2 and k+4. In this example, our variable dictionary
will take the form of a well-labeled figure.
k+4
k
k+2
Second Edition: 2012-2013
2.5. APPLICATIONS
105
2. Set up an Equation. The perimeter of the triangle is the sum of the three
sides. If the perimeter is 537 feet, then:
k + (k + 2) + (k + 4) = 537
3. Solve the Equation. To solve for k, first simplify the left-hand side of the
equation by combining like terms.
k + (k + 2) + (k + 4) = 537
3k + 6 = 537
3k + 6 − 6 = 537 − 6
3k = 531
3k
531
=
3
3
k = 177
Original equation.
Combine like terms.
Subtract 6 from both sides.
Simplify.
Divide both sides by 3.
Simplify.
4. Answer the Question. Thus, the first side has length 177 feet. Because
the next two consecutive odd integers are k+2 = 179 and k+4 = 181, the
three sides of the triangle measure 177, 179, and 181 feet, respectively.
5. Look Back. An image helps our understanding. The three sides are
consecutive odd integers.
181 feet
177 feet
179 feet
Note that the perimeter (sum of the three sides) is:
177 feet + 179 feet + 181 feet = 537 feet
(2.3)
Thus, the perimeter is 537 feet, as it should be. Our solution is correct.
29. We follow the Requirements for Word Problem Solutions.
1. Set up a variable dictionary. Let M represent the marked price of the
article.
Second Edition: 2012-2013
CHAPTER 2. SOLVING LINEAR EQUATIONS
106
2. Solve the equation. Because the store offers a 14% discount, Yao pays
86% for the article. Thus, the question becomes “86% of the marked
price is \$670.8.” This translates into the equation
86% × M = 670.8,
or equivalently,
0.86M = 670.8
Use a calculator to help divide both sides by 0.86.
0.86M
670.8
=
0.86
0.86
M = 780
3. Answer the question. Hence, the original marked price was \$780.
4. Look back. Because the store offers a 14% discount, Yao has to pay 86%
for the article. Check what 86% of the marked price will be.
86% × 780 = 0.86 × 780
= 670.8
That’s the sales price that Yao paid. Hence, we’ve got the correct solution.
31. We follow the Requirements for Word Problem Solutions.
1. Set up a variable dictionary. Let k represent the smallest of three consecutive even integers.
2. Set up an equation. Because k is the smallest of three consecutive even
integers, the next two consecutive even integers are k + 2 and k + 4.
Therefore, the statement “the sum of three consecutive even integers is
−486” becomes the equation:
k + (k + 2) + (k + 4) = −486
3. Solve the equation. First, combine like terms on the left-hand side of the
equaton.
k + (k + 2) + (k + 4) = −486
3k + 6 = −486
Second Edition: 2012-2013
2.5. APPLICATIONS
107
Subtract 6 from both sides, then divide both sides of the resulting equation by 3.
3k + 6 − 6 = −486 − 6
3k = −492
3k
−492
=
3
3
k = −164
4. Answer the question. The smallest of three consecutive even integers is
−164.
5. Look back. Because the smallest of three consecutive even integers is
−164, the next two consecutive even integers are −162, and −160. If we
sum these integers, we get
−164 + (−162) + (−160) = −486,
so our solution is correct.
33. We follow the Requirements for Word Problem Solutions.
1. Set up a variable dictionary. Let M represent the amount invested in the
mutal fund.
2. Set up the equation. We’ll use a table to help summarize the information
in this problem. Because the amount invested in the certificate of deposit
is \$3,500 more than 6 times the amount invested in the mutual fund,
we represent the amount invested in the certificate of deposit with the
expression 6M + 3500.
Investment
Mutual fund
Certificate of deposit
Totals
Amount invested
M
6M + 3500
45500
The second column of the table gives us the needed equation. The two
investment amounts must total \$45,500.
M + (6M + 3500) = 45500
Second Edition: 2012-2013
```
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# Many of the convenience foods on the market today, like dry
Author Message
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4284
Kudos [?]: 544 [0], given: 0
Many of the convenience foods on the market today, like dry [#permalink]
### Show Tags
04 Apr 2004, 09:25
00:00
Difficulty:
(N/A)
Question Stats:
100% (02:27) correct 0% (00:00) wrong based on 1 sessions
### HideShow timer Statistics
11. Many of the convenience foods on the market today, like dry cereals, have less nutrients than natural foods, which were dominant a decade or two ago. Many nutritionists claim that dry cereal gives less nourishment than natural foods like eggs or bacon. Opponents of nutritionists' views state that examination of grade-school students show less nutritional deficiency than in their parents' time.
Which of the following, If true, would tend to strengthen the opponents' view?
(A) Grade school children reported eating no breakfast at all.
(B) Fewer convenience foods were available to the parents.
(C) Adults claim to eat convenience foods as well as natural foods.
(D) Convenience foods can be digested just as quickly as natural foods.
(E) Consumers are not likely to sacrifice convenience for nutrition.
_________________
Best Regards,
Paul
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Director
Joined: 03 Jul 2003
Posts: 651
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### Show Tags
04 Apr 2004, 09:31
This is another tough one!
I'll go with B
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Joined: 04 Apr 2004
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### Show Tags
04 Apr 2004, 12:11
Paul wrote:
11. Many of the convenience foods on the market today, like dry cereals, have less nutrients than natural foods, which were dominant a decade or two ago. Many nutritionists claim that dry cereal gives less nourishment than natural foods like eggs or bacon. Opponents of nutritionists' views state that examination of grade-school students show less nutritional deficiency than in their parents' time.
Which of the following, If true, would tend to strengthen the opponents' view?
(A) Grade school children reported eating no breakfast at all.
(B) Fewer convenience foods were available to the parents.
(C) Adults claim to eat convenience foods as well as natural foods.
(D) Convenience foods can be digested just as quickly as natural foods.
(E) Consumers are not likely to sacrifice convenience for nutrition.
Quote:
(B) Fewer convenience foods were available to the parents.
parents' time:
few conv. foods available=>more nutriens=>less nutritional def.
de facto "grade-school students show less nutritional deficiency than in their parents' time"
=> nutritionists are wrong
all other answers are out of scope.
Marat
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### Show Tags
04 Apr 2004, 18:54
Everyone is on the money OA is B
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Paul
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# Many of the convenience foods on the market today, like dry
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https://open.kattis.com/contests/pgj7cu/problems/candlebox | 1,696,285,888,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511021.4/warc/CC-MAIN-20231002200740-20231002230740-00668.warc.gz | 486,797,133 | 8,283 | Hide
# Problem CCandle Box
Rita loves her Birthday parties. She is really happy when blowing the candles at the Happy Birthday’s clap melody. Every year since the age of four she adds her birthday candles (one for every year of age) to a candle box. Her younger daydreaming brother Theo started doing the same at the age of three. Rita’s and Theo’s boxes look the same, and so do the candles.
One day Rita decided to count how many candles she had in her box:
– No, no, no! I’m younger than that!
She just realized Theo had thrown some of his birthday candles in her box all these years. Can you help Rita fix the number of candles in her candle box?
Given the difference between the ages of Rita and Theo, the number of candles in Rita’s box, and the number of candles in Theo’s box, find out how many candles Rita needs to remove from her box such that it contains the right number of candles.
## Input
The first line of the input has one integer $D$, corresponding to the difference between the ages of Rita and Theo.
The second line has one integer $R$, corresponding to the number of candles in Rita’s box.
The third line has one integer $T$, corresponding to the number of candles in Theo’s box.
## Constraints
$1$ $\leq$ $D$ $\leq$ $20$ Difference between the ages of Rita and Theo $4$ $\leq$ $R$ $<$ $1\, 000$ Number of candles in Rita’s box $0$ $\leq$ $T$ $<$ $1\, 000$ Number of candles in Theo’s box
## Output
An integer representing the number of candles Rita must remove from her box such that it contains the right number of candles.
Sample Input 1 Sample Output 1
2
26
8
4
Hide | 394 | 1,610 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2023-40 | latest | en | 0.915518 |
https://www.mersenneforum.org/showthread.php?s=801398e8226b4a2ca61ddac865ef29de&t=15557&goto=nextoldest | 1,638,080,771,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358469.34/warc/CC-MAIN-20211128043743-20211128073743-00514.warc.gz | 982,081,848 | 9,619 | mersenneforum.org Post Lots and Lots of Top-5000 Primes Here
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2010-03-17, 22:24 #1 Kosmaj Nov 2003 E2616 Posts Post Lots and Lots of Top-5000 Primes Here After 2000 posts in our first prime posting thread, and more than 2300 reported primes, we are moving to a new thread. Please post your primes here! And happy hunting
2010-03-25, 11:10 #2 pb386 Nov 2003 23×163 Posts 1389*2^809811-1 (243781 digits)
2010-03-26, 10:20 #3 Beyond Dec 2002 44410 Posts 2655*2^853543-1 (256946 digits)
2010-03-26, 11:04 #4 pb386 Nov 2003 23×163 Posts 1285*2^809975-1 (243830 digits)
2010-03-27, 14:14 #5 pb386 Nov 2003 23×163 Posts 1065*2^961019-1 (289299 digits)
2010-03-28, 12:43 #6 pb386 Nov 2003 23×163 Posts 1119*2^957013-1 (288093 digits) 1069*2^811021-1 (244145 digits) 1023*2^811532-1 (244299 digits)
2010-03-28, 13:48 #7 pb386 Nov 2003 23·163 Posts 1095*2^960229-1 (289061 digits)
2010-03-29, 21:48 #8 Beyond Dec 2002 44410 Posts 251*2^1269198-1 (382070 digits)
2010-03-30, 21:15 #9
Cruelty
May 2005
22×11×37 Posts
Quote:
Originally Posted by Beyond 251*2^1269198-1 (382070 digits)
Nice one! Congratulations!
2010-04-01, 07:31 #10 VBCurtis "Curtis" Feb 2005 Riverside, CA 5×1,013 Posts I am the King of small megabit primes: 281*2^1009502-1 is prime. (303893 digits) That makes 3 megabits under 1020k. As we say in poker, I run good. -Curtis
2010-04-02, 14:59 #11 pb386 Nov 2003 23·163 Posts 7th drive 2295*2^867724-1 (261215 digits)
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A copy of the license is included in the FAQ. | 880 | 2,418 | {"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-2021-49 | latest | en | 0.582188 |
https://puzzling.stackexchange.com/questions/123100/sudoku-variants-1-schr%C3%B6dingers-grave-6x6-grid | 1,702,187,300,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101195.85/warc/CC-MAIN-20231210025335-20231210055335-00237.warc.gz | 371,264,236 | 41,659 | # Sudoku variants #1: Schrödinger's Grave (6x6 grid)
Not in conjunction with my Minesweeper puzzles - that would be concerning!
Inspired by Cracking The Cryptic's video "The Graveyard Of Schrödinger" (original gimmick by fjam)
How this gimmick works:
Normal sudoku rules almost apply. Place the digits 0-6 into cells such that each digit appears once in every row, column and region. To accommodate this, one cell in each row, column and region is a Schrödinger cell, which contains two digits. In graves (cages), digits must sum to the day, month or year of the date on the grave. Digits cannot repeat within graves. Escape the graveyard by carefully plotting an orthogonally connected path between the green and red cells. The path may not cross the highest or lowest digits in a grave.
## Some things to clarify (as best as I can):
1. What happens when there is only a grave with two cells?
Here's my take on this. Say we have this table (ripped directly from the video) (the dates mean that the numbers in the cage add up to either the day, month, or year listed on it, for example R1C2 and R1C3 add up to either 2 (the month), 13 (the day), or 20 (the year)):
Green cell 13/02/20 13/02/20
14/01/23 31/08/30 31/08/30
14/01/23 31/08/30 31/08/30
Which Simon from Cracking the Cryptic solved this box as such:
3 7 4/9
8 1 2
6 5 0
With a line going through R1C1, R1C2, R2C2, R2C3 before going into Box 2.
This is valid because we have:
1. All digits 0-9 ✓
2. Orthogonal line not going through the highest and lowest digits in each "grave" ✓
3. The digits in each "grave" adding up to either the year, month, or day for the grave listed ✓
Which brings us to our second question: How is this valid? Because
sure, we have$$7+4\overset{✓}=13$$but what we also have is$$7+9\ne13,2,20$$
I honestly think this is because of our "Schrödinger" cell in box 1, so I think that's an exception to the rule "digits must sum to the day, month or year of the date on the grave" however if someone could give me a more in depth explanation on this that would be good.
# The puzzle (only 6x6)
How the 6 boxes are going to be split up:
1. Box 1: R1C1, R1C2, R1C3, R2C1, R2C2, R2C3
2. Box 2: R1C4, R1C5, R1C6, R2C4, R2C5, R2C6
3. Box 3: R3C1, R3C2, R3C3, R4C1, R4C2, R4C3
4. Box 4: R3C4, R3C5, R3C6, R4C4, R4C5, R4C6
5. Box 5: R5C1, R5C2, R5C3, R6C1, R6C2, R6C3
6. Box 6: R5C3, R5C5, R5C6, R6C4, R6C5, R6C6
$$\color{green}✓$$ 9/12/14 10/15/16 10/15/16 10/15/16 10/15/16
8/13/25 9/12/14 9/12/14 10/15/16 12/06/67 12/06/67
8/13/25 12/24/87 12/24/87 11/27/28 11/27/28 8/23/28
8/13/25 12/24/87 12/24/87 11/27/28 11/27/28 8/23/28
1/7/27 11/13/16 11/13/16 8/11/27 5/11/12 5/11/12
1/7/27 1/7/27 11/13/16 8/11/27 5/11/12 $$\color{red}✓$$
Note: Please give me any feedback you have about this puzzle, I have never done something like this before and I think that this is honestly a really fun variant of sudoku.
• I assume 8/23/25 is a typo and should be 8/13/25. Also, unless I'm missing something, boxes 4 and 5 are broken - the numbers in them need to sum to 21, but the gravestones that cover them precisely do not have combinations of numbers that can sum to 21. Nov 6 at 16:43
• @bobajob 8/23/25 is a typo, also I don't think boxes 4/5 are broken (since 0 exists in this puzzle, read the bit about the Schrödinger cells to understand what is up with that) Nov 6 at 16:47 | 1,233 | 3,368 | {"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": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2023-50 | latest | en | 0.910978 |
https://codereview.stackexchange.com/questions/tagged/graph?tab=newest&page=8 | 1,571,134,791,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986657949.34/warc/CC-MAIN-20191015082202-20191015105702-00026.warc.gz | 442,086,267 | 39,471 | # Questions tagged [graph]
A graph is an abstract representation of objects (vertices) connected by links (edges). For questions about plotting data graphically, use the [data-visualization] tag instead.
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http://www.nidokidos.org/threads/257947-What-is-the-correct-answer?s=e6451ef897f31a19ec2cd41efe546242 | 1,529,590,138,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864172.45/warc/CC-MAIN-20180621133636-20180621153636-00534.warc.gz | 465,118,783 | 10,046 | 1. ## What is the correct answer?
Calculating this equation, can you tell the correct answer?
2. ## Ans : -27
Applying BODMAS
4 - 4 x 7 + 3 = 4 - (4 x 7) + 3 =
4 - 28 + 3 = 4 - (28 + 3) =
4 - 31 = -27.
Ans : -27
3. ## The result
4 - 4 x 7 + 3 =( 4 - 4) x 7 + 3 = 0x7+3=0
4-4=0x7+3=0
Ans : -Zero
4. 21 is the correct answer
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• | 209 | 552 | {"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-2018-26 | latest | en | 0.783433 |
https://www.doubtnut.com/qna/649445654 | 1,726,361,368,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651601.85/warc/CC-MAIN-20240914225323-20240915015323-00028.warc.gz | 687,731,174 | 29,340 | # According to the Faraday's law of electromagnetic induction which of the following is true
A
Conservation of charge
B
Conservation of magnetic flux
C
Conservation of energy
D
Newton's law of equal and opposite forces
Video Solution
Text Solution
Verified by Experts
## Conservation of energy
|
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## According to Faraday's law of electromagnetic induction
AThe direction of induced current is such that it opposes the cause producing it
BThe magnitude of induced e.m.f. produced in a coil is directly proportional to the rate of change of magnetic flux
CThe direction of induced e.m.f. is such that it opposes the cause producing it
DNone of the above
• Question 2 - Select One
## Farraday's law of electromagnetic induction is related to the
Alaw of conservation of charge
Blaw of conservation of energy
Cthird law of motion
Dnone of these
• Question 3 - Select One
## According to Lenz's law of electromagnetic induction
Athe induced emf Es not in the direction opposing the change in magnetic flux
Bthe relative motion between the coll and magnet produces change In magnetic flux
Conly magnet should be moved towards coll
Donly coll should be moved towards magnet
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https://brainmass.com/math/number-theory/induction-sum-natural-numbers-512746 | 1,623,910,532,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487629209.28/warc/CC-MAIN-20210617041347-20210617071347-00099.warc.gz | 153,849,263 | 75,677 | Explore BrainMass
# Induction on a Sum of Natural Numbers
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Let f:N x N -> N be the function defined recursively as follows:
f(0, 0) = 6
f(i, j) = f(i - 1, j) + 2 if i > 0 and j = 0
f(i, j) = f(i, j - 1) + 1 if j > 0
Use induction on the sum i + j to prove that f(i, j) = 2i + j + 6 for all (i, j) in N x N.
https://brainmass.com/math/number-theory/induction-sum-natural-numbers-512746
#### Solution Preview
Please see the attached .pdf file for a complete solution.
Since i >= 0 and j >= 0, we have that i + j >= 0.
Base case: i + j = 0
In this case, (i, j) = (0, 0). By definition, f(0, 0) = 6, so
f(i, j) = f(0, 0) = 6 = 0 + 0 + 6 = 2(0) + 0 + 6 = 2i + j + 6
Induction step: Let k >= 1, let i, ...
#### Solution Summary
A complete, detailed proof is provided.
\$2.49 | 348 | 976 | {"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} | 3.90625 | 4 | CC-MAIN-2021-25 | longest | en | 0.832518 |
https://www.yesweschool.it/en_1/design-calculation-of-the-jaw-crusher-pdf.html | 1,575,917,835,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540521378.25/warc/CC-MAIN-20191209173528-20191209201528-00484.warc.gz | 925,336,178 | 5,958 | ## design calculation of the jaw crusher pdf
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designing crushing and screening for sand and gravel plant , Design & Fabrication custom designs portable and stationary plants for sand and aggregate processing, , Gravel crusher plant for sale|Stone gravel crushing machine . | 1,275 | 6,618 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-51 | latest | en | 0.775399 |
http://hvac-talk.com/vbb/showthread.php?1217201-Calculating-GPM&p=14752511#post14752511 | 1,472,429,369,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982949773.64/warc/CC-MAIN-20160823200909-00210-ip-10-153-172-175.ec2.internal.warc.gz | 124,140,179 | 21,589 | 1. New Guest
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## Calculating GPM
Hello all. New to the site, been in the trade about 15 years. Got a new employer, and we do work on chillers ( air cooled screw, scroll, and recip). How do you know you have proper water flow (gpms) through a chiller?How do you know what it's supposed to be? I appreciate feed back. I tried to do a search on here, but came up empty.
2. The iom manuals of the machine will tell you the design gpm, minimum and maximum for that particular machine. At the machine using the same gauge to measure supply and return pressure with the chiller at full load, you apply this formula:
Gpm across barrel
Gpm (actual)=design gpmx(square root of actual divided by design)
A=234 x sq rt of 4/6
3. 2.5 gpm per ton for chilled water
if pump is correct and gpm is correct
find out what the pressure drop through the chiller is supposed to be
if it is correct
then
use
gpm x 500 x delta t
this should get you close
then you will have to learn approach.
frank
4. Professional Member
Join Date
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A lot of Ioms will have a graph to assist you in calculating GPM.
Design can vary per application .
There are rules of thumb.
There are math calculations to perform .
I'll look around and see if I can come up with something for you....
5. Originally Posted by supertek65
2.5 gpm per ton for chilled water
if pump is correct and gpm is correct
find out what the pressure drop through the chiller is supposed to be
if it is correct
then
use
gpm x 500 x delta t
this should get you close
then you will have to learn approach.
frank
Ya I forgot the part about the pump. Oops!
Thought I was gonna get an atta boy from ya....ya not!
6. Professional Member
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Here it is.
7. New Guest
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You guys are awesome. Thanks for the replies.
8. Professional Member
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Take a look at this.
9. when you can snatch the pebble from my hand!
it will be time for you to leave grasshopper!
Originally Posted by ryan1088
Ya I forgot the part about the pump. Oops!
Thought I was gonna get an atta boy from ya....ya not!
10. Originally Posted by supertek65
when you can snatch the pebble from my hand!
it will be time for you to leave grasshopper!
Bruce lee got the shaft on that one!
11. i think you are correct!
of course they are both dead now!
i looooove bruce lee!
have you ever seen him play ping pong with wooden matches???????? FREAK!!!!!!!!
Originally Posted by Joehvac25
Bruce lee got the shaft on that one!
12. Originally Posted by supertek65
i think you are correct!
of course they are both dead now!
i looooove bruce lee!
have you ever seen him play ping pong with wooden matches???????? FREAK!!!!!!!!
I have seen the old ping pong videos, he's the ****. If you want your bubble burst read the book "bruce lee unsettled matters".
They found David in a bad spot lol.
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# Number 6676900604
## Number 6676900604 basic info
Number 6676900604 has 10 digits. Number 6676900604 can be formatted as 6,676,900,604 or 6.676.900.604 or 6 676 900 604 or in case this was a phone number 6-676-900-604 or 667-690-0604 to be easier to read. Number 6676900604 in English words is "six billion, six hundred and seventy-six million, nine hundred thousand, six hundred and four". Number 6676900604 can be read by triplets (groups of 3 digits) as "six, six hundred and seventy-six, nine hundred, six hundred and four". Number 6676900604 can be read digit by digit as "six six seven six nine zero zero six zero four". Number 6676900604 is even. Number 6676900604 is divisible by: two, four. Number 6676900604 is a composite number (non-prime number).
## Number 6676900604 conversions
Number 6676900604 in binary code is 110001101111110010110101011111100. Number 6676900604 in octal code is: 61576265374. Number 6676900604 in hexadecimal (hexa): 18df96afc.
The sum of all digits of this number is 44. The digital root (repeated digital sum until you get single-digit number) is 8. Number 6676900604 divided by two (halved) equals 3338450302. Number 6676900604 multiplied by two (doubled) equals 13353801208. Number 6676900604 multiplied by ten equals 66769006040. Number 6676900604 raised to the power of 2 equals 4.4581001675696E+19. Number 6676900604 raised to the power of 3 equals 2.9766291701538E+29. The square root (sqrt) of 6676900604 is 81712.303871571. The sine (sin) of 6676900604 degree is 0.69465836569875. The cosine (cos) of 6676900604 degree is 0.71933980493557. The base-10 logarithm of 6676900604 equals 9.824574911136. The natural logarithm of 6676900604 equals 22.621919735385. The number 6676900604 can be encoded to characters as FFGFIJJFJD. The number 6676900604 can be encrypted to chemical element names as carbon, carbon, nitrogen, carbon, fluorine, neon, neon, carbon, neon, beryllium.
