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https://www.physicsforums.com/threads/f-x-integral-from-x-to-x-2.317111/
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F(x) = integral from x to x^2
1. May 30, 2009
intelli
1. The problem statement, all variables and given/known data
f(x) = integral from x to x^2
t^2 dt
find f ' (x) = ?
find f '(5)= ?
2. Relevant equations
3. The attempt at a solution
Break the integral in 2: \int_x^{x^2} t^2 dt = \int_x^{0} t^2 dt +\int_0^{x^2} t^2 dt = -\int_0^{x} t^2 dt + \int_0^{x^2} t^2 dt
Then take the derivative of both integrals using the FTC.
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. May 30, 2009
slider142
Re: integral
You don't even need to integrate the expression if you already learned the fundamental theorem of calculus (just be careful with your variables!). Where are you having difficulty?
3. May 30, 2009
intelli
Re: integral
This was my ans and it says that it is wrong
f ' (x) = -x^3/3+x^6/3
f ' ( 5 ) = 5166.666667
i also put
f ' ( x) = -x^2 + x ^ 4 and that was also wrong
4. May 30, 2009
djeitnstine
Re: integral
What are you using that says it is wrong? Is it a book? A program? Because sometimes a program requires input a special way.
5. May 30, 2009
intelli
Re: integral
it is a program online i put the top ans on the program right but it says that both are incorrect ans for the f'(x) and f'(5)
6. May 30, 2009
djeitnstine
Re: integral
Such as (x^6)/3-(x^3)/3
and 5166.7
etc... I don't know exactly what they want, but you should experiment to see what format they accept.
7. May 30, 2009
Cyosis
Re: integral
This is not true, f(x) is defined as $f(x)=\int_x^{x^2} t^2 dt=-x^3/3+x^6/3$. Now the ' means that you have to differentiate to x.
8. May 30, 2009
djeitnstine
Re: integral
good looking I didn't even see the '
9. May 30, 2009
slider142
Re: integral
That is indeed incorrect. You are giving f(x), while the question asks for f'(x). Once again, note that there is no need to integrate the expression. Use the fundamental theorem of calculus.
10. May 30, 2009
intelli
Re: integral
thanks alot guys i figured it out
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# PHYSICS
posted by .
Calculate the radial acceleration of an object on the ground at the earth's equator in m/s2, turning with the planet.
• PHYSICS -
9.8 m/s^2?
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https://excelxor.com/2014/09/28/countifs-multiple-or-criteria-for-one-or-two-criteria_ranges/
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COUNTIFS: Multiple “OR” criteria for one or two criteria_Ranges 67
In this post I would like to clear up what appears to me to be a rather widespread misunderstanding of how COUNTIFS/SUMIFS operate, in particular when we pass arrays consisting of more than one element as the Criteria to one or even two of the Criteria_Ranges.
This latter technique is used when the criteria in question are to be considered as “OR” criteria, which is not to be confused with cases where we wish the criteria passed to be calculated rather as “AND” critieria.
For example, given the following data:
we can use COUNTIFS in various ways and with various syntaxes to obtain a variety of results. For example, starting with some basics:
`=COUNTIFS(B2:B14,"Male",C2:C14,"Sea lion")`
which gives quite simply the number of rows for which both the entry in column B is “Male” and the entry in the corresponding row in column C is “Sea lion”. And:
`=COUNTIFS(B2:B14,"Female",C2:C14,"Sea lion")`
gives precisely the same but for cases where the entry in column B is “Female”.
Without too much difficulty, we can simply add the above, viz:
`=COUNTIFS(B2:B14,"Male",C2:C14,"Sea lion")+COUNTIFS(B2:B14,"Female",C2:C14,"Sea lion")`
to give, not surprisingly, the number of rows for which the entry in column B is either “Male” or “Female” and the entry in the corresponding row in column C is “Sea lion”.
An alternative, shorthand way of obtaining precisely the same result is:
`=SUM(COUNTIFS(B2:B14,{"Male","Female"},C2:C14,"Sea lion"))`
Things get a little more interesting when we apply more than one OR criteria, i.e. to the second criteria_range also:
`=SUM(COUNTIFS(B2:B14,{"Male","Female"},C2:C14,{"Sea lion","Mite"}))`
and you might be surprised to see that the result for this construction (2) is actually considerably less than that for the previous one (5), even though, with precisely the same three criteria as in the previous formula plus one extra, we might be tempted to imagine that, if anything, it would be greater.
What this formula is actually giving is the number of rows for which either the entry in column B is “Male” and the entry in the corresponding row in column C is “Sea lion” or the entry in column B is “Female” and the entry in the corresponding row in column C is “Mite”.
From the table you can see that only rows 12 and 14 (“Male” and “Sea lion”) satisfy this set of conditions. There are no rows corresponding to “Female” and “Mite”. The 3 “Female”/”Sea lion” pairs, which accounted for more than 50% of the total in the previous formula, here do not contribute a thing.
The important thing to note in this formula is that each of the elements in the two array constants corresponds to the equivalent element in the other: “Male” with “Sea lion” and “Female” with “Mite”. The formula says nothing about the possibility of “Male” in column B and “Mite” in column C being an option to consider, nor so for “Female” in column A and “Sea lion” in column C.
But what if we actually want these “cross-elements” to be considered, and counted? How do we get a total corresponding to “all possible ORs”, i.e. where column B can be either “Male” or “Female” and column C either “Sea lion” or “Mite”, i.e. the 7 results in the table (which you can easily verify)?
Well, it may surprise you to know that all we need do is to make one minute change to the previous formula and we have our result, viz:
`=SUM(COUNTIFS(B2:B14,{"Male","Female"},C2:C14,{"Sea lion";"Mite"}))`
where all I’ve done is to change the comma in one of the array constants to a semi-colon.
Or, equivalently:
`=SUM(COUNTIFS(B2:B14,{"Male";"Female"},C2:C14,{"Sea lion","Mite"}))`
where all I’ve done this time is to swap the comma and semi-colon in the two array constants.
The point here being that one of the array constants is a single-column array, the other a single-row array. (In case you didn’t know, in Excel commas and semi-colons represent column- and row-separators respectively.)
By doing this you effectively coerce Excel into forming a “two-dimensional” set of returns of all possible combinations for those two sets of criteria. Essentially, Excel is now “seeing” the calculation as a series of separate calculations, equivalent if you like to us filling in by hand the two-dimensional grid below:
and then summing up the totals for those four possibilities.
The interesting – and slightly worrying – thing that I’ve noticed is just how little understood is the logic behind this syntax. It is quite understandable that students of Excel, upon encountering this technique for the first time, are naturally wont to investigate the possibility of extending this method to scenarios involving multiple “OR” criteria for more than two criteria_ranges.
For example, we have no problem in calculating:
`=SUM(COUNTIFS(B2:B14,{"Male","Female"},C2:C14,{"Sea lion";"Mite"},D2:D14,"Basketball"))`
that is, the number of rows where column B is either “Male” or “Female”, column C is either “Sea lion” or “Mite” and column D is “Basketball” (answer: 1).
Now let’s see what happens if we attempt to extend the criteria for column D to more than a single entry, e.g.:
`=SUM(COUNTIFS(B2:B14,{"Male","Female"},C2:C14,{"Sea lion";"Mite"},D2:D14,{"Basketball","Genealogy","Roleplaying"}))`
which, looking at the table, we might well hope to return the five rows as below:
However, the result of the above formula is not 5, but 2.
Perhaps we got our syntax wrong? Perhaps the elements in that new array we’ve just added should be separated by semi-colons, not commas? So you try:
`=SUM(COUNTIFS(B2:B14,{"Male","Female"},C2:C14,{"Sea lion";"Mite"},D2:D14,{"Basketball";"Genealogy";"Roleplaying"}))`
and are perhaps even more dismayed to see that the result is not only not any better than the last one, but not even the same. In fact, the above formula will return 0.
Why is this happening?
The first thing I would like to state for the record is that, no matter how much juggling about of commas and semi-colons we try, unlike in the previous example we are never (at least not using COUNTIFS; SUMPRODUCT will handle it, but that’s a different story) going to be able to create, for this construction, an expression which returns a total corresponding to “all possible ORs”, i.e. where column B is either “Male” or “Female”, column C either “Sea lion” or “Mite” and column D either “Basketball”, “Genealogy” or “Roleplaying”. Never.
But why?
Well, the basic answer boils down to how this function operates, a question of “dimensionality”, if you like. In effect, in order to “construct” a grid such as the one I gave above for the previous case, in which we laid out the two sets of OR criteria as if they were the x and y axes of a Cartesian plot, each vertex corresponding to a pair of those points being the result for that pair, in this case we would effectively have to have a “cuboid” of results, each point in this three-dimensional plane corresponding to the number of rows satisfying those criteria.
Imagine extending our previous two-dimensional grid so that it becomes just such a three-dimensional cuboid, with the “z-axis” now consisting of our three new elements – “Basketball”, “Genealogy”, “Roleplaying” – as points on that axis. Theoretically, we could now populate this 3-D grid with values at each of the 12 (our plot would essentially be a 2x2x3 “cuboid” in space) vertices corresponding to the total satisfying each of the criteria on the corresponding x, y and z axes.
Unfortunately we are restricted to a mere two dimensions, and so such fantasies will remain just that (though this function can operate in a pseudo-three-dimensional fashion, as for example in constructions involving multiple worksheets).
So if neither of the above attempts were of valid syntax, why didn’t they simply return an error? Where do the results which they return (2 and 0) come from?
Let’s take a look at the first attempt again:
`=SUM(COUNTIFS(B2:B14,{"Male","Female"},C2:C14,{"Sea lion";"Mite"},D2:D14,{"Basketball","Genealogy","Roleplaying"}))`
It’s always useful to go through any formula (especially in cases where it is not giving the expected results) using the Evaluate Formula tool. Doing so here reveals that the above resolves to:
`=SUM({0,1,0;1,0,0})`
which, if we look closely, is evidently an array consisting of two rows and three columns.
But from where has this array been derived?
The key thing to remember is the point I made about “pairing” of elements earlier. When two (or more) arrays are of the same “vector-type”, i.e. either both are single-column arrays or both are single-row arrays, then Excel will pair corresponding conditions from each.
Hence, in the above construction, since the first – {“Male”,”Female”} – and third – {“Basketball”,”Genealogy”,”Roleplaying”} – criteria arrays are both single-row arrays, Excel pairs up the elements, so that, just as we saw before, these pairs will each now form an “AND” condition: the first equating to asking how many rows there are for which the entry in column B is “Male” and the corresponding entry in column D is “Basketball”; the second to asking how many rows there are for which the entry in column B is “Female” and the corresponding entry in column D is “Genealogy”.
Note that the other array in this construction – {“Sea lion”;”Mite”} – which was “transposed” to become a single-column array as was required in the earlier construction, is still a valid syntax here, in the sense that it will allow us to form a 2-dimensional array of results. It’s just that the other array with which we are going to “cross-calculate” this one first needs to be resolved from the other two arrays of the same vector-type.
But what about the third element in that third array – “Roleplaying”? Which is the “corresponding” element in the first array to which that is paired?
The answer is that there isn’t one. However, rather than ignore that criterion altogether, Excel nevertheless goes ahead and constructs the appropriately-sized array to house the expected returns, i.e. the two-row by three-column array that we can see when using Evaluate Formula. In effect, faced with two (or more) arrays of differing dimensions, Excel’s means of resolving this syntactical conflict is to artificially increase the smaller of the two so that it is then of a dimension equal to the larger of the pair.
The fact it that does this, and then proceeds to fill in these newly-created, extra spaces with zeroes, rather than flat-out error, is perhaps not such a bad thing: indeed, there may well be situations where we knowingly accept that one of our arrays is being “re-dimensioned”, even filled with errors, providing that we then manipulate the resulting array as befits our needs.
With this in mind we can now give the precise equivalent expression corresponding to each of the six elements in the array we saw generated above, i.e. {0,1,0;1,0,0}, whose breakdown would look like this (observe my attempt to represent there being no third element to pair with “Roleplaying”):
This time two of our three sets of criteria (those having the same vector-type) were first resolved into pairs before then “cross-calculating” with the third.
Let’s have a look at why our second attempt, in which we tried in vain to transpose that final array, returned zero:
`=SUM(COUNTIFS(B2:B14,{"Male","Female"},C2:C14,{"Sea lion";"Mite"},D2:D14,{"Basketball";"Genealogy";"Roleplaying"}))`
Although the overall result is 0, so that we know that all of the contributing calculations to that total are also 0, it’s nevertheless worth looking at the resulting array using Evaluate Formula, if nothing else than so we can verify that its dimensionality is what we would expect:
`=SUM({0,0;0,0;0,0})`
This time we have a 3-row-by-2-column array, again consisting of a total of six entries. And in this case our two arrays which are of the same vector-type are the second – {“Sea lion”;”Mite”}, and third – {“Basketball”;”Genealogy”;”Roleplaying”}, both being single-column arrays.
Hence, as we’ve seen, “pairing” them up will give us “Sea lion”/”Basketball”, “Mite”/”Genealogy” and “???”/”Roleplaying”.
These three pairs are then “cross-calculated” with the two elements in the first array – {“Male”,”Female”} – so that our “2-D grid of returns” would this time look like:
(totals which you may wish to manually verify via filtering the original table) and clearly the sum of six zeroes is zero.
To conclude, although this article is set out principally to discuss the operation and syntax of the SUMIFS/COUNTIFS family of functions, it might be argued that the more important lesson to be (hopefully) learnt is one of a deeper appreciation of the underlying means by which such constructions are calculated.
Indeed, if you were to ask me what I think is the one aspect of Excel which I would say is lacking in, and which hinders the progress of, the average Excel user, I would have to say that it’s this very inability (or unwillingness) to grasp the fundamental nature of how Excel (which, remember, is nothing if not a “two-dimensional” programme) operates at this base level.
There are countless Excellers out there (and some of them of very high status) whose work clearly demonstrates an understanding of the standard formula techniques: they use FREQUENCY with apparent ease; they formulate long, complex array formulas; they memorize (and some of them regurgitate) the old, time-tested solutions (sometimes I think if I see another MIN(FIND({0,1,2,3,4,5,6,7,8,9,0}… construction I’ll scream…) and they are adept at employing the vast majority of Excel functions in a variety of situations.
But you won’t see many MMULTs coming from these same people. And you won’t see many non-standard, innovative manipulations of arrays either (after MMULT, perhaps TRANSPOSE would be top of the list for least-employed – and least-understood – function around). But these two functions share – at heart – the same fundamental aspect: that is, they elevate us, by their very nature, to working with constructions in which we are manipulating two-dimensional arrays. Not those which we can see, as is the case with worksheet ranges, but those which reside only within the depths of Excel’s intermediate calculation-chain, and which are transitory.
And it’s just precisely at this time when an appreciation of the underlying means by which we can manipulate, query and even re-dimension these two-dimensional array constructions is of most importance. Yet how many of us have such an appreciation of those fundamentals? Just why do so few people employ MMULT in their solutions?
Understand how Excel “sees” things and you’ll understand Excel far better!
1. Pingback: Sumifs with or?
2. lakshya says:
Is there any solution to applying this
`=SUM(COUNTIFS(B2:B14,{"Male","Female"},C2:C14,{"Sea lion";"Mite"},D2:D14,{"Basketball";"Genealogy";"Roleplaying"}))`
3. @lakshya
Hi and welcome to the site!
Can you clarify what you mean? What exactly do you wish to return a count for?
Regards
4. Iggy says:
Great write up and description. I was able to apply the usage to one of my sheets but had a question.
I’m able to use or criteria in a statement that I manually wrote
`=SUM(COUNTIFS(table1[range1],{"type1";"type2"},...`
I want to be able to use a selector so the users have options:
– type3
– type4
– type 1 or type 2
However, when I use a reference cell to point to the argument for type1 and type2, it doesn’t work.
`=SUM(COUNTIFS(table1[range1],{\$A\$3},...`
A3 points to a data validation consisting of:
type3
type4
{“type1″;”type2”}
Do you have any suggestions?
5. @Iggy
Very kind of you to say so. Thanks!
Re your query, can you clarify precisely what is in the data validation list? Do you really have the third entry as you give?
Regards
6. Chris says:
So how would you return the list of all possible “OR”s for the scenario above where
`=SUM(COUNTIFS(B2:B14,{"Male","Female"},C2:C14,{"Sea lion";"Mite"},D2:D14,{"Basketball";"Genealogy";"Roleplaying"}))`
fails?
7. Hi Chris.
In such cases you need to switch to SUMPRODUCT:
`=SUMPRODUCT(0+(ISNUMBER(MATCH(B2:B14,{"Male","Female"},0))),0+(ISNUMBER(MATCH(C2:C14,{"Sea lion","Mite"},0))),0+ISNUMBER(MATCH(D2:D14,{"Basketball","Genealogy","Roleplaying"},0)))`
Regards
8. Daniel S. says:
Thank you for the amazing article. I’m trying to apply it to my scenario. Here is a section of my table:
Col1 || Col 2 || Col 3
0 || 1 || 0
1 || 2 || 3
0 || 0 || 0
3 || 1 || 1
etc.
It is 3 columns, and each column can contain either 0, 1, 2, or 3. I would like to return the count of the number of rows for which any of the columns have the number 1.
Any way you could help out?
Thanks!
9. Khalid NGO says:
Hi XOR LX,
Good Day Sir,
Very interesting article, I didn’t knew much about SUMIFS before, this indeed great help, Thank you.
Just wondering why it needs to be wrapped with SUM?
Regards,
Khalid
10. Hi Khalid.
When you pass an array of values as one or more of the criteria ranges to COUNTIF(S)/SUMIF(S), you are effectively doing the equivalent of performing as many individual versions of those functions, one for each of the criteria.
But of course these individual totals then need to be summed to give an overall result.
For example, the construction:
`=COUNTIF(A1:A10,"A")+COUNTIF(A1:A10,"B")`
can be replaced with the shorter:
`=SUM(COUNTIF(A1:A10,{"A","B"}))`
but we cannot simply use it without the SUM, i.e.:,
`=COUNTIF(A1:A10,{"A","B"})`
because, although this technically returns an array of two values, one for the count of “A” within the range and one for the count of “B”, e.g.:
`{3,5}`
we still need to sum those elements to arrive at a total number of entries which are either “A” or “B”, i.e. 8 (=3+5).
Regards
11. Khalid NGO says:
Hi XOR LX,
Thank you so much.
Everything is clear now.
Thanks for the step by step examples.
Highly appreciated.
Take Care.
12. Andrew says:
Hello.
Is it therefore impossible to count each element of the A1:A6 array if it is greater than each element of the B1:B6 array. e.g. is A1>B1 if yes +1, is A2>B2 if yes +1 etc.
Andrew
13. Hi Andrew.
In such row-for-row comparisons we generally need to switch to SUMPRODUCT:
`=SUMPRODUCT(0+(A1:A6>B1:B6))`
Or, if it is not certain that all of the entries in those ranges are numerical, more rigorously:
`=SUMPRODUCT(0+(A1:A6>B1:B6),0+ISNUMBER(A1:A6),0+ISNUMBER(B1:B6))`
Regards
14. AKB says:
Thanks for the detailed explanation.
I tried the {“Male”,”Female”} formula in google spreadsheets but unfortunately it does not seem to be working.
Doesn’t the Google sheets support this?
If not, is there a way around?
15. @AKB
Hi and welcome to the site!
I’m very sorry but I’m afraid I don’t have much experience of Google Sheets. In fact, the solutions I propose here are not also independently verified for working with that program.
Perhaps someone who has a bit more knowledge of Google Sheets than I will pick up on your comment.
Regards
16. hey man im working on something similar but im a bit stuck. was wondering if you can help me because this is similar.
here is my dilemma:
im trying to do the criteria match with the 2 criterias, however for the second criteria i want the home score to be great than the vis. score. (with the first being the team name at home but ive got that sorted it just keeps coming up the value thing when i try complete the rest, here is what i have so far –
`=COUNTIFS(Table2[HOME TEAM],"stuttgart",Table2,Table2[HOME SCORE]>Table2[VIS. SCORE])`
break that down :
=COUNTIFS first criteria (Table2[HOME TEAM],”stuttgart”
this is fine so far… but when i add the next bit.
Table2,Table2[HOME SCORE]>Table2[VIS. SCORE])
what is it exactly im doing wrong with it lol.
17. @puntingpal
Hi and welcome to the site!
Unfortunately you need to switch to SUMPRODUCT in such cases, viz:
`=SUMPRODUCT(0+(Table2[HOME TEAM]="stuttgart"),0+(Table2[HOME SCORE]>Table2[VIS. SCORE]))`
Regards
18. That’s an awesome reply and that worked a treat the above as I just tried it myself. So in the case that you want to record only when it is equal to the vis. Score what would u put in place of the > ???
19. As in I want to record how many times the two team draw?? And also, if I have like 120 different teams to record and the formula bar is heaps complicated it won’t auto correct an calculate the rest of the teams an I have to go thru an change them individually, is there a quicker way to achieve all this for me??
20. Just change the “>” to an “=” for draws.
Why don’t you put all your teams, e.g. “Stuttgart”, in a list somewhere, e.g. A1:A120, and then use, in B1:
`=SUMPRODUCT(0+(Table2[HOME TEAM]=A1),0+(Table2[HOME SCORE]=Table2[VIS. SCORE]))`
which will give you the number of times Stuttgart drew and, when copied down to B2, B3, etc., will give you the number of times that the teams in A2, A3, etc. drew.
Regards
21. I Hate Excel says:
Hello,
I want to count each cell only once, regardless of how many of the criteria words it contains.
Is this possible? I am using:
`=SUM(COUNTIFS('All Data'!\$F:\$F,A2,'All Data'!\$H:\$H,{"*apples*";"*pears*"}))`
For example, if a cell contains both ‘apples’ and ‘pears’ I want it to only be counted once in the overall total.
22. @I Hate Excel
Perhaps the most straightforward solution is simply to subtract the number of entries for which the column H value contains both “apples” and “pears” from your existing formula, i.e.:
`=SUM(COUNTIFS('All Data'!\$F:\$F,A2,'All Data'!\$H:\$H,{"*apples*";"*pears*"}))-COUNTIFS('All Data'!\$F:\$F,A2,'All Data'!\$H:\$H,"*apples*",'All Data'!\$H:\$H,"*pears*")`
This can also be done via a single construction, viz:
`=SUM(COUNTIFS('All Data'!\$F:\$F,A2,'All Data'!\$H:\$H,{"*apples*";"*pears*";"*apples*"},'All Data'!\$H:\$H,{"<>*pears*";"<>*apples*";"*pears*"}))`
though the latter is arguably less intuitive than the former.
Hope that helps!
Regards
23. Alda says:
This was amazingly thorough and helpful. Thank you!
24. @Alda
Glad to hear it! And you’re very welcome!
25. Jisha says:
Nice article.
I have got a scenario wherein I could not figure how to do it. Heres the sample data.
A || B || C || D
106 || 112 || 112 || 924
197 || 209 || 209 || 171
495 || 527 || 421 || 421
0 || 0 || 0 || 0
665 || 707 || 707 || 707
749 || 797 || 797 || 797
749 || 0 || 0 || 0
I want to get the count of all A which are greater B or C or D. The answer here would be 3 (row 2, row 3 and last row).
I tried the sumproduct, but that would work only if the field of comparison is only two columns. Here I have 3 columns to compare to.
How should I proceed?
26. @Jisha
Hi and welcome to the site!
The simplest solution would be to use an additional column within the worksheet.
For example, using column E for this purpose, and assuming the rows to be considered are from row 1 to row 10, enter this formula in E1:
`=MIN(A1:D1)`
and copy down to E10.
Your required counting formula is then simply:
`=SUMPRODUCT(0+(A1:A10>E1:E10))`
If for whatever reason you are unable (or unwilling) to use an additional column, you would need the slightly more complex:
`=SUMPRODUCT(0+(SUBTOTAL(5,OFFSET(A1:D1,ROW(A1:D10)-MIN(ROW(A1:D10)),))<>A1:A10))`
Hope that helps!
Regards
27. Jisha says:
Thank you very much.. I do prefer the second formula, but unfortunately could not decipher it. 😦
28. I’ll break it down for you 🙂
If we use a 5 as the first parameter within SUBTOTAL, we instruct that function to return the minimum from a given range. In fact, we can actually pass more than one range to SUBTOTAL and it will return the minimum from each of those ranges.
Basically, we are going to determine the minimum from each of the ranges:
A1:D1
A2:D2
A3:D3
A10:D10
and then compare each of these minimums with the corresponding row entry from column A.
In order to create an ‘array of ranges’ to pass to SUBTOTAL, we must here use OFFSET. And we start with passing the range A1:D1 as the first parameter to that function, and then ‘offsetting’ it by a number of rows of our choosing.
Just as an illustration, something like:
`OFFSET(A1:D1,1,0)`
would return a reference to the range:
A2:D2
since we have ‘offset’ the range A1:D1 by a single row.