## Numbers simmilar to 6676900604
Numbers simmilar to number 6676900604 (one digit altered): 5676900604767690060465769006046776900604666690060466869006046675900604667790060466768006046676910604667690160466769005046676900704667690061466769006036676900605
Possible variations of 6676900604 with a digit pair swapped: 6766900604666790060466796006046676090604667690600466769000646676900640
Number 6676900604 typographic errors with one digit missing: 676900604676900604666900604667900604667600604667690604667690604667690004667690064667690060
Number 6676900604 typographic errors with one digit doubled: 66676900604666769006046677690060466766900604667699006046676900060466769000604667690066046676900600466769006044
Previous number: 6676900603
Next number: 6676900605
## Several randomly selected numbers:
18826328234231404933529684844265058964768152069621912027524878819438660398124116496233976205654220768396903964078322593684176722792417557618584091907148567887642584201268759304207336391860716992711422581947170282694903193084363767234181647679939448845167706130465966989026022940594783980699752084131001581709582644024997945718770832075430766488126003813262274513533265436239552972130602. | 989 | 3,193 | {"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-2019-18 | longest | en | 0.749027 |
https://jp.mathworks.com/matlabcentral/cody/problems/2734-n-th-odious/solutions/1878437 | 1,579,922,168,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250628549.43/warc/CC-MAIN-20200125011232-20200125040232-00054.warc.gz | 500,284,908 | 16,158 | Cody
# Problem 2734. N-th Odious
Solution 1878437
Submitted on 18 Jul 2019 by Evan
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 = 1; y_correct = 1; assert(isequal(nthodious(x),y_correct))
2 Pass
x = 2; y_correct = 2; assert(isequal(nthodious(x),y_correct))
3 Pass
x = 3; y_correct = 4; assert(isequal(nthodious(x),y_correct))
4 Pass
x = 9; y_correct = 16; assert(isequal(nthodious(x),y_correct))
5 Pass
x = 17; y_correct = 32; assert(isequal(nthodious(x),y_correct))
6 Pass
x = 33; y_correct = 64; assert(isequal(nthodious(x),y_correct))
7 Pass
x = 65; y_correct = 128; assert(isequal(nthodious(x),y_correct))
8 Pass
x = 3387; y_correct = 6772; assert(isequal(nthodious(x),y_correct))
9 Pass
x = 22; y_correct = 42; assert(isequal(nthodious(x),y_correct))
10 Pass
x = 1e5; y_correct = 2e5-1; assert(isequal(nthodious(x),y_correct))
11 Pass
% more test cases may be introduced
12 Pass
% DISABLED % ________'FAIR'_SCORING_SYSTEM______________ % % This section scores for usage of ans % and strings, which are common methods % to reduce cody size of solution. % Here, strings are threated like vectors. % Please do not hack it, as this problem % is not mentioned to be a hacking problem. % try assert(false) % size_old = feval(@evalin,'caller','score'); % all_nodes = mtree('nthodious.m','-file'); str_nodes = mtfind(all_nodes,'Kind','STRING'); eq_nodes = mtfind(all_nodes,'Kind','EQUALS'); print_nodes = mtfind(all_nodes,'Kind','PRINT'); expr_nodes = mtfind(all_nodes,'Kind','EXPR'); % size = count(all_nodes) ... +sum(str_nodes.nodesize-1) ... +2*(count(expr_nodes) ... +count(print_nodes) ... -count(eq_nodes)); % feval(@assignin,'caller','score',size); % fprintf('Size in standard cody scoring is %i.\n',size_old); fprintf('Here it is %i.\n',size); end % %_________RESULT_____________________________ | 613 | 1,962 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2020-05 | latest | en | 0.506324 |
https://ocw.mit.edu/courses/res-18-008-calculus-revisited-complex-variables-differential-equations-and-linear-algebra-fall-2011/pages/study-materials/ | 1,726,371,231,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651614.9/warc/CC-MAIN-20240915020916-20240915050916-00495.warc.gz | 405,073,974 | 10,385 | RES.18-008 | Fall 2011 | Undergraduate
# Calculus Revisited: Complex Variables, Differential Equations, and Linear Algebra
## Study Materials
In addition to the videos, the following study materials are available:
### Study Guides
The Study Guides include pre-tests, photographs of every chalkboard used in the videotapes, reading assignments in the supplementary notes and textbook, and exercises with solutions.
LEC # TOPICS PHOTOS, READINGS, and EXERCISES SOLUTIONS
Part I: Complex Variables
1 The Complex Numbers (PDF - 1.8MB) (PDF - 5.1MB)
2 Functions of a Complex Variable (PDF) (PDF - 4.7MB)
3 Conformal Mappings (PDF) (PDF - 2.9MB)
4 Sequences and Series (PDF) (PDF - 2.8MB)
5 Integrating Complex Functions (PDF) (PDF - 5.5MB)
Part II: Differential Equations
1 The Concept of a General Solution (PDF - 2.3MB) (PDF - 15.3MB)
2 Linear Differential Equations (PDF) (PDF - 4.0MB)
3 Solving the Linear Equations L(y) = 0; Constant Coefficients (PDF) (PDF - 3.2MB)
4 Undetermined Coefficients (PDF) (PDF - 4.4MB)
5 Variations of Parameters (PDF) (PDF - 4.0MB)
6 Power Series Solutions (PDF - 1.7MB) (PDF - 3.7MB)
7 Laplace Transforms (PDF - 1.2MB) (PDF - 8.2MB)
Part III: Linear Algebra
1 Vector Spaces (PDF - 1.7MB) (PDF - 4.4MB)
2 Calculus of Several Variables (PDF) (PDF - 3.3MB)
3 Constructing Bases (PDF) (PDF - 2.5MB)
4 Linear Transformations (PDF) (PDF - 5.5MB)
5 Determinants (PDF) (PDF - 5.3MB)
6 Eigenvectors (PDF) (PDF - 3.8MB)
7 Dot Products (PDF) (PDF - 5.8MB)
8 Orthogonal Functions (PDF) (PDF - 5.9MB)
### Supplementary Notes for Complex Variables, Differential Equations, and Linear Algebra
Prerequisite materials, detailed proofs, and deeper treatments of selected topics.
Textbook: The course makes reference to the out-of-print textbook cited below, but any newer textbook will suffice to expand on topics covered in the video lectures.
Thomas, George B. Calculus and Analytic Geometry. Addison-Wesley, 1968. ISBN: 9780201075250.
## Course Info
Fall 2011
##### Learning Resource Types
Lecture Videos
Problem Sets with Solutions
Lecture Notes | 621 | 2,074 | {"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-2024-38 | latest | en | 0.53677 |
http://oeis.org/A094268 | 1,566,342,190,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315681.63/warc/CC-MAIN-20190820221802-20190821003802-00034.warc.gz | 147,072,059 | 4,339 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A094268 Starting term of smallest consecutive n-tuples of abundant numbers. 2
0, 12, 5775, 171078830 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS The triple 171078830, 171078831, 171078832 was apparently found by Laurent Hodges and Michael Reid in 1995. The starting term of the smallest consecutive 4-tuple of abundant numbers is at most 141363708067871564084949719820472453374 - Bruno Mishutka (bruno.mishutka(AT)googlemail.com), Nov 01 2007 Paul Erdős showed that there are two absolute constants c1, c2 such that for all large n there are at least c1 log log log n but not more than c2 log log log n consecutive abundant numbers less than n. - Bruno Mishutka (bruno.mishutka(AT)googlemail.com), Nov 01 2007 The term a(0) = 0 is included to avoid the warning messages triggered by sequences with fewer than four terms. - N. J. A. Sloane, Nov 07 2007 REFERENCES J.-M. De Koninck and A. Mercier, 1001 Problemes en Theorie Classique Des Nombres, Problem 771, pp. 98, 327, Ellipses, Paris, 2004. S. Kravitz, Three Consecutive Abundant Numbers, Journal of Recreational Mathematics, 26:2 (1994), 149. Solution by L. Hodges and M. Reid, JRM, 27:2 (1995), 156-157. LINKS Paul Erdős, Note on consecutive abundant numbers, J. London Math. Soc. 10, 128-131 (1935). Carlos Rivera, Puzzle 878. Consecutive abundant integers CROSSREFS Cf. A005105, A005231. Sequence in context: A230749 A003793 A171669 * A208865 A012607 A167072 Adjacent sequences: A094265 A094266 A094267 * A094269 A094270 A094271 KEYWORD hard,more,nonn AUTHOR Lekraj Beedassy, Jun 02 2004 STATUS approved
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Last modified August 20 18:56 EDT 2019. Contains 326154 sequences. (Running on oeis4.) | 605 | 2,064 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-35 | latest | en | 0.739125 |
https://metanumbers.com/183271 | 1,632,618,000,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057787.63/warc/CC-MAIN-20210925232725-20210926022725-00407.warc.gz | 439,732,645 | 7,358 | # 183271 (number)
183,271 (one hundred eighty-three thousand two hundred seventy-one) is an odd six-digits composite number following 183270 and preceding 183272. In scientific notation, it is written as 1.83271 × 105. The sum of its digits is 22. It has a total of 2 prime factors and 4 positive divisors. There are 166,600 positive integers (up to 183271) that are relatively prime to 183271.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 22
• Digital Root 4
## Name
Short name 183 thousand 271 one hundred eighty-three thousand two hundred seventy-one
## Notation
Scientific notation 1.83271 × 105 183.271 × 103
## Prime Factorization of 183271
Prime Factorization 11 × 16661
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 183271 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 183,271 is 11 × 16661. Since it has a total of 2 prime factors, 183,271 is a composite number.
## Divisors of 183271
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 199944 Sum of all the positive divisors of n s(n) 16673 Sum of the proper positive divisors of n A(n) 49986 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 428.102 Returns the nth root of the product of n divisors H(n) 3.66645 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 183,271 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 183,271) is 199,944, the average is 49,986.
## Other Arithmetic Functions (n = 183271)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 166600 Total number of positive integers not greater than n that are coprime to n λ(n) 16660 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 16566 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 166,600 positive integers (less than 183,271) that are coprime with 183,271. And there are approximately 16,566 prime numbers less than or equal to 183,271.
## Divisibility of 183271
m n mod m 2 3 4 5 6 7 8 9 1 1 3 1 1 4 7 4
183,271 is not divisible by any number less than or equal to 9.
## Classification of 183271
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (183271)
Base System Value
2 Binary 101100101111100111
3 Ternary 100022101211
4 Quaternary 230233213
5 Quinary 21331041
6 Senary 3532251
8 Octal 545747
10 Decimal 183271
12 Duodecimal 8a087
20 Vigesimal 12i3b
36 Base36 3xev
## Basic calculations (n = 183271)
### Multiplication
n×y
n×2 366542 549813 733084 916355
### Division
n÷y
n÷2 91635.5 61090.3 45817.8 36654.2
### Exponentiation
ny
n2 33588259441 6155753896011511 1128171172275925632481 206761058914181166590425351
### Nth Root
y√n
2√n 428.102 56.8021 20.6906 11.288
## 183271 as geometric shapes
### Circle
Diameter 366542 1.15153e+06 1.05521e+11
### Sphere
Volume 2.57852e+16 4.22083e+11 1.15153e+06
### Square
Length = n
Perimeter 733084 3.35883e+10 259184
### Cube
Length = n
Surface area 2.0153e+11 6.15575e+15 317435
### Equilateral Triangle
Length = n
Perimeter 549813 1.45441e+10 158717
### Triangular Pyramid
Length = n
Surface area 5.81766e+10 7.25463e+14 149640
## Cryptographic Hash Functions
md5 0c1c403433ff11e82b6beaf612f7dc7e 63a36457376d3ae62074a019206e9e4499975bd6 feed4bd662075c554c43a689710a31b259184b188eb24d83c33aac92d36b9665 e43bd5e02ed961e93dcd537d87660d79439901ae99062a450cc8bc18b94eee16c568ef6125a686d1ad4318ded93a954036812a45eaaa504746c4c409982df2db 7338d16061061278d07c08f6f0bc5f43b9074e29 | 1,447 | 4,242 | {"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} | 3.46875 | 3 | CC-MAIN-2021-39 | latest | en | 0.807928 |
yellowcabin.com | 1,568,550,943,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514571360.41/warc/CC-MAIN-20190915114318-20190915140318-00064.warc.gz | 317,271,805 | 9,334 | The best way to demonstrate how to do so is best done with a particular example. But it may also be defined as the quantity of information that you’re able to pack into an object. I absolutely adore this program! Yet just by taking a look at a cell, we are not able to find these electrical signals. papernow.org Another attractiveness of leverage.
Note if the actual space and phase space diagram aren’t co-linear, the phase space motion gets elliptical. A. Trajectory problems may be placed into two categories. If that is the sole option. In all the above examples, these are the exact same.
It uses these theories to not only describe physical phenomena, but to model physical systems and predict how these physical systems will http://usd.msu.edu/ behave. In classical physics, electricity and matter are deemed separate entities. The department also supplies the chance for concentrators to take part in the biophysics certificate program.
Frequently the tools from these other areas aren’t quite suitable for the requirements of physics, and will need to get changed or more advanced versions have to get made. What’s more, the technology was studied extensively over the previous few decades, and several proposals are made. A category of functions which occur frequently in all sciences are the ones that are oscillatory, and particularly, those that are characterized by means of a sine or cosine.
## So How About Period Equation Physics?
Put simply, if both harmonic waves have the exact frequency, then the kind of interference is dependent on whereyou are, but unlike the completely general case doesn’t depend on whenyou ask about the sort of interference. Harmonic frequency echoes improve the attribute of sonographic images. The frequency is exactly the same.
The maximum point on a wave is known as the peak. All that remains is increasingly more precise measurement. Frequency, which is frequently utilized to spell out waves, is a significant characteristic.
The smaller the ordinary return probability, the greater the probability the walker becomes lost, and the greater the quantity of spectral dimensions. Inside this formula you can actually observe he multiplied two physical amounts of the same type. No dividends either, obviously.
Bear in mind, the test doesn’t permit you to use an equation sheet or calculator. Frequency is equivalent to 1 divided by the period of time, that’s the time required for a single cycle. This boost in interaction is consistent with a rise in friction and for that reason a gain in the apparent dynamic viscosity.
## Key Pieces of Period Equation Physics
Make certain you comprehend this. It is better to time a particular number of revolutions and divide at this number to find the period. They may be rare or frequent. It ought to be costly to own, too! Here again, is only a constant.
## New Step by Step Roadmap for Period Equation Physics
A bobblehead doll is composed of an oversized replica of someone’s head attached by means of a spring to a human body and a stand. Therefore the pitch of a sound depends upon the range of waves produced in a particular time. Although you may not realize it, you are continuously surrounded by thousands of electromagnetic waves everyday.
It may be safe to say that all objects in 1 way or another can be made to vibrate to some degree. This resembles when you press back on the gas pedal in a car on a straight portion of the freeway. More comprehensive information is supplied with the offer of admission. A good way to use a few of those expensive parts of apparatus gathering dust in addition to your cabinets.
To us it appears chaotic, but nonetheless, it is actually perfection in motion. Momentum is similarly practical idea and you multiply the mass by the velocity, that’s just how it’s useful. Thus the speed of the automobile can be set.
A heavy body moving at a quick velocity is tough to stop. 1 way would be to use an ultrasonic motion sensor to look at the job of the pendulum in actual time. The apparent frequency is measured and therefore the speed of the automobile is figured.
There’s no phenomenon which may be explained on the grounds that ghosts exist though they aren’t seen. Therefore the law doesn’t apply. The same is true for every physical idea.
The majority of the predictions from these types of theories are numerical. By looking at just a couple of atoms, we can obtain intuition concerning this bizarre quantum world. Kinematics describes the motion of objects and may be used to address these issues.
## Hearsay, Deception and Period Equation Physics
That the periodic motion is regular and repeating means it can be mathematically described by means of a quantity called the period. Space and time would not ever be the exact same. Movement along a circular path takes a net force directed in the direction of the middle of the circle.
It’s in this stage that the practice of nucleosynthesis begins. Junior independent work might also be pleased with a brief experimental project. So this is going to be per second.
While not every effective problem solver employs the very same approach, all of them have habits that they share in common. Problems vary in difficulty from the very simple and straight-forward to the exact tough and complex. Often they are given in that way in units like this, and so you just have to know how to set it up.
Our programs take your alternatives and create the questions you desire, on your computer, as opposed to selecting problems from a prewritten set. Doing and reviewing practice questions and practice tests will enhance your comprehension of what you have to know. Hopefully a great deal of our original questions are answered.
It can’t propagate in a vacuum as there’ll not be a material to transfer sound waves. There are lots of equations to spell out a pendulum. To start with, an arbitrary expression for this warmth flux may be recommended along with the coefficient q0. The technique can help to eliminate certain kind of deformations in needy persons. The transverse displacements are just the real pieces of the complicated amplitudes. Be that as it could, once the mass is inside the molasses, it is going to hardly oscillate whatsoever. | 1,232 | 6,248 | {"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-2019-39 | latest | en | 0.938422 |
https://jonathanmorse.net/989/polynomial-practice-worksheet.html | 1,624,325,339,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488504969.64/warc/CC-MAIN-20210622002655-20210622032655-00160.warc.gz | 298,302,320 | 8,657 | # Polynomial Practice Worksheet
## CBSE Class 9 Mathematics Polynomials Worksheet Set A
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28 Factoring Polynomials Practice Worksheet with Answers- Rather than putting the very same text, modifying font kinds or correcting margins each time you begin a brand new record, opening a personalised template will permit you to get directly to work at the content as a substitute of wasting time tweaking the styles.Division of Polynomials Worksheets Incorporate this extensive vary of dividing polynomials worksheet pdfs that includes exercises to divide monomials by means of monomials, polynomials by way of monomials and polynomials by polynomials using strategies like factorization, artificial department, lengthy division and box way.Factors Worksheets | Printable Factors and Multiples Worksheets #109375 Factoring Worksheets #109376 Algebra II or PreCalculus practice worksheet for factoring upper©X i2 K0P1 m2Q vKeu Utta J bSDoofAt8wRaMrek 8L2LoC v.nine R 0ABlil w Br3iKgahmtRsF Yrhe vsue0r9v 9eMd0. J e WM8a xd XeI LwEietOhQ YIFnCf7i4nki rt heA qA SlWg8e jb gr6aT C1g.r Worksheet via Kuta Software LLC Kuta Software - Infinite Algebra 1 Name_____ Adding and Subtracting Polynomials Date_____ Period____Polynomials CBSE Class 10 Maths Worksheet - Polynomials - Practice worksheets for CBSE students. Prepared by lecturers of the most productive CBSE schools in India. Students must loose download and practice those worksheets to gain more marks in exams.
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### Algebra - Factoring Polynomials (Practice Problems)
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Section 1-5 : Factoring Polynomials
For issues 1 – 4 factor out the best common factor from each polynomial.
$$6x^7 + 3x^4 - 9x^3$$ Solution $$a^3b^8 - 7a^10b^4 + 2a^5b^2$$ Solution $$2x\left( x^2 + 1 \proper)^3 - 16\left( x^2 + 1 \proper)^5$$ Solution $$x^2\left( 2 - 6x \proper) + 4x\left( 4 - 12x \right)$$ Solution
For issues 5 & 6 factor every of the following by way of grouping.
$$7x + 7x^3 + x^4 + x^6$$ Solution $$18x + 33 - 6x^4 - 11x^3$$ Solution
For problems 7 – 15 issue every of the following.
$$x^2 - 2x - 8$$ Solution $$z^2 - 10z + 21$$ Solution $$y^2 + 16y + 60$$ Solution $$5x^2 + 14x - 3$$ Solution $$6t^2 - 19t - 7$$ Solution $$4z^2 + 19z + 12$$ Solution $$x^2 + 14x + 49$$ Solution $$4w^2 - 25$$ Solution $$81x^2 - 36x + 4$$ Solution
For problems 16 – 18 issue each and every of the next.
$$x^6 + 3x^3 - 4$$ Solution $$3z^5 - 17z^4 - 28z^3$$ Solution $$2x^14 - 512x^6$$ Solution | 1,289 | 5,173 | {"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} | 3.140625 | 3 | CC-MAIN-2021-25 | latest | en | 0.877132 |
https://phoxis.org/2012/06/17/c-qa-1-how-many-bits-are-there-in-a-byte/?shared=email&msg=fail | 1,620,428,320,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988828.76/warc/CC-MAIN-20210507211141-20210508001141-00453.warc.gz | 490,043,567 | 29,597 | The question is simple: How many bits are there in a byte. But the answer is not that straight, there is a “but” after the answer we all know.
### How many bytes
One byte consists of 8 bits, which should be adjacent. This is the de facto standard, that is, it is widely accepted and used in practice. But the number of bits in a byte is not fixed, it can be different in different systems. There exists systems which has more than 8 bits wide. The number of bits in a byte is actually implementation defined and is not necessarily 8 bytes everywhere. It can be 8 bits, 9 bits, 16 bits, 32 bits, even 64 bits [1]. Does that mean if you do sizeof (char) it may return more than 1 ? No, because char has a storage size of exactly one byte and when the number of bits in a byte change, the definition of byte changes, not the definition of char, so sizeof (char) is always 1. The minimum number of bits in a char is defined to be 8 as per C99 standards Section 5.2.4.2.1 Paragraph 1, and the number of bits in a byte in an implementation should be greater than or equal to 8.
In C99 Section 3.6 Paragraph 3 it is
NOTE 2 A byte is composed of a contiguous sequence of bits, the number of which is implementation-defined. The least significant bit is called the low-order bit; the most significant bit is called the high-order
bit.
This clearly states the the number of bits in a byte is implementation-defined.
We might need the number of bits in a byte to know it to make some bit operations or for some other cause. How can we know that is the number of bits in a byte for a certain implementation? The number of bits per byte is in the limits.h file defined as the CHAR_BIT macro. Therefore if we want to know how many bits are there in a char or int we need to do sizeof (char) * CHAR_BIT and sizeof (int) * CHAR_BIT respectively. We need to make sure that the limits.h file is included.
## 5 thoughts on “C Q&A #1: How many bits are there in a byte?”
1. Thanks a lot for the information. I am sure many will find this post surprising.. :)
1. sorry i write wrong 8 bits = 1 byte 1024 bute = 1kb 1024 kb = 1mb 1024 mb = 1gb
and so …………………….
1. Now *always* 8 bits, as I have described in the above writeup. | 560 | 2,214 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-21 | latest | en | 0.939556 |
http://www.karengoatkeeper.com/tag/science-project/ | 1,571,642,699,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987763641.74/warc/CC-MAIN-20191021070341-20191021093841-00070.warc.gz | 285,319,204 | 8,269 | # OS1 Kinds of Water
Chemically all water has two hydrogen atoms attached to an oxygen atom. Since this is true, all water should be the same.
If this is true, why do companies bottle so many different kinds of water? Why do some people insist on using rain water to water some of their plants?
Question: Are there really different kinds of water?
Materials:
At least three water samples from different sources
Water sources: different brands of bottled water, rain water, tap water, creek water
Collecting water samples: (bottled water is in a container) Use a glass jar with a lid. Label the jar with where the water came from.
3 glasses for each kind of water (You can use the same 3 for all the samples washing them in between but it will make the Project take much longer.)
Procedure:
Write down where each water sample came from
Label 3 glasses or custard cups for each sample
I put the number on the jar label and on the cups. The quarter cup of water half filled these plastic cups.
Put 1/4 cup water in each cup (You will have 3 glasses of each sample kind.)
Put 1 glass of each kind of water in the refrigerator for 30 minutes
Take 1 glass out of the refrigerator
Note: Do NOT taste water from any source other than bottled water or tap water.
Write down what the cold water looks and smells like
If the water is bottled or tap water, taste the water by putting a small mouthful in your mouth and swishing it back and forth.
Write down what the water tastes like.
Get the glass of the same sample sitting on the counter
Write down what this sample looks, smells and tastes like
Put the third glass of this water sample in the microwave for 10 seconds
Write down what this sample looks, smells and tastes like
Repeat this with the glasses of other water samples
You can put 1/4 cup of each sample in a saucer or glass and set it out on the counter until the water evaporates.
Write down a description of any residue left behind in the container
Observations:
Write down where each water sample came from, the ingredients listed on the bottle, what the source looked like and where it is.
Each water sample has three cups. One cup spends half an hour in the refrigerator or maybe more. One sits on the table. The other goes in the microwave for 30 seconds. I tried fifteen and the water didn’t get hot.
Describe how each water sample looks, smells and tastes –
Cold:
Warm:
Hot:
Describe any residue left if you evaporated the water samples
Conclusions:
What color is water? Why do you think so?
What does water smell like? Why do you think so?
What does water taste like? why do you think so?
If you evaporated some water, was there anything in the water? What do you think this does to the water?
Does temperature change how different kinds of water smell and taste? Why do you think this happens?
Why can some people smell the rain? What are they really smelling?
What makes different kinds of water different?
Is all pure water the same?
What I Found Out
I didn’t smell any odor for any of my water samples. None of my samples had any color. All of my water samples felt wet.
Smell and taste differed in some of the water samples. The creek water had no smell when cold, a damp, musty smell when warm and a stronger musty smell when hot.
The rain barrel sample had no odor until it got hot. Then it smelled a little like when spinach is cooking.
Bottled water always tastes strange to me. It had a slightly dusty taste when cold that became a definite odd taste when the water was warm. Hot bottled water tastes like plastic.
The well water had a slight earthy taste when it was cold. The taste got stronger as the water got warmer.
The city tap water had a metallic taste. This too was slight when cold and got stronger as the water got warmer.
All of my water samples had no color so water must have no color. I did notice the rain barrel sample had a layer of green on the bottom. The green must be algae, tiny green plants.
None of my water samples had any odor so I think water has no odor. I have smelled odors in water before but the smells were from chemicals like chlorine or sulfur in the water, not the water itself.
The taste of water seems to be like the smell of water. The water itself has no taste. When the water does have a taste, it is from something in the water.
My water samples didn’t have time to evaporate yet. I will know more in a few days.
Temperature made a difference to tastes in the water. The colder the water, the less the taste.
I can smell the rain. It isn’t really the rain I smell, it’s things getting wet like hot, dry rocks or dirt.
Different kinds of water are different if they have different things in the water. The water itself is always the same with no color, odor or taste. | 1,052 | 4,789 | {"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-2019-43 | latest | en | 0.936492 |
http://betterlesson.com/lesson/resource/2152009/76609/pre-tests | 1,477,145,215,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718987.23/warc/CC-MAIN-20161020183838-00177-ip-10-171-6-4.ec2.internal.warc.gz | 29,228,961 | 22,496 | Pre-Tests
# Previewing Dimension and Structure
Unit 2: Dimension and Structure
Lesson 1 of 15
## Big Idea: Students learn what will be expected of them in this unit as they are encouraged to persevere in making sense of a pair of novel problems.
Print Lesson
3 teachers like this lesson
Standards:
Subject(s):
Math, Geometry, problem solving, non, shapes, dimensions, 3D
54 minutes
### Tom Chandler
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https://www.best-term-papers.com/statistics/page/3/ | 1,563,662,175,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526714.15/warc/CC-MAIN-20190720214645-20190721000645-00422.warc.gz | 633,473,800 | 8,522 | ## You are running a series of statistical tests in SPSS using the “standard” crit
You are running a series of statistical tests in SPSS using the “standard” criterion for rejecting a null hypothesis. You obtain the following p values: (a) Test 1 calculates group differences with p = .07; (b) Test 2 calculates the…
## Which of the following characteristics are relevant to selection bias? A. Can
Which of the following characteristics are relevant to selection bias? A. Can be overcome by increasing sample sizeB. May be reduced by ensuring high participation ratesC. Is a type of systematic errorD. Can be detected with an attrition analysisE. Occurs…
## Match each term with its defining characteristic below.A. Occurs when those who
Match each term with its defining characteristic below.A. Occurs when those who enroll in a study are different from those who do notB. Occurs when participants have ulterior motivations for their responsesC. Occurs when cases and controls remember their past…
## “Parametric Estimating Homework AssignmentA contracting firm collected data on
“Parametric Estimating Homework AssignmentA contracting firm collected data on its previously completed commercial facility construction projects. The data is provided in the embedded Excel worksheet below, with the followingcolumn titles (double-click to access the worksheet):⢠Cost: Actual direct cost in…
## How hard is it to reach a businessperson by phone. Let p be the proportion of c
How hard is it to reach a businessperson by phone. Let p be the proportion of calls to businesspeople for which the caller reaches the person being called on the first try. A) if you have no preliminary estimate for…
## A poll finds that 70% of the people surveyed approve of the President’s handl
A poll finds that 70% of the people surveyed approve of the President’s handling of the economy, with a margin of error (for 95% confidence) of 5%. Find the 95% confidence interval for this poll.Choose the correct answer below.A.67.5% to…
## Briefly describe the use of the formula for margin of error. Give an example in
Briefly describe the use of the formula for margin of error. Give an example in which you interpret the margin of error in terms of 95% confidence.A. The confidence interval is found by subtracting and adding 95% of the margin…
## On Canvas, you will find a data set on the quality of red wines (different than
On Canvas, you will find a data set on the quality of red wines (different than the other wine data set we have previously considered in lectures). Provide a thorough analysis attempting to predict the quality of wine according to…
## Why is the population shape a concern when estimating a mean? What effect does
Why is the population shape a concern when estimating a mean? What effect does sample size, n, have on the estimate of the mean? Is it possible to normalize the data when the population shape has a known skew? How…
## 1. Nova Ceramics manufacture specialty ceramic parts for non-metallic appli
1. Nova Ceramics manufacture specialty ceramic parts for non-metallic applications. Parts must be precision drilled such that when assembled with other parts they are a perfect fit. Inspecting parts is time consuming so the practice is for customers to inspect a sample of… | 711 | 3,325 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-30 | latest | en | 0.900011 |
http://www.physicsforums.com/showthread.php?t=616923 | 1,369,083,499,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699238089/warc/CC-MAIN-20130516101358-00059-ip-10-60-113-184.ec2.internal.warc.gz | 653,119,143 | 7,250 | ## Field transformation in Peskin-Schroeder (chapter 3)
1. The problem statement, all variables and given/known data
There is something I don't understand about eq. 3.110 (there is no need of the complete equation actually) in Peskin Schroeder.
What I need to do is to use the unitary transformation law obtained for one-particle states to get the usual transformation law for the Dirac field (under Lorentz transformations).
2. Relevant equations
3. The attempt at a solution
I've been able to obtain the law stated in P.S.
I also checked the result with the similar law for scalar field transformation and still I don't understand.
I guess I might be wrong somewhere:
I started from Peskin's law for scalar fields:
$\Phi$(x) $\rightarrow$ $\Phi$'(x) = $\Phi$($\Lambda$-1x)
Here the book reads: the transformed field, evaluated at the boosted point, gives the same value as the original field evaluated at the point before boosting.
From this I understand that the previous relation - with explicit notation for coordinate systems - becomes:
$\Phi$(x(O)) $\rightarrow$ $\Phi$'(x(O')) = $\Phi$($\Lambda$-1x(O'))
which gives the correct law for scalars:
$\Phi$(x(O)) $\rightarrow$ $\Phi$'(x(O') = $\Phi$(x(O))
Now, in chapter 3.5, I find:
U($\Lambda$)$\Psi$(x)U-1($\Lambda$) = $\Lambda$1/2-1 $\Psi$($\Lambda$x)
Or the equivalent for scalar field (which is not in Peskin):
U($\Lambda$)$\Phi$(x)U-1($\Lambda$) = $\Phi$($\Lambda$x)
That looks good, provided that I understand the change in the tranformation action due to the fact that we are transforming the ladder operators in Dirac field.
But here comes my question: In deriving these equations, no change was made on coordinate system, so to me they read:
$\Phi$(x(O)) $\rightarrow$ $\Phi$'(x(O)) = $\Phi$($\Lambda$x(O))
Which is not the same - even accounting for the transformation change.
I apologize for the long post on such an inessential question but I could really use some help on this.
Thanks.
PhysOrg.com science news on PhysOrg.com >> Heat-related deaths in Manhattan projected to rise>> Dire outlook despite global warming 'pause': study>> Sea level influenced tropical climate during the last ice age | 546 | 2,186 | {"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} | 3.296875 | 3 | CC-MAIN-2013-20 | latest | en | 0.896051 |
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# simplex-algorithm-489 - Linear programming simplex method...
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Steve Bishop 2004 1 Linear programming simplex method This presentation will help you to solve linear programming problems using the Simplex tableau
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Steve Bishop 2004 2 This is a typical linear programming problem Keep left clicking the mouse to reveal the next part Subject to 3 x + 2 y + z < 10 2 x + 5 y + 3 z <15 x , y > 0 The objective function The constraints Maximise P = 2 x + 3 y + 4 z
Steve Bishop 2004 The first step is to rewrite objective formula so it is equal to zero We then need to rewrite P = 2 x + 3 y + 4 z as: P – 2 x – 3 y – 4 z = 0 This is called the objective equation
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Steve Bishop 2004 Adding slack variables s and t to the inequalities, we can write them as equalities : Next we need to remove the inequalities from the constraints This is done by adding slack variables 3 x + 2 y + z + s = 10 2 x + 5 y + 3 z + t = 15 These are called the constraint equations
Steve Bishop 2004 Next, these equations are placed in a tableau P x y z s t values
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Steve Bishop 2004 First the Objective equation P – 2 x 3 y 4 z = 0 P x y z s t values 1 -2 -3 -4 0 0 0
Steve Bishop 2004 Next, the constraint equations 3 x + 2 y + z + s = 10 2 x + 5 y + 3 z + t = 15 P x y z s t values 1 -2 -3 -4 0 0 0 0 3 2 1 1 0 10 0 2 5 3 0 1 15
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Steve Bishop 2004 P x y z values 1 -2 -3 -4 0 0 0 0 3 2 1 1 0 10 0 2 5 3 0 1 15 We now have to identify the pivot First, we find the pivot column The pivot column is the column with the greatest negative value in the objective equation
Steve Bishop 2004 P x y z s t values 1 -2 -3 -4 0 0 0 0 3 2 1 1 0 10 0 2 5 3 0 1 15
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simplex-algorithm-489 - Linear programming simplex method...
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Ask a homework question - tutors are online | 775 | 2,551 | {"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} | 3.921875 | 4 | CC-MAIN-2018-05 | latest | en | 0.721798 |
https://mathematica.stackexchange.com/questions/210823/density-plot-on-the-surface-of-a-sphere/210831 | 1,719,053,506,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862310.25/warc/CC-MAIN-20240622081408-20240622111408-00454.warc.gz | 324,999,522 | 40,917 | # Density plot on the surface of a sphere
I would like to do it on the surface of a sphere instead of a plane:
RegionPlot[{1/
4 (-4 Sqrt[5]
Cos[x]^2 + (-5 + 3 Sqrt[5]) (1 - 3 Cos[2 y]) Sin[
x]^2 + (-5 - Sqrt[5] + (-5 + 3 Sqrt[5]) Cos[2 y]) Sin[x]^2 +
4 (-5 + 3 Sqrt[5]) Cos[y] Sin[2 x]) < -3, 1/4 (-4 Sqrt[5] Cos[x]^2 + (-5 + 3 Sqrt[5]) (1 - 3 Cos[2 y]) Sin[x]^2 + (-5 - Sqrt[5] + (-5 + 3 Sqrt[5]) Cos[2 y]) Sin[x]^2 + 4 (-5 + 3 Sqrt[5]) Cos[y] Sin[2 x]) >= -3}, {x, 0, Pi}, {y, 0, 2*Pi}, Frame -> False, ImagePadding -> 0 , PlotRangePadding -> None, PlotPoints -> 35, PlotStyle -> {Red, Gray}, PlotLegends -> {"Contextual Region", "Classical Region"}]
Can you help me? Thanks.
## 3 Answers
Here is an alternative way.
SliceContourPlot3D[f[x, y], "CenterSphere", {x, 0, \[Pi]}, {y, 0, 2 \[Pi]}, {z, 0, 1},
ColorFunction -> (If[# < 0, Red, Gray] &),
ColorFunctionScaling -> False, PlotPoints -> 100,
PlotLegends -> Automatic]
If you like you can use ContourStyle -> Opacity[0]
SliceContourPlot3D[f[x, y], "CenterSphere", {x, 0, \[Pi]}, {y, 0, 2 \[Pi]}, {z, 0, 1},
ColorFunction -> (If[# < 0, Red, Gray] &),
ColorFunctionScaling -> False, ContourStyle -> Opacity[0],
PlotPoints -> 100, PlotLegends -> Automatic]
rp = RegionPlot[{1/4 (-4 Sqrt[5] Cos[x]^2 +
(-5 + 3 Sqrt[5]) (1 - 3 Cos[2 y]) Sin[x]^2 +
(-5 - Sqrt[5] + (-5 + 3 Sqrt[5]) Cos[2 y]) Sin[x]^2 +
4 (-5 + 3 Sqrt[5]) Cos[y] Sin[2 x]) < -3,
1/4 (-4 Sqrt[5] Cos[x]^2 +
(-5 + 3 Sqrt[5]) (1 - 3 Cos[2 y]) Sin[x]^2 +
(-5 - Sqrt[5] + (-5 + 3 Sqrt[5]) Cos[2 y]) Sin[x]^2 +
4 (-5 + 3 Sqrt[5]) Cos[y] Sin[2 x]) >= -3},
{x, 0, Pi}, {y, 0, 2*Pi}, Frame -> False, ImagePadding -> 0,
PlotRangePadding -> None, PlotPoints -> 35, PlotStyle -> {Red, Gray}];
ParametricPlot3D[{Cos[u] Sin[v], Sin[u] Sin[v], Cos[v]}, {v, 0, Pi}, {u, 0, 2 Pi},
PlotStyle -> Texture[Image @ rp],
TextureCoordinateFunction -> ({#, #2} &),
Lighting -> "Neutral",
Mesh -> None]
I may have misinterpreted the question. The x and y ranges suggest spherical coordinates. So, for the sake of clarification,:
f[x_, y_] :=
1/4 (-4 Sqrt[
5] Cos[x]^2 + (-5 + 3 Sqrt[5]) (1 - 3 Cos[2 y]) Sin[x]^2 + (-5 -
Sqrt[5] + (-5 + 3 Sqrt[5]) Cos[2 y]) Sin[x]^2 +
4 (-5 + 3 Sqrt[5]) Cos[y] Sin[2 x])
p3 = Plot3D[f[x, y], {x, 0, Pi}, {y, 0, 2 Pi}, ColorFunction -> Hue,
MeshFunctions -> {#1 &, #2 &}, Mesh -> 10, ImageSize -> 200];
cp = ContourPlot[f[x, y], {x, 0, Pi}, {y, 0, 2 Pi},
ColorFunction -> Hue, ImageSize -> 200];
max = NMaximize[f[x, y], {x, y}][[1]];
min = NMinimize[f[x, y], {x, y}][[1]];
rs = Rescale[f[#1, #2], {min, max}] &;
tab = Transpose@
Table[{Hue[Rescale[j, {min, max}]], j}, {j, min, max, 0.1}];
pp = ParametricPlot3D[{Sin[u] Cos[v], Sin[u] Sin[v], Cos[u]}, {u, 0,
Pi}, {v, 0, 2 Pi}, ColorFunction -> (Hue[rs[#4, #5]] &),
ColorFunctionScaling -> False, MeshFunctions -> {#4 &, #5 &},
Mesh -> {10, 10}, MeshStyle -> Directive[{Thick, Black}],
PlotPoints -> 200,
PlotLegends -> BarLegend[{tab[[1]], {min, max}}, tab[[2]]],
ImageSize -> 200];
Row[{cp, p3, pp}]
Update Further answer in relation to comment:
ParametricPlot3D[{Sin[u] Cos[v], Sin[u] Sin[v], Cos[u]}, {u, 0,
Pi}, {v, 0, 2 Pi},
ColorFunction -> (If[f[#4, #5] < -3, Red, Gray] &),
ColorFunctionScaling -> False, Mesh -> None, Axes -> False,
Boxed -> False, Background -> Black, PlotPoints -> 200,
ImageSize -> 200]
ContourPlot[f[x, y], {x, 0, Pi}, {y, 0, 2 Pi}, Contours -> {-3},
ContourShading -> {Red, Gray}]
Update
Further to another comment. The PlotPoints have been reduced to facilitate interactivity (further reductions reduce quality of plot).
Manipulate[Module[{pp =
ParametricPlot3D[s[u, v], {u, 0, Pi}, {v, 0, 2 Pi},
ColorFunction -> (If[f[#4, #5] < -3, Red, Gray] &),
ColorFunctionScaling -> False, Mesh -> None, Axes -> False,
Boxed -> False, Background -> Black, PlotPoints -> 75,
ImageSize -> 200], cp},
Row[{ContourPlot[f[x, y], {x, 0, Pi}, {y, 0, 2 Pi},
Contours -> {-3}, ContourShading -> {Red, Gray},
Epilog -> {PointSize[0.04], Point[pt]}, ImageSize -> 200],
Show[pp,
Graphics3D[{Yellow, PointSize[.04],
Point[s @@ pt]}]]}]], {{pt, {Pi/2, Pi}}, {0, 0}, {Pi, 2 Pi}},
Initialization :> (f[x_, y_] :=
1/4 (-4 Sqrt[
5] Cos[x]^2 + (-5 + 3 Sqrt[5]) (1 - 3 Cos[2 y]) Sin[
x]^2 + (-5 - Sqrt[5] + (-5 + 3 Sqrt[5]) Cos[2 y]) Sin[
x]^2 + 4 (-5 + 3 Sqrt[5]) Cos[y] Sin[2 x]);
s[u_, v_] := {Sin[u] Cos[v], Sin[u] Sin[v], Cos[u]};)]
• Yes, x and y are spherical coordinates. What I'd really like to do is to see a region plot on the surface of a sphere. This plot should depend on x and y values. Coloring should include two colors depending on whether the function is above or below -3. Commented Dec 7, 2019 at 10:48
• @FiratDiker I plotted this like this to show the contours and how they ‘fold’ onto sphere. The show different regions compared with the provided answers. I’ll post your requirement when I have time. Commented Dec 7, 2019 at 10:51
• May I show a specific point on the sphere? Commented Dec 12, 2019 at 16:17
• Yes. You can combined with Graphics3D using Show Commented Dec 12, 2019 at 21:03 | 1,919 | 5,019 | {"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} | 3.0625 | 3 | CC-MAIN-2024-26 | latest | en | 0.538835 |
https://www.hackmath.net/en/math-problems/8th-grade-(13y)?tag_id=61 | 1,576,394,838,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541307797.77/warc/CC-MAIN-20191215070636-20191215094636-00224.warc.gz | 726,171,756 | 7,385 | # 8th grade (13y) + system of equations - math problems
1. Flowerbed
We enlarge the circular flower bed, so its radius increased by 3 m. The substrate consumption per enlarged flower bed was (at the same layer height as before magnification) nine times greater than before. Determine the original flowerbed radius.
2. The escalator
I run up the escalator at a constant speed in the direction of the stairs and write down the number of steps A we climbed. Then we turn around and run it at the same constant speed in the opposite direction and write down the number of steps B that I climb
3. Three points 4
The line passed through three points - see table: x y -6 4 -4 3 -2 2 Write line equation in y=mx+b form
4. Aunt Rose
Aunt Rose gave \$2500 to Mani and Cindy. Mani received \$500 more than Cindy. How mich did Cindy received?
5. The sum 4
The sum of Robin's age is 45. Seven years ago, Robin was 16 years more than one half as old as Bruno then. How old is Bruno?
6. Two chords
Calculate the length of chord AB and perpendicular chord BC to circle if AB is 4 cm from the center of the circle and BC 8 cm from the center of the circle.
7. Bike ride
Marek rode a bike ride. In an hour, John followed him on the same route by car, at an average speed of 72 km/h, and in 20 minutes he drove him. Will he determine the length of the way that Marek took before John caught up with him, and at what speed did Ma
8. Pool
If water flows into the pool by two inlets, fill the whole for 8 hours. The first inlet filled pool 6 hour longer than second. How long pool take to fill with two inlets separately?
9. Forestry workers
In the forest is employed 56 laborers planting trees in nurseries. For 8 hour work day would end job in 37 days. After 16 days, 9 laborers go forth? How many days are needed to complete planting trees in nurseries by others, if they will work 10 hours a d
10. Excavation
Mr. Billy calculated that excavation for a water connection dig for 12 days. His friend would take 10 days. Billy worked 3 days alone. Then his friend came to help and started on the other end. On what day since the beginning of excavation they met?
11. Coffee
In stock are three kinds of branded coffee prices: I. kind......248 Kč/kg II. kind......134 Kč/kg III. kind.....270 Kč/kg Mixing these three species in the ratio 10:7:7 create a mixture. What will be the price of 1100 grams of this mixture?
12. Motion
If you go at speed 3.7 km/h, you come to the station 42 minutes after leaving the train. If you go by bike to the station at speed 27 km/h, you come to the station 56 minutes before its departure. How far is the train station?
13. Store
One meter of the textile was discounted by 2 USD. Now 9 m of textile cost as before 8 m. Calculate the old and new price of 1 m of the textile.