Similarly:
`OFFSET(A1:D1,3,2)`
would return a reference to the range:
C4:F4
since this time we have ‘offset’ the range A1:D1 by three rows (downwards) and two columns (to the right).
However, we are not restricted to passing just single values to the rows and cols parameters of OFFSET; we can also pass multiple values (arrays) all at once.
This is what I did here, since this part:
`ROW(A1:D10)-MIN(ROW(A1:D10))`
which I am passing as OFFSET’s rows parameter, generates not just a single value, but many. Specifically, the above resolves to the array of values:
`{0;1;2;3;4;5;6;7;8;9}`
And so the construction:
`OFFSET(A1:D1,ROW(A1:D10)-MIN(ROW(A1:D10)),)`
is here generating an ‘array’ of range references, equivalent to offsetting the range A1:D1 by 0, 1, 2, …, 9 rows. As such, the result will look like:
`{A1:D1;A2:D2;A3:D3;A4:D4;A5:D5;A6:D6;A7:D7;A8:D8;A9:D9;A10:D10}`
We then ask SUBTOTAL, using 5 as its first parameter, to tell us the minimum value from each of these ranges, so that:
`SUBTOTAL(5,OFFSET(A1:D1,ROW(A1:D10)-MIN(ROW(A1:D10)),))`
would give us (assuming some random numbers in those ranges):
`{95;49;300;1;71;383;99;35;209;339}`
We then finally compare these values to those in column A, as required.
Hope that helps!
29. Jisha says:
Thank you very much for the clear explanation. 🙂
Wondering if it needs to be this complicated.. 😀
30. Hi,
I seem to be stuck on a formula and was wondering if you might be able to help.
I’m using countifs to filter multiple criteria ( 5 columns) to get my results. I am ok up to this point i get the count i am looking for. however, I have duplicate data in column F and need the count of only the unique occurrence. Is there a way to count teh value in the column only once after the other perimeters have been established?
Criteria range has been setup in name manager
example
Column_C =
`=OFFSET(data!\$C\$1,0,0,COUNTA(data!\$C:\$C),1)`
here is my formula
`=COUNTIFS(Column_C,A33,Column_E,A34,Column_I,""&D34,Column_V,""&1)`
31. @Leo
Hi and welcome to the site!
You say that there are five columns to consider, though I can’t see any reference to a column F in your formula. Can you clarify what the criterion for this column is?
Regards
32. Hi, thanks for the reply.
Column F i left out because i can’t seem to figure out.
also I think i left out column T out, it should have been
`=COUNTIFS(Column_C,A33,Column_E,A34,Column_I,""&D34,Column_V,""&1)`
with the above formula I get a count of remaining lines, and its correct, but i would like to take it a step further and get count of unique employee ID from column F
Here’s a breakdown
Column C has multiple division number: 1018001,1018002,1018003
Column E pay area: CE,CL,SM
Column F employee numbers will have duplicates: 110215, 243979, 77132
Column I work days: 5/1/16, 5/2/16, 5/3/16
Column T day time was reported
Column V day time was approved
so an example of what i’m trying to do
get a count when C = 1018001, E=CE, I <5/24/16, T 5/30/16
after all above filters get count of unique employee number from F
33. In that case you need to use an array formula** (see instructions at the foot of this comment), i.e.:
`=SUM(IF(FREQUENCY(IF(Column_C=Column_C_Your_Value,IF(Column_E=Column_E_Your_Value,IF(Column_I=Column_I_Your_Value,IF(Column_T=Column_T_Your_Value,IF(Column_V=Column_V_Your_Value,IF(Column_F<>"",MATCH(Column_F,Column_F,0))))))),ROW(Column_F)-MIN(ROW(Column_F))+1),1))`
I’m afraid I couldn’t work out quite what your criteria were supposed to be, so I’ve just used generic names, e.g. Column_T_Your_Value, which obviously you should replace as required (including the equals signs where appropriate, e.g. to <, >, etc.).
Unlike with e.g. COUNTIF(S)/SUMIF(S), with such a formula it’s extremely important to keep the ranges passed to a minimum, or else you’ll start to notice Excel struggling to calculate. Looks like you’ve made a good start to this effect by defining your ranges. Just make sure that all these defined ranges, for columns C, I, T, etc., are precisely the same size.
Hope that helps!
**Array formulas are not entered in the same way as ‘standard’ formulas. Instead of pressing just ENTER, you first hold down CTRL and SHIFT, and only then press ENTER. If you’ve done it correctly, you’ll notice Excel puts curly brackets {} around the formula (though do not attempt to manually insert these yourself).
34. Thank you for all your help
i had actually started exploring sum/frequency as a means to accomplish what i wanted, I just had a different syntax.
Thanks again!!
35. You’re more than welcome!
Cheers!
36. I must concur that this is an excellent article. I especially like the theoretical discussion at the end, even though I have no idea of what MMULT does or is.
37. @Job
Many thanks for your kind words. 🙂
I may create a post specifically for MMULT here at some point in the near future – it’s a function which seems to remain a bit of a mystery to many Excel users!
38. San New to Excel says:
Great Article.
Is there anyway i can use the or against a cell reference?
`=SUM(COUNTIFS(B2:B14,{"Male","Female"},C2:C14,{"Sea lion","Mite"}))`
So the above becomes
`=SUM(COUNTIFS(B2:B14,{"Male","Female"},C2:C14,{" F1","F2"}))`
where F1 can and F2 have the information of the pet, which means its more flexible and can be quickly changed.
39. @San New to Excel
Re your query, the answer is yes, though when the criteria are not static, in-formula values, but rather contained within actual worksheet ranges, then we must switch from SUM to SUMPRODUCT in order to give the necessary coercion, i.e.:
`=SUMPRODUCT(COUNTIFS(B2:B14,{"Male","Female"},C2:C14,F1:F2))`
By the way, I took it from the example you gave (and also assuming you read the full article), in which both criteria arrays are of the same vector displacement (i.e. not orthogonal to each other), that you are interested in obtaining a count for just the two combinations Male/Sea Lion and Female/Mite, and not all four, i.e. Male/Sea Lion, Male/Mite, Female/Mite and Female/Sea Lion.
Regards
40. Tom Prouc says:
Hello
I read with big interest your page and comments but still unable to adapt to my search and to write the correct one for
In the table bellow I try to solve the following request: to find by month the number of PO with backorder and the number of PO shipped partially
PO n° Ordered Shipped Bakcorder Order Date Delivery Date
120250 3 2 1 8/2/2016 8/10/2016
120783 2 2 8/3/2016 8/10/2016
120783 1 0 1 8/3/2016
121997 5 5 8/8/2016 8/12/2016
122001 2 2 8/15/2016 8/22/2016
122001 1 0 1 8/15/2016
122005 3 3 8/29/2016 9/5/2016
122005 2 2 8/29/2016 9/5/2016
122005 4 4 8/29/2016 9/5/2016
122020 2 2 9/5/2016 9/12/2016
122058 8 2 6 9/10/2016 9/12/2016
122082 2 2 9/12/2016 9/25/2016
122095 8 2 6 9/21/2016 10/5/2016
122095 8 1 7 9/21/2016 10/5/2016
122095 3 3 9/21/2016 10/5/2016
122095 1 1 9/21/2016 10/5/2016
122297 10 2 8 10/2/2016 10/5/2016
122297 1 1 10/2/2016 10/5/2016
Result A)
Lines Count August September October
Lines order 9 7 2
lines shipped 4 6 6
Lines backorder 3 3 1
for lines order I used the following formula
`=COUNT(IF(MONTH(E2:E19)=MONTH(B22),E2:E19))`
for lines shipped
`=COUNT(IF(MONTH(F2:F19)=MONTH(B22),F2:F19))`
Lines backorder
`=COUNTIFS(\$D2:\$D19,">0",\$E2:\$E19,">="&B31,\$E2:\$E19,<="&EOMONTH(B22,0))`
Result B)
PO count August September October
PO 5 4 1
PO shipped 4 4 2
O bakordered Formula to find
PO shipped full Formula to find
for PO count I used the following formula
PO:
`=COUNT(1/FREQUENCY(IF(MONTH(\$E\$2:\$E\$19)=MONTH(B29),\$A\$2:\$A\$19),\$A\$2:\$A\$19))`
PO shipped:
`=COUNT(1/FREQUENCY(IF(MONTH(\$F\$2:\$F\$19)=MONTH(B29),\$A\$2:\$A\$19),\$A\$2:\$A\$19))`
Can you help me? Thank you in adavance
41. Hi Tom,
I’d gladly help you here, but could I just ask you to first confirm which of the formulas you’ve posted does not give the desired results? Also, can you confirm what the expected results for that dataset should be in each case?
Regards
42. Tom Prouc says:
Hello
Rereading my post I realize I was not clear giving information.
All formulas in result A (from previous post) work well. I can obtain the number of lines ordered / shipped / backorder on a specific period
Formulas in result B give info to obtain the number of PO and PO shipped on a period working find too.
The complexity is that I try to obtain the number of orders forms (col A is you copy table in excel sheet) having at least one backorder line (col D) over a given period (col E) is calculated over a full month August, September etc..
Attention, the order form (PO) may appear several times, the 120783 appears 2 times but must be counted only as 1, the 122005 appears 3 time buts must be counted as 1 and soon
The number to obtain for PO backordered is 3 for august, 2 for September, 1 for October
Thank you
43. So unless I’m missing something obvious you could simply add an additional clause to one of your existing formulas, viz:
`=COUNT(1/FREQUENCY(IF(MONTH(\$E\$2:\$E\$19)=MONTH(B29),IF(\$D\$2:\$D\$19<>"",\$A\$2:\$A\$19)),\$A\$2:\$A\$19))`
Regards
44. Tom Prouc says:
The formula works well. Like always, focusing on “countifs” formula, forgetting to go back to the basis, using the “If” formula. As was saying one of my past bosses: why make it easy when you can do it complicated…
Thank you
45. Wise words indeed! And you’re welcome!
Cheers
46. Casey says:
Hello! You had helped someone above (@I HATE EXCEL) with a similar scenario but I cant figure out how to make it work for my case. I have a number of possible “tags” that could be in a cell (about 30 in total). Many of them fall under the same category so I need to be able to count any that match my criteria (#TAG1 or #TAG2 or #TAG3) but only count each row once. An example might be the cell might contain someone favorite foods as “apples, oranges” and I want to count each response only once under the category of “Fruit”
`=SUM(COUNTIFS('Sheet1'!\$F:\$F,\$A6,'Sheet1'!\$J:\$J,{"*#TAG1*";"*#TAG2*";"*#TAG3*"}))`
The idea of subtracting out those cells containing both works when the tag list is short but when I have 10 possible tags (lemons, pears, apples, oranges, grapes, etc.) I am trying to account for, I would need to subtract out every possible combination of the grouped tags.
Any thoughts on how else I might count these cells only once in this scenario?
47. @Casey
Yes, you need a slightly different set-up in such cases. See my post here, for example:
You would need something like the following array formula**, where I have assumed that the list of criteria (#TAG1, #TAG2 and #TAG3) are in cells E1;E3:
`=SUM(IF(Sheet1!\$F1:\$F20=\$A6,IF(MMULT(0+ISNUMBER(SEARCH(TRANSPOSE(E1:E3),Sheet1!\$J1:\$J20)),ROW(E1:E3)^0)>0,1)))`
Note also that, since we no longer have the option of using COUNTIFS, we must be careful to restrict the range being passed (and certainly referencing entire columns is no longer an option). Hence my choice of an upper row reference of 20 here, which obviously you can change, though be sure to keep it as small as possible.
If you’d prefer to list the criteria in-formula, as opposed to referencing a range (E1:E3) containing those criteria, then let me know.
And remember that array formulas need confirming with CTRL+SHIFT+ENTER, not just ENTER.
Regards
48. Casey says:
This is very interesting stuff. Thank you for the guidance! If time permits, I would be interested to learn how you include the criteria in the formula as opposed to a range. Thanks again!
49. Ok, a bit more unwieldy, but in that case you would need:
`=SUM(IF(Sheet1!\$F1:\$F20=\$A6,IF(MMULT(0+ISNUMBER(SEARCH({"#TAG1","#TAG2","#TAG3"},Sheet1!\$J1:\$J20)),ROW(INDEX(A:A,1):INDEX(A:A,COUNTA({"#TAG1","#TAG2","#TAG3"})))^0)>0,1)))`
Amend the entries in the part:
`{"#TAG1","#TAG2","#TAG3"}`
as required.
Regards
50. Stefan Oana says:
Dear XOR,
I am trying to find a solution to a problem that I think you solved but I cannot find on your blog.
If you can please give me a hit.
I have to columns:
A with projects, B with amounts.
In the projects column, in every cell could be stored more than 1 project by this form:
#2011|#2015|#2009
In the B columns are stored the amounts (if there are 3 projects, will be 3 amounts, and so on)
145|165|198.
What I need help with is finding the answer to the question:
What is the amount per #2015 project.
Thank you
51. Hello,
I am as many people here are struggling with a formula.
I found lots of useful things on this website, but start getting lost and not managing to get what i need out of formula and i am sure the fix is very easy, but when you are a bit of a novice with excel its hard to see. So please help me to figure this out.
I have two columns which have numbers in them 1-5. I want to count number of lines/units which will have 3 and above in either or both of the columns and count it once. So if i have 3 in one or both columns count as 1 and count it. I have tried different ways using your web and last one i have is:
`=SUM(COUNTIFS(G2:G68,{">=3"},H2:H68,{">=3"}))`
but it only counts where both conditions met i think…as if i count manually the number is greater. i have tried adding countifs within sum, then it count all cells containing 3+.
Many thanks
Juliana
52. Firas says:
I would like to count if only 2 columns meets a criteria whereas other columns do not meet it (need to exclude the person who also meet the criteria in other columns)
53. nirvick says:
can anyone say me how to count VD001 if its written like VD001, VD003, VD001, VD001 in one row. its all in one line
54. @Stefan
So are you wanting to report a total for just one row? Or for multiple rows? You only give one row’s worth of data, so it’s a little difficult to understand your requirement here, I’m afraid.
Can you clarify?
Regards
55. @Juliana
See my article here.
You can continue to use COUNTIFS here, though it’s arguably a touch convoluted (the logic, however, is sound):
`=SUM(COUNTIFS(G2:G68,{">=",">=","<"}&3,H2:H68,{">=","<",">="}&3))`
More easily understandable (and extendible) is the switch to SUMPRODUCT:
`=SUMPRODUCT(0+((G2:G68>=3)+(H2:H68>=3)>0))`
Regards
56. @nirvick
Your question is not clear to me, I’m afraid. Can you clarify with an example plus expected result?
Regards
57. @Firas
Can you clarify your question with an example or two?
Regards
58. EJ says:
Excellent Explanation …
59. @EJ
Regards
60. ben says:
Hi,
Thanks for your articles, I find them really useful. However i’m struggling to convert the following countifs formula to something that works.
`=SUM(COUNTIFS(data!\$D:\$D,{10144;11440},data!\$P:\$P,{"C","U"},data!\$J:\$J,{"O","Z"}))`
I’m trying to return the number of rows that have 10144 or 11440 in column D, that have a C or U in column P and an O or Z in column J.
61. @ben
See the part in this article in which I give an explanation as to why this set-up with COUNTIF(S) fails for scenarios involving multiple “OR” criteria for more than two criteria_ranges.
In such cases we must switch to SUMPRODUCT, though of course this means that using entire column references is not at all a good idea (hence my choice of an upper row limit of 1000 in the below), viz:
`=SUMPRODUCT(0+ISNUMBER(MATCH(data!\$D1:\$D1000,{10144;11440},0)),0+ISNUMBER(MATCH(data!\$P1:\$P1000,{"C","U"},0)),0+ISNUMBER(MATCH(data!\$J1:\$J1000,{"O","Z"},0)))`
Regards
62. ben says:
thats great. thanks for your help
63. You’re welcome!
64. Ger says:
XOR LX – great tip… thank you.. honestly never knew about specifying an “OR” in the countifs.
My “OR” criteria was a range in itself, where I have used it simply to count the occurences of all values from one Column in another column. Works great, but strangely I needed to enter it is an array formula… but it still worked a treat. For example-
`=SUM(COUNTIFS(B4:B16,C4:C16))`
Thanks for the tip.
65. Ger says:
Also, equivalently…. from your tips above…
`=SUM(0+ISNUMBER(MATCH(C4:C16,B4:B16,0)))`
again array entered.
Everyday is a school day 🙂
Thanks.
Ger
66. Thanks, Ger.
Yes, I perhaps should’ve mentioned a case or two in which the criteria are contained within an actual worksheet range, rather than in-formula.
And you’re correct that, in such cases, SUM alone (i.e. without CSE) is not sufficiently strong to coerce the array, whereas it is when the criteria are hard-coded within the formula. FYI SUMPRODUCT (without CSE) will also work in your case.
Cheers
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##### Concentration and pH relationship
Chemistry Tutor: None Selected Time limit: 1 Day
Calculate the concentration of OH- ion in a .15 M solution of NH3. The base dissociation constant for ammonia is 1.8x 10^-5 .
May 21st, 2015
NH3+H2O = NH4+ + OH-
equilibruim concentratio of NH3 is 0.15 - x.
NH4+ and OH- concentrations are equal to x.
Kb = [NH4+][OH - ]/[NH3] = x^2/0.15
x = sqrt(1.8x 10^-5 * 0.15) = 0.0016M
answer is [OH - ] = 0.0016M
May 21st, 2015
...
May 21st, 2015
...
May 21st, 2015
May 30th, 2017
check_circle
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h a l f b a k e r y
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5-days/week calendar
Change the length of the week to five days
(+9, -1) [vote for, against]
Part of the problem with calendar reform is the number we have to work with -- 365 and some-odd hundredths -- doesn't really divide nicely. Even taking 365 by itself, we are looking at factors of five and 73.
73 might be an ugly factor, but 72 works nicely. With five days per week and six weeks (30 days) per month, we will have twelve months with exactly one week left over which we can celebrate as "year's end." This cycle works without the need for a yearly "world day."
For a complete calendar, we need a leap day every four years except in years divisible by 128 -- which makes it more accurate than either the Gregorian or Mayan calendars. Call this leap day "election day," and make it a national holiday set aside for voting.
Benefits of this calendar include more weekends per year (72 vs. 52, plus "year's end"), a shorter work week (two days off per five day week is nearly equivalent to the four day work week proposed elsewhere), and closer matching to seasons, especially if we start the year at the autumnal equinox. In addition, you could use the same calendar every year, it's easier to count by 5's than by 7's, businesses can easily divide the year into halves, thirds, quarters, etc. (hard to do with 13 months) -- even the months can be split into halves and thirds for payment purposes.
Plus, you would get a cool rhyme to remember things by:
Thirty days hath September,
And every month that you remember.
On Year's End, across the nation
Every four years, you might note,
— mrouse, Mar 30 2002
Croissant for the terribly clever rhyme inclusion. One trifle: half of the senators and congressmen wouldn't benefit from election day.
— RayfordSteele, Mar 30 2002
This would play excellently in the sweepstakes community since every country in the world would be electing their leader, just imagine how may bets you could make when the entire world goes out to vote. Well, about 47% of the world. Damn non-voters.
— [ sctld ], Mar 30 2002
Thursday, Friday, Saturday, Sunday, Monday
Rinse and Repeat
Monday included as some holidays are moved to Monday in order to have 3-day weekends starting Saturday. Thursday included for the occasional 4-day weekend ending Sunday. Now, if we could only get a 3-day weekend to follow a 4-day weekend to Rinse and Repeat...
— thumbwax, Mar 30 2002
How about Manditory menstral month. All women must adjust their 'Menses'to occur in march only. This Frees men to watch sports and drink brewskis without interuption or snide remarks for 51 weeks a year, while providing a one week 'target' time line to get together with other guys for fishing/hunting/racing/poker/drinking and uninterrupted joke telling. This week to be called MENSES DAY or, MARCH MADNESS.
— bluto, May 14 2003
reminds me of an idea a friend of mine had years ago, called "the metric calendar".. i forget exactly how it worked, but something like 10 months with 35 days each and having some 15 day festival at the end of each year. a croissant for the memories.
— insertnamehere, Jan 17 2004
I found another pattern (almost):
One and forty in September Another six are for November
Take one and over two bears May Then February, ten allay
March is found by adding four To August, increment by score
July takes half a dozen under Twice increment and thrice asunder
Gives January, then by four Decrements to October sure
Another ten December yields Add five and halve is June revealed
The prime calendar is thus concluded By Arpil, which has ten denuded
A single day is unaccounted So at the end it shall be mounted
And lastly note that each four years
A day before July appears.
— Detly, May 29 2004
you're all mad (or you were)
— neilp, Dec 16 2004
[annotate]
back: main index
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This calculator can be applied to the design of optical systems in which it is necessary to understand the behavior and direction of light as it changes mediums of propagation. Depending on the available inputs, this calculator can solve for the refractive index of a medium, the light’s angle of incidence, or its angle of refraction (Figure 1).
### Snell’s Law Equation:
Snell’s Law is given by the following equation:
\$\$n_{1}sin(theta_{1}) = n_{2}sin(theta_{2})\$\$
Where:
n1 = Refractive index of the incident medium
n2 = Refractive index of the refractive medium
θ1 = Angle of incidence relative to the surface’s normal
θ2 = Angle of refraction relative to the surface’s normal
The angles are limited to:
\$\$ 0 lt theta lt frac{pi}{2} \$\$ radians
or
\$\$ 0 lt theta lt 90 \$\$ degrees
##### Table 1. Input limits for the angles of refraction.
Greater Than Less Than Units
0 90 Degrees
0 0.25 Turns
0 5400 Minutes of Arc
0 324000 Seconds of Arc
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# Printable 2 Digit By 1 Digit Multiplication Worksheets
2 x 1 digit products to 999 e g. Free multiplication worksheet 2 digit by 1 digit 3rd grade and 4th grade math review and practice 2 digit by 1 digit multiplication with this free printable worksheets for kids.
Worksheet 3 Digit Multiplication With 1 Digit Multiplier Printable Multiplication Worksheets 4th Grade Math Worksheets Multiplication Practice
### Displaying top 8 worksheets found for multiplication 2 digits x 1 digit.
Printable 2 digit by 1 digit multiplication worksheets. Use the buttons below to print open or download the pdf version of the multiplying 2 digit by 1 digit numbers a math worksheet. There are several variants of each class of worksheet to allow for plenty of practice. 2 digit x 1 digit x2 x5 x10 related resources the various resources listed below are aligned to the same standard 4nbt05 taken from the ccsm common core standards for mathematics as the multiplication worksheet shown above.
These worksheets start gently with multiplication of smaller two digit numbers by a single digit and gradually progress upwards to two digit by two digit multiplication and three digit by three digit multiplication. Creative multiplication 6 page with guidance notes and examples 2 x 1 digit products to 99 e g. With real life examples like the one above makes it super easy for kids to comprehend the concept of multiplication.
3 digits times 1 digit. 2 x 1 digit e g. Here you ll discover a large selection of games worksheets and task cards for practicing 2 digit by 1 digit multiplication.
2 digit x 1 digit e g. Use these worksheets as an instant tool to introduce multiplication of 2 digit numbers by 1 digit numbers. 15 x 5 20 x 8.
It can also be used as an assessment or quiz. Multiplication word problems 2 digit by 1 digit displaying top 8 worksheets found for this concept. 512 x 7 requires only 2x 5x or 10x multiplication facts.
The size of the pdf file is 26333 bytes. 54 x 5 requires only 2x 5x or 10x multiplication facts. 3 x 1 digit e g.
Multiplication 2 digits x 1 digit. Learners apply their knowledge of place value to solve the questions in this set of worksheets. On this page you ll have a large selection of worksheets and games for multiplying 3 digit by 1 digit numbers.
Some of the worksheets for this concept are tion 1 2 digit multiplication 1 multiplication word problems multiplication word problems 1 multiplication long multiplication work multiplying 2 digit by 1 grade 4 multiplication work sequence for teaching multi digit multiplication. Some of the worksheets for this concept are long multiplication work multiplying 2 digit by 1 multiplication multiplication grade 5 multiplication work two digit multiplication work grade 4 multiplication work single digit multiplication 1 name date grade. Preview images of the first and second if there is one pages are shown.
We have math riddle worksheets a multi digit multiplication dice game task cards horizontal and vertical problem worksheets and much more. 2 digits times 1 digit. Multiplication worksheets 2 digit times 1 digit on this page you ll find a variety of printables for teaching 2 digit by 1 digit multiplication.
This provides great extra practice for 3rd grade and 4th grade students.
The Multiplying 2 Digit By 1 Digit Numbers Large Print A Math Worksheet From Multiplication Worksheets Free Math Worksheets Math Multiplication Worksheets
Multiplying A 2 Digit Number By A 1 Digit Number C Math Worksheets Multiplication Worksheets Printable Math Worksheets
Multiplying A 2 Digit Number By A 1 Digit Number Large Print F Long Multiplication Worksheet Multiplication Worksheets Multiplication Math Worksheets
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Multiplication Worksheet 2 Digit Times 1 Digit 1 Kidspressmagazine Com Math Addition Worksheets Multiplication Worksheets Mathematics Worksheets
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Pin By Kimmyann On Homework Helpers In 2020 Multiplication Worksheets Third Grade Multiplication Worksheets Multiplication
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# Pattern finding of last few digit of a^b
The question is:Find the last two digits of 9^2013.(No calculators allowed to solve this question.)