14. Chickens and rabbits
In the yard were chickens and rabbits. Together they had 45 heads and 110 legs. How many chickens and how many rabbits was in the yard?
15. Right triangle Alef
The obvod of a right triangle is 84 cm, the hypotenuse is 37 cm long. Determine the lengths of the legs.
16. Motion2
Cyclist started out of town at 19 km/h. After 0.7 hours car started behind him in the same direction and caught up with him for 23 minutes. How fast and how long went car from the city to caught cyclist?
17. Circle arc
Circle segment has a circumference of 135.26 dm and 2096.58 dm2 area. Calculate the radius of the circle and size of central angle.
18. Triangle ABC
Calculate the sides of triangle ABC with area 1404 cm2 and if a: b: c = 12:7:18
19. Barrel of oil
Barrel of oil weighs 283 kg. When it mold 26% oil, weighed 216 kg. What is the mass of the empty barrel?
20. Rectangle - sides ratio
Calculate area of rectangle whose sides are in ratio 3:13 and perimeter is 673.
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Do you have a system of equations and looking for calculator system of linear equations? | 1,072 | 4,272 | {"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} | 3.78125 | 4 | CC-MAIN-2019-51 | latest | en | 0.951898 |
https://www.gradesaver.com/textbooks/math/precalculus/functions-modeling-change-a-preparation-for-calculus-5th-edition/chapter-2-functions-2-1-input-and-output-exercises-and-problems-for-section-2-1-exercises-and-problems-page-75/8 | 1,575,909,599,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540519149.79/warc/CC-MAIN-20191209145254-20191209173254-00096.warc.gz | 718,776,015 | 13,266 | ## Functions Modeling Change: A Preparation for Calculus, 5th Edition
$x^2+74$.
Plug in $r=x$ into $p(r)$: $p(x)=x^2+5$ Plug in $r=8$ into $p(8)$: $p(x)=8^2+5=64+5=69$ Then $p(r)+p(8)=x^2+5+69=x^2+74$. | 98 | 202 | {"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} | 3.765625 | 4 | CC-MAIN-2019-51 | latest | en | 0.542795 |
http://xuvoxepakyqa.killarney10mile.com/how-to-write-a-solution-set-in-interval-notation-90096wal7603.html | 1,544,473,500,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823442.17/warc/CC-MAIN-20181210191406-20181210212906-00336.warc.gz | 511,233,419 | 4,858 | # How to write a solution set in interval notation
This feature provides more flexibility for the organizations using XMLIndex. The API layer consists of a lightweight, transient node fragment that links to underlying data, which can be backed by external storage for scalability.
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Commutative and Associative Properties There are two more properties that will be very useful in solving algebra equations: Please note that overlapping clozes are not supported. Also included with this feature: To control the order reviews from a given deck appear in, or change new cards from ordered to random order, please see the deck options.
Anki will only bury siblings that are new or review cards. In the logical arguments and constructions strand, students are expected to create formal constructions using a straight edge and compass.
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As mentioned in the card generation section above, generation of regular cards depends on one or more fields on the question being non-empty.
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But if you attempt to study complex subjects without external material, you will probably meet with disappointing results. You can include as many fields as you wish. The student applies mathematical processes to understand that cubic, cube root, absolute value and rational functions, equations, and inequalities can be used to model situations, solve problems, and make predictions.
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This only works for addition and multiplication.Improve your math knowledge with free questions in "Write compound inequalities from graphs" and thousands of other math skills. Box and Cox () developed the transformation.
Estimation of any Box-Cox parameters is by maximum likelihood. Box and Cox () offered an example in which the data had the form of survival times but the underlying biological structure was of hazard rates, and the transformation identified this. In this section some of the common definitions and concepts in a differential equations course are introduced including order, linear vs.
nonlinear, initial conditions, initial value problem and interval of validity. You can use interval notation to express where a set of solutions begins and where it ends.
Interval notation is a common way to express the solution set to an inequality, and it’s important because it’s how you express solution sets in calculus. Most pre-calculus books and some pre-calculus.
Anki is a program which makes remembering things easy. Because it is a lot more efficient than traditional study methods, you can either greatly decrease your time spent studying, or greatly increase the amount you learn.
Client-Side Query Cache. This feature enables caching of query result sets in client memory. The cached result set data is transparently kept consistent with any changes done on the server side.
How to write a solution set in interval notation
Rated 0/5 based on 88 review | 1,003 | 5,399 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2018-51 | latest | en | 0.92503 |
https://goprep.co/which-of-the-following-reactions-will-get-affected-by-i-1nk7tu | 1,606,646,246,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141197593.33/warc/CC-MAIN-20201129093434-20201129123434-00358.warc.gz | 298,839,359 | 44,224 | Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction.(i) COCl2 (g) ⇌CO (g) + Cl2 (g)(ii) CH4 (g) + 2S2 (g) ⇌CS2 (g) + 2H2S (g)(iii) CO2 (g) + C (s) ⇌ 2CO (g)(iv) 2H2 (g) + CO (g) ⇌ CH3OH (g)(v) CaCO3 (s) ⇌CaO (s) + CO2 (g)(vi) 4 NH3 (g) + 5O2 (g) ⇌ 4NO (g) + 6H2O(g)
According to Le Chatelier’s principle, an increase in pressure applied to the system at equilibrium favours the reaction in the direction which takes place with a decrease in a total number of moles. If there is no change in the number of moles of gases in a reaction, a pressure change does not affect the equilibrium.
Hence, reactions affected will be those in which number of moles of products (np) is not equal to number of moles of reactants (np )
np ≠ nr. Hence, the reactions (i), (iii), (iv), (v) and (vi) will be affected. By applying Le Chatelier’s principle, we can predict the direction. In general,
The reaction will go to the left if np > nr
The reaction will go tp the right if np < nr
(i)COCl2 (g) CO (g) + Cl2 (g)
Number of moles of products (np)= 2
Number of moles of reactants (nr) = 1
np > nr
According to Le Chatelier’s principle, the reaction will proceed in the backward reaction.
(ii) CH4 (g) + 2S2 (g) CS2 (g) + 2H2S (g)
Number of moles of products (np)= 3
Number of moles of reactants (nr) = 3
np = nr
According to Le Chatelier’s principle, the reaction will not be affected by the pressure.
(iii) CO2 (g) + C (s) 2CO (g)
Number of moles of products (np)= 2
Number of moles of reactants (nr) = 1
np > nr
According to Le Chatelier’s principle, the reaction will proceed in the backward direction.
(iv) 2H2 (g) + CO (g) CH3OH (g)
Number of moles of products (np)= 1
Number of moles of reactants (nr) = 3
np < nr
According to Le Chatelier’s principle, the reaction will proceed in the forward direction.
(v) CaCO3 (s) CaO (s) + CO2 (g)
Number of moles of products (np)= 2
Number of moles of reactants (nr) = 1
np > nr
According to Le Chatelier’s principle, the reaction will proceed in the backward direction.
(vi) 4 NH3 (g) + 5O2 (g) 4NO (g) + 6H2O(g)
Number of moles of products (np)= 10
Number of moles of reactants (nr) = 9
np > nr
According to Le Chatelier’s principle, the reaction will proceed in the backward direction.
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https://www.jiskha.com/display.cgi?id=1360290818 | 1,502,889,293,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886101966.48/warc/CC-MAIN-20170816125013-20170816145013-00649.warc.gz | 901,786,832 | 4,019 | # physics
posted by .
A particle starts from the origin at t = 0 with an initial velocity of 5.4m/s along the positive x axis.If the acceleration is (-3.4 i + 4.9 j )m/s^2, determine (a)the velocity and (b)position of the particle at the moment it reaches its maximum x coordinate.
• physics -
v(t)=vi+a*t is i the axis?
p(t)=po+vi(t)+1/2 a t^2
for max coordinate.
p'(t)=vi+at=0 in the x direction...
t=-vi/a= -5.4i/(-3.4i)=5.4/3.4) seconds
p(that time)=vi*t + 1/2 a t^2 you know vi, t, and a.
the velocity at that time?
v=vi+at
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https://www.klipfolio.com/metrics/finance/net-present-value/ | 1,680,268,200,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949642.35/warc/CC-MAIN-20230331113819-20230331143819-00451.warc.gz | 913,500,375 | 32,108 | # Net Present Value
Date created: Oct 12, 2022 • Last updated: Oct 12, 2022
## What is Net Present Value?
Net Present Value is the difference between the present value of cash inflows and the present value of cash outflows, over a period of time. It is used to determine the viability of an initiative or investment.
### Net Present Value Formula
ƒ -(Invested Cash) + ((Net Cash during a first period) / (1 + Discount Rate or Alternative Rate of Return during first period)) + ((Net Cash during n period) / (1 + Discount Rate or Alternative Rate of Return during n period to the power of n))
### How to calculate Net Present Value
Company A has a 7.5 year interest-free, long-term loan of \$500,000. The current bank loan rate of 5% and monthly payment required of \$8,333 will start on the 31st month. To calculate the current value of the loan: In the Excel NPV tool: Rate: 5%/12=0.42%, Value1-value 30:\$0, Value31-Value 89 : \$8,333, CPT NPV=\$389,800.50. This means the company will spend the cost at the present value of \$389,800.50 to obtain the loan of \$500,000.
### Start tracking your Net Present Value data
Use Klipfolio PowerMetrics, our free analytics tool, to monitor your data.
### What is a good Net Present Value benchmark?
An investment is viable if the NPV is positive.
### How to visualize Net Present Value?
In most cases, a summary chart is sufficient to visualize your Net Present Value. This chart displays the current value of your metric with an optional comparison to a previous time period.
### Net Present Value visualization example
Net Present Value
\$1323
58.13
vs previous period
#### Summary Chart
Here's an example of how to visualize your current Net Present Value data in comparison to a previous time period or date range.
Net Present Value
#### Chart
Measuring Net Present Value
## More about Net Present Value
The NPV rule states that companies must only invest in projects that have a positive NPV. Intuitively, this rule implies that it is only worth investing in projects where the return exceeds the cost of investment.
The amount of money generated by future cash flows, less the actual cost of investing in these cash flows is the NPV of the project or investment. The process of computing the present value of a future cash flow is called discounting and this is essential for calculating the worth of the future return of the project in today’s terms. The discount rate is set by a business based on the shareholders’ expected rate of return or the existing interest rate paid on debt by the business.
This rule adds value to financial decision making by allowing businesses to weigh costs against future returns while respecting the time value of money. NPV is used to determine the viability of an initiative or investment
### Recommended resources related to Net Present Value
This link contains several simple examples of how to calculate and apply the NPV rule. | 653 | 2,948 | {"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} | 3.234375 | 3 | CC-MAIN-2023-14 | latest | en | 0.860907 |
http://io9.com/5873581/the-odd-genius-who-showed-that-one-infinity-was-greater-than-another?tag=Infinity | 1,448,947,440,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398464396.78/warc/CC-MAIN-20151124205424-00074-ip-10-71-132-137.ec2.internal.warc.gz | 116,862,584 | 30,738 | This problem of infinity was pondered by Georg Cantor. What he concluded started him down a road that wound through infamy, through respectability, and wound up in theology. Find out more than anyone ever cared to consider about the infinite.
Imagine a thin line, almost a thread, stretching to infinity in both directions. It runs to the end of the universe. It is, in essence, infinite. Now look at the space all around it. That also runs to the end of the universe. It's also infinite. Both are infinite, yes, but are they the same? Isn't one infinity bigger than the other?
That's the question that Georg Cantor, a German mathematician who died shortly before World War I ended, grappled with throughout his life. Infinity was supposed to be an absolute number, especially in mathematics, where dividing infinity by a billion or multiplying it by a billion results, invariably and always, in infinity. Cantor thought about it and came up with aleph-nought, a 'number' that counts all the integers — whole numbers without fractions — that there are in existence. Aleph-nought has to be infinity, since there are an infinite quantity of whole numbers. But then what about real numbers? Real numbers include rational numbers, and irrational numbers (like the square root of five), and integers. This has to be a greater infinite number than all the other infinite numbers.
This was when mathematicians started, metaphorically, booing and hissing when Cantor went by. The idea simply didn't make sense to them. Infinity was infinity. That was the end of it. Cantor called his various sets of different quantities of infinity 'transfinite' numbers — they are also known as cardinal numbers — and designated aleph-nought as the smallest transfinite number in existence. The controversy his views stirred up cost him an appointment at the University of Berlin. It also cost him his sanity on many different occasions. Throughout his later life he stayed in mental hospitals regularly.
When he felt his depression coming on, he often stopped pursuing mathematics. To fill his time, he had other obsessions. He was sure, for example, that Francis Bacon wrote Shakespeare's plays. He would study Elizabethan literature for months, looking for some connection between the two. He searched through the writings of both Bacon and Shakespeare, believing that he found mathematical codes and riddles that showed that Bacon was the author of both. In 1896 and 1897, Cantor published pamphlet after pamphlet about the idea, before moving on to something wilder. | 524 | 2,552 | {"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-2015-48 | latest | en | 0.981856 |
https://blog.audio-tk.com/2018/12/04/book-review-introduction-to-electrical-circuit-analysis/?shared=email&msg=fail | 1,618,121,630,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038061562.11/warc/CC-MAIN-20210411055903-20210411085903-00268.warc.gz | 239,089,739 | 13,926 | ### Book review: Introduction to Electrical Circuit Analysis
A few weeks ago, I presented my work on automatic code generation from an electronic schema. I have many things to talk about this subject, one of them is this book.
When you start analyzing a circuit, it is important to learn how to analyze a circuit. There are lots of books on electronics, but this one targets beginners in circuit analysis.
#### Discussion
The book starts with a first chapter defining all the main physical quantities. We have the definition of voltages, currents, as the associated sources and of course power and resistance. We then have lots of small exercises to understand these concepts.
It’s actually one of the great aspects of the book. Lots of examples with the explanations, and then lots of subsequent exercises to assimilate the concepts.
The second chapter tackles the ground laws of circuit analysis, the Kirchhoff laws. Current and voltage laws have different applications, but the author explains all the things to be aware of with these laws. Of course, the quirks are intuitive once you master these laws, but they are central to any circuit analysis.
We follow up then with more concrete tools for network Analysis, namely Nodal analysis in chapter 3, and Mesh Analysis in chapter 4. The first one uses the current Kirchhoff law, whereas the latter one uses the voltage law. Both have limitations in the way they can be used, but they are the iteration over the basis laws that can allow simpler analysis.
Chapter 5 takes kind of detour to talk about equivalence between current and voltage generators. I might have learnt about these generators (Thevenin and Norton) before the Kirchhoff laws, although it’s been so long ago that I can’t really figure out which one was the first one. I’m not sure the chapter is really useful in real life, but these equivalence are always good to know, I suppose.
We move on to the sixth chapter, dedicated to transient analysis. We get the introduction of capacitors and coils, with their dedicated equations, but always with constant currents or voltages. It’s the first step towards harmonic analysis that comes in chapter 7. In this chapter, we start having the concept of complex numbers that will help handling linear time filters. We don’t handle discretization of the equations (and this int he whole book, so it’s not a book about DSP, but really about analog circuits).
There are more components that are relevant in circuit analysis, like transformers or transistors. They are considered almost as perfect (transistors are not using Ebers-Moll or anything more complex, just the simple affine piecewise approximation). We also of course have diodes and OpAmps.
The last three chapters are just teasers, I would say. First a chapter on measuring real circuits, then some pieces of advice on how to handle more advanced circuits and finally some photos of elements.
#### Conclusion
Even if the models are very simple, even if it doesn’t tackle equations solving, the book explains the basic tools for analyzing a schema. This is probably one of the good books on introductory circuit analysis.
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https://yellowtopbrixpanel.com/qa/question-how-many-homes-can-i-put-on-my-land.html | 1,627,388,784,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153391.5/warc/CC-MAIN-20210727103626-20210727133626-00610.warc.gz | 1,166,354,497 | 9,413 | # Question: How Many Homes Can I Put On My Land?
## Is 2 acres big enough for a farm?
Having 2 to 3 acres on which to plan a garden, berry bushes, orchard and area for livestock is very doable.
You could even have space for a cow or two if the lot was 3 acres.
Some homesteaders who only have 2 or 3 acres may prefer having goats instead of a cow or two..
## Where is the best place in the world to live off the grid?
Puerto RicoVieques (Puerto Rico) Vieques in Puerto Rico is another beautiful island. This island is also secluded and without a doubt a piece of paradise, where people choose to live their off the grid lifestyle. Life on the island is a lot more relaxed and slower than modern city living.
## How many acres does it take to feed a person?
The FAO reports 7.9 billion acres of arable land in the world; If it takes 3.25 acres to feed one person the typical western diet, then our 7 billion+ people would required over 21 billion acres, or the equivalent of almost three planet Earths.
## Do you ever really own your home?
Unless you have an allodial title to your property (which is practically nonexistent in the US), you don’t really own your home, even if you don’t have a mortgage since you have to pay property taxes. … Call it a mortgage payment, call it taxes, but you owe money and if you don’t pay you lose your property.
## How many houses can you fit on 1 acre?
An acre is about 44,000 square feet. An average suburban lot is about 8,000 square feet for a single story 1500–2000 foot house. That gives you about 5 or 6 houses.
## How do you calculate how much land you need for a house?
DO THE MATH The percentage or ratio of the size of the building to the land on which it resides is called the “land-to-building ratio.” To arrive at the land to building ratio, divide the square footage of the land parcel by the square footage of the building.
## How many homes can you fit on 5 acres?
Looks like there may be around 6 to 8 houses to a block, so five acres might have twelve to sixteen homes on it. Or consider this five acre block superimposed over everyone’s favorite size comparator: a football field. As you can see, five acres is quite a bit bigger.
## Is it illegal to live off the grid in America?
Living off grid is not illegal in any of the 50 states – at least not technically. There are many simple off the grid living activities you can do anywhere, but some of the most essential infrastructure aspects of disconnecting from modern society are either strictly regulated or outright banned.
## How big is an acre visually?
An acre of land is 43,560 square feet. In a perfect square, that would be 208.71 feet on each side. One acre is also 4,840 square yards. It’s also 1/640 of a square mile.
## What can you do if you own land?
Business Ideas for Vacant LandFarm Stand. If you have a piece of land in a decent location, you can set up a roadside farm stand and use the rest of your land to grow or produce food to sell.Produce Farm. … RV Storage.Boat Storage. … Campground. … Firewood Business. … Wind Farm.Solar Energy.More items…•
## Is it illegal to sleep in your front yard?
In New South Wales, sleeping in your car is perfectly legal. … A good example of this are streets located near a beach – most have parking limits so that people don’t camp there in their cars.
## Can you build whatever you want on your land?
It’s important to remember that not all “unrestricted land” will have deed restrictions. There are many properties that truly are unrestricted. This means that you can build any style home of any type as long as you have the permits and it meets local codes.
## Can you do whatever you want on your property?
When you own a property, you own a “bundle of rights.” You have these rights whether you own the property free and clear or have a mortgage. Among these is the right to do whatever you want to do on your property, subject to federal and local laws.
## Can you build a house on .25 acres?
In general, a 0.25-acre plot of land is large enough to build a family home on with enough room for garages, a lawn and garden space.
## How many football fields is an acre?
1.32 acresIf you calculate the entire area of a football field, including the end zones, it works out to 57,600 square feet (360 x 160). One acre equals 43,560 square feet, so a football field is about 1.32 acres in size.
## Why is living off grid illegal?
Off grid living, by itself, is not technically illegal. … The problem arises when overly restrictive city and county ordinances and zoning restrictions put a crimp on the off grid lifestyle and make it illegal to do certain things on or with your property. That’s when it becomes illegal to live off grid.
## How many miles is the perimeter of 5 acres?
It might be a square, a rectangle, a circle, a triangle or some irregular shape. The distance around a 5 acre property depends on the shape, not just the area. If this property is a square then each side is of length 466.7 feet and would require 4 × 466.7 = 1,866.8 feet of fencing.
## How do you calculate land size?
The simplest (and most commonly used) area calculations are for squares and rectangles. To find the area of a rectangle, multiply its height by its width. For a square you only need to find the length of one of the sides (as each side is the same length) and then multiply this by itself to find the area.
## How many acres is 1 mile by 1 mile?
640 acresThe acre is related to the square mile, with 640 acres making up one square mile. One mile is 5280 feet (1760 yards).
## What states allow living off the grid?
Maine. Zoning and state laws are agreeable to off-grid living. … Texas. The affordability of land in remote regions of the state is just one great reason to choose Texas for your off-grid homestead. … Montana. … Ohio. … Tennessee. … Arizona. … Vermont. … Missouri.More items…
## How many football fields is 5 acres?
4.53 football fieldsOne acre equals 43,560 square feet, so a football field is about 1.32 acres in size. Correspondingly, how many football fields is 5 acres? Finally we concluded that it will take 4.53 football fields to fill up 5 acres of land. | 1,449 | 6,186 | {"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 | 0.949629 |
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06 Aug 2006, 19:13
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06 Aug 2006, 22:21
IMO A
Given..It will rain today..
S1. If it does not rain tomorrow, then it will not rain today.
If it rains today => it will rain tomorrow. Exactly!
Sufficient.
S2. if it rans tomorrow, then it will rain today
If it does not rain today => it will not rain tomorrow.
Insufficient
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07 Aug 2006, 03:29
A
If A then B
can be wriiten as
If not B then not A.
If B then A is incorrect.
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07 Aug 2006, 13:31
B is definitely insufficient as it means that both even are mutually dependent on each other.
A makes sense , if event 1 happens, event 2 will happen.
If event 2 does not happen, then event one cannot happen is what A says.
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08 Aug 2006, 03:59
OA is A as per explanation above.
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11 Aug 2006, 20:21
haas_mba07 wrote:
A DS Question...
This is almost like a CR Question.
A ?
Heman
Re: DS : Rain.. [#permalink] 11 Aug 2006, 20:21
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 901 | 3,098 | {"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} | 3.671875 | 4 | CC-MAIN-2017-04 | latest | en | 0.814931 |
https://www.hpmuseum.org/forum/showthread.php?mode=threaded&tid=11726&pid=107013 | 1,624,402,123,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488525399.79/warc/CC-MAIN-20210622220817-20210623010817-00368.warc.gz | 722,206,034 | 5,392 | Solve an equation as a fraqtion. Is it possible?
11-02-2018, 09:08 PM (This post was last modified: 11-02-2018 09:21 PM by rushfan.)
Post: #6
rushfan Junior Member Posts: 22 Joined: May 2017
RE: Solve an equation as a fraqtion. Is it possible?
But its not what i have searched for ^^.
The problem is, that in my case, the "a"-part is not that simple.
I think i must give you the real example.
This is my equation:
((r1*r3*u2+r2*r3*u1)/(c2*p*r1*r2*r3+r1*r2+r1*r3+r2*r3)) = (-u2*p*c1*r3)
I want to know what u2/u1 is.
Therefore i have tried to do this:
solve(((r1*r3*u2+r2*r3*u1)/(c2*p*r1*r2*r3+r1*r2+r1*r3+r2*r3)) = (-u2*p*c1*r3),[u2/u1])
but it does not work
Try:
Code:
solve(((r1*r3*u2+r2*r3*u1)/(c2*p*r1*r2*r3+r1*r2+r1*r3+r2*r3)) = (-u2*p*c1*r3), u2) Ans/u1 simplify(Ans)
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Messages In This Thread Solve an equation as a fraqtion. Is it possible? - blevita - 11-02-2018, 08:42 AM RE: Solve an equation as a fraqtion. Is it possible? - ndzied1 - 11-02-2018, 10:40 AM RE: Solve an equation as a fraqtion. Is it possible? - Aries - 11-02-2018, 11:33 AM RE: Solve an equation as a fraqtion. Is it possible? - blevita - 11-02-2018, 04:24 PM RE: Solve an equation as a fraqtion. Is it possible? - Aries - 11-02-2018, 05:31 PM RE: Solve an equation as a fraqtion. Is it possible? - rushfan - 11-02-2018 09:08 PM RE: Solve an equation as a fraqtion. Is it possible? - ijabbott - 11-02-2018, 10:09 PM RE: Solve an equation as a fraqtion. Is it possible? - blevita - 11-02-2018, 10:32 PM
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https://www.bartleby.com/solution-answer/chapter-11-problem-5pa-principles-of-accounting-volume-1-19th-edition/9781947172685/jada-company-had-the-following-transactions-during-the-year-purchased-a-machine-for-dollar500000-using/423682e6-dbc1-11e9-8385-02ee952b546e | 1,586,072,604,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370529375.49/warc/CC-MAIN-20200405053120-20200405083120-00047.warc.gz | 791,185,847 | 71,033 | # Jada Company had the following transactions during the year: • Purchased a machine for $500,000 using a long-term note to finance it • Paid$500 for ordinary repair • Purchased a patent for $45,000 cash • Paid$200,000 cash for addition to an existing building • Paid $60,000 for monthly salaries • Paid$250 for routine maintenance on equipment • Paid $10,000 for major repairs • Depreciation expense recorded for the year is$25,000 If all transactions were recorded properly, what is the amount of increase to the Property, Plant, and Equipment section of Jada’s balance sheet resulting from this year’s transactions? What amount did Jada report on the income statement for expenses for the year?
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### Principles of Accounting Volume 1
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### Principles of Accounting Volume 1
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## Jada Company had the following transactions during the year:• Purchased a machine for $500,000 using a long-term note to finance it• Paid$500 for ordinary repair• Purchased a patent for $45,000 cash• Paid$200,000 cash for addition to an existing building• Paid $60,000 for monthly salaries• Paid$250 for routine maintenance on equipment• Paid $10,000 for major repairs• Depreciation expense recorded for the year is$25,000If all transactions were recorded properly, what is the amount of increase to the Property, Plant, and Equipment section of Jada’s balance sheet resulting from this year’s transactions? What amount did Jada report on the income statement for expenses for the year?
To determine
Introduction:
Financial statements are the financial performance indicator of a company, which discloses the profits and losses of the company and the position of assets and liabilities of the company at the end of the year.
To compute:
The increase in amount of property, plant and equipment to be shown in balance sheet and increased amount to be charged to income statement for the year by Company J.
### Explanation of Solution
Calculation of increase in amount of property, plant and equipment to be disclosed in the balance sheet for the year by Company J:
Particulars Amount ($) Machinery purchased for using a long-term note 500,000 Add: Purchased a patent 45,000 Paid cash for addition to an existing building 200,000 Paid cash for extraordinary repairs 10,000 Less: Depreciation expense for the year ($25,000) Total increase in amount of property, plant and equipment to be shown in balance sheet
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# 221_2010_Makeup_Spring - (b Find the zero input response of...
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North South University Spring 2010 Makeup Mid 1 Course: ETE/EEE-221/383 Secs. 1+2 Time: 75 min Marks: 25 Answer all of the following questions 1. Consider a power signal x ( t ) = 1 . 5 * rect ( t - 1 2 ) whose period is 2 π . Sketch the signal and calculate its power. 3+4=7 2. (a) Consider a system described by the ODE ( D 3 + 5 D 2 + 8 D + 4) y ( t ) = x ( t ) Find the characteristic modes of the system.
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Unformatted text preview: (b) Find the zero input response of the system if y (0) = 1 , ˙ y (0) = 0 , ¨ y (0) =-1. (c) What would be the zero state input of the system if it is driven by an input x ( t ) = 4 δ ( t-2)? 3+4+5=12 3. Sketch the convolution of the signals x ( t ) = u ( t + 1)-2 u ( t ) + u ( t-1) with rect ( t/ 2). 6 1...
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## This note was uploaded on 06/08/2010 for the course ETE ETE 221 taught by Professor Adm during the Spring '10 term at North South University.
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# Indians out there...got to read this !!!
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13 Jan 2008, 08:50
This is not really related to MBA but it does provide much needed moments of levity....for anyone Indian or following up with news in India
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13 Jan 2008, 20:44
Hilarious !
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haha!
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Hi,
Nice buddy, fantastic really make me laugh.
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lol
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"Vijay Mallya says that he will buy the Blackburn Rovers and rename it to 'Team India'"
Re: Indians out there...got to read this !!! [#permalink] 12 Mar 2008, 10:54
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http://www.fixedpointtheoryandapplications.com/content/2010/1/126192 | 1,386,699,488,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164023039/warc/CC-MAIN-20131204133343-00016-ip-10-33-133-15.ec2.internal.warc.gz | 336,832,335 | 15,523 | Research Article
# Existence of Solution and Positive Solution for a Nonlinear Third-Order -Point BVP
Jian-Ping Sun* and Fan-Xia Jin
Author Affiliations
Department of Applied Mathematics, Lanzhou University of Technology, Lanzhou, Gansu 730050, China
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Fixed Point Theory and Applications 2010, 2010:126192 doi:10.1155/2010/126192
Received: 5 November 2010 Accepted: 14 December 2010 Published: 19 December 2010
© 2010 Jian-Ping Sun and Fan-Xia Jin.
This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
In this paper, we are concerned with the following nonlinear third-order -point boundary value problem: , , , , . Some existence criteria of solution and positive solution are established by using the Schauder fixed point theorem. An example is also included to illustrate the importance of the results obtained.
### 1. Introduction
Third-order differential equations arise in a variety of different areas of applied mathematics and physics, for example, in the deflection of a curved beam having a constant or varying cross-section, a three-layer beam, electromagnetic waves, or gravity-driven flows and so on [1].
Recently, third-order two-point or three-point boundary value problems (BVPs) have received much attention from many authors; see [210] and the references therein. In particular, Yao [10] employed the Leray-Schauder fixed point theorem to prove the existence of solution and positive solution for the BVP
(11)
Although there are many excellent results on third-order two-point or three-point BVPs, few works have been done for more general third-order -point BVPs [1113]. It is worth mentioning that Jin and Lu [12] studied some third-order differential equation with the following -point boundary conditions:
(12)
The main tool used was Mawhin's continuation theorem.
Motivated greatly by [10, 12], in this paper, we investigate the following nonlinear third-order -point BVP:
(13)
Throughout, we always assume that and . The purpose of this paper is to consider the local properties of on some bounded sets and establish some existence criteria of solution and positive solution for the BVP (1.3) by using the Schauder fixed point theorem. An example is also included to illustrate the importance of the results obtained.
### 2. Main Results
Lemma 2.1.
Let . Then, for any , the BVP
(21)
has a unique solution
(22)
where
(23)
Proof.
If is a solution of the BVP (2.1), then we may suppose that
(24)
By the boundary conditions in (2.1), we know that
(25)
Therefore, the unique solution of the BVP (2.1)
(26)
In the remainder of this paper, we always assume that . For convenience, we denote
(27)
The following theorem guarantees the existence of solution for the BVP (1.3).
Theorem 2.2.
Assume that is continuous and there exist and such that
(28)
Then the BVP (1.3) has one solution satisfying
(29)
Proof.
Let be equipped with the norm , where . Then is a Banach space.
Let , . Then the BVP (1.3) is equivalent to the following system:
(210)
Furthermore, it is easy to know that the system (2.10) is equivalent to the following system:
(211)
Now, if we define an operator by
(212)
where
(213)
then it is easy to see that is completely continuous and the system (2.11) and so the BVP (1.3) is equivalent to the fixed point equation
(214)
Let . Then is a closed convex subset of . Suppose that . Then and . So,
(215)
(216)
which implies that
(217)
From (2.16) and , we have
(218)
On the other hand, it follows from (2.17) that
(219)
In view of (2.18) and (2.19), we know that
(220)
which shows that . Then it follows from the Schauder fixed point theorem that has a fixed point . In other words, the BVP (1.3) has one solution , which satisfies
(221)
On the basis of Theorem 2.2, we now give some existence results of nonnegative solution and positive solution for the BVP (1.3).
Theorem 2.3.
Assume that , , , , is continuous, and there exist and such that
(222)
Then the BVP (1.3) has one solution satisfying
(223)
Proof.
Let
(224)
Then is continuous and
(225)
Consider the BVP
(226)
By Theorem 2.2, we know that the BVP (2.26) has one solution satisfying
(227)
Since , we get
(228)
In view of (2.28) and , we have
(229)
which implies that
(230)
It follows from (2.28), (2.30), and the definition of that
(231)
Therefore, is a solution of the BVP (1.3) and satisfies
(232)
Corollary 2.4.
Assume that all the conditions of Theorem 2.3 are fulfilled. Then the BVP (1.3) has one positive solution if one of the following conditions is satisfied:
(i);
(ii);
(iii), .
Proof.
Since it is easy to prove Cases (ii) and (iii), we only prove Case (i). It follows from Theorem 2.3 that the BVP (1.3) has a solution , which satisfies
(233)
Suppose that . Then for any , we have
(234)
which shows that is a positive solution of the BVP (1.3).
Example 2.5.
Consider the BVP
(235)
where , .
A simple calculation shows that and . Thus, if we choose and , then all the conditions of Theorem 2.3 and (i) of Corollary 2.4 are fulfilled. It follows from Corollary 2.4 that the BVP (2.35) has a positive solution.
### Acknowledgment
This paper was supported by the National Natural Science Foundation of China (10801068).
### References
1. Greguš, M: Third Order Linear Differential Equations, Mathematics and its Applications (East European Series),p. xvi+270. Reidel, Dordrecht, The Netherlands (1987)
2. Anderson, DR: Green's function for a third-order generalized right focal problem. Journal of Mathematical Analysis and Applications. 288(1), 1–14 (2003). Publisher Full Text
3. Bai, Z: Existence of solutions for some third-order boundary-value problems. Electronic Journal of Differential Equations. 25, 1–6 (2008)
4. Feng, Y, Liu, S: Solvability of a third-order two-point boundary value problem. Applied Mathematics Letters. 18(9), 1034–1040 (2005). Publisher Full Text
5. Guo, L-J, Sun, J-P, Zhao, Y-H: Existence of positive solutions for nonlinear third-order three-point boundary value problems. Nonlinear Analysis. Theory, Methods & Applications. 68(10), 3151–3158 (2008). PubMed Abstract | Publisher Full Text
6. Hopkins, B, Kosmatov, N: Third-order boundary value problems with sign-changing solutions. Nonlinear Analysis. Theory, Methods & Applications. 67(1), 126–137 (2007). PubMed Abstract | Publisher Full Text
7. Ma, R: Multiplicity results for a third order boundary value problem at resonance. Nonlinear Analysis. Theory, Methods & Applications. 32(4), 493–499 (1998). PubMed Abstract | Publisher Full Text
8. Sun, J-P, Ren, Q-Y, Zhao, Y-H: The upper and lower solution method for nonlinear third-order three-point boundary value problem. Electronic Journal of Qualitative Theory of Differential Equations. 26, 1–8 (2010)
9. Sun, Y: Positive solutions for third-order three-point nonhomogeneous boundary value problems. Applied Mathematics Letters. 22(1), 45–51 (2009). Publisher Full Text
10. Yao, Q: Solution and positive solution for a semilinear third-order two-point boundary value problem. Applied Mathematics Letters. 17(10), 1171–1175 (2004). Publisher Full Text
11. Du, Z, Lin, X, Ge, W: On a third-order multi-point boundary value problem at resonance. Journal of Mathematical Analysis and Applications. 302(1), 217–229 (2005). Publisher Full Text
12. Jin, S, Lu, S: Existence of solutions for a third-order multipoint boundary value problem with -Laplacian. Journal of the Franklin Institute. 347(3), 599–606 (2010). Publisher Full Text
13. Sun, J-P, Zhang, H-E: Existence of solutions to third-order -point boundary-value problems. Electronic Journal of Differential Equations. 125, 1–9 (2008) | 2,042 | 7,852 | {"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-2013-48 | latest | en | 0.922964 |
https://www.physicsforums.com/threads/volume-and-pressure-cylinders.158631/ | 1,548,063,084,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583771929.47/warc/CC-MAIN-20190121090642-20190121112642-00527.warc.gz | 925,768,416 | 12,224 | # Volume and Pressure Cylinders
1. Feb 28, 2007
### metalmagik
1. The problem statement, all variables and given/known data
Problem comes straight from a 2001 AP Physics B test
http://img151.imageshack.us/img151/6159/physicscylinderszq4.png [Broken]
2. Relevant equations
PV = nRT
KE = 3/2KT
deltaU = Q+W
3. The attempt at a solution
Okay, for the first two (a and b) I arrived at an answer of double the initial pressure of State 1 and half of the initial volume of State 1. This was because after I drew myself a little P vs. V graph, I knew the two States had to be on the same isotherm, so pressure had to increase in order for volume to decrease. Also, I just realized as I write this, if T is constant, in PV = nRT that means nRT is a constant and as P increases, V must decrease.
I'm pretty sure that's right, if it's wrong, I would appreciate clarification.
For c I got Isobaric, simply because isothermal and adiabatic do not work.
For d I got Yes, 4 --> 1 is isobaric simply because I drew a P versus V graph and since it has to go back to the same isotherm State 2 was on, but volume had to decrease, it had to be isobaric.
I do not understand e. I'm having a hardtime knowing what the weight does...do you have to use P = F/A?
Thanks for any help at all on this, any corrections if im wrong and any elaboration on any topic is greatly appreciated.
Last edited by a moderator: May 2, 2017 | 384 | 1,412 | {"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} | 3 | 3 | CC-MAIN-2019-04 | latest | en | 0.965649 |
https://au.mathworks.com/matlabcentral/cody/problems/40-reverse-run-length-encoder/solutions/163636 | 1,585,443,363,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493121.36/warc/CC-MAIN-20200328225036-20200329015036-00524.warc.gz | 365,426,810 | 15,699 | Cody
# Problem 40. Reverse Run-Length Encoder
Solution 163636
Submitted on 18 Nov 2012 by Evan
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 5 1 2 4 1 1 3]; correct = [5 5 2 1 1 1 1 3]; assert(isequal(correct, RevCountSeq(x)));
ans = 2 1 4 1 5 2 1 3 ans = 5 5 2 1 1 1 1 3
2 Pass
%% x = [1 9]; correct = [9]; assert(isequal(correct, RevCountSeq(x)));
ans = 1 9 ans = 9
3 Pass
%% x = [9 1]; correct = ones(1,9); assert(isequal(correct, RevCountSeq(x)));
ans = 9 1 ans = 1 1 1 1 1 1 1 1 1
4 Pass
%% x = [1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9]; correct = 1:9; assert(isequal(correct, RevCountSeq(x)));
ans = 1 1 1 1 1 1 1 1 1 1 2 3 4 5 6 7 8 9 ans = 1 2 3 4 5 6 7 8 9 | 372 | 809 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-16 | latest | en | 0.577556 |
http://designalyze.com/vidhan-sabha-yfr/keith-number-program-in-java-4d56f7 | 1,720,962,609,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514580.77/warc/CC-MAIN-20240714124600-20240714154600-00239.warc.gz | 6,517,440 | 6,594 | 124; by SMC ACADEMY. August 16, 2015 at 1:59 pm. Write a program to input a number and display whether the number is a Smith number or not. 26. Java Programs Friday, 25 October 2013. The sequence goes like this: 1, 9, 7, 17, 33, 57, 107, 197, 361, ... import java.io. Keith Number //Write a program to accept a number and check it is a Keith number or not. Then continue the sequence, where each subsequent term is the sum of the previous n terms. *; class keith_number The happy number can be defined as a number which will yield 1 when it is replaced … Keith numbers were introduced by Mike Keith in 1987. Program to determine whether a given number is a happy number Explanation. A Smith Number is a composite number whose sum of digits is equal to the sum of digits in its prime factorization. June 17, 2019 at 2:04 pm. As an example, consider the 3-digit number N = 197. 01-01-2018 27212 times. If a Fibonacci sequence (in which each term in the sequence is the sum of the previous t terms) is formed, with is the first ‘t’ terms being the decimal digits of the number n, then n … ; In this step, we are creating three strings. harish. If N = 187, then n=3, the first three numbers of the sequence are 1, 8, 7 and n=3. Examples: Input : n = 4 Output : Yes Prime factorization = 2, 2 and 2 + 2 = 4 Therefore, 4 is a smith number Input : n = 6 Output : No Prime factorization = 2, 3 and 2 + 3 is not 6. Vansh Rathour. Menu Based Program in Java Using Switch ... 05-05-2018 39447 times. A Smith number is a composite number, the sum of whose digits is the sum of the digits of its prime factors obtained as a result of prime factorization (excluding 1). In this program, we need to determine whether a given number is a happy number or not. By definition, N is a Keith number if N appears in the sequence thus constructed. Explanation of the above java buzz number program : The commented numbers in the above program denote the step numbers below : Define one integer variable no to store the user input value and one Scanner variable sc to read all user inputs. If you seed it with the three digit number 197, then 197 will (possibly coincidentally) be seen as a Keith number. In number theory, a Keith number or repfigit number (short for repetitive Fibonacci-like digit) is a natural number in a given number base with digits such that when a sequence is created such that the first terms are the digits of and each subsequent term is the sum of the previous terms, is part of the sequence. Java programs java python Python programs. Note: A Keith Number is an integer N with ‘d’ digits with the following property: If a Fibonacci-like sequence (in which each term in the sequence is the sum of the ‘d’ previous terms) is … Units of Measurement CGS FPS MKS SI. Blue Java Program For Keith Number. It is full of simple and easy programmes. ex- 197 is a keith number Sep, 2018. //A Keith number is an integer n with d digits with the following property. Happy number. Sample data: ... Java program to find the future date. 7 comments . Keith Number /*To determine whether an n-digit number N is a Keith number, create a Fibonacci-like sequence that starts with the n decimal digits of N, putting the most significant digit first. This is very helpful.for my course. If you seed it with 187, then 197 won't be seen as a Keith number. Write a Program in Java to input a number and check whether it is a Keith Number or not. if a fibonacci like sequence //is formed with the first d terms. 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If a fibonacci like sequence //is formed with the first d terms with first! | 2,943 | 11,884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-30 | latest | en | 0.879886 |
https://math.stackexchange.com/questions/2632051/what-can-be-said-about-statement-xy4-sinx-siny-ge-0 | 1,571,732,077,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987813307.73/warc/CC-MAIN-20191022081307-20191022104807-00457.warc.gz | 576,556,145 | 32,357 | # What can be said about statement $x+y+4\sin(x)\sin(y)\ge 0$
## Problem
What can be said about following statement when dot $(x,y)$ is close enough to origin.
$$x+y+4\sin(x)\sin(y)\ge 0$$
## Attempt to solve
I mark the given expression as:
$$f(x,y)=x+y+4\sin(x)\sin(y)$$
Now i simply examine the situation where $f(x,y)\ \ge 0$ and $$\lim_{(x,y)\rightarrow (0,0)}f(x,y)$$
I could start off my examining the gradient at point $(0,0)$
$$\nabla f(x,y)=\begin{bmatrix} 4\cos(x)\sin(y)+1 \\ \cos(y)4\sin(x)+1 \end{bmatrix}$$
$$\nabla f(0,0)=\begin{bmatrix} 1 \\ 1 \end{bmatrix}$$
It seems $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ would be increasing direction. Now my intuition would tell me that if $\nabla f(x,y)\neq0$ wouldn't necessary mean that critical point couldn't exist in this particular point. If $\nabla f(x,y)=0$ i would say this is certainly critical point.
Since it's uncertain if this is critical point or not. Computing a hessian matrix would result in more definite answer to this question.
$$H(x,y)=\begin{bmatrix} -4\sin(x)\sin(y) & 4\cos(x)\cos(y) \\ \cos(y)4\cos(x) & - 4\sin(x)\sin(y) \end{bmatrix}$$
$$H(0,0) = \begin{bmatrix} 0 & 4 \\ 4 & 0 \end{bmatrix}$$ egeinvalues are $\lambda_1=-4,\lambda_2=4$ so this would mean hessian is indefinite matrix which means point $(0,0)$ is saddle point.
At this point it would be probably a good idea to make plot of this function since i don't know other good methods from now on.
The plot confirms that $\nabla f(x,y)$ at point $(0,0)$ gives the increasing direction. The $- \nabla f(x,y)$ is also increasing direction even though negative gradient is usually interpreted as the largest decreasing direction. In this the interpretation would be incorrect. We can see that:
$$\hat{d_1}= \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \hat{d_2}= \begin{bmatrix} -1 \\ -1 \end{bmatrix}$$ are increasing direction and the decreasing direction would be:
$$\hat{d_3}= \begin{bmatrix} 1 \\ -1 \end{bmatrix},\hat{d_4}= \begin{bmatrix} -1 \\ 1 \end{bmatrix}$$
Now what happens if we try to approach point $(0,0)$. When $(x,y)$ values approach zero the variation in $f(x,y)$ values decreases. Also values of $f(x,y)$ become smaller and smaller until it reaches point $(0,0)$ (if we approach the critical point using a circle and start decreasing radius by some constant).
Now if i was able to prove this circle can get arbitrary close to the point (0,0) (arbitrary small radius) this would prove the existence of limit at point $(0,0)$ ? I don't know how formally existence of a limit can be proven in multivariable case but this sort of approach would sound like it could work. Now these things would indicate that following statement is true:
$$\lim_{(x,y)\rightarrow (0,0)}f(x,y)=0$$
Also when we can evaluate this function's value at point $(0,0)$ meaning it is defined in this particular point. Limit would automatically exist in this point since it would be already defined which wouldn't require any proof in the first place ?
I've understood that the reason behind why there is such thing called as "limit " is attended to use in cases where we have undefined point and we want to know that can we get arbitrary close to the point without evaluating it's value at that point. If the answer is yes then we would have limit at this particular point.