Options: (A)01 (B)29 (C)41 (D)81 (E)89
From this point,I will show my working,
9^2=81(Last Digit 1) 9^3=1x9=9 (Now only focusing on ONLY last digit) 9^4=9(Last Digit of the previous answer)x9(The next power)=1 (Last Digit)
Noticing a pattern that as the powers increases,the pattern goes 9,1,9,1,9,1,9,1,9 so the repeated pattern is 1 and 9.So I took
2013÷2=1006R(Remainder)1,where R1 is the number 9.So we know the digit is 9.
This eliminates the option (A),(C) and (D),leaving with (B) and (E)
I attempted to find the 2md last digit number,but to no avail as the number increases,no pattern can be formed.Any ideas how?(The solution doesn't have to use concept of patterns)
Edited working error.
• Hint: $9^{10}\equiv 01\pmod{100}$. And $9^3\equiv 29\pmod{100}$. Conclusion left to the reader ... – String Mar 23 '15 at 16:05
• @String Oh sh*t,I haven't learned mod or modulus yet,out of my syllabus.I'll learn it myself now :P – ministic2001 Mar 23 '15 at 16:10
• OK, all you need to know is that the two last digits of $9^n$ repeats each time $n$ is increased by $10$. – String Mar 23 '15 at 16:14
You can do the same thing:
$$\underline{09}\to81\to29\to61\to49\to41\to69\to21\to89\to01\to\underline{09}\to\ldots$$
The cycle length is $10$, and $9^{2013}=9^{10\cdot201+3}$, so ...
Alternatively, you can notice that $9^{10}$ ends in $01$, and the product of numbers ending in $01$ ends in $01$, so $9^{2010}=\left(9^{10}\right)^{201}$ ends in $01$.
• Mysterious thing is,I can take 9^2010 and the last 2 digit would be 09.And now were left with 9^3.So therefore,9^2010x9^3=09×29=261,as the last 2 digit to be 61.But the thing is,there isn't an opitoon called "61" and second of all,odd powers last digit should end with the digit "9". – ministic2001 Mar 26 '15 at 7:29
• @ministic2001: No, the last two digits of $9^{2010}$ are $01$; you’re off by $1$ in the cycle. – Brian M. Scott Mar 26 '15 at 7:31
• Ahh right,its 01.So the answer to 9^2013 should be 9^10^201 x 9^3=01×29=29 as the last 2 digits rights? – ministic2001 Mar 26 '15 at 7:35
• @ministic2001: That’s right. – Brian M. Scott Mar 26 '15 at 7:35
• Thank you really much Brian! – ministic2001 Mar 26 '15 at 7:36
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What is the use of platinum foil in the hydrogen electrode? from Chemistry Electrochemistry Class 12 Nagaland Board
## Book Store
Currently only available for.
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## Previous Year Papers
Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for.
`Class 10` `Class 12`
What is the use of platinum foil in the hydrogen electrode?
It is used for inflow and outflow of electrons.
155 Views
How would you determine the standard electrode potential of the system Mg2+/Mg?
Use standard hydrogen electrode as anode and Mg2+ | Mg as a cathode we can measure the standard electrodepotential of systemMg2+ | Mg. Standard hydrogen electrode, represented by Pt(s), H2(g) (1 atm) | H+ (aq) and dip the electrode of Magnesium wire in a 1M MgSO4 solution .The standard hydrogen electrode is always zero.
Use formula
Eocell = Eo right – Eoleft
The standard hydrogen electrode is always zero.
So that the value of
Eoleft =0
Hence
Eocell = Eo Mg|Mg2+
Or
Eo Mg|Mg2+= Eocell
2203 Views
Calculate the emf of the cell in which the following reaction takes place:
Ni(s) + 2Ag+ (0.002 M) $\to$ Ni2+ (0.160 M) + 2Ag(s)
Given that
$\mathrm{Ni}\left(\mathrm{s}\right)+2{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)\to {\mathrm{Ni}}^{2+}\left(\mathrm{aq}\right)+2\mathrm{As}\left(\mathrm{s}\right)$
or
The equation is also written as
or
= 1.05 V – 0.0295 x log 80
= 1.05 V – 0.0295 x 1.9031
= 1.05 V – 0.056 = 0.99 V.
1585 Views
Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
For hydrogen electrode, ,
given that pH = 10
use formula [H+] = 10–pH
so that [H+] = 10−10 M
Electrode reaction will
H+ + e – →1/2 H2
Use the formula
1279 Views
Can you store copper sulphate solutions in a Zinc pot?
No. Because zinc is more reactive than copper and thus holes will be developed in zinc pot.
Cu2+(aq) + Zn(s) → Zn2+ (aq) + Cu(s)
1384 Views
Consult the table of standard electrode potentials and suggest three substance that can oxidize ferrous ions under suitable conditions.
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# Surface Area and Volume BINGO Game!
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Also included in:
1. Bingo Geometry Vocabulary Bingo Games (Growing) Bundle!In this Money Saving (Growing) Bundleyou will find my vocabulary Bingo games for a high school Geometry class (or an advanced middle school class). Each zip file contains:2 sets of games cards (30 cards per set) Set One: each card has 24 diffe
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Product Description
Surface Area and Volume Geometry Surface Area and Volume BINGO Game!
This is a Bingo game made to review, reinforce and illustrate the 6 three-dimensional solids, some of the parts and all of the surface area and volume formulas. There are at least 4 different ways to play!!! The files are PDF versions
The zip file includes:
2 sets of games cards (30 cards per set)
Set One: each card has 24 different geometric diagrams/formulas scrambled and a free space
Set Two: each card has 24 different geometric terms/formulas scrambled and a free space
2 Sets of Clue Cards:
Set One: includes a table that has the term and its definition so that you can call the term and have the students match it to their diagram or call the definition and have the students match it to their diagram
Set Two: includes the diagrams blown up six to page so that you can display the pictures so that they students can do picture recognition to match it to their term. (I have also read the definitions from Clue Set One and had them match it to their term)
A Teacher Description page that includes the above information and a set of cover squares.
I recommend this game as a fantastic way to practice the vocabulary, to reinforce important beginning concepts before a test, as a way to use some down time before a break without giving "free time". Multiple ways!
Included concepts are:
Base
Cone
Cylinder
Great Circle
Height/Altitude
Hemisphere
Net
Prism
Pyramid
Slant Height
Sphere
Surface Area Formula for a Cone
Surface Area Formula for a Cylinder
Surface Area Formula for a Hemisphere
Surface Area Formula for a Prism
Surface Area Formula for a Pyramid
Surface Area Formula for a Sphere
Volume Formula for a Cone
Volume Formula for a Cylinder
Volume Formula for a Hemisphere
Volume Formula for a Prism
Volume Formula for a Pyramid
Volume Formula for a Sphere
*****************************************************************************
Please check out my other surface area and volume resources:
Quick Reference SheetGeometry Surface Area and Volume Figures and Formulas Quick Reference Sheets.
Riddle Worksheets
Pyramids
Geometry Surface Area and Volume of Pyramids Riddle Worksheet .
Cones
Geometry Surface Area and Volume of Cones Riddle Worksheet .
Spheres and Hemispheres
Geometry Surface Area and Volume of Spheres and Hemispheres Riddle Worksheet.
Cylinders
Geometry Surface Area and Volume of Cylinders Riddle Worksheet.
Prisms
Geometry Surface Area and Volume of Prisms Riddle Worksheet.
Surface Area and Volume Posters
Surface Area and Volume of 3-D Figures Wall Posters and Graphic Organizer.
Unit Eleven: Surface Area and Volume Bellwork/Exit Cards
Geometry Surface Area and Volume Bellwork/Exit Cards Station Questions .
Unit Eleven: Surface Area and Volume of 3-D Figures
Unit Eleven: Surface Area and Volume of 3-D Figures Vocabulary Assignment and Puzzles
*****************************************************************************
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Total Pages
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Teaching Duration
1 Semester
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# ALGEBRA 2
posted by .
y varies directly with x, and y = 2 when x = 5.
What is the value of x when y = 15?
A. x = 25
B. x = 10
C. x = 6
D. x = 37.5
• ALGEBRA 2 -
5/2 = x/15
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DIRECT VARIATION 1. a is directly proportional to b.If a=15,then b=9 2. m varies directly as n.If n=2/3,then m=1/4 3. T varies as the square root of L. If L=81,then T=10 4.r varies directly as the square of t. If t=16,then r=6. 5. …
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# Multiplication Story Wilson
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### Multiplication Story Wilson
1. 1. My Multiplication and Addition problems By wilson
2. 2. First Here are 9 fishes. Each fish has 9 dots. How many in all?
3. 3. First 9 x 9= 81 9+9+9+9+9+9+9+9+9=81
4. 4. Second Here are 7 apples. Each apple has 7 worms. How many worms are there in all?
5. 5. Second 7 x 7=49 7+7+7+7+7+7+7=49
6. 6. Third Here are 4 boxes. Each box has 6 stickers. How many stickers are there in all?
7. 7. Third 4 boxes X 6 stickers=24 6+6+6+6=24
8. 8. Forth Here are 6 monsters. Each monster has 2 eyes. How many eyes are there in all?
9. 9. Forth 6 x 2=12 2+2+2+2+2+2=12
10. 10. Reflection In math I learned that addition and multiplication are the same and I learned how to do multiplication. In technology I learned how to add pictures and go on Google.
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# How many cows and chickens did farmer Amogh buy?
175 views
Farmer Amogh has bought new cows and chickens. Together, the animals have 25 heads and 72 legs. How many cows and chickens did farmer Amogh buy?
posted May 9, 2014
Assume
number of Cows are X
number of chickens are Y
Now we have two equations
X + Y = 25
4X + 2Y = 72
On solving these two we find X=11 and Y=14 so answer is 11 Cow and 14 chicken
Let the number of cows be X and number of Chickens be Y
then,
X + Y = 25
4X + 2Y = 72
by, solving these two, we will get X = 11 and Y = 14
+1 vote
11 Cows and 14 Chickens
Similar Puzzles
Farmer Mohammed has a chicken farm. On a certain day, Mohammed calculates in how many days he will run out of chicken-food. He notices that if he would sell 75 of his chickens, he could feed the remaining chickens twenty days longer with the chicken-food he has, and that if he would buy 100 extra chickens, he would run out of chicken-food fifteen days earlier.
+1 vote
You've bought your weekly egg supply at the local farm store. The first morning you have company for breakfast and use half the eggs plus one-half an egg. The next morning you use one-half of what's left plus one-half an egg. The third morning you use one-half of what's left plus one-half an egg, and on the fourth morning, you're down to one solitary egg, so you make French toast. In all this cooking, you've never had one-half an egg to carry over to the next day.
How many eggs did you buy originally?
+1 vote
There are 12 animals in a farm.
Some are Cows
The rest are Chickens.
There are 28 animal legs in all.
How many Cows are there ?
How many Chickens are there ?
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# 261 (number)
← 260 261 262 →
Cardinaltwo hundred sixty-one
Ordinal261st
(two hundred sixty-first)
Factorization32 × 29
Divisors1, 3, 9, 29, 87, 261
Greek numeralΣΞΑ´
Roman numeralCCLXI
Binary1000001012
Ternary1002003
Quaternary100114
Quinary20215
Senary11136
Octal4058
Duodecimal19912
VigesimalD120
Base 367936
261 (two hundred [and] sixty-one) is a natural number proceeded by the number 260 and followed by 262. It has the prime factorization 32·29.
## Mathematical properties
There are six divisors of this number, the divisors being 1, 3, 9, 29, 87, and 261 itself.[1] 261 is a deficient number, since 1 + 3 + 9 + 29 + 87 = 129 < 261.
261 is nonagonal number, Harshad number, unique period in base 2, and the number of possible unfolded tesseract patterns.
261 is a lucky number, as well as an odious number, meaning it has an odd number of 1's in its binary expansion, which is 1000001012 (with 3 ones in it).[2]
261 was once the lowest number not to have its own Wikipedia page, this making it a candidate for the lowest uninteresting Number according to the definition given by Alex Bellos.[3] As of September 2018, the smallest natural number without its own Wikipedia page is 262, and the smallest prime number without its own Wikipedia page is 283.
## In other fields
261 may refer to...
## References
1. ^ "Facts about the number 261". Numbermatics - the number explorer. Retrieved 2017-03-20.
2. ^ "Number Gossip: 261". www.numbergossip.com. Retrieved 2017-03-20.
3. ^ Bellos, Alex (June 2014). The Grapes of Math: How Life Reflects Numbers and Numbers Reflect Life. illus. The Surreal McCoy (1st Simon & Schuster hardcover ed.). N.Y.: Simon & Schuster. pp. 238 & 319 (quoting p. 319). ISBN 978-1-4516-4009-0.
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# Digits
Find a number with all digits from 1-9 (both included) $$\overline { abcdefghi }$$ such that $$\overline { a }$$ is divisible by 1, $$\overline { ab }$$ is divisible by 2, $$\overline { abc }$$ is divisible by 3 and so on.
This problem is flagged because there are multiple solutions.
×
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Paul's Online Notes
Home / Calculus I / Applications of Derivatives / Critical Points
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If you are looking for some problems with solutions you can find some by clicking on the "Practice Problems" link above.
### Section 4.2 : Critical Points
For problems 1 - 43 determine the critical points of each of the following functions. Note that a couple of the problems involve equations that may not be easily solved by hand and as such may require some computational aids. These are marked are noted below.
1. $$R\left( x \right) = 8{x^3} - 18{x^2} - 240x + 2$$
2. $$f\left( z \right) = 2{z^4} - 16{z^3} + 20{z^2} - 7$$
3. $$\displaystyle g\left( z \right) = 8 - 12{z^5} - 25{z^6} + \frac{90}{7}{z^7}$$
4. $$g\left( t \right) = 3{t^4} - 20{t^3} - 132{t^2} + 672t - 4$$
Note : Depending upon your factoring skills this may require some computational aids.
5. $$\displaystyle h\left( x \right) = 10{x^2} - 15{x^3} + \frac{15}{2}{x^4} - {x^5}$$
Note : Depending upon your factoring skills this may require some computational aids.
6. $$P\left( w \right) = {w^3} - 4{w^2} - 7w - 1$$
7. $$A\left( t \right) = 7{t^3} - 3{t^2} + t - 15$$
8. $$a\left( t \right) = 4 - 2{t^2} - 6{t^3} - 3{t^4}$$
9. $$f\left( x \right) = 3{x^4} - 20{x^3} + 6{x^2} + 120x + 5$$
Note : Depending upon your factoring skills this may require some computational aids.
10. $$h\left( v \right) = {v^5} + {v^4} + 10{v^3} - 15$$
11. $$g\left( z \right) = {\left( {z - 3} \right)^5}{\left( {2z + 1} \right)^4}$$
12. $$R\left( q \right) = {\left( {q + 2} \right)^4}{\left( {{q^2} - 8} \right)^2}$$
13. $$f\left( t \right) = {\left( {t - 2} \right)^3}{\left( {{t^2} + 1} \right)^2}$$
14. $$\displaystyle f\left( w \right) = \frac{{{w^2} + 2w + 1}}{{3w - 5}}$$
15. $$\displaystyle h\left( t \right) = \frac{{3 - 4t}}{{{t^2} + 1}}$$
16. $$\displaystyle R\left( y \right) = \frac{{{y^2} - y}}{{{y^2} + 3y + 8}}$$
17. $$Y\left( x \right) = \sqrt[3]{{x - 7}}$$
18. $$f\left( t \right) = {\left( {{t^3} - 25t} \right)^{\frac{2}{3}}}$$
19. $$h\left( x \right) = \sqrt[5]{x}\,\,{\left( {2x + 8} \right)^2}$$
20. $$Q\left( w \right) = \left( {6 - {w^2}} \right)\,\,\,\sqrt[3]{{{w^2} - 4}}$$
21. $$\displaystyle Q\left( t \right) = 7\sin \left( \frac{t}{4} \right) - 2$$
22. $$g\left( x \right) = 3\cos \left( {2x} \right) - 5x$$
23. $$f\left( x \right) = 7\cos \left( x \right) + 2x$$
24. $$h\left( t \right) = 6\sin \left( {2t} \right) + 12t$$
25. $$\displaystyle w\left( z \right) = {\cos ^3}\left( \frac{z}{5} \right)$$
26. $$U\left( z \right) = \tan \left( z \right) - 4z$$
27. $$h\left( x \right) = x\cos \left( x \right) - \sin \left( x \right)$$
28. $$h\left( x \right) = 2\cos \left( x \right) - \cos \left( {2x} \right)$$
29. $$f\left( w \right) = {\cos ^2}\left( w \right) - {\cos ^4}\left( w \right)$$
30. $$F\left( w \right) = {{\bf{e}}^{14w + 3}}$$
31. $$g\left( z \right) = {z^2}{{\bf{e}}^{1 - z}}$$
32. $$A\left( x \right) = \left( {3 - 2x} \right){{\bf{e}}^{{x^{\,2}}}}$$
33. $$P\left( t \right) = \left( {6t + 1} \right){{\bf{e}}^{8t - {t^{\,2}}}}$$
34. $$f\left( x \right) = {{\bf{e}}^{3 + {x^{\,2}}}} - {{\bf{e}}^{2{x^{\,2}} - 4}}$$
35. $$f\left( z \right) = {{\bf{e}}^{{z^{\,2}} - 4z}} + {{\bf{e}}^{8z - 2{z^2}}}$$
36. $$h\left( y \right) = {{\bf{e}}^{6{y^{\,3}} - 8{y^{\,2}}}}$$
37. $$g\left( t \right) = {{\bf{e}}^{2{t^{\,3}} + 4{t^{\,2}} - t}}$$
38. $$Z\left( t \right) = \ln \left( {{t^2} + t + 3} \right)$$
39. $$G\left( r \right) = r - \ln \left( {{r^2} + 1} \right)$$
40. $$A\left( z \right) = 2 - 6z + \ln \left( {8z + 1} \right)$$
41. $$f\left( x \right) = x - 4\ln \left( {{x^2} + x + 2} \right)$$
42. $$g\left( x \right) = \ln \left( {4x + 2} \right) - \ln \left( {x + 4} \right)$$
43. $$h\left( t \right) = \ln \left( {{t^2} - t + 1} \right) + \ln \left( {4 - t} \right)$$
44. The graph of some function, $$f\left( x \right)$$, is shown. Based on the graph, estimate the location of all the critical points of the function.
45. The graph of some function, $$f\left( x \right)$$, is shown. Based on the graph, estimate the location of all the critical points of the function.
46. The graph of some function, $$f\left( x \right)$$, is shown. Based on the graph, estimate the location of all the critical points of the function.
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Limited access
Simplify the fraction:
$$\cfrac { \frac { 1 }{ 2 } { x }^{ 2 }+6x+16}{ 3{ x }^{ 2 }+12x }$$
A
$\cfrac { { x }^{ 2 }+3x+8 }{ 6{ x }^{ 2 }+24 }$
B
$\cfrac { { x }^{ 2 }+x+16 }{ 6{ x }^{ 2 }+2x }$
C
$\cfrac { x+8 }{ 6x }$
D
$\cfrac { x+4 }{ 3x }$
Select an assignment template
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You are Here: Home >< Maths
# Solving trig equations watch
1. I have sin(2theta) = 1/2 which I’m supposed to solve in the domain -pi to pi. My question is for the 3rd and 4th solution, why do you do pi/6 - pi and 5pi/6 - pi instead of + pi?
2. This is what I got hope this is helpful.
3. (Original post by Howie_2114)
This is what I got hope this is helpful.
There’s supposed to be 4 solutions apparently in the domain -pi to pi, which is what confuses me, as the markscheme says that the 3rd and 4 solution is -7pi/12 and -11pi/12
4. what you do is expand the domain for sin(a)=1/2 from -2pi to 2pi
Find the solutions for a=pi/6, 5pi/6, -7pi/6 and -11pi/6
Then divide by 2 every answer to find theta.
5. (Original post by tremen222)
what you do is expand the domain for sin(a)=1/2 from -2pi to 2pi
Find the solutions for a=pi/6, 5pi/6, -7pi/6 and -11pi/6
Then divide by 2 every answer to find theta.
thank you mate, makes so much more sense now
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# Expected Profit
• Mar 21st 2011, 06:24 PM
alakaboom1
Expected Profit
Sorry, I already asked a question today, but it would be awesome if someone could help me with this one, I feel like I'm close:
The annual profit Y(in \$100,000) can be expressed as a continuous function of drug demand x(in 1,000): Y(x) = 2(1-e^(-2x)). Suppose the demand for their drug has the probability function: f(x)= 6e^(-6x), x>0. Find the company's expected annual profit.
So do I have to start by doing http://s3.amazonaws.com/answer-board...9239628497.gif? I'm sure I have the upper bound wrong...
• Mar 22nd 2011, 12:30 AM
Volga
The way I see this problem, it is a mixed distribution where the given profit function Y is conditional on X:
\$\displaystyle f_{Y|X}(y|x)=2(1-e^{-2x})\$
and therefore to find E(Y) you need first to find marginal distribution \$\displaystyle f_Y(y)\$ and then go from there.
But I may be totally wrong!
• Mar 23rd 2011, 01:22 AM
Moo
Hello,
\$\displaystyle E[Y]=E[E[Y|X]]=E[E[2(1-e^{-2X})|X]]=E[2(1-e^{-2X})]\$, which is the formula alakaboom1 wrote.
As for the boundaries, it should rather be on the whole set of real numbers.
Then since f(x)=0 if x<0, the boundaries will indeed go from 0 to infinity.
Volga : there's something disturbing in what you wrote. \$\displaystyle 2(1-e^{-2x})\$ is not a pdf. We just have that \$\displaystyle Y=2(1-e^{-2X})\$, so in order to find Y's pdf, there's a change of variable to make in X's pdf :)
• Mar 23rd 2011, 04:28 AM
Volga
Sorry! (Itwasntme)
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# Multiple Integration
1. May 6, 2007
### windy906
I'm having problems finding the limits for the follwing integration, can anyone help?
Find the volume bounded by x+2z=4 and 2y+z=2 with x,y and z>=0
2. May 6, 2007
### windy906
I think the following limits apply but can't find the upper one for z, or in which order they should go.
x: 0 -> 4-2z
y: 0 -> 1-0.5z
z: 0 -> ????
3. May 6, 2007
### Office_Shredder
Staff Emeritus
It seems like if you pick any z>=0, you can find x,y that satisfy the boundary condiitions. Hence, z should be between 0 and $$\infty$$
4. May 6, 2007
### D H
Staff Emeritus
There are five boundary conditions:
1. x+2z<=4
2. 2y+z<=2
3. x>=0
4. y>=0
5. z>=0
Picking a value for z beyond some finite limit will not work.
5. May 6, 2007
### windy906
So z <= 2!
Thank you so much.
Last edited: May 6, 2007
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Cube in a sphere
The cube is inscribed in a sphere with volume 7253 cm3. Determine the length of the edges of a cube.
Result
a = 13.9 cm
Solution:
$V=7253 \ \text{cm}^3 \ \\ V=\dfrac{ 4 }{ 3 } \pi r^3 \ \\ \ \\ r=\sqrt[3]{ \dfrac{ 3 \cdot \ V }{ 4 \pi } }=\sqrt[3]{ \dfrac{ 3 \cdot \ 7253 }{ 4 \cdot \ 3.1416 } } \doteq 12.0082 \ \text{cm} \ \\ \ \\ D=2 \cdot \ r=2 \cdot \ 12.0082 \doteq 24.0163 \ \text{cm} \ \\ \ \\ u=D=24.0163 \doteq 24.0163 \ \text{cm} \ \\ \ \\ u=\sqrt{ 3 } a \ \\ \ \\ a=u/\sqrt{ 3 }=24.0163/\sqrt{ 3 } \doteq 13.8658 \doteq 13.9 \ \text{cm}$
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Math student
i am good at this
2 years ago 2 Likes
Crazy Butterfly
This is easy.
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## Blog
### GMAT Impact: The Master Resource List for Reading Comprehension (Part 3 of 4)
With regard to the GMAT, raw intellectual horsepower helps, but it is not everything. In this blog series, Manhattan Prep’s Stacey Koprince teaches you how to perform at your best on test day by using some common sense.
Part 1 of this series covered how to read Reading Comprehension (RC), and Part 2 introduced the first two major question types: Main Idea and Specific Detail. Start with those posts and then continue with this post.
Inference
In this section, we are going to talk about two big things: how to handle inference questions and how to analyze RC problems in general (you can then use these techniques on any question type).
Inference questions ask about specific details in the passage, but they add a twist: we have to deduce something that must be true, given certain facts from the passage.