If someone manages to read all of this i highly appreciate your effort in doing so. Some sort of feedback on this would be great. Do my claims seem correct / false or am i missing something important ?
• The question is really vague. I would just take $\sin p \approx p$ where p is either x or y. And go from there. But again, the question is not clear to me – Yuriy S Feb 2 '18 at 0:42
• Not sure I understand, but isn't the claim false with the simple counter-example of $f(x,x) = 2x+4(\sin x)^2$? $f(x,x)$ is negative for $x<0$ (and $x$ close to $0$) and positive for $x>0$ (and $x$ close to $0$). wolframalpha.com/input/?i=plot+2x%2B4(sin+x)%5E2,+x%3D-1..1 – Clement C. Feb 2 '18 at 1:06 | 1,177 | 4,012 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2019-43 | latest | en | 0.794137 |
https://math.stackexchange.com/questions/915809/why-is-2-sin-5x-cos-5x-2-sin-x-cos-x-sin-10x-sin-2x/915816 | 1,716,607,137,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058770.49/warc/CC-MAIN-20240525004706-20240525034706-00805.warc.gz | 327,781,078 | 35,070 | # Why is (2 sin 5x cos 5x) (2 sin x cos x) = sin 10x sin 2x? [closed]
The question says it all, why? Also, those are not powers, all of them are co-efficient, thanks.
PS This is actually not a homework, was a solved example I was going through and I got confused.
• What have you tried? If you tell us this then we will be better able to help you. And it helps us feel that we are not just doing your homework for you. Sep 1, 2014 at 11:42
• This is actually not a homework, was a solved example I was going through and I got confused. Sep 1, 2014 at 12:33
• @AkshatTripathi Is is still confusing with the answer below? Sep 1, 2014 at 14:01
• @Jean-ClaudeArbaut Nono, solved, thanks for asking. :) Sep 1, 2014 at 14:32
$\sin(2x) = 2\sin(x)\cos(x)$ | 242 | 751 | {"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} | 3.640625 | 4 | CC-MAIN-2024-22 | latest | en | 0.975145 |
https://financetrainingcourse.com/store/Constructing-Volatility-Surfaces-in-EXCEL-p36848744 | 1,721,803,011,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518157.20/warc/CC-MAIN-20240724045402-20240724075402-00226.warc.gz | 209,584,933 | 13,474 | 0Constructing Volatility Surfaces in EXCEL. Study Guide - FinanceTrainingCourse.com Store
# Constructing Volatility Surfaces in EXCEL
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Volatility surface plots are a construct of the latter across moneyness (strike prices) and maturity (time to expiry). Unlike implied volatilities which are determined by backing out volatility from the Black Scholes option price equation applied to at the money options, local volatilities use implied volatilities and a one factor Black Scholes model to drive local volatility values across the surface. Calibrating volatility surfaces is one of the tweaks market practitioners use to sidestep the constant volatility assumption present in the Black Scholes equation for pricing European options.
This course begins by explaining the difference and pros and cons of various volatility estimates, from trailing volatilities, to implied volatilities to local volatilities. It then looks at the benefits of purchasing deep out of the money options, and how a change in volatility impacts the value of these options. The next lesson delves deeper into the differences between implied and local volatilities. A step-by-step guide on how to build surfaces for local volatilities for options on stock prices using Dupire’s formula in EXCEL is illustrated. Forward implied volatilities using the methodology described by Taleb are also derived.
Finally, the course extends the calculation to determine local volatility surfaces for options on commodity futures contracts. The resulting surface is plotted against moneyness (as given by strike/futures price) and maturity as well as against futures price and maturity for a given strike. Results are compared with implied and forward implied volatilities.
The course consists of:
• A 40 page PDF guide that shows how to build a volatility surface step by step in EXCEL using Dupire's formula.
• A 20 page PDF guide that shows how to calculate the volatility surface in EXCEL for options of futures contracts.
• An EXCEL spreadsheet that is used as a simple teaching template by the PDF tutorial above. The Excel sheet shows the implementation of Dupire's formula as well as the resultant volatility surface. The sheet also shows Taleb's implementation of implied forward volatility using term structure of volatility concepts.
• An EXCEL spreadsheet that shows the calculation of local volatility surfaces for options on Crude Oil WTI commodity futures contracts. The sheet also shows the calculation of the implied forward volatility and the comparison of local volatilities with implied & forward implied volatilities.
#### Learning Objectives
After taking this course you will be able to:
• Distinguish between trailing, implied and local volatilities
• Calculate implied volatility
• Calculate local volatility using Dupire’s formula
• Understand the advantages of purchasing a deep out of the money option
• Construct a volatility surface of implied volatilities in EXCEL
• Construct a volatility surface of local volatilities in EXCEL for options on stock price
• Calculate a term structure of forward implied volatilities in EXCEL
• Construct a local volatility surface in EXCEL for options on commodity futures contracts
#### Prerequisites
Familiarity with the Black Scholes equation, derivative products and pricing models for pricing European options and EXCEL.
#### Target Audience
This advanced practitioner course is aimed at professionals who deal with pricing, valuation and risk issues related to derivative transactions.
Save this product for later | 715 | 3,653 | {"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-2024-30 | latest | en | 0.878562 |
http://studyset.net/solved/?paper_id=9210834 | 1,498,367,794,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320438.53/warc/CC-MAIN-20170625050430-20170625070430-00701.warc.gz | 375,503,176 | 8,966 | #### Description of this paper
##### Part II, Problem 3 | Final Exam | CTL.SC1x Courseware | edX Part-(Answered)
Description
Question
Part II, Problem 3 | Final Exam | CTL.SC1x Courseware | edX
Part II, Problem 3
You work for an agriculture supply company in Africa. Part of your job is to procure fertilizer for local small farmers in the area. Some problems with the fertilizer are that it must be ordered before the growing season begins, it has a very long lead-time, and the farmers need it only over a very short window of time. Additionally, the fertilizer cannot be stored for the next growing season, as it will deteriorate. So, you have to make the decision of how much fertilizer to order before the farmers in the local area have even determined how much of what crop to plant.
To make matters worse, you discover that no records were kept for the amount of fertilizer demanded by the farmers in the past. You have asked your customer farmers who have been there for decades and they tell you that typically they will collectively plant 210.0 acres. The fewest number of acres planted in a season that any farmer could remember was 190.0 acres while the most was 1080.0 acres. You decide to estimate the demand using a triangle distribution. You know that each acre of planted crop will require 41.0 pounds of fertilizer.
Fertilizer costs 0.22 $/pound and you sell the fertilizer to the farmers at 0.55$/pound. Additionally, you estimate that every pound of fertilizer you are short is equivalent to costing you 1 $/pound in loss of goodwill to the farmers that might translate into future loss of business. 3.1 How many pounds of fertilizer should you order if you wish to maximize profitability? Please round your answer to the nearest 100 pound. - unanswered 3.2 What Cycle Service Level (CSL) does your answer in 3.1 correspond to? Answer as a percentage with two decimals, without the percentage symbol. For instance, if your solution is 95.554%, you answer 95.55. - unanswered 3.3 When you tell a local manager about your plans, he asks you about the left-over fertilizer. He points out that, according to a recent law, fertilizer that is not sold need to be taken to a special destruction facility. Together you estimate that the cost associated with taking it to the facility is 0.5$/pound.
If you include a destruction cost of 0.5 $/pound in your calculations, how many pounds of fertilizer should you order if you wish to maximize profitability? Please round your answer to the nearest 100 pound. (Hint: Think of the destruction cost as a negative salvage value.) - unanswered 3.4 What Cycle Service Level (CSL) does your answer in 3.3 correspond to? Answer as a percentage with two decimals, without the percentage symbol. For instance, if your solution is 95.554%, you answer 95.55. - unanswered 3.5 You do some additional research and find a not-for-profit community plantation that is happy to receive any unsold fertilizer. They are even willing to come and pick it up, so donating left-over fertilizer comes at no cost to you. Rather, you believe it comes with a long-term goodwill benefit (revenue) as you will give back to the local community. You can safely assume the plantation will pick up all unsold fertilizer so that none has to be taken to the destruction facility. After hearing about the plantation, the local manager says: "That is great news! We should aim for a 95% service level since having too much in stock is costless!" Call the goodwill benefit from donating left-over fertilizer [mathjaxinline]g[/mathjaxinline]. What size of [mathjaxinline]g[/mathjaxinline] will make a 95% Cycle Service Level the profit optimizing Cycle Service Level? Answer with two decimals. - unanswered Check your answerSave your answer You have used 0 of 3 submissions ' data-progress_detail="0/5" data-progress_status="none" data-url="/courses/course-v1:MITx+CTL.SC1x_2+1T2016/xblock/block-v1:MITx+CTL.SC1x_2+1T2016+type@problem+block@a0724d02edd245c0a95559faef557aae/handler/xmodule_handler" data-problem-id="block-v1:MITx+CTL.SC1x_2+1T2016+type@problem+block@a0724d02edd245c0a95559faef557aae"> Part II, Problem 3 (5 points possible) You work for an agriculture supply company in Africa. Part of your job is to procure fertilizer for local small farmers in the area. Some problems with the fertilizer are that it must be ordered before the growing season begins, it has a very long lead-time, and the farmers need it only over a very short window of time. Additionally, the fertilizer cannot be stored for the next growing season, as it will deteriorate. So, you have to make the decision of how much fertilizer to order before the farmers in the local area have even determined how much of what crop to plant. To make matters worse, you discover that no records were kept for the amount of fertilizer demanded by the farmers in the past. You have asked your customer farmers who have been there for decades and they tell you that typically they will collectively plant 210.0 acres. The fewest number of acres planted in a season that any farmer could remember was 190.0 acres while the most was 1080.0 acres. You decide to estimate the demand using a triangle distribution. You know that each acre of planted crop will require 41.0 pounds of fertilizer. Fertilizer costs 0.22$/pound and you sell the fertilizer to the farmers at 0.55 $/pound. Additionally, you estimate that every pound of fertilizer you are short is equivalent to costing you 1$/pound in loss of goodwill to the farmers that might translate into future loss of business.
3.1
How many pounds of fertilizer should you order if you wish to maximize profitability? Please round your answer to the nearest 100 pound.
3.2
What Cycle Service Level (CSL) does your answer in 3.1 correspond to? Answer as a percentage with two decimals, without the percentage symbol. For instance, if your solution is 95.554%, you answer 95.55.
3.3
When you tell a local manager about your plans, he asks you about the left-over fertilizer. He points out that, according to a recent law, fertilizer that is not sold need to be taken to a special destruction facility. Together you estimate that the cost associated with taking it to the facility is 0.5 $/pound. If you include a destruction cost of 0.5$/pound in your calculations, how many pounds of fertilizer should you order if you wish to maximize profitability? Please round your answer to the nearest 100 pound. (Hint: Think of the destruction cost as a negative salvage value.)
3.4
What Cycle Service Level (CSL) does your answer in 3.3 correspond to? Answer as a percentage with two decimals, without the percentage symbol. For instance, if your solution is 95.554%, you answer 95.55.
3.5
You do some additional research and find a not-for-profit community plantation that is happy to receive any unsold fertilizer. They are even willing to come and pick it up, so donating left-over fertilizer comes at no cost to you. Rather, you believe it comes with a long-term goodwill benefit (revenue) as you will give back to the local community. You can safely assume the plantation will pick up all unsold fertilizer so that none has to be taken to the destruction facility.
After hearing about the plantation, the local manager says: "That is great news! We should aim for a 95% service level since having too much in stock is costless!"
Call the goodwill benefit from donating left-over fertilizer g
Paper#9210834 | Written in 27-Jul-2016
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https://doubtnut.com/question-answer/if-90-lt-a-lt-180-180-lt-b-lt-270-and-cos-a-sqrt3-2-sin-b-3-5-then-2-tan-b-sqrt3-tan-a-cot2-a-cos-b-1810642 | 1,590,921,214,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413097.49/warc/CC-MAIN-20200531085047-20200531115047-00021.warc.gz | 318,010,332 | 57,169 | or
# If 90^@ < A < 180^@ ,180^@ < B < 270^@ and cos A=-(sqrt3)/2, sin B =-3/5 then (2 tan B+sqrt3 tan A)/(cot^2 A+cos B)=
Question from Class 12 Chapter Default
Apne doubts clear karein ab Whatsapp par bhi. Try it now. | 93 | 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} | 2.578125 | 3 | CC-MAIN-2020-24 | latest | en | 0.199121 |
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• Physics Notes On – Law Of Refraction – For W.B.C.S. Examination.
When a beam of light passes through two different media via an interface, its behaviour is governed by the laws of refraction of light. These laws are also known as Snell’s law.Continue Reading Physics Notes On – Law Of Refraction – For W.B.C.S. Examination.
The two laws followed by a beam of light traversing through two media are:
The ray of light incident to the interface, the normal trajectory of the beam (from the point of incidence), and the refracted ray must all lie in the same plane.For any two different media, the since of the angle at which the beam of light is incident is always proportional to the sine of the angle at which the refracted ray emerges.
In other words, the quotient of the sine of the angle of incidence and the sine of the angle of refraction is always constant.
It is important to note that insight into the extent to which a particular medium refracts a beam of light is given by the refractive index of that medium.
The point of refraction is created where the incident ray lands and the angle that it makes with the refracted ray not forgetting the normal line that is dropped on the plane perpendicularly.
The medium through which the rays of light are passing creates a considerable difference in refraction unlike in reflection of light. The refractive indices make the dependency on the medium apparent in Snell’s Law.
The normal on the surface is used to gauge the angles that the refracted ray creates at the contact point. n1 and n2 are the two different mediums that will impact the refraction.
The refractive index of water is 1.33 whereas the refractive index of air is 1.00029. Thus, to understand the concept of Snell’s Law let’s consider the light of wavelength 600nm that goes from water into the air. To calculate the angle made by the outgoing ray we apply the figures in the formula mentioned above.
1.33 sin 30o = 1.00029 sin x
x = 41o
When one fishes with a spear it is not as difficult as fishing with a rod as the fisherman has to encounter refraction in the latter case.
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https://www.jiskha.com/display.cgi?id=1326245477 | 1,516,510,301,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890187.52/warc/CC-MAIN-20180121040927-20180121060927-00094.warc.gz | 948,898,403 | 5,149 | # math-calculus
posted by .
evaluate
|8|
okay i know this means square root of 8 squared and that would be 8.
|-8|
this also positive because negative 8squared gives you positive.
i not get these ones
||3|-|9||
theres too many lines
and these one
-|-sqrt81|
-|cubicrt27
||5|-|-32||
• math-calculus -
• math-calculus -
First of all |8| does not mean "square root" it means
absolute value.
In simple terms it means that you want to take the positive result of the value inside the | | signs.
||3-|9||
= |3-9|
= |-6|
= 6
-|-√81|
= -|-9|
= - 9
etc.
• math-calculus -
thanks jordan and reiny :) but reiny i not understand the whole absolute value thing becuase for -|-sqrt81| you get -9 and that not positive.
• math-calculus -
suppose it wasn't there, then the answer would be +9
But it is there, so .....
don't confuse a negative inside the || with one outside of the ||
• math-calculus -
okay so everything in || will always be positive? and can u please explain how did u get rid of the two extra lines here ||3-|9||.
• math-calculus -
work it like you had brackets.
start with the innermost and work towards the outside ones, only drop the || signs if you have a single digit left inside.
e.g. the last one:
||5|-|-32||
= | 5 - 32| , I worked on the inside ones
= | -27|
= 27
(perhaps in your notebook you could use different colours for each set,
e.g.
use red for the inside, and blue for the inside )
• math-calculus -
okay those bracket lines really confusing. how you get |5-32| because that negative sign be separated from f and -32. sorry i just real confused right now
• math-calculus -
i not meant f i mean 5.
• math-calculus -
Did you not read what I said before the
||5|-|-32|| ??
• math-calculus -
okay wait i think i sort of get it now. and absolute value mean how far you are from 0 ?
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1) For a reaction delta Go is more negative than delta Ho. What does this mean? | 873 | 3,085 | {"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} | 4.03125 | 4 | CC-MAIN-2018-05 | latest | en | 0.911095 |
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rational expressions typically contain a variable in the denominator. after multiplying both sides of the previous example by the lcd, we were left with a linear equation to solve. this is not always the case; sometimes we will be left with a quadratic equation. to solve it, rewrite it in standard form, factor, and then set each factor equal to 0. up to this point, all of the possible solutions have solved the original equation. multiplying both sides of an equation by variable factors may lead to extraneous solutionsa solution that does not solve the original equation., which are solutions that do not solve the original equation. here the result is a quadratic equation.
in this case, choose the factored equivalent to check: here −2 is an extraneous solution and is not included in the solution set. sometimes all potential solutions are extraneous, in which case we say that there is no solution to the original equation. to clear the fractions, multiply by the lcd, (x−4)(x+5). if we multiply the expression by the lcd, x(2x+1), we obtain another expression that is not equivalent. solution: the goal is to isolate x. assuming that y is nonzero, multiply both sides by y and then add 5 to both sides. solution: in this example, the goal is to isolate c. we begin by multiplying both sides by the lcd, a⋅b⋅c, distributing carefully. 81. explain how we can tell the difference between a rational expression and a rational equation.
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Now, let us continue where we left off in the previous part in part 3b. In the bonus that we will calculate the voltage quantities. Now in the previous section for part 3b, we calculated the phase current quantities for a SLGF on a 13.8 kV system.
Now, this is the bonus step, in this step, we will calculate the phase voltage quantities at the point of the fault on a 13.8 kV system. Now, we do that with a very very similar process – right, and what we will do first is we will go back to our faulted sequence network and redraw it again.
In this figure, we want to focus on the positive, negative, and zero sequence voltages that are shown on the Low Voltage 13.8kV fictitious bus on the positive, negative, and zero sequence networks… and we want to determine the sequence voltage quantities by hand and from the sequence voltage quantities, we will calculate the actual phase voltage quantities.
So this is our objective in this section.
Now lets continue where we left off in the previous part in part 3b. In the bonus step we will calculate the voltage quantities. Now in the previous section for part 3b we calculated the phase current quatities for a single line to ground fault on 13.8 kV system. Now this is a bonus step in this step we will calculate the phase voltage quantities at the point of the fault on the 13.8 kV system. Now we do tht with a very very similar process right and what we will do first is that we will go back to a faulted sequence network and redraw it again. In this figure we want to focus on the positive negative and zero sequence voltages that are shown on the low voltage 13.8 kV fictitious bus on the positive negative zero sequence network. And we want to determine the sequence voltage quantities by hand and from the sequence voltages quantities we will calculate the actual phase voltage quantities so this is our objective in this section.
Now, before we move forward, let’s first decide a base value of voltage for the calculations and because the voltage at the point of the fault is 13.8 KV, we will select that as our base value below.
Vbase is equal to 13.8 KV divided by root 3.
V_Base=13.8kV/√3
Now, to calculate positive sequence voltage we will now apply KVL in the positive sequence loop. So here we go,
The positive sequence voltage will equal to the sum of the voltages across the voltage source minus the voltage drop due to the positive sequence impedance, that is 1 angle 0 minus j0.15 (which is the sum of the positive sequence generator and transformer impedance) multiplied by the per unit positive sequence current that is flowing to the 13.8kV fictitious bus which is –j2.5 pu.
The –j2.5 pu was calculated in the previous part 3b section. That gives us 0.625 pu
V_a^((1) )=1 ∠0°-(j0.15)(-j2.5)=0.625 pu
Now, it needs to be converted back to the actual voltage quantities by multiplying with the base value of 13.8 KV divided by square root 3.
V_A^((1) )=(0.625 pu)* 13.8kV/√3=4.98 kV∠0°
So that gives us 4.98 kV at the angle of 0 degrees.
Similarly, now that we have calculated the positive sequence voltage quantities, now we will calculate the negative sequence voltage quantity.
Similarly, we will apply KVL again in the negative sequence and zero sequence networks to calculate the per unit voltage and then multiply by the voltage base to get the actual voltage quantities.Since there is no voltage source in the negative sequence loop, it will be considered 0.
So the negative sequence voltage is equal to zero minus j0.15 pu (which is the transformer and generator impedance) times –j2.5 pu (which is the current that is going through the 13.8 KV fictitious bus – negative sequence. That is equal to -0.375 pu. Okay.
V_a^((2) )=0-(j0.15)(-j2.5)=-0.375 pu
In terms of the actual voltage quantities that’s equal to 2.988 kV at the angle of 180 degrees.
And that voltage will entirely consist of the voltage drop across the negative sequence impedance that is j0.15 * -j2.5, again that gives us -0.375 per unit and we will multiply accordingly to determine the 2.988 kV at the angle of 180 degrees. This is the negative sequence voltage.
V_a^((2) )=(-0.375pu)*(13.8kV/√3)=2.988 kV∠180°
For Zero sequence voltage, we do the following:
Zero sequence voltage is equal to 0 minus j0.10 times the –j2.5 which is equal to –j0.25 pu
V_a^((0) )=0-(j0.10)(-j2.5)=-j0.25 pu
Here, the impedances will be –j0.10 only because the transformer pu impedance is only considered and the generator impedance which j0.05 must be excluded from the loop due to the open delta… and we already know that zero sequence current will NOT flow through the HV side which strengthen this point even more.
Furthermore, like the negative sequence, we don’t have a voltage source for the zero sequence that is why the voltage source equals zero… and again, we get -0.25 pu for the zero sequence voltage quantity. And then we multiplied with the base value to get the actual voltage quantities which is 1.992 kV at the angle of 180 degrees. That’s a zero sequence voltage quantity
V_a^((0) )=(-0.25)*(13.8kV/√3)=1.992 kV∠180°
Now, so we have calculated the positive sequence voltage quantity, the negative sequence voltage quantity and the zero sequence voltage quantity. Now, let us determine using our familiar transformation equations to get the phase voltage quantities at the point of the fault on a 13.8 KV bus.
Plugging in these calculated values of the three sequence components and the “a” operator. Here’s what we will get. Solving the math with a scientific calculator will lead us to the following answers.
The voltage at the point of the fault on phase A is equal to 0 kV
V_a= 0
The voltage at the point of the fault on phase B is equal to 7.52 kV at the angle of negative 113 degrees.
V_a=7.52kV∠-113°
The voltage at the point of the fault on phase C will equal to 7.52 kV at the angle of positive 113 degrees.
V_c=7.52kV ∠113°
Now, this is an interesting result. Now let’s redraw the original drawing and let’s talk about this.
This is very interesting, logical and very intuitive result, because we know that the fault was one line to ground fault on phase A. Now the point of the fault is very clear that the phase A voltage will drive to zero and which is something very common and that is something, we should totally expect.
However, phase B and phase C voltage magnitudes will have a slight depression in voltage.
During normal and balanced operating conditions, we should expect phase voltages to be 13.8kV divided by square root of 3 which equals 7.96 kV.