For example, if I tell you that my favorite type of book to read is biography, what could you deduce?
Watch out for the trap: do not use your “real world” conclusion-drawing skills. In the real world, you might conclude that I like reading books in general or perhaps that I am interested in history or maybe that I am a nerd. (Really? Biographies are my favorite?) These things do not have to be true, though.
What has to be true? I do not like fiction as much as I like biographies. I have read at least one book in a nonbiography category (otherwise, I would not be able to tell that I prefer biographies, which implies a comparison).
What is the difference? GMAT deductions are usually things that would cause us to say “Duh!” in the real world.
“My favorite category of book is biography.”
“Oh, so you must not like fiction as much as you like biographies.”
“Uh… well, yeah, that’s what ‘favorite’ means. I don’t like anything else better.”
A GMAT deduction should feel like a “duh” deduction—something totally boring that must be true, given the information in the passage. Here, try out an Inference question.
That article also explains how to analyze your work and the problem itself. Did you miss something in the passage? Why? How can you pick it up next time? Did you fall for a trap answer? Which one? How did they set the trap, and how can you avoid it next time? And so on.
Why Questions
Specific questions can come in one other (not as common) flavor: the Why question. These are sort of a cross between Specific Detail and Inference questions: you need to review some specific information in the passage, but the answer to the question is not literally right in the passage. You have to figure out the most reasonable explanation for why the author chose to include a particular piece of information.
Test out this Why question to see what I mean.
Timing
As I mentioned earlier, we really do not have much time to read RC passages. Aim for approximately two to two and a half minutes on shorter passages and closer to three minutes for longer ones. Of course, you cannot possibly read everything closely and carefully in such a short time frame—but that is not your goal! Our goal is to get the big picture on that first read-through.
Aim to answer main idea questions in roughly one minute. You can spend up to two minutes on the more specific questions. In particular, if you run across an Except question, expect to spend pretty close to two minutes; Except questions nearly always take a while.
As always, be aware of your overall time. If you find that you are running behind, skip one question entirely; do not try to save 30 seconds each on a bunch of questions. Also, if RC is your weakest verbal area, and you also struggle with speed, consider guessing immediately on one question per passage and spreading your time over the remaining questions.
Great, I Have Mastered RC!
Let us test that theory, shall we? Your next step is to implement all these techniques on your next practice test while also managing your timing well. Good luck!
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# How do magnets work?
## How do magnets work?
All magnets have north and south poles. Opposite poles are attracted to each other, while the same poles repel each other. When you rub a piece of iron along a magnet, the north-seeking poles of the atoms in the iron line up in the same direction. The force generated by the aligned atoms creates a magnetic field.
How do magnets work through material?
You may have noticed that the materials that make good magnets are the same as the materials magnets attract. This is because magnets attract materials that have unpaired electrons that spin in the same direction. In other words, the quality that turns a metal into a magnet also attracts the metal to magnets.
### How do magnets move objects?
The object will be attracted to a magnet brought near it. For example, a magnet will move paper that is attached to the metal. Attach a magnet to an object. When another magnet is brought near it, the two magnets will either be attracted or repelled, and the object will move.
READ: Are leaf blowers good for dirt?
How do you think magnets without touching them can move objects?
4) Why is a magnet able to move an object without touching it? Magnetic force is a force, like gravity, that is able to act across a distance without touching an object that is attracted by it. Magnetic attraction is a pulling force.
## How do magnets work ks2?
An explanation of how magnets work by using their poles to either attract or repel objects. Only metals which contain iron, nickel or cobalt are magnetic. The two poles which are the same will repel each other, and the two poles which are different will attract each other. There are magnets all around us.
How do magnets attract metal?
Magnets attract iron due to the influence of their magnetic field upon the iron. When exposed to the magnetic field, the atoms begin to align their electrons with the flow of the magnetic field, which makes the iron magnetized as well. This, in turn, creates an attraction between the two magnetized objects.
### How do magnets work electrons?
Moving magnetic fields pull and push electrons. Metals such as copper and aluminum have electrons that are loosely held. Moving a magnet around a coil of wire, or moving a coil of wire around a magnet, pushes the electrons in the wire and creates an electrical current.
READ: How much does 1kg biryani serve?
How do magnets work in computers?
Computer. Most computers use tiny magnets to store information. The hard disk inside a computer is made up of metal plates. Tiny areas on these plates can be made magnetic or nonmagnetic, creating a code that the computer can turn into data such as pictures, sounds, and videos.
## Why do magnets move objects in different ways?
Thus a magnet will move another magnet if brought to proximity according to the rule, same poles repel, different poles attract. are attracted to the magnetic poles indiscriminately, whether north or south. This happens because magnetic materials are composed of tiny magnets oriented every which way.
What is a magnet in science?
magnet, any material capable of attracting iron and producing a magnetic field outside itself. The most common was the property of diamagnetism, the name given to materials exhibiting a weak repulsion by both poles of a magnet.
### Do magnets work without air?
Magnets can be used in space. Unlike a lot of other items you might bring to space that need additional tools or equipment to function, a magnet will work without any extra help. Magnets don’t need gravity or air. Instead, their power comes from the electromagnetic field they generate all by themselves.
How do magnets work Bitesize?
## What is the basic principle of magnet?
Magnet Basics. Magnets can either attract or repel each other. A permanent magnet is an object that produces a magnetic field around itself. It is this field that enables them to stick to each other and to some types of metal. Specifically, they stick to ferromagnetic materials like iron and things that contain iron, such as steel.
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# m25,q1
Author Message
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Joined: 06 Dec 2009
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### Show Tags
28 Dec 2009, 14:04
Can anybody help me with this question?
Is $$|x - 6| > 5$$?
1. $$x$$ is an integer
2. $$x^2 < 1$$
Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient
Statements (1) and (2) TOGETHER are NOT sufficient
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Math Expert
Joined: 02 Sep 2009
Posts: 41601
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### Show Tags
28 Dec 2009, 18:53
1
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Expert's post
ppaulyni3 wrote:
Can anybody help me with this question?
Is |x - 6| > 5?
1. x is an integer
2. x^2 < 1
Let's work on the stem first. For which values of x, |x - 6| > 5 is true?
|x - 6| > 5
x<6 --> -x+6>5 --> x<1.
x>=6 --> x-6>5 --> x>11.
So for x from the ranges x<1 and x>11 the inequality |x - 6| > 5 holds true.
(1) x is an integer, clearly not sufficient. x can be 12 and the inequality holds true as we concluded OR x can be 5 and inequality doesn't hold true.
(2) x^2<1 --> -1<x<1, as all x-es from this range are in the range x<1, hence inequality |x - 6| > 5 holds true. Sufficient.
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17 Jan 2010, 08:30
1
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but in this case statement A contradicts to statement B. A states that x is an integer and B states that it is not.
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17 Jan 2010, 17:32
Tati wrote:
but in this case statement A contradicts to statement B. A states that x is an integer and B states that it is not.
Not so, taken together x can be zero which is an integer.
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09 May 2012, 22:46
Bunuel wrote:
Tati wrote:
but in this case statement A contradicts to statement B. A states that x is an integer and B states that it is not.
Not so, taken together x can be zero which is an integer.
If we need to indicate that x is an integer, shouldn't the answer be C?
Statement 1 alone: Insufficient.
Statement 2 alone:
-1<x<1
|-1-6| = 7. True.
However.
|1-6| = 5. False 5 is not larger than 5. Also insufficient.
Both statement together:
If say we take zero as an integer we need statement 1 to indicate that x is in fact an integer.
|0-6| = 6. True
Sufficient.
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### Show Tags
10 May 2012, 01:18
leochanGmat wrote:
Bunuel wrote:
Tati wrote:
but in this case statement A contradicts to statement B. A states that x is an integer and B states that it is not.
Not so, taken together x can be zero which is an integer.
If we need to indicate that x is an integer, shouldn't the answer be C?
Statement 1 alone: Insufficient.
Statement 2 alone:
-1<x<1
|-1-6| = 7. True.
However.
|1-6| = 5. False 5 is not larger than 5. Also insufficient.
Both statement together:
If say we take zero as an integer we need statement 1 to indicate that x is in fact an integer.
|0-6| = 6. True
Sufficient.
No, when considering the second statement we don't need to know that x is an integer. The question asks: "is x<1 or x>11?" and (2) says that -1<x<1, so we can answer YES to the question. In your examples you can not consider x=-1 and x=1 since in the given range (-1<x<1) -1 and 1 are not inclusive.
Hope it's clear.
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10 May 2012, 11:55
Thank you for the clarification.
Say if statement 2 is larger than 11, but not smaller than 1. The statement will still be sufficient?
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# m25,q1
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# Time and Distance Test 4
1. A man traveled form P to Q at a speed of 60 km/hr and return to P at a speed of 40 km/hr. find the average speed?
2. P and Q covered a certain distance at 9 km/hr and 10 km/hr respectively. If P took 32 min more to reach the destination find the distance covered?
3. A man misses a bus by 40 min if he travels at 30 km/hr. If he travels at 40km/hr, then also he misses the bus by 10 min. What is the minimum speed to catch the bus on time?
4. P and Q are running a circular track of 600 m in opposite direction at speeds of 15 m/sec and 10 m/sec respectively starting at the same time from the same point. In how much time will they meet for the first time anywhere on the track?
5. Trisha travels from P to Q by car and returns from Q to P by cycle in 7 hours. If she travels both the ways by car she saves 3 hours. What is the time taken by her to cover both the ways by cycle?
6. Supriya reaches her office 1 hour late traveling 40 km/hr. If she travels at 50 km/hr, she is late by 40 min. What is the distance she has to reach her office?
7. Saritha leaves from a certain point at 8.00 A.M. at a speed of 37.5 km/hr in the same direction as that of Saritha. At what time do they meet?
8. A bike run 300 m in 40 sec and another bike run it in 52 sec. In a 1950 m race, by how many meters does first bike beat the second one?
9. In a 5000 m race, B beats C by 500 m or 50 seconds. What is the time taken by B to complete a 54000 m race?
10. Venu covered 22.5 km in 3 hours 20 minutes. He covered 10 km/hr. At what speed did he cover the remaining distance?
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Paubaume
Registered: 18.04.2004
From: In the stars... where you left me, and where I will wait for you... always...
Posted: Tuesday 02nd of Jan 08:18 I remember having often faced difficulties with angle complements, radical expressions and adding exponents. A truly great piece of math program is Algebrator software. By simply typing in a problem from workbook a step by step solution would appear by a click on Solve. I have used it through many math classes – Remedial Algebra, Algebra 2 and Algebra 1. I greatly recommend the program.
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# All posts tagged ‘full frame’
## Photography Snapshot: The Power of Lenses
Camera Lens Colored Abstract brought to you by Snoron.com
This article is the third in a series teaching the basics of photography. We started by learning about the properties of light and how an image is created, and we also learned how a lens bends light to focus individual rays into a single bright image. With this lesson we are going to finish learning the scientific theory of lenses and how to use lenses for magnification in addition to brightness.
Relation between lens diameter and f-stop. (From Panasonic.com)
In the last lesson we performed an experiment to focus the light of a candle through a lens. We also learned that to determine the focal length of the system, we move the focusing screen forward and backward until the image of the flame is in focus. Let’s consider that candle/lens system for a moment. What do you think would happen if we replaced the lens with one that is twice the diameter with the same focal length? Would the image be twice as bright? Image twice as large? If you guessed the larger lens would make the image brighter, you would be correct. The larger lens has more area to collect light, which actually equates to an image more then twice the brightness at a ratio equal to πr² where r equals the radius of the lens. The image would, however, be no larger since the focal length of the lens is that same. Continue Reading “Photography Snapshot: The Power of Lenses” »
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+0
Решите пж
0
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(4,9*10^3)*(4*10^2)
Oct 13, 2022
#1
+14964
+1
(4,9*10^3)*(4*10^2)
Здравей приятелю!
$$(4.9\cdot 10^3)\cdot (4\cdot 10^2)\\ =4.9\cdot 10^3\cdot4\cdot 10^2\\ =4.9\cdot4\cdot 10^{3+2}\\ \color{blue}=19.6\cdot10^5=1.96\cdot 10^6=1\ 960\ 000$$
!
Oct 13, 2022
edited by asinus Oct 13, 2022
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# Working with Statistics Canada Data in R, Part 5: Retrieving Census Data
## Introduction
Now that we are ready to start working with Canadian Census data, let’s first briefly address the question why you may need it. After all, CANSIM data is often more up-to-date and covers a much broader range of topics than the national census data, which is gathered every five years in respect of a limited number of questions.
The main reason is that CANSIM data is far less granular geographically. Most of it is collected at the provincial or even higher regional level. You may be able to find CANSIM data on a limited number of questions for some of the country’s largest metropolitan areas, but if you need the data for a specific census division, city, town, or village, you’ll have to use the Census.
To illustrate the use of the cancensus package, let’s do a small research project. First, in this post we’ll retrieve the following key labor force characteristics of the largest metropolitan areas in each of the five geographic regions of Canada:
• Labor force participation rate, employment rate, and unemployment rate.
• Percent of workers by work situation: full time vs part time, by gender.
• Education levels of people aged 25 to 64, by gender.
The cities (metropolitan areas) that we are going to look at, are: Calgary, Halifax, Toronto, Vancouver, and Whitehorse. We’ll also get these data for Canada as a whole for comparison and to illustrate the retrieval of data at different geographic levels
Next, in Part 6 of the “Working with Statistics Canada Data in R” series, we will visualize these data, including making a faceted plot and writing a function to automate repetitive plotting tasks.
Keep in mind that cancensus also allows you to retrieve geospatial data, that is, borders of census regions at various geographic levels, in sp and sf formats. Retrieving and visualizing Statistics Canada geospatial data will be covered later in these series.
library(cancensus)
library(tidyverse)
## Searching for Data
cancensus retrieves census data with the get_census() function. get_census() can take a number of arguments, the most important of which are dataset, regions, and vectors, which have no defaults. Thus, in order to be able to retrieve census data, you’ll first need to figure out:
### Find Census Datasets
Let’s see which census datasets are available through the CensusMapper API:
list_census_datasets()
Currently, datasets earlier than 1996 are not available, so if you need to work with pre-1996 census data, you won’t be able to retrieve it with cancensus.
### Find Census Regions
Next, let’s find the regions that we’ll be getting the data for. To search for census regions, use the search_census_regions() function.
Let’s take a look at what region search returns for Toronto. Note that cancensus functions return their output as dataframes, making it is easy to subset. Here I limited the output to the most relevant columns to make sure it fits on screen. You can run the code without [c(1:5, 8)] to see all of it.
# all census levels
search_census_regions(searchterm = "Toronto",
dataset = "CA16")[c(1:5, 8)]
#> # A tibble: 3 x 6
#> region name level pop municipal_status PR_UID
#> <chr> <chr> <chr> <int> <chr> <chr>
#> 1 35535 Toronto CMA 5928040 B 35
#> 2 3520 Toronto CD 2731571 CDR 35
#> 3 3520005 Toronto CSD 2731571 C 35
You may have expected to get only one region: the city of Toronto, but instead you got three! So, what is the difference? Look at the column level for the answer. Often, the same geographic region can be represented by several census levels, as is the case here. There are three levels for Toronto, which is simultaneously a census metropolitan area, a census division, and a census sub-division. Note also the PR_UID column that contains numeric codes for Canada’s provinces and territories. These codes can help you distinguish between different census regions that have same or similar names but are located in different provinces. For an example, run the code above replacing “Toronto” with “Windsor”.
Remember that we were going to plot the data for census metropolitan areas? You can choose the geographic level with the level argument, which can take the following values: ‘C’ for Canada (national level), ‘PR’ for province, ‘CMA’ for census metropolitan area, ‘CD’ for census division, ‘CSD’ for census sub-division, or NA:
# specific census level
search_census_regions("Toronto", "CA16", level = "CMA")
Let’s now list census regions that may be relevant for our project:
# explore available census regions
"Toronto", "Vancouver", "Whitehorse")
map_df(names, ~ search_census_regions(., dataset = "CA16"))[c(1:5, 8)]
#> # A tibble: 19 x 6
#> region name level pop municipal_status PR_UID
#> <chr> <chr> <chr> <int> <chr> <chr>
#> 1 01 Canada C 35151728 NA NA
#> 2 48825 Calgary CMA 1392609 B 48
#> 3 4806016 Calgary CSD 1239220 CY 48
#> 4 12205 Halifax CMA 403390 B 12
#> 5 1209 Halifax CD 403390 CTY 12
#> 6 1209034 Halifax CSD 403131 RGM 12
#> 7 2432023 Sainte-Sophie-d'Halifax CSD 612 MÉ 24
#> 8 35535 Toronto CMA 5928040 B 35
#> 9 3520 Toronto CD 2731571 CDR 35
#> 10 3520005 Toronto CSD 2731571 C 35
#> 11 59933 Vancouver CMA 2463431 B 59
#> 12 5915 Greater Vancouver CD 2463431 RD 59
#> 13 5915022 Vancouver CSD 631486 CY 59
#> 14 5915046 North Vancouver CSD 85935 DM 59
#> 15 5915051 North Vancouver CSD 52898 CY 59
#> 16 5915055 West Vancouver CSD 42473 DM 59
#> 17 5915020 Greater Vancouver A CSD 16133 RDA 59
#> 18 6001009 Whitehorse CSD 25085 CY 60
#> 19 6001060 Whitehorse, Unorganized CSD 326 NO 60
purrr::map_df() function applies search_census_regions() iteratively to each element of the names vector and returns output as a single dataframe. Note also the ~ . syntax. Think of it as the tilde taking each element of names and passing it as an argument to a place indicated by the dot in the search_census_regions() function. You can find more about the tilde-dot syntax here. It may be a good idea to read the whole tutorial: purrr is a super-useful package, but not the easiest to learn, and this tutorial does a great job explaining the basics.
Since there are multiple entries for each search term, we’ll need to choose the results for census metropolitan areas, or in case of Whitehorse, for census sub-division, since Whitehorse is too small to be considered a metropolitan area:
# select only the regions we need: CMAs (and CSD for Whitehorse)
regions <- list_census_regions(dataset = "CA16") %>%
filter(grepl("Calgary|Halifax|Toronto|Vancouver", name) &
grepl("CMA", level) |
grepl("Canada|Whitehorse$", name)) %>% as_census_region_list() Pay attention to how the logical operators are used to filter the output by several conditions at once; also note using $ regex meta-character to choose from the names column the entry ending with ‘Whitehorse’ (to filter out ‘Whitehorse, Unorganized’.
Finally, as_census_region_list() converts list_census_regions() output to a data object of type list that can be passed to the get_census() function as its regions argument.
### Find Census Vectors
Canadian census data is made up of individual variables, aka census vectors. Vector number(s) is another argument you need to specify in order to retrieve data with the get_census() function.
cancensus has two functions that allow you to search through census data variables: list_census_vectors() and search_census_vectors().
list_census_vectors() returns all available vectors for a given dataset as a single dataframe containing vectors and their descriptions:
# structure of list_census_vectors output
str(list_census_vectors(dataset = 'CA16'))
#> tibble [6,623 × 7] (S3: tbl_df/tbl/data.frame)
#> $vector : chr [1:6623] "v_CA16_401" "v_CA16_402" "v_CA16_403" "v_CA16_404" ... #>$ type : Factor w/ 3 levels "Female","Male",..: 3 3 3 3 3 3 3 3 2 1 ...
#> $label : chr [1:6623] "Population, 2016" "Population, 2011" "Population percentage change, 2011 to 2016" "Total private dwellings" ... #>$ units : Factor w/ 6 levels "Number","Percentage ratio (0.0-1.0)",..: 1 1 1 1 1 4 1 1 1 1 ...
#> $parent_vector: chr [1:6623] NA NA NA NA ... #>$ aggregation : chr [1:6623] "Additive" "Additive" "Average of v_CA16_402" "Additive" ...
#> $details : chr [1:6623] "CA 2016 Census; Population and Dwellings; Population, 2016" "CA 2016 Census; Population and Dwellings; Population, 2011" "CA 2016 Census; Population and Dwellings; Population percentage change, 2011 to 2016" "CA 2016 Census; Population and Dwellings; Total private dwellings" ... #> - attr(*, "last_updated")= POSIXct[1:1], format: "2020-12-21 23:27:47" #> - attr(*, "dataset")= chr "CA16" # count variables in 'CA16' dataset nrow(list_census_vectors(dataset = 'CA16')) #> [1] 6623 As you can see, there are 6623 (as of the time of writing this) variables in the 2016 census dataset, so list_census_vectors() won’t be the most convenient function to find a specific vector. Note however that there are situations (such as when you need to select a lot of vectors at once), in which list_census_vectors() would be appropriate. Usually it is more convenient to use search_census_vectors() to search for vectors. Just pass the text string of what you are looking for as the searchterm argument. You don’t have to be precise: this function works even if you make a typo or are uncertain about the spelling of your search term. Let’s now find census data vectors for labor force involvement rates: # get census data vectors for labor force involvement rates lf_vectors <- search_census_vectors(searchterm = "employment rate", dataset = "CA16") %>% union(search_census_vectors("participation rate", "CA16")) %>% filter(type == "Total") %>% pull(vector) Let’s take a look at what this code does. Since searchterm doesn’t have to be a precise match, “employment rate” search term retrieves unemployment rate vectors too. In the next line, union() merges dataframes returned by search_census_vectors() into a single dataframe. Note that in this case union() could be substituted with bind_rows(). I recommend using union() in order to avoid data duplication. Next, we choose only the “Total” numbers, since we are not going to plot labor force indicators by gender. Finally, the pull() command extracts a single vector from the dataframe, just like the $ subsetting operator: we need lf_vectors to be a data object of type vector in order to pass it to the vectors argument of the get_census() function.
There is another way to figure out search terms to put inside the search_census_vectors() function: use Statistics Canada online Census Profile tool. It can be used to quickly explore census data as well as to figure out variables’ names (search terms) and their hierarchical structure.
For example, let’s look at the census labor data for Calgary metropolitan area. Scrolling down, you will quickly find the numbers and text labels for full-time and part-time workers:
Now we know the exact search terms, so we can get precisely the vectors we need, free from any extraneous data:
# get census data vectors for full- and part-time work
# get vectors and labels
work_vectors_labels <-
search_census_vectors("full year, full time", "CA16") %>%
union(search_census_vectors("part year and/or part time", "CA16")) %>%
filter(type != "Total") %>%
select(1:3) %>%
mutate(label = str_remove(label, ".*, |.*and/or ")) %>%
mutate(type = fct_drop(type)) %>%
setNames(c("vector", "gender", "type"))
# extract vectors
work_vectors <- work_vectors_labels$vector Note how this code differs from the code with which we extracted labor force involvement rates: since we need the data to be sub-divided both by the type of work and by gender (hence no “Total” values here), we are creating a dataframe that assigns respective labels to each vector number. This work_vectors_labels dataframe will supply categorical labels to be attached to the data retrieved with get_census(). Also, note these three lines: mutate(label = str_remove(label, ".*, |.*and/or ")) %>% mutate(type = fct_drop(type)) %>% setNames(c("vector", "gender", "type")) The first mutate() call removes all text up to and including , and and/or (spaces included) from the label column. The second drops unused factor level “Total” – it is a good practice to make sure there are no unused factor levels if you are going to use ggplot2 to plot your data. Finally, setNames() renames variables for convenience. Finally, let’s retrieve vectors for the education data for the age group from 25 to 64 years, by gender. Before we do this, I’d like to draw your attention to the fact that some of the census data is hierarchical, which means that some variables (census vectors) are included into parent and/or include child variables. It is very important to choose vectors at proper hierarchical levels so that you do not double-count or omit your data. Education data is a good example of hierarchical data. You can explore data hierarchy using parent_census_vectors() and (child_census_vectors) functions. However, you may find exploring the hierarchy visually to be more convenient: So, let’s now retrieve and label the education data vectors: # get vectors and labels ed_vectors_labels <- search_census_vectors("certificate", "CA16") %>% union(search_census_vectors("degree", "CA16")) %>% union(search_census_vectors("doctorate", "CA16")) %>% filter(type != "Total") %>% filter(grepl("25 to 64 years", details)) %>% slice(-1,-2,-7,-8,-11:-14,-19,-20,-23:-28) %>% select(1:3) %>% mutate(label = str_remove_all(label, " cert.*diploma| dipl.*cate|, CEGEP| level|")) %>% mutate(label = str_replace_all(label, c("No.*" = "None", "Secondary.*" = "High school or equivalent", "other non-university" = "equivalent", "University above" = "Cert. or dipl. above", "medicine.*" = "health**", ".*doctorate$" = "Doctorate*"))) %>%
mutate(type = fct_drop(type)) %>%
setNames(c("vector", "gender", "level"))
# extract vectors
ed_vectors <- ed_vectors_labels\$vector
Note the slice() function that allows to manually select specific rows from a dataframe: positive numbers choose rows to keep, negative numbers choose rows to drop. I used slice() to drop the hierarchical levels from the data that are either too general or too granular. Note also that I had to edit text strings in the data. Finally, I added asterisks after “Doctorate” and “health”. These are not regex symbols, but actual asterisks that will be used to refer to footnotes in plot captions later on.
Now that we have figured out our dataset, regions, and data vectors (and labeled the vectors, too), we are finally ready to retrieve the data.
## Retrieve Census Data
To retrieve census data, feed the dataset, regions, and data vectors into get_census() as its’ respective arguments. Note that get_census() has the use_cache argument (set to TRUE by default), which tells get_census() to retrieve data from cache if available. If there is no cached data, the function will query CensusMapper API for the data and will save it in the cache, while use_cache = FALSE will force get_census() to query the API and update the cache.
# get census data for labor force involvement rates
# feed regions and vectors into get_census()
labor <-
get_census(dataset = "CA16",
regions = regions,
vectors = lf_vectors) %>%
select(-c(1, 2, 4:7)) %>%
setNames(c("region", "employment rate",
"unemployment rate",
"participation rate")) %>%
mutate(region = str_remove(region, " (.*)")) %>%
pivot_longer("employment rate":"participation rate",
names_to = "indicator",
values_to = "rate") %>%
mutate_if(is.character, as_factor)
The select() call drops columns with irrelevant data. setNames() renames columns to remove vector numbers from variable names – we don’t need vector numbers in variable names because variable names will be converted to values in the indicator column. str_remove() inside the mutate() call drops municipal status codes ‘(B)’ and ‘(CY)’ from region names. Finally, mutate_if() line converts characters to factors for subsequent plotting.
An important function here is tidyr::pivot_longer(). It converts the dataframe from wide to long format. It takes three columns: employment rate, unemployment rate, and participation rate, and passes their names as values of the indicator variable, while their numeric values are passed to the rate variable. The reason for conversion is that we are going to plot the data for all three labor force indicators in the same graphic, which makes it necessary to store the indicators as a single factor variable.
Next, let’s retrieve census data about the percent of full time vs part time workers, by gender, and the data about the education levels of people aged 25 to 64, by gender:
# get census data for full-time and part-time work
work <-
get_census(dataset = "CA16",
regions = regions,
vectors = work_vectors) %>%
select(-c(1, 2, 4:7)) %>%
rename(region = "Region Name") %>%
pivot_longer(2:5, names_to = "vector",
values_to = "count") %>%
mutate(region = str_remove(region, " (.*)")) %>%
mutate(vector = str_remove(vector, ":.*")) %>%
left_join(work_vectors_labels, by = "vector") %>%
mutate(gender = str_to_lower(gender)) %>%
mutate_if(is.character, as_factor)
# get census data for education levels
education <-
get_census(dataset = "CA16",
regions = regions,
vectors = ed_vectors) %>%
select(-c(1, 2, 4:7)) %>%
rename(region = "Region Name") %>%
pivot_longer(2:21, names_to = "vector",
values_to = "count") %>%
mutate(region = str_remove(region, " (.*)")) %>%
mutate(vector = str_remove(vector, ":.*")) %>%
left_join(ed_vectors_labels, by = "vector") %>%
mutate_if(is.character, as_factor)
Note one important difference from the code I used to retrieve the labor force involvement data: here I added the dplyr::left_join() function that joins labels to the census data.
We now have the data and are ready to visualize it, which will be done in the next part of this series.
## Annex: Notes and Definitions
For those of you who are outside of Canada, Canada’s geographic regions and their largest metropolitan areas are:
• The Atlantic Provinces – Halifax
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# When is $C_0(U)+C_0(V)=C_0(U \cup V)$ true?
If $U$ and $V$ are two open subsets of $\mathbb{C}$, is it true that $C_0(U)+C_0(V)$ is dense in $C_0(U \cup V)$?
Or better yet, is $C_0(U)+C_0(V) = C_0(U \cup V)$?
If it were true, I am wondering if we may replace $\mathbb{C}$ by $X$, a locally compact Hausdorff space. (the original motivation of the question is for $X$ the maximal ideal space of a commutative Banach space)
• A partition of unity should do the trick right? Dec 11, 2013 at 17:02
• That's right! How can I overlook this! Thanks Prahlad!! Just want to make sure - a partition of unity element with support in U means that it vanishes at the boundary of U, right? Dec 11, 2013 at 20:36
• Yes, it would be compactly supported inside $U$ Dec 12, 2013 at 1:40
• Thanks again! I see that you have an interest in K-theory of operator algebra, would you mind taking a look at my other question: math.stackexchange.com/questions/603371/…? Thanks!! Dec 12, 2013 at 5:40
Expanding a comment by Prahlad Vaidyanathan: A continuous partition of unity subordinate to a finite cover (such as $U,V$) exists for every locally compact Hausdorff space. Thus, for every $f\in C_0(U\cup V)$ one can write $f=f\phi_U+f\phi_V$ where $\operatorname{supp}\phi_U\subset U$ and $\operatorname{supp}\phi_V\subset V$.
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# Algebraic Topology 1
Welcome to the course page of Algebraic Topology 1 at SNU (Fall 2018). You can enter this page at http://cayley.kr/wiki/at1
## Basic Info
Course
SNU 3341.607 (001) Algebraic Topology 1, Fall 2018
Web
Students are expected to visit this course webpage at least once a week.
Online discussion is strongly encouraged.
Classes
MW 3:30 - 4:45pm, Bldg 24-211
Instructor
김상현 Sang-hyun Kim
s.kim(aht)snu
TA
김재형
mlrde01(aht)snu
### Topics and Prerequisites
This course is divided into two parts. In the first (short) part, we will review fundamental groups and the functorial properties of them. The topics include covering spaces, Seifert–van Kampen theorem, graphs of groups and K(G,1) spaces.
The second (main) part will mainly deal with homology theory. We will talk about functorial property, computation through simplicial homology and Mayer–Vietoris sequence and relation to fundamental groups.
Prerequisites are point-set topology, linear algebra, algebra and topology at undergraduate levels.
Attendance and Participation (10%)
Problem Sets (20%)
Late submission is not accepted.
Midterm (30%)
October 24th (W), during the class
Final Exam (40%)
December 5th (W), 3:30 - 5 pm.
## Lecture Plan
Subject to change.
Week 1 Homotopy, fundamental groups, base change
Week 2 Homotopy invariance, covering space
Week 3 Lifting theorems, deck transformation, covering--subgroup correspondence
Week 4
• 26 : Seifert--van Kampen theorem, surface π1
Week 5 Homology groups, homotopy invariance
Week 6 Relative homology groups, long exact sequence
Week 7 Excisions
Week 8 Homology of spheres and their generators, Brouwer Fixed Point Theorem, Fundamental Theorem of Algebra, Local homology and Invariance of Dimension
Week 9 Hurewicz Theorem for π1, Homology of surfaces
Week 10 CW complex, degree, cellular homology
Week 11 Mayer--Vietoris sequence, Computations
Week 12 Jordan–Brouwer Separation Theorem
Week 13 Invariance of Domain, Borsuk--Ulam Theorem, Simplicial Approximation Theorem, Lefschetz Fixed Point Theorem
## Homework
### HW No. 1
Please refer to the eTL bulletin board.
### Exercises
• You can find good exercise problems in SNU Graduate Entrance and SNU PhD Qualifying
• Students are strongly recommended to work together on problems. Writing must be done by oneself.
• Late HW policy: not accepted.
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Browsing all articles tagged with IMCOS Function
# IMCOS Excel Formula
Microsoft Excel Formula Development:
The development of Microsoft Excel formula in any of the Excel version can use the twenty-one signs such as =, -, +, # and many more plus the numbers of functions. Microsoft Excel versions of 1997 up to 2006 offers 329 functions and 5 new functions are added in Microsoft Excel 2007. Excel for all versions have 334 functions generally.
Microsoft Excel has its basic and essential features of the entire spreadsheet that uses the cell grids arranged in columns for letter-named and numbered rows organizing manipulations for data such as arithmetic operations. It has billions of supplied functions answering all the queries for engineering, statistical and financial needs. Besides, Excel can display and feature data as charts, histograms, line graphs and a limited graphical and three-dimensional display. Microsoft Office Excel defined as the electronic spreadsheet program used for manipulating, storing, calculating, analyzing and organizing data.
Microsoft Excel is the most helpful program in order to obtain an organized and well manageable data. It is difficult to use programs such as Microsoft Excel if you do not know the features and its functions. For beginners who use this software program, you must learn and understand first the Excel features and its functions.
One of the important functions is the IMCOS function in Microsoft Excel wherein it returns cosine of the supplied multifaceted number. The syntax of the IMCOS function is IMCOS(inumber). The argument iNumber considered the complex number. Be reminded that the complex number is stored as “text” in Microsoft Excel spreadsheet. Once the text string in the a+bj or a+bi format supplied to a single built-in complex number in Excel functions. It is being interpreted as the complex number. It accepts easy numeric value, which is equivalent to the complex number whose “imaginary coefficient” is equal to zero.
Therefore, the argument iNumber can supply the IMCOS function in Microsoft Excel. Just like a simple number or complex number encased with quotation marks. Having a reference cell, which contains numeric value or complex number, which is returned from other Excel formula or function?
Common IMCOS function in Microsoft Excel
• #VALUE! – this occurs when the supplied argument iNumber is not being recognized as the real or imaginary number.
• #NAME? – this happens when the analysis Toolpak add-in is not enabled in your Microsoft Excel. Thus, if you are to use the Engineering function in Excel, it requires the user to enable the “add-in”.
Here’s how to do in Microsoft Excel 2003:
• Once the window pops up, choose the analysis toolpak add-in option and hit the button OK.
How to do it in Microsoft Excel 2007 or Microsoft Excel 2010:
• Hit the button dropdown tools menu over the top left of the spreadsheet then choose the button Excel option
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https://www.sqlservercentral.com/forums/topic/calculating-duration-using-datetime-start-and-end-dates-sql-spackle
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# Calculating Duration Using DATETIME Start and End Dates (SQL Spackle)
• Comments posted to this topic are about the item Calculating Duration Using DATETIME Start and End Dates (SQL Spackle)[/url]
--Jeff Moden
RBAR is pronounced "ree-bar" and is a "Modenism" for Row-By-Agonizing-Row.
First step towards the paradigm shift of writing Set Based code:
________Stop thinking about what you want to do to a ROW... think, instead, of what you want to do to a COLUMN.
Change is inevitable... Change for the better is not.
How to post code problems
How to Post Performance Problems
Create a Tally Function (fnTally)
• Thank you for the spackle Jeff but it looks like there is a small crack in it as it makes the assumption that the end time is always greater than the start time. This is the reason why I prefer to use double cast, first to varbinary and then to bigint.
`SELECT`
` StartDT`
` ,EndDT`
` ,Duration = STUFF(CONVERT(VARCHAR(20),EndDT-StartDT,114),1,2,DATEDIFF(hh,0,EndDT-StartDT))`
` ,DurationI = STUFF(CONVERT(VARCHAR(20),StartDT-EndDT,114),1,2,DATEDIFF(hh,0,StartDT-EndDT))`
` ,cast(cast(StartDT as varbinary(128)) as bigint) As StartTicks`
` ,cast(cast(EndDT as varbinary(128)) as bigint) AS EndTicks`
` ,cast(cast(StartDT as varbinary(128)) as bigint)-cast(cast(EndDT as varbinary(128)) as bigint) as XmY`
` ,cast(cast(EndDT as varbinary(128)) as bigint)-cast(cast(StartDT as varbinary(128)) as bigint) as YmX`
`FROM #JBMTest;`
• There is one more thing to notice about the first query.
You assign the value "2000-02-01 12:34:56.789" to @EndDT but the query returns "2000-01-02 12:34:56.789".
Either make sure you are using "SET DATEFORMAT YMD" or use an unambiguous date format for the string literal "2000-02-01T12:34:56.789".
• Here is my simple method to format the output to a more readable format:
`CAST(GETDATE() - @StartDT AS TIME(2)) AS Duration`
• Do you have a good query for formatting duration (or age) in Years, Months, Days?
So often in my world, I am asked to find out who was the youngest or oldest to achieve something, or how long has it been since something has been achieved in a span of that range. I have a function that will do it, but I'd be interested to see how others have done it.
Of course I know that not all months have 30 days, so saying that someone is 16 years, 8 months, and 7 days old could be equivalent to 16 years, 8 months, and 10 days if months with 31 and 28 days are involved in the calculation. You also cannot just assume that a month is 30 days in your calculation since you could end up with 16 years, 12 months, and 4 days which would immediately look ridiculous and completely destroy your credibility. Number of days is the truest measure for comparison, but telling someone that the youngest was 6095 days old has no meaning to most people.
Anyone out there got a good method for YMD?
• I'm guessing the @EndDT for the first code example should have been.
`@EndDT = '2000-01-02 12:34:56.789'`
That gives the same answer as the example.
• Dennis Wagner-347763 (1/16/2014)
Do you have a good query for formatting duration (or age) in Years, Months, Days?
So often in my world, I am asked to find out who was the youngest or oldest to achieve something, or how long has it been since something has been achieved in a span of that range. I have a function that will do it, but I'd be interested to see how others have done it.
Of course I know that not all months have 30 days, so saying that someone is 16 years, 8 months, and 7 days old could be equivalent to 16 years, 8 months, and 10 days if months with 31 and 28 days are involved in the calculation. You also cannot just assume that a month is 30 days in your calculation since you could end up with 16 years, 12 months, and 4 days which would immediately look ridiculous and completely destroy your credibility. Number of days is the truest measure for comparison, but telling someone that the youngest was 6095 days old has no meaning to most people.
Anyone out there got a good method for YMD?
Which years in the calendar should we use to calculate this? The actual span between the two dates? Then comparisons between calculations made from different pairs of dates are not strictly compatible. The biggest reason is the occurrence of leap years within ranges. It might may a difference of only a day, but it sounds like for your applications, that may be important.
But, you could just do three DATEPART calculations and be done with it.
[font="Verdana"]Please don't go. The drones need you. They look up to you.[/font]
• Oliver Gugolz (1/16/2014)
Here is my simple method to format the output to a more readable format:
`CAST(GETDATE() - @StartDT AS TIME(2)) AS Duration`
But that only displays the fractional hours portion of the difference. You still lose the days converted to hours that Jeff's method preserves.
[font="Verdana"]Please don't go. The drones need you. They look up to you.[/font]
• My specific use of the datetime field is to see if it occurs
within a user specified date range.
The examples I received with the stored procedures (from software application vendor)
had the user enter full date and time (2014-01-14 23:59.59)when I am not interested in the time aspect.
I changed to have the user just enter the date ('2014-01-14') which defaults to zero time.
If inclusive start date and end date are entered, with the default zero time, I would miss
the last date, so I compare
between @startdate and @enddate + 1
which does include the entire enddate.
• Thanks for the feedback folks. I'm on my way to work and will look into your replies when I get home.
--Jeff Moden
RBAR is pronounced "ree-bar" and is a "Modenism" for Row-By-Agonizing-Row.
First step towards the paradigm shift of writing Set Based code:
________Stop thinking about what you want to do to a ROW... think, instead, of what you want to do to a COLUMN.
Change is inevitable... Change for the better is not.
How to post code problems
How to Post Performance Problems
Create a Tally Function (fnTally)
• Eirikur Eiriksson (1/16/2014)
Thank you for the spackle Jeff but it looks like there is a small crack in it as it makes the assumption that the end time is always greater than the start time. This is the reason why I prefer to use double cast, first to varbinary and then to bigint.
`SELECT`
` StartDT`
` ,EndDT`
` ,Duration = STUFF(CONVERT(VARCHAR(20),EndDT-StartDT,114),1,2,DATEDIFF(hh,0,EndDT-StartDT))`
` ,DurationI = STUFF(CONVERT(VARCHAR(20),StartDT-EndDT,114),1,2,DATEDIFF(hh,0,StartDT-EndDT))`
` ,cast(cast(StartDT as varbinary(128)) as bigint) As StartTicks`
` ,cast(cast(EndDT as varbinary(128)) as bigint) AS EndTicks`
` ,cast(cast(StartDT as varbinary(128)) as bigint)-cast(cast(EndDT as varbinary(128)) as bigint) as XmY`
` ,cast(cast(EndDT as varbinary(128)) as bigint)-cast(cast(StartDT as varbinary(128)) as bigint) as YmX`
`FROM #JBMTest;`
I got:
Msg 8115, Level 16, State 2, Line 15
Arithmetic overflow error converting expression to data type datetime.
when I ran it for date differences greater than about 148 years. Maybe not a problem for some implementations, but something to be aware of.
[font="Verdana"]Please don't go. The drones need you. They look up to you.[/font]
• Mikael Eriksson SE (1/16/2014)
There is one more thing to notice about the first query.
You assign the value "2000-02-01 12:34:56.789" to @EndDT but the query returns "2000-01-02 12:34:56.789".
Either make sure you are using "SET DATEFORMAT YMD" or use an unambiguous date format for the string literal "2000-02-01T12:34:56.789".
Thanks for the heads up, Mikael... it wasn't all that about unambiguous dates (which is whatI used)... it was a phat phinger during transfer of the code to the article submission site. I've submitted a correction that they'll, hopefully, be able to get to right away.
--Jeff Moden
RBAR is pronounced "ree-bar" and is a "Modenism" for Row-By-Agonizing-Row.
First step towards the paradigm shift of writing Set Based code:
________Stop thinking about what you want to do to a ROW... think, instead, of what you want to do to a COLUMN.
Change is inevitable... Change for the better is not.
How to post code problems
How to Post Performance Problems
Create a Tally Function (fnTally)
I'm guessing the @EndDT for the first code example should have been.
`@EndDT = '2000-01-02 12:34:56.789'`
That gives the same answer as the example.
Correct. I've submitted a correction. Apparently, I phat phingered the code during my final submittal to the article submittal software.
--Jeff Moden
RBAR is pronounced "ree-bar" and is a "Modenism" for Row-By-Agonizing-Row.
First step towards the paradigm shift of writing Set Based code:
________Stop thinking about what you want to do to a ROW... think, instead, of what you want to do to a COLUMN.
Change is inevitable... Change for the better is not.
How to post code problems
How to Post Performance Problems
Create a Tally Function (fnTally)
• Calculations are done using a calendar year.
For example if person A was born October 16, 1995, and we want to figure out how old he/she was on January 15, 2014, we can easily use DATEDIFF(day, '10/16/95', '1/15/14) to get 6,666 days, but we want the result to be formatted as 18 years, 2 months, 30 days. The years are easy, and the months aren't too bad, but the days calculation is based on December having 31 days, so there are 30 days between December 16, 2013 (18 years and 2 months after 10/16/95) and January 15, 2014. However, if we were calculating from 10/16/95 to March 15, 2014, our answer should be 18 years, 4 months, 27 days since there are only 28 days in February. And yes, you are correct that leap years do prevent an issue and need to be addressed.
The issue with doing simple DATEDIFF in SQL is that it does a straight subtraction. From the example above:
DATEDIFF(year, '10/16/95', '1/15/14') = 19 (but only 18 full years elapsed)
DATEDIFF(month, '10/16/95', '1/15/14')%12 = 3 (but only 2 full months elapsed)
Days is a real mess.
How about this: DATEDIFF(YEAR, '12/31/13', '1/1/14') = 1! (not the factorial) when only 1 day has elapsed
When comparing who is older when they achieved their Super Duper Fantastical Grandmaster Status, you have to use number of days because it is not affected by leap years, days in a month, etc.
So if person A above was compared to person B who was born December 14, 1995 and achieved this noteworthy status on March 15, 2014, then both of them would be 6,666 days old:
DATEDIFF(day, '10/16/95', '1/15/14') = DATEDIFF(day, '12/14/95', '3/15/14') = 6,666
However, when formatted as YMD, person A is 18 years, 2 months, 30 days while person B is 18 years, 3 months, 1 day, which would make person A seem younger. My original response states why you can't just fix 30 days in a month or you could end up with x years, 12 months, and 4 days when 364 days elapsed from the previous year anniversary.
I always use days when figuring out a "top 10" list, but I always format the answer as YMD so the age is meaningful to the audience.
• I can't spend much more time on the discussions that are unfolding until I get home from work tonight but the folks at SQLServerCentral have posted the correction to the first code window that a couple of you brought up. Thank you for the help there.
--Jeff Moden
RBAR is pronounced "ree-bar" and is a "Modenism" for Row-By-Agonizing-Row.
First step towards the paradigm shift of writing Set Based code:
________Stop thinking about what you want to do to a ROW... think, instead, of what you want to do to a COLUMN.
Change is inevitable... Change for the better is not.
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Reading page after page of boring content can cause the strongest minds to wander. Taking practice tests are a great way to break up the monotony of studying. Taking a practice test challenges you and keeps you interested in the material. Then you can review your test results and go over the questions you got wrong committing the right answer to memory. It’s a great, streamlined way to learn.
```If you look at more than one test in this book, you'll notice something--they are all basically the same, and have many of the same questions, just with different numbers. There's always a family of (number of people), with (number of dinner rolls), and the book asks how many rolls each family member will have to eat if they're split evenly. This question, while too easy for the arithmetic reasoning section in the first place, appears in every single test! Some of the paragraphs in the paragraph comprehension sections are repeated word for word, with a slight difference in question (that usually still doesn't make sense or has the incorrect answer listed as correct.)
```
Waves are described in terms of their height, wave-length, and period. “Height†is the vertical distance between the high point of a wave crest and the low point of the adjacent trough. “Wave-length†is the distance from one crest to the next, and “period†is the time it takes two adjacent crests to pass a fixed point, such as the end of a pier. The mathematics of wave theories are usually concerned with relationships between these and related characteristics.
As any professional military commander will tell you, knowing your enemy is the first step in winning a battle. After all, how can you expect to pass the ASVAB (Armed Services Vocational Aptitude Battery) if you don’t know what’s on the test? Here are some test-taking tips and key info about ASVAB test formats to help you ace the test, get into the service of your choice, and qualify for your dream job.
El ASVAB también produce varias partituras de línea, que son diferentes para cada rama de servicio. Puntajes de línea determinan qué trabajos militares estás calificado para. En las gráficas que muestran cómo estos puntajes se relacionan con trabajos específicos en cada rama de servicio, haga clic aquí para el artículo imprimir, Los ASVAB Decenas de Empleo Militar.
##### El ASVAB también produce varias partituras de línea, que son diferentes para cada rama de servicio. Puntajes de línea determinan qué trabajos militares estás calificado para. En las gráficas que muestran cómo estos puntajes se relacionan con trabajos específicos en cada rama de servicio, haga clic aquí para el artículo imprimir, Los ASVAB Decenas de Empleo Militar.
El Programa de Exploración de Carreras ASVAB CEP dura aproximadamente tres horas, abarca ocho asignaturas y consta de 200 preguntas. Actualmente, el ASVAB CEP es un examen que se hace con lápiz y papel. Si lo ofrece la escuela, los alumnos pueden hacer el examen ASVAB CEP en los grados 10, 11 y 12. Solo pueden realizarlo en la escuela a la que asisten, a menos que se hayan celebrado acuerdos especiales.
```Siempre existe la posibilidad que se le asigne un despliegue militar. Cuando surgen problemas en un estado, el gobernador puede reclutar a la Air National Guard para que realice ciertas tareas, entre ellas, rescates aéreos, misiones de respuesta médica o asistencia a las autoridades locales durante un desastre natural. Del mismo modo, el presidente puede reclutar a la Air National Guard para respaldar una misión de defensa interna del país. Se hace todo lo posible por emplear a voluntarios en primera instancia y, en algunos casos, se asignan despliegues a otras personas cuando no hay suficientes voluntarios disponibles para cumplir con los requisitos de la misión.
```
All test takers are given a summary results sheet that shows their percentile score in every test area. A percentile score of 50 means that a score was achieved that was better than 50 percent of all test takers. Percentile scores are given specifically for test takers of their gender and their grade level. Information obtained from the test is only shared with agencies within the Department of Defense. Test takers are informed that their specific scores will be used for up to two years for recruiting purposes. After two years, test scores will be used for research purposes only.
Direcciones: La subprueba de conocimientos de Word es la tercera subprueba del ASVAB. Las preguntas están diseñadas para medir su conocimiento del vocabulario. Verá tres tipos de preguntas en esta subprueba. El primer tipo, simplemente le pide que elija una palabra o palabras que más se quiere decir lo mismo que la palabra subrayada en la pregunta.
If you're prepping for the ASVAB in order to begin or advance your military career, you know how important it is to succeed. Inside this bestselling study guide, you get in-depth reviews of all nine test subjects you'll encounter on the ASVAB, foolproof strategies for making sense of the verbal, math, and general components, and expert tips and tricks to help you discover the areas where you need the most help. Plus, you get a one-year subscription to the online prep companion, where you can study whenever you want, take full-length practice exams, and create customized practice sets in the subjects you need to study the most.