>V_phase=13.8kV/√3=7.96 kV
however, during phase A to ground fault, at the point of the fault, Phase B and C voltages are 7.52 kV, which is very interesting, because there is a slight depression on the unfaulted phases.
Another interesting thing to note is that the angles of Phase B and C voltages are opposite. They are equal in magnitude but the angles are opposite which makes sense because Phase A voltage is driven to zero and again, at the point of the fault, and the remaining two phases will need to balance out.
So phase voltages are goanna be equal but the angles are goanna be opposite.
Now, In the next video we will walk through a similar calculation for the line to line fault.
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An incredible amount of time and effort is needed to develop high-quality video tutorials. Each video (Part 1 for example) takes approximately 10 hours to complete which includes learning the concept ourselves, brainstorming creative ways to teach and explain the concepts, writing the script, audio recording, video recording, and editing. It's no wonder why Hundreds-of-Thousands of people have watched, liked, subscribed, and left positive comments on Youtube channel. Your support truly makes all the difference. | 2,202 | 9,333 | {"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-2019-04 | longest | en | 0.91473 |
https://blog.ed.ted.com/2015/01/22/unlocking-fractals-an-exercise-in-pure-mathematics/ | 1,591,387,862,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348502204.93/warc/CC-MAIN-20200605174158-20200605204158-00125.warc.gz | 274,922,246 | 8,672 | ## Unlocking fractals: An exercise in pure mathematics
“This shape here just came out of an exercise in pure mathematics. Bottomless wonders spring from simple rules, which are repeated without end.’
This poetic definition of fractal geometry is the closing note of Benoit Mandelbrot’s 2010 TED Talk ‘Fractals and the art of roughness.’ And while this definition touches on the extraordinary nature of these incredibly complex and infinite patterns, there is no better way to understand fractals than by seeing them. In this TED Talk and TED-Ed Lesson pairing, we will take a closer look at the unbelievably beautiful world of fractal geometry.
Fractals are infinitely complex patterns that are self-similar across different scales, created by repeating a simple process over and over in an ongoing feedback loop.
Confused yet? We’ll start our exploration of fractals with the man who coined the term: Benoit Mandelbrot.
Studying complex dynamics in the 1970s, Benoit Mandelbrot had a key insight about a particular set of mathematical objects: that these self-similar structures with infinitely repeating complexities were not just curiosities, as they’d been considered since the turn of the century but were in fact a key to explaining non-smooth objects and complex data sets – which make up, let’s face it, quite a lot of the world. Mandelbrot coined the term “fractal” to describe these objects, and set about sharing his insight with the world.
In this TED Talk from 2010, Mandelbrot develops a theme he first discussed at TED in 1984 — the extreme complexity of roughness and the way that fractal math can find order within patterns that seem unknowably complicated.
Now that you’ve been primed on fractals, let’s take a look at them in action. When TED-Ed Educators Alex Rosenthal and George Zaidan were tasked with writing a TED-Ed Lesson on fractals (with a targeted K-12 audience), they tackled the difficult math concept with a narrative bent. Diving headfirst into a fun film noir treatment, the educators (along with animator Jeremiah Dickey) created an amazing visual world for the fractals to shine.
When asked why they decided to go for narrative, educator Alex Rosenthal says, “I go in looking for the narrative. My background is in theater and in film, and I’m always looking for those ways to enter in, especially when it comes to math and science. Not just reporting it but to tell the story. Because I think a good story and characters can help get you in and give you an empathetic, intuitive sense of the subject more than if it was just the pure information.”
For extra credit, check out one more TED Talk on The fractals at the heart of African design. >> | 589 | 2,692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-24 | longest | en | 0.943128 |
https://math.stackexchange.com/questions/869656/divergence-set-at-radius-of-convergence | 1,569,080,309,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574532.44/warc/CC-MAIN-20190921145904-20190921171904-00374.warc.gz | 575,775,526 | 30,365 | # Divergence set at radius of convergence
I came up with this question on my own while I was musing around reviewing notes. After unsuccessful Google search (thwarted by a deluge amount of webpages on basic calculus), I decided to ask here.
We know that radius of convergence characterize the behaviour of a power series at point inside and outside the radius, but not on it. Calculus book never take much of a careful look at that details. Thus the question is thus:
Does there exist a power series $s(z)=\sum\limits_{k=0}^{\infty}a_{k}z^{k}$ such that the set $$D=\{z\in\mathbb{C}:|z|=1, \;s(z) \text{ diverges } \}$$ is $\underline{not}$ a countable union of arcs (close, open or otherwise).
Hope someone can solve it. Thank you.
Here is a variant of this question asked on MathOverflow: https://mathoverflow.net/q/49395/. In particular, your question is answered by the cited result that $D$ can be any $G_\delta$ subset of the circle (such a set can be uncountable with dense complement). | 259 | 998 | {"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} | 2.625 | 3 | CC-MAIN-2019-39 | latest | en | 0.924126 |
https://www.physicsforums.com/threads/finding-the-time-constant.638187/ | 1,675,503,016,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500095.4/warc/CC-MAIN-20230204075436-20230204105436-00448.warc.gz | 970,456,156 | 13,761 | # Finding the time constant
Rombus
## Homework Statement
Find the time constant
$x(t)=4 e^{-4t} u(t)$
## Homework Equations
$\tau = \frac{1}{\lambda}$
## The Attempt at a Solution
$\tau = \frac{1}{\lambda}= \frac{1}{4} = .25$
I know the unit step is shifting the start of the decay function to zero, but I'm not sure how or if this can affect the time constant. So my question is how does the unit step affect the time constant?
Homework Helper
Dearly Missed
## Homework Statement
Find the time constant
$x(t)=4 e^{-4t} u(t)$
## Homework Equations
$\tau = \frac{1}{\lambda}$
## The Attempt at a Solution
$\tau = \frac{1}{\lambda}= \frac{1}{4} = .25$
I know the unit step is shifting the start of the decay function to zero, but I'm not sure how or if this can affect the time constant. So my question is how does the unit step affect the time constant?
It has no effect. Just recall the definition of time constant.
RGV | 261 | 937 | {"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} | 3.609375 | 4 | CC-MAIN-2023-06 | latest | en | 0.811393 |
https://courses.ansys.com/index.php/courses/intro-to-modal-based-methods/lessons/how-many-modes-to-include-modal-truncation-lesson-3/ | 1,716,447,747,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058611.55/warc/CC-MAIN-20240523050122-20240523080122-00202.warc.gz | 152,102,651 | 23,328 | How Many Modes to Include? (Modal Truncation) — Lesson 3
In linear dynamics, the mode-superposition method provides a very efficient method of determining the response of a system. Since this method uses linear combination of modes for calculating the response of a structure over a range of frequencies, an adequate number of modes needs to be carefully extracted. This video lesson will discuss the significance of modal truncation, and we will also discuss some guidelines and tools on how to extract enough modes. We have a short lecture followed by a workshop example to demonstrate the importance of modal truncation using Ansys Mechanical.
Video Highlights
01:12 — Generalized coordinates or modal methods
02:05 — Extracting an optimal number of modes
03:22 — Guideline 1: Appropriate frequency range
04:10 — Guideline 2: Ratio of effective mass to total mass
05:20 — Guideline 3: Understanding actual modes
07:20 — Improving accuracy with fewer modes
08:18 — Walkthrough Example — Clamp | 214 | 1,003 | {"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-2024-22 | latest | en | 0.824079 |
https://www.yht7.com/news/19782 | 1,642,761,995,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303356.40/warc/CC-MAIN-20220121101528-20220121131528-00400.warc.gz | 1,122,062,496 | 8,197 | #### 解决浮点运算精度不准确,BigDecimal 加减乘除
``````package com.kflh.boxApi.utils.util;
import java.math.BigDecimal;
/**
* @program: BoxApi
* @description: 计算浮点数
* @author: eterntiyz
* @create: 2019-01-17 11:10
*/
public class DoubleCalendar {
/**
* @Description: 浮点加法
* @Param: []
* @return: java.lang.Double
* @Author: tonyzhang
* @Date: 2019-01-17 11:27
*/
public static Double add(String str1,String str2) {
BigDecimal bignum1 = new BigDecimal(str1);
BigDecimal bignum2 = new BigDecimal(str2);
return bignum3.doubleValue();
}
/**
* @Description: 浮点减法
* @Param: []
* @return: java.lang.Double
* @Author: tonyzhang
* @Date: 2019-01-17 11:27
*/
public static Double subtract(String str1,String str2) {
BigDecimal bignum1 = new BigDecimal(str1);
BigDecimal bignum2 = new BigDecimal(str2);
BigDecimal bignum3 = bignum1.subtract(bignum2);
return bignum3.doubleValue();
}
/**
* @Description: 浮点乘法
* @Param: str1为分母,str2为分子
* @return: java.lang.Double
* @Author: tonyzhang
* @Date: 2019-01-17 11:26
*/
public static Double multiply(String str1,String str2) {
BigDecimal bignum1 = new BigDecimal(str1);
BigDecimal bignum2 = new BigDecimal(str2);
BigDecimal bignum3 = bignum1.multiply(bignum2);
return bignum3.doubleValue();
}
/**
* @Description: 浮点除法
* @Param: []
* @return: java.lang.Double
* @Author: tonyzhang
* @Date: 2019-01-17 11:26
*/
public static Double divide(String str1,String str2) {
BigDecimal bignum1 = new BigDecimal(str1);
BigDecimal bignum2 = new BigDecimal(str2);
//参数意义.bignum1为分母,bignum2为分子,scale保留的位数,BigDecimal.ROUND_DOWN表示不进位
BigDecimal bignum3 = bignum1.divide(bignum2,2,BigDecimal.ROUND_DOWN);
return bignum3.doubleValue();
}
public static void main(String[] args) {
System.out.println(divide("4600.0","0.6"));
}
}
`````` | 565 | 1,737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-05 | latest | en | 0.191377 |
https://www.freemathhelp.com/forum/threads/puzzle-probability-how-many-options.111478/ | 1,566,591,079,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318986.84/warc/CC-MAIN-20190823192831-20190823214831-00237.warc.gz | 833,794,823 | 10,873 | # Puzzle Probability - How many options
#### Dramine
##### New member
Hello fellow Mathematics Members. After finishing school I have spent some time working on mathematic exercises to enhance my knowledge. I have stumbled upon this exercise I cannot solve. If anyone could give me a step by step explanation (leading up to the solution), it would be great. I have spent hours on it already (including scrolling through websites) but I haven't found a clue on how to solve it:
All 4 people have the same empty (not filled) Puzzle with (consisting of) 4 different parts and they now each write their own name on all of their 4 parts. They now give one part to the other person so that in the end everyone has one set with all 4 (different) names on them. Then how many ways are there to do this?
I would really appreciate a fast answer
I wish you all a nice day/night!
Sincerely Dramine
#### lev888
##### Full Member
By puzzle you mean jigsaw puzzle? All pieces are different?
#### Dramine
##### New member
By puzzle you mean jigsaw puzzle? All pieces are different?
Yes exactly. And all four jigsaw parts are different. As far as I know the jigsaw puzzle copy everyone has, should be same though (Because it has to form one jigsaw puzzle when everyone exchanged parts).
#### Jomo
##### Elite Member
I would really appreciate a fast answer
Are you really serious? You want us to the problem for you? On top of that you want it fast. The title of the website is not free math solutions but rather free math help. It is like a game, you submit a problem that you can't do, show the work that you tried to solve the problem and then some tutor will guide you to the correct solution. It really is fun and I truly hope that you give it a try.
I'll even give you a hint. Suppose the four people are A, B, C and D, and the 4 pieces are called w, x, y and z. What might a listing of what the 4 people have look like. If you know what the form of what something should like you might be able to count all the combination.
Consider A has w from A (Aw), Bx, Cx, Dz and B has .... and C has and D has ..... This will yield just one result. | 498 | 2,141 | {"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-2019-35 | latest | en | 0.972768 |
https://www.opticsforhire.com/blog/types-of-projections-in-wide-angle-lenses-part-1/ | 1,709,061,890,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474686.54/warc/CC-MAIN-20240227184934-20240227214934-00773.warc.gz | 894,854,783 | 37,917 | Select Page
# An Overview of Wide-Angle Lens Projections
There are two primary methods of image formation in lens design.
1. Perspective projection (F-Tan Theta lenses also called Rectilinear or Orthoscopic).
2. Equidistant projection (F-Theta lenses also called Equiangular)
The Perspective projection method is most often used during the design of lenses with field of view in the 40-60 degree range. This formation method maintains straight lines in images while stretching space. For some photography types such as satellite imaging, this stretching is acceptable even at wide angles. For many applications however, it is not acceptable. The photo on the right above shows a 130 degree image with distortions.
Equidistant projection is used in wide-angle lenses like a fisheye lens. It bends straight lines but can provide more than 180 degrees of lens FOV (left photo). Moreover, this projection saves angles. This is useful in astronomy photography and many applications on earth.
There are 3 less common methods of image formation or projection these are:
1. Stereographic
2. Equisolid
3. Orthographic projections
Need assistance designing a custom optic or imaging lens ? Learn more about our lens design services here.
The key criteria used to evaluate which methods are best include:
1. Image format at given field of view depending on image construction equation
2. Space distortion
3. Object distortion
4. Light distribution on image
5. Theoretical and achievable lens field of view
1. ## Image format at given field of view depending on image construction equation
### 1.1 Types of projections and image formation equation
?- angle of object FOV
y’ – image height
f – focal length
Next table shows how image size depends on type of projection for given angular FOV
### 1.2 Image size calculation of different types of projection for 120° FOV
Different projections give different image height for given angular FOV (see below for details)
## 2. Space distortion
### 2.1 Definition
Space distortion is defined as the ratio of paraxial value of areas forming by small solid angle at image plane for given angle ω of FOV to the area forming by equal value of solid angle at center FOV (ω=0).
Derivation of values of Space distortion was implemented in [1]
Plots of space distortion depending on the type of projection are shown below.
### 2.7 Summary
– There is image stretching for Perspective, Stereographic and Equidistant projections.
– Distortion of Perspective projection limits FOV to 120-140 deg.
– Stereographic projections can provide maximal angular FOV 210-250 deg.
– Equidistant projections can provide maximal angular FOV 300-350 deg.
-There is compression space for Orthographic projection.
– Orthographic projection has 180 deg. limit of angular FOV.
– There is no space distortion for Equisolid projection.
– Equisolid projection can provide maximal angular FOV up to 360 deg.
## 3. Object distortion
Perspective, Equidistant, Equisolid and Orthographic projections give deformation of shape of small objects through the field of view.
Only stereographic projection saves shape of small objects.
Stereographic projection preserves circles.
Stereographic projection is conformal –preserves angles of intersects of two curves
## 4. Distribution of illumination
### 4.1 Light distribution in perspective projection
Maximal field of view is limited by intensity drop-off at edge of image.
A well known formula for perspective projection describes decreasing light distribution from center to edge is
It can be used for other projections if consider changing size of square in plane of image.
### 4.2 Distribution of illumination for different projections
Ratio of areas for Orthographic projection
Ratio of areas for Equisolid projection
For Equidistant
For Stereographic
– Orthographic projection has perfectly even image illumination.
– Equidistant has a very slow intensity decreasing.
– Equidistant, stereographic, and equi-solid provide very wide FOV because of slow intensity drop.
## 5. Maximal theoretical and feasible FOV
Specific properties of particular projections
## 6. Examples of lens design with different image projections
Below we present examples of lenses for each projection. The lenses with close parameters for each projection were found and then additionally optimized by Zemax to match the image formation equation for each projection.
Angular FOV is 120 deg. for all samples
## 8. Conclusion
• All image formation or projection types can be useful. Which one depends on the application
• Perspective projection is useful to preserve straight lines in an image. The maximal field of view should however be less than 140 degrees. This is widely used in photography lenses and aero photo lenses
• Stereographic projection is useful if preserving the shape of small object on image plane is required. FOV can be more than 180 deg. This is widely used in machine vision systems.
• Equidistant projection is useful if preserving angular sizes of object on image plane is required. FOV can be more than 180 deg. This is widely used in fish eye lenses and astronomical cameras.
• Equisolid projection is useful if preserving constant ration of solid angles in object and image spaces is required. FOV can be more than 180 deg. This is used in scientific photography.
• Orthographic projection is useful if evenness of illumination through the entire image plane is required. FOV can be up to 180 degrees. This is used in cheap cameras and door eye viewers.
## 9. Wide-Angle Lens Projections FAQ
### What are the different types of projections used in wide-angle lenses?
Wide-angle lenses utilize various projections like Perspective, Equidistant, Stereographic, Equisolid, and Orthographic. Each has unique characteristics suitable for different applications, ranging from photography to scientific imaging.
### How does the Perspective projection affect images in wide-angle lenses?
Perspective projection, often used in lenses with a 40-60 degree field of view, maintains straight lines in images but can cause space stretching. It’s typically utilized in photography and aero photo lenses where this distortion is acceptable.
### What are the advantages of Equidistant and Stereographic projections in wide-angle lenses?
Equidistant projection, common in fisheye lenses and astronomical cameras, preserves angular sizes on the image plane and can offer over 180 degrees of field of view. Stereographic projection, used in machine vision systems, maintains the shape of small objects and can also exceed 180 degrees in field of view.
## 10. References of theoretical materials
1. Field of View – Rectilinear and Fishye Lenses
http://www.bobatkins.com/photography/technical/field_of_view.html
2. Margaret M. Fleck. Perspective Projection: The Wrong Imaging Model. 1995, Technical report 95-01
http://mfleck.cs.illinois.edu/my-papers/stereographic-TR.pdf
3. Models for the various classical lens projections
http://michel.thoby.free.fr/Fisheye_history_short/Projections/Models_of_classical_projections.html
4. About the various projections of the photographic objective lenses
http://michel.thoby.free.fr/Fisheye_history_short/Projections/Various_lens_projection.html
## 11. References of lens design
1. U.S.Patent 3661447
2. JP #: 04-267,212
3. Imaging lens and imaging device US 20090009888 A1
4. Wide-Angle Objective. Zeiss #1058 page #0550.
5. JP Patent 4238312
## 12. Related Content
Exploring the World of Fisheye Lenses: Dive into the fascinating realm of fisheye lens design. Learn about the unique challenges and creative possibilities these wide-angle lenses offer.
Transform Your Photography with Lens Attachments: Discover how wide-angle and telephoto lens attachments can revolutionize your photographic experience, offering new perspectives and capabilities. | 1,673 | 7,906 | {"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-2024-10 | longest | en | 0.890798 |
https://www.solutionsfolks.com/ExpertAnswers/in-a-certain-island-of-the-caribbean-there-are-n-cities-numbered-from-1-to-n-for-each-ordered-pa861 | 1,708,528,950,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473518.6/warc/CC-MAIN-20240221134259-20240221164259-00027.warc.gz | 1,027,623,194 | 6,613 | Home / Expert Answers / Computer Science / in-a-certain-island-of-the-caribbean-there-are-n-cities-numbered-from-1-to-n-for-each-ordered-pa861
# (Solved): In a certain island of the Caribbean there are N cities, numbered from 1 to N. For each ordered ...
In a certain island of the Caribbean there are cities, numbered from 1 to . For each ordered pair of cities you know the cost of flying directly from to . In particular, there is a flight between every pair of cities. Each such flight takes one day and flight costs are not necessarily symmetric. Suppose you are in city and you want to get to city . You would like to use this opportunity to obtain frequent flyer status. In order to get the status, you have to travel on at least minDays consecutive days. What is the minimum total of a flight schedule that gets you from to in at least minDays days? Design a dynamic programming algorithm to solve this problem. Assume you can access for any pair in constant time. You are also given and . Hint: one way to solve this problem is using dynamic states similar to those on Bellman-Ford's algorithm. Please answer the following parts: 1. Define the entries of your table in words. E.g. or is ... 2. State a recurrence for the entries of your table in terms of smaller subproblems. Don't forget your base case(s). 3. Write pseudocode for your algorithm to solve this problem. 4. State and analyze the running time of your algorithm.
We have an Answer from Expert | 330 | 1,468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-10 | latest | en | 0.936813 |
http://clay6.com/qa/11251/if-z-1-and-z-2-are-two-complex-numbers-such-that-z-1-z-2-z-1-z-2-then-argz- | 1,480,832,228,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541214.23/warc/CC-MAIN-20161202170901-00424-ip-10-31-129-80.ec2.internal.warc.gz | 50,482,347 | 27,230 | Browse Questions
# If $z_1\:\:and\:\:z_2$ are two complex numbers such that $|z_1+z_2|=|z_1|+|z_2|$, then $argz_1-argz_2=?$
$\begin{array}{1 1}(A) \; \pi\\(B)\;\large\frac{\pi}{2}\\(C)\;-\large\frac{\pi}{2}\\(D)\; 0 \end{array}$
Let $z_1=r_1(cos\theta_1+isin\theta_1)\:\:and\:\:z_2=r_2(cos\theta_2+isin\theta2)$
$\Rightarrow\:z_1+z_2=(r_1cos\theta_1+r_2cos\theta_2)+i(r_1sin\theta_1+sin\theta_2)$
$|z_1|=r_1,\:|z_2|=r_2,\:argz_1=\theta_1,\:argz_2=\theta_2\:\:and$
$|z_1+z_2|=\sqrt {(r_1cos\theta_1+r_2cos\theta_2)^2+(r_1sin\theta_1+r_2sin\theta_2)^2}$
$=\sqrt{r_1^2+r_2^2+2r_1r_2cos(\theta_1-\theta_2)}$
If $|z_1+z_2|=|z_1|+|z_2|,$ then $cos(\theta_1-\theta_2)=1$
$\Rightarrow\:\theta_1-\theta_2=0$
$\Rightarrow\:argz_1-argz_2=0$ | 394 | 732 | {"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} | 3.6875 | 4 | CC-MAIN-2016-50 | longest | en | 0.203153 |
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# Short Questions
## Find y and dy if y x 2 1 when x changes from 3 to 3.02
dy dx
Using differential find , when x 2 2 y 2 16
dx dy
1
Use differential to approximate the value of 31 5 .
x 1
x 2 dx (Example) dx
Evaluate Evaluate
x x 1
1
Evaluate x x 1 dx Evaluate 2 x 3 2 dx
2
1 1 x2
Evaluate
d Evaluate 1 x2 dx
dx
sin
2
Evaluate dx Evaluate x dx
xa xb
1
1 cos x dx tan
2
Evaluate Evaluate x dx
a
Evaluate 1 cos 2x dx Evaluate 2 at b
dt (Example)
1 sin x dx (Example) a
x2
Evaluate Evaluate xdx (Example)
1 dx
Evaluate a x2 2
dx (Example) Evaluate x 2
4 x 13
1 sec2 x
Evaluate
x ln x
dx Evaluate tan x
dx
1 sin
Evaluate 1 x tan 2 1
x
dx Evaluate 1 cos 2
d
x e dx (Example) x sin x dx
x
Evaluate Find
ln x dx tan
1
Find Evaluate x dx
e sin x 2cos x dx x x a dx x a
2x
Evaluate Evaluate (example)
1
em tan x
Evaluate e cos x sin x dx
x
dx Evaluate
1 x2
5x 8 a b x
Evaluate x 3 2 x 1 dx Evaluate x a x b dx
3
x 3x 2 dx (Example)
4
## sec x sec x tan x dx (Example)
3
Evaluate Evaluate
1 0
2 2
x x dx (Example) x 1 dx
2
Evaluate Evaluate
1 1
0 2
1 x
Evaluate x 1
2
2
dx Evaluate
1
x 22
dx
1 2
1 2
Evaluate
1
x x x 1 dx
2
Evaluate ln x dx
(long) 1
6 5
cos d x x 2 1 dx
3
Evaluate Evaluate
0 2
3x 2 2 x 1 x2 2
3 3
Evaluate 2 x 1 x2 1 dx Evaluate 1 x 1 dx
dy
Evaluate x xy 2
dx
## Write two properties of definite integral
Find the area between the x-axis and the curve y x 2 1 from x=1 to x=2
Find the area between the x-axis and the curve y 4 x x 2
Find the area bounded by cos function from x to x
2 2
1
Find the area between the x-axis and the curve y cos x from x to .