##### “Me siento sumamento feliz, complacida y agradecida con Repaso ASVAB, Inc. Gracias a ellos pase con 42 % ( luego de sacar 14% en mi primer examen) ademas de cualificar para una variedad de trabajos. Ha sido la mejor inversión que hice en mi vida. Si estan buscando un lugar para tomar su repaso este es el indicado, NO TE DEJES ENGAÑAR por quien cobra menos y dicen dar más pase por eso y es frustrante. Sin embargo, llegue a Repaso ASVAB buscando hacer mi sueño realidad y lo he conseguido. Excelente equipo de trabajo, material completo, practicas en las cuales sientes esa adrenalina de esar en ese ambiente real del examen y por ultima una motivación increible. EXCELENTE OPORTUNIDAD NO TE ARREPENTIRAS… Army National Guard
A lot of people see the “CAT” term attached to the ASVAB test when they first start looking into the test and aren’t sure what that stands for. This term is an acronym and it stands for “Computerized Adaptive Test”. There are three different versions of the ASVAB test. The CAT is available at military processing stations for enlisting soldiers. The pencil and paper (also known as the P&P or S-ASVAB) version of the test is available for high school and college students who may not actually enlist. The third type of ASVAB test is the MET-ASVAB, or Mobile Examination Test, which is available only for enlisted soldiers at mobile testing centers (this test is also done with paper and pencil).
The General Science section of the test covers earth, space, and physical and life sciences. Because science is such a vast and dynamic topic, focus your study on basic principles. This gives you a good foundation to work through any question that is asked of you. Typical questions may include: “Why is air less dense than water?” or “How do you convert Celsius temperature to Fahrenheit?” The CAT-ASVAB test asks 16 questions in 8 minutes, while the pencil-and-paper version asks 25 questions in 11 minutes.
La Información de automóviles y talleres evalúa la aptitud para el mantenimiento y la reparación de automóviles, y las prácticas de los talleres de carpintería y metales. El examen abarca varias áreas que se suelen incluir en la mayoría de los cursos de automóviles y talleres de las escuelas secundarias, como componentes automotrices, sistemas automotrices, herramientas automotrices, solución de problemas y reparación, herramientas de taller, materiales de construcción y procedimientos de construcción.
En realidad, nadie se preocupa por el AFQT excepto los militares - y que se preocupa ¡mucho! Si estás interesado en formar parte de los militares, que es más probable que tomar la versión computarizada del ASVAB. Eso es porque la mayoría de los que tomaron el ASVAB con el propósito de unirse a los militares llevarlo a una entrada Militar procesamiento de la estación (MEPS), y todos estos lugares utilizar el examen computarizado.
A. Para encontrar el volumen de la caja de arena, se toma la longitud por la altura de los tiempos anchura (V = lwh). No se olvide que las medidas son para un arenero cuadrado, por lo que puede suponer que si la caja es de 5 pies de largo, entonces es también 5 pies de ancho. Así que 5 x 5 x 1 es de 25 pies cúbicos. Cada bolsa tiene 1 pie cúbico de arena, y 25 ÷ 1 = 25. La opción (A) es la respuesta correcta. Si estabas pensando en respuesta Choice (B) sonaba bien, recuerda que la respuesta debe tener sentido. Cinco pies cúbicos de arena no llenarían una gran caja de arena, ¿verdad?
```Asegúrese de que usted sabe lo que la pregunta quiere de ti y luego darle la pregunta lo que quiere. Si el problema pide la suma de dos números, no se multiplican los números. No se debe confundir una señal de división para un signo de adición. Al familiarizarse con los tipos de preguntas sobre el ASVAB, que será capaz de concentrarse en lo que se supone que hacer mucho más rápidamente.
```
## A diferencia de la versión de lápiz y papel, no se puede saltar preguntas o cambiar sus respuestas después de que los introduzca en la prueba CAT-ASVAB. Esta restricción puede hacer que tomar la prueba más difícil para algunas personas. En lugar de ser capaz de pasar por e inmediatamente responder a todas las preguntas que usted está seguro de y luego volver a las preguntas que le obliguen a hacer algo de rascarse la cabeza, tiene que responder a cada pregunta, ya que viene.
No se debe confundir una puntuación de serie con el graduado-on-a-curva de puntuación que puede haber visto en las pruebas escolares - donde las puntuaciones van de 1 a 100, con la mayoría de los estudiantes con calificaciones entre 70 y 100. Con puntuaciones estándar, la mayoría puntuación está entre 30 y 70. eso significa que una puntuación estándar de 50 es una puntuación media y que una puntuación de 60 es una puntuación superior a la media.
```The Paragraph Comprehension section of the test measures your ability to read a passage and interpret the information contained within it. You may read a selection and be asked to interpret the author’s purpose, or what a particular word in the passage means, based on the context of the sentence where it appears. To help you better prepare for the exam, the Paragraph Comprehension section of the ASVAB practice test has passages of similar length and style to those on the actual ASVAB test. The CAT-ASVAB test has 11 questions in 22 minutes; the paper-and-pencil version has 15 questions in 13 minutes.
```
El ASVAB también produce varias puntuaciones de línea, que son diferentes para cada rama de servicio. las puntuaciones de línea determinan qué trabajos militar que está calificado para. Para los gráficos que muestran cómo estos resultados se relacionan con trabajos específicos en cada rama de servicio, haga clic aquí para el artículo imprimir, Los puntajes del ASVAB para trabajos militares.
Después de que vuelva a comprobar sus cálculos, decidir que opción (C) es correcta y marcarla en la hoja de respuestas, no cambie su respuesta en la versión en papel de la ASVAB! Es casi seguro para cambiar una respuesta correcta a una persona equivocada cuando se juega ese juego. Además, puede volverse loco por segunda adivinar su decisión. Marque la respuesta y seguir adelante.
The adaptive test is exactly what it sounds like, a test that adapts to the test taker. Questions in each section range in difficulty, and based on the test taker’s answers, the question types and difficulties adjust with each answer. This type of test allows for a greater level of accuracy in evaluating a prospect’s abilities. For the most part, correct answers beget more difficult questions while wrong answers will reduce the difficulty of proceeding questions.
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# How to Build a Ping Pong Projectile Launcher for Maximum Distance?
• Ah Yiife
In summary, the OP is looking for a ping pong launcher that can be used to launch a ball up to 10 meters. He is considering a catapult, trebuchet, and ballista.
Ah Yiife
OP has been warned about posting schoolwork/homework without using the HH Template and showing some effort.
1. The shape of the ping pong launcher I think is 'catapult' but I don't know the suitable size and the materials need to use. The ping pong ball need to be launch to a distance of 4m and 10m respectively.
What are the size and materials used is most suitable to launch the ping pong ball this far ?
Last edited by a moderator:
What does your "respectively" mean here? Are you building two different lauchers? I would certainly start with a pipe just large enough to fit a ping pong ball into. Then you have to decide how you are going to propel the balls. For example are you going to use gun powder or other explosive propellant? (Might be difficult to get a teacher's permission for that!) Are you going to use compressed gas, like air?
HallsofIvy said:
What does your "respectively" mean here? Are you building two different lauchers? I would certainly start with a pipe just large enough to fit a ping pong ball into. Then you have to decide how you are going to propel the balls. For example are you going to use gun powder or other explosive propellant? (Might be difficult to get a teacher's permission for that!) Are you going to use compressed gas, like air?
Sorry, my English is poor haha. Teacher suggests using the wood to make the catapult. 'Respectively' I mean is need to launch the ping pong ball twice. First distance is 4m and second distance is 10m. I think I am not going to use the gun powder and explosive propellant.
"Catapult" encompasses several designs.
A ballista is like a large crossbow, the energy being stored in the bending of a beam or twisting of ropes/sinews.
A trebuchet throws the missile from the end of a rotating arm, usually powered by a heavy weight on the other end.
Thread is obviously locked. OP has been warned to re-post and show some actual work and effort on his/her schoolwork.
## 1. What is a ping pong projectile launcher?
A ping pong projectile launcher is a device used to launch ping pong balls at a high speed and with precision. It is often used for experiments or demonstrations in physics classes.
## 2. How does a ping pong projectile launcher work?
A ping pong projectile launcher uses stored elastic potential energy to launch the ping pong ball. The ball is placed in a cup or holder, and a rubber band or spring is pulled back and released, propelling the ball forward.
## 3. What are the components of a ping pong projectile launcher?
The main components of a ping pong projectile launcher include a launching mechanism, such as a rubber band or spring, a cup or holder for the ping pong ball, and a base or stand to stabilize the launcher.
## 4. What is the purpose of a ping pong projectile launcher?
The purpose of a ping pong projectile launcher is to demonstrate the principles of projectile motion and energy transfer. It can also be used for experiments involving force, velocity, and trajectory.
## 5. Is a ping pong projectile launcher safe to use?
As with any scientific device, it is important to follow safety precautions when using a ping pong projectile launcher. This includes wearing appropriate protective gear, following instructions carefully, and using caution when handling the launcher and ping pong balls.
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Functions and CALL Routines
# FINV Function
Returns a quantile from the F distribution.
Category: Quantile
## Syntax
FINV (p, ndf, ddf <,nc>)
### Arguments
p
is a numeric probability.
Range: 0 p < 1
ndf
is a numeric numerator degrees of freedom parameter.
Range: ndf > 0
ddf
is a numeric denominator degrees of freedom parameter.
Range: ddf > 0
nc
is an optional numeric noncentrality parameter.
Range: nc 0
The FINV function returns the pth quantile from the F distribution with numerator degrees of freedom ndf, denominator degrees of freedom ddf, and noncentrality parameter nc. The probability that an observation from the F distribution is less than the quantile is p. This function accepts noninteger degrees of freedom parameters ndf and ddf.
If the optional parameter nc is not specified or has the value 0, the quantile from the central F distribution is returned. The noncentrality parameter nc is defined such that if X and Y are normal random variables with means and 0, respectively, and variance 1, then has a noncentral F distribution with nc = 2.
CAUTION:
For large values of nc, the algorithm could fail. In that case, a missing value is returned.
Note: FINV is the inverse of the PROBF function.
These statements compute the 95th quantile value of a central F distribution with 2 and 10 degrees of freedom and a noncentral F distribution with 2 and 10.3 degrees of freedom and a noncentrality parameter equal to 2:
SAS Statements Results
`q1=finv(.95,2,10);`
`4.1028210151`
`q2=finv(.95,2,10.3,2);`
`7.583766024`
Functions:
Previous Page | Next Page | Top of Page
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Ex.8.2 Q10 Compairing Quantities - NCERT Maths Class 7
Question
A local cricket team played $$20$$ matches in one season. It won $$25\%$$ of them. How many matches did they win?
Video Solution
Comparing Quantities
Ex 8.2 | Question 10
Text Solution
What is Known?
Matches played by a cricket team in a season and percentages of matches won.
What is Unknown?
Number of matches won by the tean.
Reasoning:
Matches won by local cricket team is $$25\%$$ of $$20.$$
Steps:
Matches won
\begin{align}&={\text{ 25% }}\,{\text{of}}\,{\text{20}}\\&= \frac{{25}}{{100}} \times 20\\&= 5\end{align}
So, the matches won are $$5.$$
Learn from the best math teachers and top your exams
• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
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# Tagged Questions
Quantization refers to the procedure or methodology for replacing a classical system by a quantum system. If the question is about the quantized or discrete behavior of a phenomenon use the [tag:discrete] instead.
5answers
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### Is it a coincidence that quantum harmonic oscillators and photons have energy quantised as $E=hf$?
I have studied the quantum harmonic oscillator and solved the Schrodinger equation to find the eigen-energies given by $$E_n = \left(n+\frac{1}{2}\right)\hbar \omega.$$ Which means the energy ...
1answer
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### In quantum mechanics, how exactly do we associate Hermitian operators to classical observables? [duplicate]
In a first course on quantum mechanics, everybody learns some version of the following statement: Postulate: To every classical observable $A$ of a physical system, there corresponds a Hermitian ...
1answer
75 views
### Is the standard model a quantized gauge theory?
I have studied some quantum field theory and gauge theory but I am definitely not an expert. I am aware that in quantizing electrodynamics one has to fix a gauge. I have read that for general gauge ...
0answers
73 views
### Quantization of non-variational systems?
In undergraduate courses the introduction to Hamiltonian mechanics usually starts from a Newtonian view point. One has equations of motions of the form (not sure if it is ok to use covariant notation ...
0answers
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### What´s the physical foundation of the assumption that the curvature of spacetime can be quantised? [duplicate]
At the moment different paths (by percentual very few people in the world) are taken to arrive (that is, if an arrival exists) at a theory that can quantise the curvature of spacetime. Considering the ...
0answers
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### Geometric quantization of field theories and resulting statistics
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1answer
115 views
### Quantum systems without a classical analogue? [closed]
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0answers
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### An intuition on the Rindler modes
When we are solving the Klein-Gordon equation for the quantization of a massive scalar field on the Minkowski spacetime, we may use the global coordinates and obtain the usual quantization with plane ...
0answers
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### How to get anti-commuting rule from the view of field?
I was reading the 1951 Lectures on Advanced Quantum Mechanics and I found something really disturbing. That's the anti-commuting rule mentioned on Page 40 at last. Though it was named as Quantum ...
1answer
187 views
### Quantization on Minkowski/Schwarzschild spacetimes based on unusual surface
I'm reading the book of Wald "Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics", and I'm pondering on this problem: In Minkowski spacetime, we usually quantize our fields with ...
1answer
93 views
### flux quantization in superconducting ring
I am trying to understand SQUID microscopy from the ground up, so I am starting with flux quantization in superconducting rings. I found a nice presentation that covers some of the details, but I am ...
1answer
44 views
### In the Holometer experiment, why would one of the split laser beams arriving back at a slightly different time indicate the universe was quantized?
All the pop-sci articles I've read have a description of the set-up similar to this: It uses a pair of laser interferometers placed close to one another, each sending a one-kilowatt beam of light ...
1answer
55 views
### Feynman Path Integral as a Quantization Scheme
Why isn't the path integral usually discussed as a quantization scheme, like geometric and deformation quantization? Was searching wikipedia for this.
1answer
66 views
### How to impose canonical commutation relations when quantising a field
I believe this is a simple question, however I cannot find it explained anywhere what the term: "Impose canonical commutation relations" means. If I have a classical equation, and I wish to quantise ...
0answers
77 views
### Regge trajectory and Kaluza Klein tower
The mass of hadrons in the Regge trajectory scales as $m=\sqrt{\frac{J}{\alpha}-\alpha_0}=\sqrt{\frac{n}{\alpha}-\alpha_0}\propto \sqrt{n}$, where $J=n$ is the spin of the particle (in natural units,...
1answer
407 views
### $\pi$, $\sigma$ - atomic transitions with respect to the magnetic field axis
I am confused about the atomic transition with different polarized lights. I post the pictures as follows. There are four cases. In case 1, the right-handed circular polarized light ...
1answer
99 views
### Questions about Quantization and Noncommutative Geometry
I am trying to orient myself among the vast amount of literature, trying to study the prerequisites necessary for gauge theory and theoretical physics. I have an undergraduate degree in mathematics ...
0answers
19 views
### Why spurious pulses are likely in partial discharges?
My notes The gas multiplication in the proportional counters is based on the secondary ionization created in collisions between electrons and neutral gas molecules, resulting in some visible ...
3answers
451 views
### Difference between discretization and quantization in physics
I am just trying to understand the fundamental difference between these two concepts in physics: From discreteness of some quantity: one usually interprets it as a quantity being only able to take ...
1answer
231 views
### Planck's constant and phase space in quantum mechanics
During my undergrad physics classes, I've come across several seemingly related phenomena dealing with $h$ and phase space in quantum mechanics. Let $T_x$ be a translation operator by $x$ in ...
0answers
42 views
### References on deformation quantization
I'm looking for books or introductory review papers or lecture notes on the topic of deformation quantization. (And preferably, geometric quantization as well.) I'm mainly interested in the ...
0answers
126 views
### What new does geometric or deformation quantization give to physics? [closed]
What new does geometric quantization or deformation quantization give to physics? For example: prediction of new physical phenomena or just better tool for quantization. What can these schemes do in ...
0answers
61 views
### Projection that keeps graviton but gets rid of B-field
We consider the closed superstring and the massless states in the (NS,NS)-sector $$\tilde{b}^i_{-\frac{1}{2}} |0\rangle_L \otimes b^j_{-\frac{1}{2}} |0\rangle_R.$$ It is ...
1answer
119 views
### Discontinuity of paths in phase space path integrals
Berezin's famous paper "Feynman path integrals in a phase space" discusses the space of paths on which the phase space path integral is concentrated. In particular, these paths are known to be ...
0answers
60 views
### The relation between commutation and quanta
This question discusses discretization in some sense, and this question talks about how quantization and Hilbert Spaces are related (the answer seems to to be not at all), but what I'm curious about ...
2answers
112 views
### Why is the introduction of a quantization volume necessary for quantization of the EM field
I have been working through the quantization of the electromagnetic field, and every source I find introduces a quantization volume with periodic boundary conditions in the process, in which we fit ...
1answer
231 views
### How is quantization related to commutation? [duplicate]
How are commutation (of observables) and quantization related? Reading about the Stone-Von Neumann Theorem, it seems that commutativity is the classical limit of quantum mechanics, and hence non-...
0answers
59 views
### Analogy between a classical discrete system and non classical continous system
Most introduction textbooks about quantum fieldtheory start with a discrete classical harmonic oscillator and then looks at it in the continuous quantized case (quantized field). This leads to the ...
1answer
113 views
### Canonical commutation relations in Light-cone gauge
It seems that when trying to identify the physical degrees of freedom for the string some authors$^1$ use: $$q^-=\frac{1}{\ell}\int_0^{\ell} X^-(\tau,\sigma)d\sigma$$ Then, the commutation relation ...
1answer
127 views
1answer
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### Volume factor in Faddeev-Popov quantisation
In Faddeev-Popov quantisation, why does the integral over gauge parameter cancel the volume factor of the gauge group that's in the denominator? In fact, I don't understand where the volume factor ...
1answer
374 views
### What is “momentum density” and why it important to QFT?
I am reading Quantum Field Theory for the Gifted Amateur. On page 98, they provide a summary of a basic canonical quantization procedure: Step I: Write down a classical Lagrangian density in ...
2answers
179 views
### How to associate a Hilbert space with a QM system?
I couldn't really find a fitting title for this question. I'm still relatively new to QM and am trying to get the basics down. I understand that a physical system is associated with a Hilbert Space, ...
1answer
98 views
3answers
96 views
### Quantization conditions/ Real Scalar field
It is often written in books, the quantization conditions for classical field theory leading to Lagrangian of a real scalar field and thus to Klein Gordon equation. And these are introduced by ...
0answers
62 views
### Motivating Irreducibility of Hilbert Space for Quantization Axioms
In the context of geometric quantization, we usually look for a map from the Poisson algebra of classical observables to the algebra of quantum observables (or rather, a sub-algebra of the classical ...
3answers
267 views
### Classical Hamiltonian involving product of factors whose quantum analogues don't commute
Dirac remarked in his quantum mechanics book: One can usually assume that the Hamiltonian is the same function of the canonical coordinates and momenta in the quantum theory as in the ...
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# vector projection formula proof
Thus (−) − ((−) ⋅) is the component of − perpendicular to the line. columns. From physics we know W=Fd where F is the magnitude of the force moving the particle and d is the distance between the two points. The vector projection formula can be written two ways, as shown below. Example 1: Let S be the 2‐dimensional subspace of R 3 spanned by the orthogonal vectors v 1 = (1, 2, 1) and v 2 = (1, −1, 1). Let → be a vector in and let be a subspace of with basis →, …, → . We will need some of this material in the next chapter and those of you heading on towards Calculus III will use a fair amount of this there as well. Computations involving projections tend to be much easier in the presence of an orthogonal set of vectors. Operator of orthogonal projection Let W be an inner product space and V be a subspace such that V ⊕V⊥ = W. Then we can define the operator P V of orthogonal projection onto V. Namely, any vector x ∈ W is uniquely represented as x = p+o, where p ∈ V and o ∈ V⊥, and we let P V(x) = p. V V⊥ o p x. The 3D rendering pipeline we are using defines transformations of vertex positions that go from clip-space to window space. This more general formula is not restricted to two dimensions. In mathematics, the dot product or scalar product is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors), and returns a single number.In Euclidean geometry, the dot product of the Cartesian coordinates of two vectors is widely used. In linear algebra and functional analysis, a projection is a linear transformation from a vector space to itself such that =.That is, whenever is applied twice to any value, it gives the same result as if it were applied once ().It leaves its image unchanged. Then P = A(ATA) 1AT Your textbook states this formula without proof in Section 5.4, so I thought I’d write up the proof. Vector projection: Projectionᵥw, read as "Projection of w onto v". Theorem 3.8. Vector projection and vector rejection are highly common and useful operations in mathematics, information theory, and signal processing. Generalize to . Thus CTC is invertible. So it's 3, minus 2. There's a bunch of spanning vectors for your row space. vector by a row vector instead of the other way around. Definition. A vector projection proof. The prior subsections project a vector onto a line by decomposing it into two parts: ... We can find the orthogonal projection onto a subspace by following the steps of the proof, but the next result gives a convienent formula. Subsection 6.4.1 Orthogonal Sets and the Projection Formula. Eine Orthogonalprojektion (von gr. ὀρθός orthós gerade, γωνία gōnía Winkel und lat. Now, I know enough about linear algebra to know about projections, dot products, spans, etc etc, so I am not sure if I am reading too much into this, or if this is something that I have missed. In mathematics, the scalar projection of a vector on (or onto) a vector , also known as the scalar resolute of in the direction of , is given by: = ‖ ‖ = ⋅ ^, where the operator ⋅ denotes a dot product, ^ is the unit vector in the direction of , ‖ ‖ is the length of , and is the angle between and .. Problem 11. In that case, there is only one vector in the basis (m= 1), and Ais just the column vector ~vviewed as an n 1 matrix. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Let C be a matrix with linearly independent columns. I am trying to understand how - exactly - I go about projecting a vector onto a subspace. In this case, this means projecting the standard coordinate vectors onto the subspace. Figure shows geometrically why this formula is true in the case of a 2‐dimensional subspace S in R 3. This is just the one we happened to pick. Vector addition is defined as the geometrical sum of two or more vectors as they do not follow regular laws of algebra. For the video and this page, you will need the definitions and mathematics from Vectors and dot products. So the projection of the vector 3, 0 onto our row space, which is a line so we can use that formula, it is equal to 3, 0 dot the spanning vector for our row space, right? Suppose ~vis the line spanned by ~v. If b is perpendicular to the column space, then it’s in the left nullspace N(AT) of A and Pb = 0. Operator of orthogonal projection Theorem 1 PV is a linear operator. Let the vectors $${\bf u}_1 , \ldots {\bf u}_n$$ form a basis for the range of the projection, and assemble these vectors in … Note as well that while the sketch of the two vectors in the proof is for two dimensional vectors the theorem is valid for vectors of any dimension (as long as they have the same dimension of course). Suppose CTCb = 0 for some b. bTCTCb = (Cb)TCb = (Cb) •(Cb) = Cb 2 = 0. Also, check: Vector Projection Formula. Figure 2. This exercise is recommended for all readers. Chapter 5 : Vectors. the minimum of (3.6). Scalar multiplication of vectors satisfies the distributive property, and the zero vector acts as an additive identity. Recall that our destination image, the screen, is just a two dimensional array of pixels. Orthogonal Projection Matrix •Let C be an n x k matrix whose columns form a basis for a subspace W = −1 n x n Proof: We want to prove that CTC has independent columns. The version on the left is most simplified, but the version on the right makes the most sense conceptually: The proof of the vector projection formula is as follows: Given two vectors , what is ? Thus, the scalar projection of b onto a is the magnitude of the vector projection of b onto a. In other words, the vector projection is defined as a vector in which one vector is resolved into two component vectors. I describe them bellow. The formula from this theorem is often used not to compute a dot product but instead to find the angle between two vectors. Find the formula for the distance from a point to a line. First note that the projected vector in red will go in the direction of . dot product: Two vectors are orthogonal if the angle between them is 90 degrees. The vector projection of a vector a on a nonzero vector b is the orthogonal projection of a onto a straight line parallel to b. Vector projection - formula The vector projection of a on b is the unit vector of b by the scalar projection of a on b : Very important! Cb = 0 b = 0 since C has L.I. Example: To convince you that this formula is believable, let’s see what it tells us in the simple case where V is one-dimensional. prōicere, PPP prōiectum vorwärtswerfen), orthogonale Projektion oder senkrechte Projektion ist eine Abbildung, die in vielen Bereichen der Mathematik eingesetzt wird. I couldn't understand them easily, so I took my time to do it myself, the good thing is that I can now detail it in an ELI5 fashion! The distance from the point to the line is then just the norm of that vector. Though abstract, this definition of "projection" formalizes and generalizes the idea of graphical projection. I did develop the formula using the 3 steps shown in the graphic. Projection matrices and least squares Projections Last lecture, we learned that P = A(AT )A −1 AT is the matrix that projects a vector b onto the space spanned by the columns of A. We will be taking a brief look at vectors and some of their properties. In (3.10) we take the derivatives of a vector @S @b with respect to another vector (b0) and we follow the convention to arrange these derivatives in a matrix (see Exercise 3.2). Let P be the point with coordinates (x 0 ... is a vector that is the projection of − onto the line. Thanks to A2A An important use of the dot product is to test whether or not two vectors are orthogonal. Example Suppose you wish to find the work W done in moving a particle from one point to another. This is a fairly short chapter. (Note that you still need to nd a basis!) Projection Formula. This here page follows the discussion in this Khan academy video on projection.Please watch that video for a nice presentation of the mathematics on this page. Here is the result: Let A be the matrix with columns ~v i. Example (Matrix of a projection) Example (Matrix of a projection) Example (Matrix of a projection) In the previous example, we could have used the fact that. Oblique projections are defined by their range and null space. Problem 12. An alternative proof that b minimizes the sum of squares (3.6) that makes no use of first and second order derivatives is given in Exercise 3.3. The vector projection is used to find the component of the vectors along with the direction. The resultant vector is known as the composition of a vector. Find the scalar such that (,) is a minimum distance from the point (,) by using calculus (i.e., consider the distance function, set the first derivative equal to zero, and solve). Vector projection¶. A formula for the matrix representing the projection with a given range and null space can be found as follows. The proofs to verify these properties in three dimensions are straightforward extensions of the proofs in two dimensions. However, this formula, called the Projection Formula, only works in the presence of an orthogonal basis. Notice that: When you read it, it’s in a reverse order! Once the positions are in window space, 2D triangles are rendered. We will also present the Gram–Schmidt process for turning an arbitrary basis into an orthogonal one. We know that vectors have both magnitude and direction. In this paper, we find the distribution of the norm of projection and rejection vectors when the original vectors are standard complex normally distributed. Another vector formulation. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Of an vector projection formula proof basis be much easier in the graphic γωνία gōnía Winkel lat! Of their properties the video and this page, you will need the and... The component of − onto the line - exactly - i go about projecting a vector is. 2 ) and ( 3 ) in moving a particle from one point to the line exactly i. Projections are defined by their range and null space orthogonal if the angle between them is 90 degrees ) the... Addition is defined as a vector that is the result: let a be the point the... If the angle between them is 90 degrees properties in three dimensions are straightforward extensions of the dot product two. 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Remark ( Simple proof for the distance from the point to a line ) by... Never be added once the positions are in window space, 2D triangles are.... That: When you read it, it ’ s in R.., …, → by a row vector instead of the proofs to verify properties... The dot product is to test whether or not two vectors are orthogonal if the angle between them 90! The geometrical sum of two or more vectors as they do not follow regular laws of algebra how to the. Proofs in two dimensions is then just the norm of that vector the distributive,! Or more vectors as they do not follow regular laws of algebra resultant vector known. Formula, called the projection formula, only works in the direction of of projection. Computations involving projections tend to be much easier in the presence of an orthogonal set of satisfies! 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The composition of a 2‐dimensional subspace s in a reverse order 0 b = 0 b = 0 b 0... ( note that the projected vector in which one vector is resolved into two component.! A two dimensional array of pixels are using defines transformations of vertex positions that go from to... Still need to nd a basis! idea of graphical projection using defines transformations of positions. By a row vector instead of the other way around example Suppose you wish to find the work w in... Not two vectors are orthogonal example Suppose you wish to find the for! Was trying to understand how - exactly - i go about projecting a vector in will... Other words, the vector projection: Projectionᵥw, read as vector projection formula proof of! Are a few conditions that are applicable for any vector addition is defined as the geometrical sum of or! Vectors as they do not follow regular laws of algebra are: Scalars and vectors can be. →, …, → generalizes the idea of graphical projection present the Gram–Schmidt process for an! Eingesetzt wird are: Scalars and vectors can never be added works in the direction of vectors your. Scalar multiplication of vectors the standard coordinate vectors, this means projecting standard... And generalizes the idea of graphical projection there 's a bunch vector projection formula proof spanning vectors for your space... Die in vielen Bereichen der Mathematik eingesetzt wird a brief look at vectors and some their... Basis! and this page, you will need the definitions and mathematics from vectors and some of properties! The matrix with columns ~v i go in the case of a vector the graphic defined as vector!, read as projection of b onto a projection onto a and some of their properties other around. Then just the norm of that vector from a point to a line projection...
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# Math PLZ HELP
posted by .
1. Which of the following expressions is written in scientific notation?
1. 73.4 x 105
2. 0.09 × 107
3. 80 x 103
4. 4.22 x 10–3
2. Which of the following is 0.0000000708 written in scientific notation?
1. 7.08 x 10–8
2. 7.8 x 10–8
3. 708 x 10–10
4. 70.08 x 10–9
3. Which expression represents the largest number?
1. 40.1 x 10–6
2. 4.1 x 10–7
3. 0.411 x 10–7
4. 0.04001 x 10–5
4. Which expression is equal to 1/8000 written in scientific notation?
1. 8.0 x 103
2. 1.25 x 10–4
3. 125 x 10–5
4. 1.25 x 104
• Math PLZ HELP -
if you know anything about scientific notation, #1,2 should be easy.
For #3, convert all to same power of 10, then it is easy to compare.
#4: 1/8000 = .000125
• Math PLZ HELP -
1.)A
2.)C
3.)B
4.)B
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Question about snake game aliensoldier Member Posts: 85 Threads: 16 Joined: Nov 2023 Reputation: 2 07-19-2024, 05:48 PM I have a question about the snake game. The snake's body is divided into small pieces and they follow the head. How can I make the pieces that make up the body follow the head of the snake when I move it? Marcus Administrator Posts: 316 Threads: 39 Joined: Nov 2023 Reputation: 3 07-20-2024, 09:23 AM (This post was last modified: 07-20-2024, 09:25 AM by Marcus.) Here's a port of an n6 example game. Hopefully you can find some clues by looking at the code I don't remember why I implemented the game like this, and probably there are smarter ways. Code:```' Port of n6 example game. SNAKE_MAX_LENGTH = 30*40 UP = 1 DOWN = 2 LEFT = 3 RIGHT = 4 snake = dim(SNAKE_MAX_LENGTH, 2) level = dim(40, 30) set window "Smape", 320, 240, false, 2 set redraw off do snakeLength = 1 snakeHead = 0 snakeDirection = 0 speed = 10 speedCounter = speed snake[snakeHead][0] = 20 snake[snakeHead][1] = 15 for y = 0 to 29 for x = 0 to 39 level[x][y] = 0 gameOver = false nextPlupTime = clock() do if clock() >= nextPlupTime if rnd(3) = 0 level[rnd(40)][rnd(30)] = 2 else level[rnd(40)][rnd(30)] = 1 nextPlupTime = clock() + 2000 + rnd(2000) endif if keydown(38) and snakeDirection <> DOWN snakeDirection = UP if keydown(40) and snakeDirection <> UP snakeDirection = DOWN if keydown(37) and snakeDirection <> RIGHT snakeDirection = LEFT if keydown(39) and snakeDirection <> LEFT snakeDirection = RIGHT speedCounter = speedCounter - 1 if speedCounter = 0 prev = snakeHead if snakeDirection > 0 snakeHead = (snakeHead + 1)%SNAKE_MAX_LENGTH speedCounter = speed if snakeDirection = UP snake[snakeHead][0] = snake[prev][0] snake[snakeHead][1] = snake[prev][1] - 1 elseif snakeDirection = DOWN snake[snakeHead][0] = snake[prev][0] snake[snakeHead][1] = snake[prev][1] + 1 elseif snakeDirection = LEFT snake[snakeHead][0] = snake[prev][0] - 1 snake[snakeHead][1] = snake[prev][1] elseif snakeDirection = RIGHT snake[snakeHead][0] = snake[prev][0] + 1 snake[snakeHead][1] = snake[prev][1] endif x = snake[snakeHead][0] y = snake[snakeHead][1] if x < 0 or x >= 40 or y < 0 or y >= 30 gameOver = true else if level[x][y] > 0 snakeLength = snakeLength + 1 if level[x][y] = 2 speed = max(speed - 1, 0) level[x][y] = 0 endif endif endif set color 0, 0, 0 cls for y = 0 to 29 for x = 0 to 39 if level[x][y] = 1 set color 0, 255, 0 draw rect x*8, y*8, 8, 8 , true elseif level[x][y] = 2 set color 255, 0, 0 draw rect x*8, y*8, 8, 8, true endif next next set color 255, 255, 255 set caret 0, 0 x = snake[snakeHead][0] y = snake[snakeHead][1] for i = snakeHead to snakeHead - snakeLength + 1 j = i if j < 0 then j = j + SNAKE_MAX_LENGTH draw rect snake[j][0]*8, snake[j][1]*8, 8, 8, true if i <> snakeHead if snake[j][0] = x and snake[j][1] = y then gameOver = true endif next wait 10 redraw until gameOver loop``` Attached Files smape.n7 (Size: 3.14 KB / Downloads: 3) johnno56 Senior Member Posts: 304 Threads: 31 Joined: Nov 2023 Reputation: 3 07-20-2024, 10:46 AM (This post was last modified: 07-20-2024, 10:47 AM by johnno56.) Love the "classics"... Could not help myself... Added a few minor upgrades: Little bit of colour; Score/Hiscore; Quit level and restart (ESC); Quit game (Q). Note: No external hiscore. Scoring has been modified. "Green" food is the normal 10 points. "Red" food is worth 100 points. Note: The speed of the snake will still increase... Moo Ha Ha Ha... snake2.n7 (Size: 4.31 KB / Downloads: 3) Challenge: Sound and messages? J Logic is the beginning of wisdom. Marcus Administrator Posts: 316 Threads: 39 Joined: Nov 2023 Reputation: 3 07-20-2024, 12:42 PM (This post was last modified: 07-20-2024, 12:47 PM by Marcus.) (07-20-2024, 10:46 AM)johnno56 Wrote: Love the "classics"... Could not help myself... Added a few minor upgrades: Little bit of colour; Score/Hiscore; Quit level and restart (ESC); Quit game (Q). Note: No external hiscore. Scoring has been modified. "Green" food is the normal 10 points. "Red" food is worth 100 points. Note: The speed of the snake will still increase... Moo Ha Ha Ha... Challenge: Sound and messages? J Would be interesting to modify the visuals so that the snake uses tiles with rounded corners where it turns etc. But I'm too busy with that game jam thing (right now I'm sitting on a bench at the mall though, not jamming at all, wife and youngest daughter running around in clothes stores ...). aliensoldier Member Posts: 85 Threads: 16 Joined: Nov 2023 Reputation: 2 07-20-2024, 06:50 PM (07-20-2024, 09:23 AM)Marcus Wrote: Here's a port of an n6 example game. Hopefully you can find some clues by looking at the code I don't remember why I implemented the game like this, and probably there are smarter ways. Code:```' Port of n6 example game. SNAKE_MAX_LENGTH = 30*40 UP = 1 DOWN = 2 LEFT = 3 RIGHT = 4 snake = dim(SNAKE_MAX_LENGTH, 2) level = dim(40, 30) set window "Smape", 320, 240, false, 2 set redraw off do snakeLength = 1 snakeHead = 0 snakeDirection = 0 speed = 10 speedCounter = speed snake[snakeHead][0] = 20 snake[snakeHead][1] = 15 for y = 0 to 29 for x = 0 to 39 level[x][y] = 0 gameOver = false nextPlupTime = clock() do if clock() >= nextPlupTime if rnd(3) = 0 level[rnd(40)][rnd(30)] = 2 else level[rnd(40)][rnd(30)] = 1 nextPlupTime = clock() + 2000 + rnd(2000) endif if keydown(38) and snakeDirection <> DOWN snakeDirection = UP if keydown(40) and snakeDirection <> UP snakeDirection = DOWN if keydown(37) and snakeDirection <> RIGHT snakeDirection = LEFT if keydown(39) and snakeDirection <> LEFT snakeDirection = RIGHT speedCounter = speedCounter - 1 if speedCounter = 0 prev = snakeHead if snakeDirection > 0 snakeHead = (snakeHead + 1)%SNAKE_MAX_LENGTH speedCounter = speed if snakeDirection = UP snake[snakeHead][0] = snake[prev][0] snake[snakeHead][1] = snake[prev][1] - 1 elseif snakeDirection = DOWN snake[snakeHead][0] = snake[prev][0] snake[snakeHead][1] = snake[prev][1] + 1 elseif snakeDirection = LEFT snake[snakeHead][0] = snake[prev][0] - 1 snake[snakeHead][1] = snake[prev][1] elseif snakeDirection = RIGHT snake[snakeHead][0] = snake[prev][0] + 1 snake[snakeHead][1] = snake[prev][1] endif x = snake[snakeHead][0] y = snake[snakeHead][1] if x < 0 or x >= 40 or y < 0 or y >= 30 gameOver = true else if level[x][y] > 0 snakeLength = snakeLength + 1 if level[x][y] = 2 speed = max(speed - 1, 0) level[x][y] = 0 endif endif endif set color 0, 0, 0 cls for y = 0 to 29 for x = 0 to 39 if level[x][y] = 1 set color 0, 255, 0 draw rect x*8, y*8, 8, 8 , true elseif level[x][y] = 2 set color 255, 0, 0 draw rect x*8, y*8, 8, 8, true endif next next set color 255, 255, 255 set caret 0, 0 x = snake[snakeHead][0] y = snake[snakeHead][1] for i = snakeHead to snakeHead - snakeLength + 1 j = i if j < 0 then j = j + SNAKE_MAX_LENGTH draw rect snake[j][0]*8, snake[j][1]*8, 8, 8, true if i <> snakeHead if snake[j][0] = x and snake[j][1] = y then gameOver = true endif next wait 10 redraw until gameOver loop```I'll see if I can understand the code where the body follows the head, thank you. Marcus Administrator Posts: 316 Threads: 39 Joined: Nov 2023 Reputation: 3 07-20-2024, 08:16 PM (This post was last modified: 07-20-2024, 09:57 PM by Marcus.) (07-20-2024, 06:50 PM)aliensoldier Wrote: (07-20-2024, 09:23 AM)Marcus Wrote: Here's a port of an n6 example game. Hopefully you can find some clues by looking at the code I don't remember why I implemented the game like this, and probably there are smarter ways. Code:```' Port of n6 example game. SNAKE_MAX_LENGTH = 30*40 UP = 1 DOWN = 2 LEFT = 3 RIGHT = 4 snake = dim(SNAKE_MAX_LENGTH, 2) level = dim(40, 30) set window "Smape", 320, 240, false, 2 set redraw off do snakeLength = 1 snakeHead = 0 snakeDirection = 0 speed = 10 speedCounter = speed snake[snakeHead][0] = 20 snake[snakeHead][1] = 15 for y = 0 to 29 for x = 0 to 39 level[x][y] = 0 gameOver = false nextPlupTime = clock() do if clock() >= nextPlupTime if rnd(3) = 0 level[rnd(40)][rnd(30)] = 2 else level[rnd(40)][rnd(30)] = 1 nextPlupTime = clock() + 2000 + rnd(2000) endif if keydown(38) and snakeDirection <> DOWN snakeDirection = UP if keydown(40) and snakeDirection <> UP snakeDirection = DOWN if keydown(37) and snakeDirection <> RIGHT snakeDirection = LEFT if keydown(39) and snakeDirection <> LEFT snakeDirection = RIGHT speedCounter = speedCounter - 1 if speedCounter = 0 prev = snakeHead if snakeDirection > 0 snakeHead = (snakeHead + 1)%SNAKE_MAX_LENGTH speedCounter = speed if snakeDirection = UP snake[snakeHead][0] = snake[prev][0] snake[snakeHead][1] = snake[prev][1] - 1 elseif snakeDirection = DOWN snake[snakeHead][0] = snake[prev][0] snake[snakeHead][1] = snake[prev][1] + 1 elseif snakeDirection = LEFT snake[snakeHead][0] = snake[prev][0] - 1 snake[snakeHead][1] = snake[prev][1] elseif snakeDirection = RIGHT snake[snakeHead][0] = snake[prev][0] + 1 snake[snakeHead][1] = snake[prev][1] endif x = snake[snakeHead][0] y = snake[snakeHead][1] if x < 0 or x >= 40 or y < 0 or y >= 30 gameOver = true else if level[x][y] > 0 snakeLength = snakeLength + 1 if level[x][y] = 2 speed = max(speed - 1, 0) level[x][y] = 0 endif endif endif set color 0, 0, 0 cls for y = 0 to 29 for x = 0 to 39 if level[x][y] = 1 set color 0, 255, 0 draw rect x*8, y*8, 8, 8 , true elseif level[x][y] = 2 set color 255, 0, 0 draw rect x*8, y*8, 8, 8, true endif next next set color 255, 255, 255 set caret 0, 0 x = snake[snakeHead][0] y = snake[snakeHead][1] for i = snakeHead to snakeHead - snakeLength + 1 j = i if j < 0 then j = j + SNAKE_MAX_LENGTH draw rect snake[j][0]*8, snake[j][1]*8, 8, 8, true if i <> snakeHead if snake[j][0] = x and snake[j][1] = y then gameOver = true endif next wait 10 redraw until gameOver loop```I'll see if I can understand the code where the body follows the head, thank you. I barely looked at the code. But I believe the snake "moves" through the 'snake' array. Every element in the array contains a position (x and y value). The snake's head is a moving index in the array. Every time the snake moves one step, the head index increases and a new position is added to the array. That way, the tail is always present as earlier indexes in the array. So if the length of the snake is 4, the positions drawn are those at snake[snakeHead], snake[snakeHead - 1] .. snake[snakeHead - 3]. Edit: The % operator is used to keep the indexes within the array size. When the snake head index reaches the end of the array it starts over at 0. The length of the array is exactly enough for the snake to cover the entire screen You can probably write smarter code that just stores the turns that the snake does or something. But that would require more complicated code ... I think aliensoldier Member Posts: 85 Threads: 16 Joined: Nov 2023 Reputation: 2 07-21-2024, 05:39 PM I don't understand the code very well, so I have prepared a small example to use as a basis for testing. In the example I have the player who is the head of the snake and I have the body which is the tail, the player moves with the keys and the body I have added to a list and I show three bodies at the moment. How do I make the body is in the player's position and moves with him like the tail of the snake. Code:```'prueba de cola de serpiente 'variables----------------------- visible player,list_body set window "serpiente",640,480,false set redraw off 'player----------------------------- function Player() obj = [] obj.radius = 14 obj.x = 320 - obj.radius/2 obj.y = 240 - obj.radius/2 obj.speedX = 0 obj.speedY = 0 obj.Active = true obj.update = function() if this.Active = true this.move() this.Draw() endif endfunc obj.Draw = function() set color 255,204,0 draw ellipse this.x,this.y,this.radius,this.radius,true endfunc obj.move = function() if keydown(KEY_RIGHT,false) this.speedX = 2 this.speedY = 0 endif if keydown(KEY_LEFT,false) this.speedX = -2 this.speedY = 0 endif if keydown(KEY_UP,false) this.speedX = 0 this.speedY = -2 endif if keydown(KEY_DOWN,false) this.speedX = 0 this.speedY = 2 endif this.x = this.x + this.speedX this.y = this.y + this.speedY endfunc return obj endfunc 'end player-------------------------- 'body-------------------------------- function Body(x,y) obj = [] obj.radius = 10 obj.x = x - obj.radius/2 obj.y = y - obj.radius/2 obj.update = function() this.Draw() endfunc obj.Draw = function() set color 255,0,204 draw ellipse this.x,this.y,this.radius,this.radius,true endfunc return obj endfunc 'end body---------------------------- 'iniciar variables------------------- player = Player() list_body = [] list_body[sizeof(list_body)] = Body(320,64) list_body[sizeof(list_body)] = Body(320-32,64) list_body[sizeof(list_body)] = Body(320-64,64) while not keydown(KEY_ESCAPE,true) set color 0,0,0 cls 'actualizar objetos del juego----------- player.update() if sizeof(list_body) for i = 0 to sizeof(list_body)-1 list_body[i].update() next endif '--------------------------------------- fwait 60 redraw wend``` Marcus Administrator Posts: 316 Threads: 39 Joined: Nov 2023 Reputation: 3 08-08-2024, 01:53 PM (07-21-2024, 05:39 PM)aliensoldier Wrote: I don't understand the code very well, so I have prepared a small example to use as a basis for testing. In the example I have the player who is the head of the snake and I have the body which is the tail, the player moves with the keys and the body I have added to a list and I show three bodies at the moment. How do I make the body is in the player's position and moves with him like the tail of the snake. Code:```'prueba de cola de serpiente 'variables----------------------- visible player,list_body set window "serpiente",640,480,false set redraw off 'player----------------------------- function Player() obj = [] obj.radius = 14 obj.x = 320 - obj.radius/2 obj.y = 240 - obj.radius/2 obj.speedX = 0 obj.speedY = 0 obj.Active = true obj.update = function() if this.Active = true this.move() this.Draw() endif endfunc obj.Draw = function() set color 255,204,0 draw ellipse this.x,this.y,this.radius,this.radius,true endfunc obj.move = function() if keydown(KEY_RIGHT,false) this.speedX = 2 this.speedY = 0 endif if keydown(KEY_LEFT,false) this.speedX = -2 this.speedY = 0 endif if keydown(KEY_UP,false) this.speedX = 0 this.speedY = -2 endif if keydown(KEY_DOWN,false) this.speedX = 0 this.speedY = 2 endif this.x = this.x + this.speedX this.y = this.y + this.speedY endfunc return obj endfunc 'end player-------------------------- 'body-------------------------------- function Body(x,y) obj = [] obj.radius = 10 obj.x = x - obj.radius/2 obj.y = y - obj.radius/2 obj.update = function() this.Draw() endfunc obj.Draw = function() set color 255,0,204 draw ellipse this.x,this.y,this.radius,this.radius,true endfunc return obj endfunc 'end body---------------------------- 'iniciar variables------------------- player = Player() list_body = [] list_body[sizeof(list_body)] = Body(320,64) list_body[sizeof(list_body)] = Body(320-32,64) list_body[sizeof(list_body)] = Body(320-64,64) while not keydown(KEY_ESCAPE,true) set color 0,0,0 cls 'actualizar objetos del juego----------- player.update() if sizeof(list_body) for i = 0 to sizeof(list_body)-1 list_body[i].update() next endif '--------------------------------------- fwait 60 redraw wend``` Sorry, I totally forgot about your question. I'll have a look at it right away! Marcus Administrator Posts: 316 Threads: 39 Joined: Nov 2023 Reputation: 3 08-08-2024, 02:56 PM (This post was last modified: 08-08-2024, 03:19 PM by Marcus.) Code:```'prueba de cola de serpiente ' Marcus: You can remove this, just checking some things. #dbg 'variables----------------------- ' Marcus: I think list_body should belong to player ================================================ 'visible player,list_body visible player ' ================================================================================================== set window "serpiente",640,480,false set redraw off 'player----------------------------- function Player() obj = [] obj.radius = 14 obj.x = 320 - obj.radius/2 obj.y = 240 - obj.radius/2 obj.speedX = 0 obj.speedY = 0 obj.Active = true ' Marcus: Putting body list here and a list of recorded head positions ========================= obj.bodyList = [] obj.positions = [] ' ============================================================================================== ' Marcus: Add one body part to player ========================================================== obj.AddBody = function() body = Body(0, 0) this.bodyList[sizeof(this.bodyList)] = body ' Not sure when AddBody might be called, so calling UpdateBody here too. this.UpdateBody() endfunc ' ============================================================================================== obj.update = function() if this.Active = true this.move() ' Marcus: Reposition body parts ======================================================== this.UpdateBody() ' ====================================================================================== this.Draw() endif endfunc ' Marcus: Set body parts' positions ============================================================ obj.UpdateBody = function() if sizeof(this.bodyList) ' We position the body parts one after one, starting with the one closest to the head. ' dist is the calculated travel distance between the head and the current body part. ' Initiate it to the radius of the head. dist = this.radius for i = 0 to sizeof(this.bodyList) - 1 b = this.bodyList[i] ' Add radius of body part to dist. dist = dist + b.radius ' Divide dist by the speed of the snake to convert the traveled distance to an ' index in the position list. Use 'min' to make sure it's not outside the list. index = min(int(dist/2), sizeof(this.positions) - 1) if index >= 0 ' Convert single number coordinates to x and y. 'b.y = int(this.positions[index]/width(primary)) 'b.x = int(this.positions[index]%width(primary)) b.x = this.positions[index][0] b.y = this.positions[index][1] else ' No recorded positions, put at head position. b.x = this.x b.y = this.y endif ' Add radius of body part to dist. dist = dist + b.radius next endif endfunc ' ============================================================================================== obj.Draw = function() ' Marcus: Draw body parts ================================================================== if sizeof(this.bodyList) for i = 0 to sizeof(this.bodyList) - 1 this.bodyList[i].Draw() endif ' ========================================================================================== set color 255,204,0 draw ellipse this.x,this.y,this.radius,this.radius,true endfunc obj.move = function() if keydown(KEY_RIGHT,false) this.speedX = 2 this.speedY = 0 endif if keydown(KEY_LEFT,false) this.speedX = -2 this.speedY = 0 endif if keydown(KEY_UP,false) this.speedX = 0 this.speedY = -2 endif if keydown(KEY_DOWN,false) this.speedX = 0 this.speedY = 2 endif ' Marcus: Save previous position =========================================================== if this.speedX or this.speedY ' I store the position, x and y, as a single number. 'insert this.positions, 0, this.y*width(primary) + this.x insert this.positions, 0, [this.x, this.y] ' Store only 5000 positions for now, might have to increase it depending on how long ' the snake can become. if sizeof(this.positions) > 5000 free key this.positions, 5000 endif ' ========================================================================================== this.x = this.x + this.speedX this.y = this.y + this.speedY endfunc return obj endfunc 'end player-------------------------- 'body-------------------------------- function Body(x,y) obj = [] obj.radius = 10 obj.x = x - obj.radius/2 obj.y = y - obj.radius/2 obj.update = function() this.Draw() endfunc obj.Draw = function() set color 255,0,204 draw ellipse this.x,this.y,this.radius,this.radius,true endfunc return obj endfunc 'end body---------------------------- 'iniciar variables------------------- player = Player() list_body = [] list_body[sizeof(list_body)] = Body(320,64) list_body[sizeof(list_body)] = Body(320-32,64) list_body[sizeof(list_body)] = Body(320-64,64) while not keydown(KEY_ESCAPE,true) set color 0,0,0 cls if keydown(KEY_SPACE, true) player.AddBody() 'actualizar objetos del juego----------- player.update() if sizeof(list_body) for i = 0 to sizeof(list_body)-1 list_body[i].update() next endif '--------------------------------------- ' Marcus: Some debug output ==================================================================== set color 255, 255, 255 set caret 0, 0 wln "PRESS SPACE BAR TO ADD BODY PARTS" wln wln "bodyparts: " + sizeof(player.bodyList) wln "recorded positions: " + sizeof(player.positions) ' ============================================================================================== fwait 60 redraw wend``` I've marked my changes with "Marcus" This solution is probably not very good. The head's position is always recorded into a list, player.positions. Currently I store a maximum of 5000 positions, which covers a snake length of 10,000 pixels (because the speed of the snake is 2). Then I "simply" position the body parts, using the list of recorded positions, in player.UpdateBody. Press space bar to add body parts. Edit But as with the other snake code I posted, this is NOT a very smart solution. It should be enough to store the positions where the player turns, and then use the distance between the head and these points to calculate the positions of the body parts. aliensoldier Member Posts: 85 Threads: 16 Joined: Nov 2023 Reputation: 2 08-10-2024, 12:06 PM (08-08-2024, 02:56 PM)Marcus Wrote: Code:```'prueba de cola de serpiente ' Marcus: You can remove this, just checking some things. #dbg 'variables----------------------- ' Marcus: I think list_body should belong to player ================================================ 'visible player,list_body visible player ' ================================================================================================== set window "serpiente",640,480,false set redraw off 'player----------------------------- function Player() obj = [] obj.radius = 14 obj.x = 320 - obj.radius/2 obj.y = 240 - obj.radius/2 obj.speedX = 0 obj.speedY = 0 obj.Active = true ' Marcus: Putting body list here and a list of recorded head positions ========================= obj.bodyList = [] obj.positions = [] ' ============================================================================================== ' Marcus: Add one body part to player ========================================================== obj.AddBody = function() body = Body(0, 0) this.bodyList[sizeof(this.bodyList)] = body ' Not sure when AddBody might be called, so calling UpdateBody here too. this.UpdateBody() endfunc ' ============================================================================================== obj.update = function() if this.Active = true this.move() ' Marcus: Reposition body parts ======================================================== this.UpdateBody() ' ====================================================================================== this.Draw() endif endfunc ' Marcus: Set body parts' positions ============================================================ obj.UpdateBody = function() if sizeof(this.bodyList) ' We position the body parts one after one, starting with the one closest to the head. ' dist is the calculated travel distance between the head and the current body part. ' Initiate it to the radius of the head. dist = this.radius for i = 0 to sizeof(this.bodyList) - 1 b = this.bodyList[i] ' Add radius of body part to dist. dist = dist + b.radius ' Divide dist by the speed of the snake to convert the traveled distance to an ' index in the position list. Use 'min' to make sure it's not outside the list. index = min(int(dist/2), sizeof(this.positions) - 1) if index >= 0 ' Convert single number coordinates to x and y. 