2
Find the area between the x-axis and the curve y sin 2 x from x=0 to x
3
dy y 2 1
Solve the differential equation x
dx e
dy
Solve the differential equation sin y cos ecx 1
dx
dy 1 x
Solve the differential equation
dx y
## Solve the differential equation xdy y( x 1)dx 0
Solve the differential equation sec2 x tan ydx sec2 y tan xdy 0
dy
Solve the differential equation 1 cos x tan y 0
dx
dy
Find the general solution of the equation x xy 2 Also find the particular solution if y=1
dx
when x=0.
Long Questions
1. Show that
dx
x a
2 2
ln x x 2 a 2 c (Gujranwala Board 2013)
a2 x x 2
2. Show that a 2 x 2 dx
2
sin 1
a 2
a x 2 c (Gujranwala Board 2012)
dx
3. Evaluate 1 3
sin x cos x
2 2
1 x
4. Evaluate dx
1 x
e x 1 sin x
5. Evaluate 1 cos x
dx
2
6. Evaluate
## 7. Evaluate 4 5x dx (Gujranwala board2016)
2
x4
8. Evaluate x 3
3x 2 4
dx
cos sin
4 4
cos sin
9. Evaluate d or d
0
2 cos 2 0
cos 2 1
4
sec
10. Evaluate sin cos d (Gujranwala Board 2014, 2015)
0
4
cos
4
11. Evaluate t dt (Gujranwala Board 2015)
0
3
## x 3x 2 dx (Example) (Gujranwala Board 2016)
3
12. Evaluate
1
13. Find the area between the x-axis and the curve y 2ax x 2 when a>0
dy dy
14. Solve the differential equation y x 2 y 2 (Gujranwala Board 2012)
dx dx | 1,146 | 2,880 | {"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} | 2.828125 | 3 | CC-MAIN-2019-43 | latest | en | 0.503912 |
http://www.nag.com/numeric/fl/nagdoc_fl23/examples/source/f07ugfe.f90 | 1,438,381,996,000,000,000 | text/plain | crawl-data/CC-MAIN-2015-32/segments/1438042988312.76/warc/CC-MAIN-20150728002308-00239-ip-10-236-191-2.ec2.internal.warc.gz | 605,536,944 | 1,282 | PROGRAM f07ugfe ! F07UGF Example Program Text ! Mark 23 Release. NAG Copyright 2011. ! .. Use Statements .. USE nag_library, ONLY : dtpcon, nag_wp, x02ajf ! .. Implicit None Statement .. IMPLICIT NONE ! .. Parameters .. INTEGER, PARAMETER :: nin = 5, nout = 6 CHARACTER (1), PARAMETER :: diag = 'N', norm = '1' ! .. Local Scalars .. REAL (KIND=nag_wp) :: rcond INTEGER :: i, info, j, n CHARACTER (1) :: uplo ! .. Local Arrays .. REAL (KIND=nag_wp), ALLOCATABLE :: ap(:), work(:) INTEGER, ALLOCATABLE :: iwork(:) ! .. Executable Statements .. WRITE (nout,*) 'F07UGF Example Program Results' ! Skip heading in data file READ (nin,*) READ (nin,*) n ALLOCATE (ap(n*(n+1)/2),work(3*n),iwork(n)) ! Read A from data file READ (nin,*) uplo IF (uplo=='U') THEN READ (nin,*) ((ap(i+j*(j-1)/2),j=i,n),i=1,n) ELSE IF (uplo=='L') THEN READ (nin,*) ((ap(i+(2*n-j)*(j-1)/2),j=1,i),i=1,n) END IF ! Estimate condition number ! The NAG name equivalent of dtpcon is f07ugf CALL dtpcon(norm,uplo,diag,n,ap,rcond,work,iwork,info) WRITE (nout,*) IF (rcond>=x02ajf()) THEN WRITE (nout,99999) 'Estimate of condition number =', & 1.0E0_nag_wp/rcond ELSE WRITE (nout,*) 'A is singular to working precision' END IF 99999 FORMAT (1X,A,1P,E10.2) END PROGRAM f07ugfe | 428 | 1,236 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2015-32 | latest | en | 0.475664 |
http://www.maplesoft.com/support/help/Maple/view.aspx?path=SumTools/Hypergeometric/IsHypergeometricTerm | 1,462,152,208,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860121418.67/warc/CC-MAIN-20160428161521-00057-ip-10-239-7-51.ec2.internal.warc.gz | 654,078,475 | 23,180 | SumTools[Hypergeometric] - Maple Help
Home : Support : Online Help : Mathematics : Discrete Mathematics : Summation and Difference Equations : SumTools : SumTools/Hypergeometric/IsHypergeometricTerm
SumTools[Hypergeometric]
IsHypergeometricTerm
test if a given expression is a hypergeometric term
Calling Sequence IsHypergeometricTerm(H, n, certificate)
Parameters
H - function of n n - variable certificate - (optional) name; assigned the computed certificate
Description
• The IsHypergeometricTerm(H,n) command returns true if $H\left(n\right)$ is a hypergeometric term of n. Otherwise, it returns false.
A function H is hypergeometric of n if $\frac{H\left(n+1\right)}{H\left(n\right)}=R\left(n\right)$, a rational function of n. $R\left(n\right)$ is the certificate of $H\left(n\right)$. If the third optional argument is included, it is assigned the certificate $R\left(n\right)$.
Examples
> $\mathrm{with}\left(\mathrm{SumTools}[\mathrm{Hypergeometric}]\right):$
> $H≔\frac{\left({n}^{2}-1\right)\left(3n+1\right)!}{\left(n+3\right)!\left(2n+7\right)!}$
${H}{:=}\frac{\left({{n}}^{{2}}{-}{1}\right){}\left({3}{}{n}{+}{1}\right){!}}{\left({n}{+}{3}\right){!}{}\left({2}{}{n}{+}{7}\right){!}}$ (1)
> $\mathrm{IsHypergeometricTerm}\left(H,n,'\mathrm{certificate}'\right)$
${\mathrm{true}}$ (2)
> $\mathrm{certificate}$
$\frac{{3}}{{2}}{}\frac{\left({3}{}{n}{+}{2}\right){}\left({3}{}{n}{+}{4}\right){}{n}{}\left({n}{+}{2}\right)}{\left({n}{-}{1}\right){}\left({2}{}{n}{+}{9}\right){}{\left({n}{+}{4}\right)}^{{2}}}$ (3) | 529 | 1,545 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 12, "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-2016-18 | longest | en | 0.426349 |
http://math.stackexchange.com/users/37833/stumbleine75?tab=activity | 1,454,988,912,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701156448.92/warc/CC-MAIN-20160205193916-00045-ip-10-236-182-209.ec2.internal.warc.gz | 144,013,858 | 12,191 | Stumbleine75
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Jul 13 accepted Question from book 'Indra's Pearls' about limit set arising from infinite words (compositions of maps) Jul 12 asked Question from book 'Indra's Pearls' about limit set arising from infinite words (compositions of maps) Jan 7 accepted Proving a function is chaotic on an interval Jan 6 comment Proving a function is chaotic on an interval I hope I'm interpreting your answer correctly. So you utilize a representation scheme for numbers in $[0, 1]$ to create a mapping between each $x$ in $[0, 1]$ and a sequence $s$. From this you deduce that the $n$th number in the sequence for the $k$th iteration of seed $x$ is $k$th number in the sequence of $x$ itself times the $n + k$th number of $x$'s sequence. Why is that deduction true? Jan 6 revised Proving a function is chaotic on an interval added 3 characters in body Jan 6 asked Proving a function is chaotic on an interval Dec 11 awarded Popular Question Sep 24 awarded Autobiographer Sep 19 accepted Solving $\log(x) = vx^α$ for $x$ via Lambert W function Sep 19 awarded Commentator Sep 19 comment Solving $\log(x) = vx^α$ for $x$ via Lambert W function Yeah I see it now, and feel bad for missing it. $y$ as you defined it completes the identity and then it is trivial to manipulate it into the answer you gave. Now, can I ask how you knew to define y the way you did? Sep 19 comment Solving $\log(x) = vx^α$ for $x$ via Lambert W function How do you get to $ye^y = -v\alpha$? Does it have to do with $W(z)e^{W(z)} = z$? Sep 19 revised Solving $\log(x) = vx^α$ for $x$ via Lambert W function edited title Sep 19 asked Solving $\log(x) = vx^α$ for $x$ via Lambert W function Aug 15 awarded Scholar Aug 15 accepted The operation $\langle a, b\rangle = \{\{∅, a\}, \{\{∅\}, b\}\}$ creates distinct sets from distinct pairs $(a,b)$ Aug 14 revised The operation $\langle a, b\rangle = \{\{∅, a\}, \{\{∅\}, b\}\}$ creates distinct sets from distinct pairs $(a,b)$ clarification Aug 14 awarded Supporter Aug 14 asked The operation $\langle a, b\rangle = \{\{∅, a\}, \{\{∅\}, b\}\}$ creates distinct sets from distinct pairs $(a,b)$ Mar 13 revised Help understanding this 'Fractal' I've just made? added 3 characters in body | 634 | 2,261 | {"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} | 3.734375 | 4 | CC-MAIN-2016-07 | latest | en | 0.874558 |
https://codegolf.stackexchange.com/questions/249107/ro1000an-nu1000era50-en100o501ng | 1,716,185,602,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058222.5/warc/CC-MAIN-20240520045803-20240520075803-00600.warc.gz | 148,016,650 | 55,387 | # ro1000an nu1000era50 en100o501ng
In the hovertext of this xkcd:
There's an "encoding" based on replacing runs of roman numerals with their sums. To do this:
• Find all runs of at least 1 roman numeral (IVXLCDM)
• Replace them with their sums, as numbers
The Roman Numerals are IVXLCDM, corresponding respectively to 1, 5, 10, 50, 100, 500, 1000. You should match them in either case.
For example, if we take encoding, c is the roman numeral 100, so it gets replaced with 100, leaving en100oding. Then, di are both roman numerals, respectively 500 and 1, so it gets replaced with their sum, 501, leaving en100o501ng.
Your challenge is to implement this, given a string. This is , shortest wins!
Note: the xkcd implements some more complex rules regarding consecutive characters like iv. This can create some ambiguities so I'm going with this simpler version.
## Testcases
xkcd -> 10k600
encodingish -> en100o501ng1sh
Hello, World! -> He100o, Wor550!
Potatoes are green -> Potatoes are green
Divide -> 1007e
bacillicidic -> ba904
• Do we have to handle either case, or can we assume, eg, all lowercase? Jun 27, 2022 at 2:49
• @Jonah I'm gonna say either case Jun 27, 2022 at 3:50
• based on the same XKCD: Not-Roman-Numeral Addition Jun 27, 2022 at 13:22
• That's Numberwang! Jun 27, 2022 at 14:12
• Shouldn't xkcd be 10k400 ?
– vsz
Jun 28, 2022 at 5:35
# Vyxals, 9 bytes
ƛøṘ∨;⁽-ḊṠ
ƛøṘ∨;⁽-ḊṠ
ƛ # Map, and for each character:
øṘ # Convert from roman numeral to a number, or 0 if invalid
∨ # Logical OR with the character: if it's 0, then use the character instead
; # Close map
Ḋ # Group results where the same result appears consecutively...
⁽- # ...with subtracting it by itself. Subtracting a number by itself is 0,
# but subtracting a string by itself is "". These are two different values, so it works as a type check.
Ṡ # Sum each. Summing a list of characters will turn it into a string, but lists of numbers will simply sum.
# The s flag joins the list together into a string before outputting.
• Wow this is clever. Jun 27, 2022 at 0:26
# Factor + grouping.extras math.unicode roman, 100 99 bytes
[ 1 group [ [ roman> ] try ] map [ real? ] group-by values [ [ Σ >dec ] try ] map flatten concat ]
Needs modern Factor for >dec. Here's a version that runs on TIO for 3 more bytes, though: Try it online!
#### How?
This answer relies on try in order to process heterogeneous arrays relatively succinctly, as roman> blows up on non- roman numbers and sum blows up on arrays of strings.
! "encodingish"
1 group ! { "e" "n" "c" "o" "d" "i" "n" "g" "i" "s" "h" }
[ [ roman> ] try ] map ! { "e" "n" 100 "o" 500 1 "n" "g" 1 "s" "h" }
[ real? ] group-by values ! { V{ "e" "n" } V{ 100 } V{ "o" } V{ 500 1 } V{ "n" "g" } V{ 1 } V{ "s" "h" } }
[ [ Σ >dec ] try ] map ! { V{ "e" "n" } "100" V{ "o" } "501" V{ "n" "g" } "1" V{ "s" "h" } }
flatten ! { "e" "n" "100" "o" "501" "n" "g" "1" "s" "h" }
concat ! "en100o501ng1sh"
• "Language: Factor + the kitchen sink". This is why I love Factor. What other language randomly comes with a "to Roman" function built-in in some standard library somewhere? Jun 27, 2022 at 22:08
• @SilvioMayolo A lot of golfing languages do. But they include it because they anticipate Roman numerals in golf questions. I think Factor has it because there is a small, but enthusiastic group of people who like writing stuff in Factor. Incidentally, the roman vocabulary is touted as being a good introductory example of Factor code, as it is well-written and includes many language features such as metaprogramming. Jun 27, 2022 at 22:21
# JavaScript (Node.js), 91 bytes
s=>s.replace(/[ivxlcdm]+/gi,s=>Buffer(s).map(c=>t-=~[9,4,,99,499,49,999][c%32%24%7],t=0)|t)
Try it online!
# R, 146 bytes
\(s){!=is.na
T=tapply
r=as.roman(t<-el(strsplit(s,"")))
a=T(r,i<-rep(seq(a=x<-rle(!r)$l),x),sum) a[!a]<-T(t,i,g<-\(x)Reduce(paste0,x))[!a] g(a)} Attempt This Online! Definitely not the best tool for the job, despite built-in roman numerals. ### Explanation outline: 1. Split string to characters (stored in t; this part could be omitted if we took input as a vector of characters). 2. Convert to type roman - successful for Roman digits, otherwise NA (stored in r). 3. Build index i for spitting on consecutive runs of NAs or not NAs. 4. Tapply (split+sapply) sum on Roman numerals. 5. Fill NAs with chunks from original string. 6. Collapse vector to one string. # Ret1na 0.8.2, 45 bytes i(m dd d 5$*c
c
ll
l
5$*x x vv v 5$*i
i+
$.& Try 1t on51ne! 51nk 1n150u500es test 100ases. E10p50anat1on: i( Run the who50e s100r1pt 100ase-1nsens1t6e50y. m dd 1000 = 2 * 500. d 5$*c
500 = 5 * 100.
c
ll
100 = 2 * 50.
l
5$*x 50 = 5 * 10. x vv 10 = 2 * 5. v 5$*i
5 = 5 * 1.
i+
\$.&
100on5ert to 500e1101a50.
• It took me a while to figure out 500e1101a50 :P Jun 27, 2022 at 8:44
• @emanresuA Too much decimal?
– Neil
Jun 27, 2022 at 8:45
# 05AB1E, 21 bytes
.γu.vd}εDSu.v©àdi®O]J
Or alternatively, with I/O as character array:
u.vøεΣa}н}.γd}ε¤diO]S
Can definitely be golfed a bit. But despite having a Roman Number builtin, the functionality of some builtins are nowhere near as versatile as the Vyxal answer.. Why you might ask?
1. The Roman Number builtin only works on uppercase letters (probably a bug..), so preserving the casing costs quite a few bytes.
2. Sum given an error on letters in the new 05AB1E version, so explicit checks before summing are unfortunately necessary.
Explanation:
.γ # Adjacent group by the (implicit) input-string:
u # Convert to uppercase
.v # Convert from letter to Roman Number
d # Check if this is a (non-negative) integer
}ε # After the adjacent group-by: map over each group:
D # Duplicate the current group
S # Convert it to a list of characters
u # Uppercase each
.v # Convert it to a Roman Number
# (it will remain an uppercase character if invalid)
© # Store this list in variable ® (without popping)
àdi # If this list contains an integer:
®O # Push and sum list ®
] # Close the if-statement and map
J # Join everything together to a single string
# (which is output implicitly as result)
u # Uppercase each character in the (implicit) input-list
.v # Convert each to a Roman Number
# (it will remain an uppercase character if invalid)
ø # Pair each with the characters of (implicit) input-list
ε # Map over each pair:
Σ # (Stable) sort the pair by:
a # Check if it's a letter (0 if number; 1 if letter)
}н # After the sort-by: pop and push the first character
}.γ # After the map: adjacent group by:
d # Is it a (non-negative) integer
}ε # After the adjacent group-by: map over each group:
¤ # Push the first item (without popping the group list)
di # If it's a (non-negative) integer:
O # Sum the group together
] # Close the if-statement and map
S # Convert the numbers and remaining groups to a flattened list
# of characters
# (which is output implicitly as result)
# Charcoal, 52 bytes
≔⁰θＦ⊞Ｏ⪪Ｓ¹ω«≔⌕⪪IVXLCDM¹↥ιη¿⊕η≧⁺×Ｘχ÷η²⊕×⁴﹪η²θ«⁺∨θωι≔⁰θ
Try it online! Link is to verbose version of code. Explanation:
≔⁰θ
Start with a running total of 0.
Ｆ⊞Ｏ⪪Ｓ¹ω«
Loop over the input characters, but include a trailing empty string.
≔⌕⪪IVXLCDM¹↥ιη
Get the index of the uppercase character in the list of Roman numerals.
¿⊕η
If this is a Roman numeral, then...
≧⁺×Ｘχ÷η²⊕×⁴﹪η²θ«
... use the index to convert it to decimal and add it to the running total, otherwise:
⁺∨θωι
Prefix the running total, if any, to the character before printing it.
≔⁰θ
Clear the running total.
# Python, 130122 117 bytes
-8 bytes by porting Arnauld's formula
-5 bytes thanks to movatica
lambda s:re.sub('[ivxlcdm]+',lambda M:str(sum(1+[9,4,0,99,499,49,999][ord(i)%32%24%7]for i in M[0])),s,0,2)
import re
Try it online!
Matches runs of roman numerals and replaces them with their sum. The 2 at the end is equivalent to re.IGNORECASE.
• M[0] instead of M.group() for 117 bytes Jun 30, 2022 at 14:11
• @movatica Nice! Thanks Jun 30, 2022 at 14:51
# C (gcc), 150159152137133 126 bytes
• +9 to handle either case
• -7 thanks to emanresu A
• -22 thanks to ceilingcat
• -4 thanks to Neil
Processes a string, looking for Roman numerals. If it finds one, the value is added to the total consecutive run. Once a non-numeral is found, the total is printed if it is greater than zero (and the total is reset), then the non-numeral is printed. Finally, the total is printed if the end of string is encountered in case the final character was a Roman numeral.
char*n="ivxlcdm",*d;t;f(char*s){for(t=0;t=!(d=index(n,*s|32))?t&&printf("%d",t),*s&&!putchar(*s):t+L"\1\5\n2dǴϨ"[d-n],*s++;);}
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• You can save 7 bytes by or with 32 and loweracse. Jun 28, 2022 at 1:54
• 135 bytes but feels like it should be shorter.
– Neil
Jun 30, 2022 at 0:08 | 2,950 | 9,103 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-22 | latest | en | 0.86376 |
https://communities.sas.com/t5/SAS-Procedures/how-to-get-a-when-statement-to-work-with-negative-numbers/td-p/201279?nobounce | 1,532,225,455,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592875.98/warc/CC-MAIN-20180722002753-20180722022753-00139.warc.gz | 627,255,909 | 30,750 | ## how to get a when statement to work with negative numbers
Solved
Regular Contributor
Posts: 240
# how to get a when statement to work with negative numbers
a HI I have an if statement
data ms;
set ms;
actual = datedif (date,today(), 'act/act');
put actual= ;
gets me number of days from date to today date...
data ms;
set ms;
select;
when (actual=.) copy= " ";
when (actual <0 or actual > -15 ) Copy "pre_15";
when ( actual >=1 or actual <=90) copy " post_90";
end;
run;
I Want to Id what's within 15 days an that number is a negative reason its a -15...
i Want to ID post 90 days I'm using today date to get the datediff and date. Thanks
Message was edited by: gilbert arredondo
Accepted Solutions
Solution
08-17-2015 11:26 PM
Super User
Posts: 23,776
## Re: how to get if statement to work with negative numbers
You need AND not OR In your conditions.
As Chris has indicated you need an equal in your assignment statement for the copy variable.
All Replies
PROC Star
Posts: 2,370
## Re: how to get if statement to work with negative numbers
1- There is no IF in your code; what with the title?
2- The copy keyword is invalid syntax as used in your code
3- The last sentences make little sense
If you want to be helped the least you can do is present a properly explained and presented case.
Regular Contributor
Posts: 240
## Re: how to get if statement to work with negative numbers
it was an if statement I change it before I asked the question....Chris if it doesn't make sense to you need to respond I try to make it as clear as possible
Solution
08-17-2015 11:26 PM
Super User
Posts: 23,776
## Re: how to get if statement to work with negative numbers
You need AND not OR In your conditions.
As Chris has indicated you need an equal in your assignment statement for the copy variable.
Regular Contributor
Posts: 240
## Re: how to get if statement to work with negative numbers
Thanks Reeza ... I will take your suggestion...It a snippe of code that works for me on other code but it's character not number
Regular Contributor
Posts: 240
## Re: how to get if statement to work with negative numbers
HI reeza
i Got it
data table;
set table;
if actual ge 0 and actual Le 90 then copy = "post_90";
thanks
🔒 This topic is solved and locked. | 591 | 2,298 | {"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-2018-30 | latest | en | 0.880553 |
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