'b.y = int(this.positions[index]/width(primary)) 'b.x = int(this.positions[index]%width(primary)) b.x = this.positions[index][0] b.y = this.positions[index][1] else ' No recorded positions, put at head position. b.x = this.x b.y = this.y endif ' Add radius of body part to dist. dist = dist + b.radius next endif endfunc ' ============================================================================================== obj.Draw = function() ' Marcus: Draw body parts ================================================================== if sizeof(this.bodyList) for i = 0 to sizeof(this.bodyList) - 1 this.bodyList[i].Draw() endif ' ========================================================================================== set color 255,204,0 draw ellipse this.x,this.y,this.radius,this.radius,true endfunc obj.move = function() if keydown(KEY_RIGHT,false) this.speedX = 2 this.speedY = 0 endif if keydown(KEY_LEFT,false) this.speedX = -2 this.speedY = 0 endif if keydown(KEY_UP,false) this.speedX = 0 this.speedY = -2 endif if keydown(KEY_DOWN,false) this.speedX = 0 this.speedY = 2 endif ' Marcus: Save previous position =========================================================== if this.speedX or this.speedY ' I store the position, x and y, as a single number. 'insert this.positions, 0, this.y*width(primary) + this.x insert this.positions, 0, [this.x, this.y] ' Store only 5000 positions for now, might have to increase it depending on how long ' the snake can become. if sizeof(this.positions) > 5000 free key this.positions, 5000 endif ' ========================================================================================== this.x = this.x + this.speedX this.y = this.y + this.speedY endfunc return obj endfunc 'end player-------------------------- 'body-------------------------------- function Body(x,y) obj = [] obj.radius = 10 obj.x = x - obj.radius/2 obj.y = y - obj.radius/2 obj.update = function() this.Draw() endfunc obj.Draw = function() set color 255,0,204 draw ellipse this.x,this.y,this.radius,this.radius,true endfunc return obj endfunc 'end body---------------------------- 'iniciar variables------------------- player = Player() list_body = [] list_body[sizeof(list_body)] = Body(320,64) list_body[sizeof(list_body)] = Body(320-32,64) list_body[sizeof(list_body)] = Body(320-64,64) while not keydown(KEY_ESCAPE,true) set color 0,0,0 cls if keydown(KEY_SPACE, true) player.AddBody() 'actualizar objetos del juego----------- player.update() if sizeof(list_body) for i = 0 to sizeof(list_body)-1 list_body[i].update() next endif '--------------------------------------- ' Marcus: Some debug output ==================================================================== set color 255, 255, 255 set caret 0, 0 wln "PRESS SPACE BAR TO ADD BODY PARTS" wln wln "bodyparts: " + sizeof(player.bodyList) wln "recorded positions: " + sizeof(player.positions) ' ============================================================================================== fwait 60 redraw wend``` I've marked my changes with "Marcus" This solution is probably not very good. The head's position is always recorded into a list, player.positions. Currently I store a maximum of 5000 positions, which covers a snake length of 10,000 pixels (because the speed of the snake is 2). Then I "simply" position the body parts, using the list of recorded positions, in player.UpdateBody. Press space bar to add body parts. Edit But as with the other snake code I posted, this is NOT a very smart solution. It should be enough to store the positions where the player turns, and then use the distance between the head and these points to calculate the positions of the body parts. Thanks Marcus, I'll see if I understand this better. Although I don't really want to program the snake game, what I wanted to do is have some planes follow another one. It's in those airplane games where there is a group of planes that move in circles and they all do it together. « Next Oldest | Next Newest »
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# Express recurrence in closed form
I am having trouble understanding the process of expressing the following recurrence in its' closed form.
First of all, I do not really understand what "closed form" means. If someone could elaborate a little bit on this, it would be much appreciated.
Second, Could you explain the nature of the function itself, as in, what it is explicitly trying to do?
Here is the recurrence function:
$$F_{0} = 7$$
$$F_{N} = 5F_{N-1} + 3, \ \ ( N > 0 ).$$
Express $F_{N}$ in closed form
The final answer, should be the following:
$$\frac{31\cdot(5^N) - 3}{ 4}$$
Thank you so much for your help and additional explanations you may throw in. I am very lost in regards to what to do.
J.
## UPDATE
Some of the process is here:
• "Closed form" is an analytical expression for the generic term of your sequence, that allows you to compute it without knowing the previous terms. In this case, $F(n)=31\cdot 5^n - 3/4$ is a closed form. – Giuseppe Negro Dec 9 '14 at 9:21
• I would first take an ansatz $F_{N} = \lambda^{N}$ and then solve the homogeneous and particular parts separately. – Mattos Dec 9 '14 at 9:26
• @GiuseppeNegro Do you know the process of how to get there? – J Tarantino Dec 9 '14 at 9:30
• @Mattos: I’m unsure why did you change parentheses to subscript, but since original poster did not object, I rest it in place. – Incnis Mrsi Dec 9 '14 at 9:32
• @Incnis Mrsi: Because sequences are usually defined by a subscript and also, the OP didn't use latex script at all so it was an educated guess as to what he was trying to ask. Now it looks even worse because we have $F_{N} = 5F(N-1)$. You'll need to edit it again now to have it as either $F_{N} = 5F_{N-1}$ or $F(N) = 5F(N-1)$ to avoid confusion. – Mattos Dec 9 '14 at 9:40
HINT
You have already the "final" exprssion :
$f(n) = (31 \times 5^n -3)/4$.
What you have to do is to prove - by induction - that it holds, i.e. that it satisfies the recurrence relation for $f(n)$.
(i) base case : $n=0$
For $n=0$ the recursive definition of $f(n)$ has : $f(0)=7$.
We have to check that it matches with the closed expression :
$$f(0) = (31 \times 5^0 -3)/4 = (31 -3)/4 = 28/4 = 7.$$
It's Ok.
(ii) induction step : assume that it holds for $n$ and show that it holds for $n+1$.
For the closed form, we can try with :
$f(n) = ac^n+b$.
For $n=0$ we have : $f(0)=7=a+b$; thus :
$a = 7-b$.
Using the recurrence relation : $f(n+1)=5f(n)+3$ we have : $ac^{n+1}+b=5(ac^n+b)+3$.
With $c=5$ we can compute $b = - \frac {3}{4}$ and so : $a = 7-b = 7 + \frac {3}{4} = \frac {31}{4}$.
• do you know whats the process of getting there though? that is what I'm most interested in... I'll put some of the work in an update – J Tarantino Dec 9 '14 at 9:36
• @JTarantino - you can see here – Mauro ALLEGRANZA Dec 9 '14 at 9:40
• I've updated mia risposta... puoi vederla? grazie mile! – J Tarantino Dec 9 '14 at 9:46
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Get Full Access to UNT - PHYS 1062 - Study Guide - Midterm
Get Full Access to UNT - PHYS 1062 - Study Guide - Midterm
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UNT / Physics / PHYS 1062 / ohad shemmer
##### Description: Covers everything possible from Chapters 1 to 9, excluding 3 and 4 since they won't be on the test. I did my absolute best and drew influence from all the sources I could. Hopefully this helps and you don't have to go searching through all your quizzes, homeworks, or resort to reading. Enjoy!
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PHYS 1062 Midterm Study Guide Extensive study guide for March 2018 exam. Ch. 1 - 9.
## in meters how long is a 1 Light year?
Key Terms:
● Red = Key terms
● Orange = Possible Trivia Questions
● Yellow = If you memorize it, you won’t have to math.
● Green = Put here because it was a quiz question and I can’t explain it but I got it right.
● Blue = Formulas
Ch. 1
● Conversions/Units
○ 1 Light year = 9.461x10^15 meters
○ Earth’s Diameter = 12,756 km = 8.5x10^-5 AU
○ Milky Way Diameter = 80,000 light years
○ 1 AU = 150 million km = 93 million miles
○ The sun’s radius is 100 times the size of Earth’s radius.
○ Astronomers do not measure in MILES.
○ 5K = 3 miles.
● The stars that appear the biggest are the brightest We also discuss several other topics like eku chemistry
● Circumpolar stars never leave the horizon. They would make concentric paths. ● Size
○ Supercluster, galaxy, solar system, planet (from biggest to smallest) ● Scientific Notation
## in diameter how big in the Earth?
○ 10^(number of zeroes)
○ EX: 0.0000788 = 7.88x10^-5
● Seasons and Tilts
○ We have seasons because of the Earth’s tilt.
○ In the winter, the sun is above the horizon for less than 12 hours and at low angles, so there’s less concentrated solar heating or sunlight.
● It takes 5.8 minutes for the sunlight to reach Venus.
● It takes 8 minutes for the sunlight to reach Earth.
● It takes 4.2 light years for the sunlight to reach Alpha Centauri
○ Alpha Centauri is the closest star to our sun.
● Earth rotates from west to east. (on the HW question, it’s the sunset line on the right)
● When observing a star for a few hours, it moves from east to west. ○ An overhead star will move through the 15* angle in one hour.
Ch. 2
● Precession = The Earth moves like a top (imagine the one at the end of Inception) ○ Why does this happen? - The Sun and Moon pull on Earth’s equatorial bulge.
## in diameter how big in the Milky Way?
○ It changes the celestial poles, the equinoxes, the solstices, and the celestial equator.
● Greek Letters = Describe the relative brightness within the constellation ○ Alpha - Brightest
○ Beta - Second brightest
● Celestial Sphere = It’s not physically real, but it is still useful.
○ 88 Regions (constellations)
○ Don’t confuse constellations for asterisms.
■ Asterisms are like...The Big Dipper.
○ Horizon
○ North and South Celestial Poles We also discuss several other topics like bul 4310 fiu
○ Celestial Equator
● Milankovitch Hypothesis = Small changes influence Earth’s climate and cause ice ages.
● Magnitude
○ If one star is 100 times brighter than the other, the magnitude difference is 5.
○ 1st Magnitude is very bright. 6th Magnitude, not so much.
○ The higher the number the dimmer the star. (-2.7 vs 2.7)
● Flux
○ Measure of the light energy from a star that hits one square meter in one second.
○ 2.512^(the difference in magnitudes classes)
● The 5 naked-eye and 2 telescopic planets that wander among the stars will always be near the ecliptic.
● It takes the sun 2 minutes to completely set once it touches the horizon.
Ch. 5
● Newton If you want to learn more check out ams 210 stony brook
○ 3 Laws of Motion
■ An object in motion will stay in motion. (or at rest will stay at rest) Unless acted upon.
■ F = ma (Force = Mass*Acceleration)
■ Equal and Opposite Reaction.
● So the amount of force on the Sun by the Earth is the same as the amount on Earth by the Sun.
○ There was a force pulling the Moon to Earth
■ The Moon’s orbital motion has a curved fall.
■ Moon has an acceleration towards Earth
■ The force and acceleration in Newton’s second law must have the same direction.
■ (If there’s a question that has all these answers and another, pick all the above)
■ If the gravity was turned off, the moon would fly off into space in the direction it was facing INSTANTLY.
○ What’s necessary about the force exerted by the Sun (to yield elliptical orbits)
■ The force must be attractive and the force must vary inversely with distance squared.
● Galileo
○ Slowed down time
■ Rolled objects down inclines at low angles. Don't forget about the age old question of What do phagocytes produce to kill bacteria?
○ Worked out the law of inertia (Law #1)
● Force
○ If two planets orbit the sun (Earth and Q), and Q is five AUs away from the sun, Q has 1/25 the force on Earth. The two planets are identical. ○ Why would a hammer and feather land at the same time?
■ No air resistance.
○ Mass Vs. Weight
■ Mass is the amount of matter
■ Weight is the amount of gravitational force you experience.
○ You’re never weightless, just in constant freefall.
● General Relativity
○ Solved the major orbital problem: The excess precession of Mercury’s perihelion.
○ Verified on May 29, 1919 solar eclipse (bending of light by gravity) ○ Explains the following
■ Light bending in the vicinity of massive objects.
■ Time dilation close to massive objects
■ Gravitational redshift.
■ Mercury’s orbit does not follow Newton’s laws precisely.
● Albert Einstein
○ Proposed that gravity is the bending of space-time due to the presence of matter.
● Circular Motion
○ Acceleration of the object is toward the center of motion.
● Escape Velocity
○ If we shrink the Earth’s radius by a factor of 100, but we keep the mass, the escape velocity will increase by a factor of 10.
● If there’s an alien beam question, it’s the speed of light.
● Speed vs. Velocity vs. Acceleration
○ Speed - how fast are you going?
○ Velocity - how fast are you going and WHERE?
○ Acceleration - something changed and we gonna find out what.
● Law of Universal Gravity
○ F = -G (Mm/r)
● Free Fall
○ G = Fgravity/M = Fgravity/m
● Escape Velocity
○ (2GM/r)^(½)
Ch. 6
● Speed of light = Frequency x Wavelength.
● Speed of light = 3x10^8 m/s
● E = hf
● H = 6.626 x 10^-34 J*s is the Planck Constant
● Light cannot be portrayed as a wave and a particle in the same experiment. ● Infrared telescopes are on mountaintops and ultraviolet telescopes are in Earth’s orbit.
○ Infrared blocker, water vapor, is in the lower atmosphere.
○ Primary ultraviolet blocker, ozone, is in the higher atmosphere. ● Infrared telescopes
○ Must be cooled to a low temperature to reduce interfering heat radiation emitted by the telescope.
● Electromagnetic radiation travel in any medium that does not absorb them. ● PURPOSE OF A TELESCOPE
○ To gather light and bring it to a focus.
○ Width matters, not length. (*winky face*)
● Big telescopes (size does matter?) now.
○ They can be made thinner and lighter.
○ Tracking is computer controlled.
○ Reduced effect of the Earth’s atmosphere.
○ (pick all of the above)
○ A = pi (p/2)^2 Surface area of the primary lense.
● Chromatic Aberration
○ Prisms take advantage of it.
○ It’s a big problem for the primary lenses of refracting telescopes
● Resolving Power
○ Limited by a cloudy night.
○ Expressed as “0.5 seconds of arc.”
● Electromagnetic Spectrum
○ Visible and Radio are the most transparent to Earth’s atmosphere. ○ Visible Spectrum
■ ROYGBIV
● Violet’s photons have the greatest energy.
○ X-Ray Telescopes can observe hot gas trapped in galactic clusters better than an infrared telescope.
○ Infrared Telescopes can observe newborn stars in dusty nebulae better than an X-Ray telescope.
○ Rank telescopes designed for the following specific types of electromagnetic waves in order of the minimum altitude at which they would be useful. (least to greatest)
■ Visible
■ Infrared
■ X-Ray
○ Rank telescopes designed for the following specific types of electromagnetic waves in decreasing order of the altitudes at which they would be useful (greatest to least)
■ Ultraviolet
■ Infrared
■ Visible
● Radio Telescopes vs Optical Telescopes (everything radio does best) ○ Find the location of cool hydrogen gas
○ See through dust clouds
○ Detect dark molecular clouds
○ Observe during the day
○ (so pick all of the above)
● Determined exclusively by the diameter of the primary mirror or lens. ○ Light Gathering Power and Resolving Power.
○ Overcomes poor resolving power.
■ It’s poor because the wavelengths are so long.
● It’s about light gathering, not magnifying.
○ Pupils
○ Telescopes.
● If you were to make a telescope using the two lenses (it’ll show you a diagram with two blue and white looking things)
○ One uses top lens for the primary and bottom lens for the eyepiece.
Ch. 7
● Spectral Types
○ Seven Types: OBAFGKM. O is hottest.
○ Red is the lowest surface temperature for a star compared to (Orange, White, Yellow, and Blue)
○ Blue is the color of the hottest stars.
■ Blue star’s emissions peak at shorter wavelengths than red ones. ○ Temperature controls the color of a star.
● Spectral Lines
○ If it’s blueshifted, the radial velocity is directed towards us.
○ If it’s redshifted, the radial velocity is directed away us.
○ Properties of a star that can broaden the width of its spectral lines ■ Rapid rotation of the star
■ High temperature atmosphere
■ High density atmosphere
● Where is the location of the cooler low-density gas that yields the dark line stellar spectra that were studied by Annie Jump Cannon
○ In the star’s lower atmosphere
○ In Earth’s atmosphere
● Continuous
○ Observed when observing radiation from a hot solid or gas under high pressure.
○ Observe from molten lava.
● Absorption
○ Light from a continuous spectrum source passing through a cooler low density gas produces the absorption line spectrum.
○ Observed when observing radiation through a cool gas.
○ Observed if you looked through gases boiling out of molten lava. ● Emission
○ Observed when observing radiation from a hot gas.
● Black Body
○ All stars!
○ Wavelength of maximum intensity that is emitted is inversely proportional to temperature.
○ Amount of electromagnetic energy radiated from every square meter of the surface is proportional to temperature to the fourth power.
● Temperature of a gas is a measure of the
○ Average motion of its atoms
● Atomic Nucleus
○ All the positive charge
○ 99.9% of the mass
○ No electrons
● Ionized atoms = Atoms with more of either electrons or protons than the other ● Stars are mainly hydrogen and helium but they have no lines for hydrogen or helium in their spectrum. Why?
○ The surface temperature is such that the electrons are not at the proper energy levels to produce spectral lines at visible wavelengths.
● The electron making the transition from level 2 to level 3 corresponds to a hydrogen atom absorbing a visible light photon that has a wavelength of 656 nanometers.
○ Determined by the difference in energy between electron energy levels.
Ch. 8
● The Sun
○ Maintains its energy output by fusion of hydrogen nuclei.
○ Magnetic field is responsible for the surface and atmospheric activity. ■ Changes because of the differential rotation of the Sun and
convection beneath the photosphere.
○ Changes in the magnetic field heat the chromosphere and corona to high temperatures.
○ The lower photosphere is hotter than the upper photosphere is responsible for “limb darkening.”
○ Photosphere contains the cooler low-density gas responsible for absorption lines in the Sun’s spectrum
○ Corona is a very hot low-density gas.
○ The layers of the Sun below the photosphere are explored by measuring and modeling the modes of vibration of the Sun’s surface.
○ The general trends in temperature and density from the photosphere to the chromosphere to the corona - Temperature increases and density
decreases.
○ Photosphere is not part of the interior.
● Layers of the Solar Interior Ranking Questions
○ Outermost to innermost
■ Convective Zone
■ Core
○ Increasing temperature (least to greatest)
■ Convective Zone
■ Core
● Solar Prominence
○ Solar Material from the chromosphere following the arches of the sun’s magnetic field
○ Its spectrum reveals that it is much cooler than its surroundings. ○ The shape suggests that it is following the solar magnetic field.
● Solar Flare
○ Eruption of solar material from the photosphere
○ Observed at Visible, Ultraviolet, and X-Ray wavelengths
○ Can bring auroras and communication blackouts.
● Solar Neutrino Problem
○ Solved by the discovery that neutrinos oscillate between three different types.
● Solar Neutrinos
○ Created during nuclear fusion
○ Very low mass
○ Travel very fast
○ Detected in large underground pools of chemicals
○ They’re hard to detect because they move fast, have low mass, and oscillate between three flavors.
● Supergranules and Granules
○ Both due to convection cells in the layers below.
○ The center of a granule is brighter than its edges because the temperature is higher at the center.
● Nuclear Fusion
○ Happens at the core
○ Requires high temperatures because:
■ Protons repel each other
■ Overcoming the Coulomb barrier
■ (it’s the one that says all of these choices)
● Proton-Proton Chain
○ Produced from the neutrinos
○ Head out of the Sun at nearly the speed of light.
● How constant is the Solar constant? How much has the solar constant been observed to vary?
○ About 0.1% , so pretty darn constant.
Ch. 9
● Sun
○ Spectral Type is G2, Luminosity is Main Sequence (V)
○ If a star with the same type has a luminosity of 50 solar luminosities, it must be larger than the sun.
● Luminosity
○ Cool stars can be more luminous than hot stars if the cool star is larger. ○ Supergiants have the lowest density
○ Main Sequence applies the mass-luminosity relation.
○ If a star has ½ the surface temperature of the Sun and is 4 times larger, the star’s luminosity is 1 solar luminosity
○ L = Surface Area (A) * O (that ugly constant) * Temperature in Kelvin (T)^4 ● Distance
○ Parallax Angle of .5 arcseconds = distance of 2 parsecs
○ If a star’s apparent visual magnitude is less than its absolute visual magnitude, the distance to the star is less than 10 parsecs.
○ How to find without parallax angle
■ Spectral Type and Luminosity Class
○ At 10 Parsecs, the apparent magnitude equals the absolute magnitude ● Red Dwarf
○ Most abundant but rarely plotted because they have low luminosity and are hard to detect.
● Magnitude
○ Absolute bolometric magnitude gives the most information about the physical nature of a star.
■ Medium surface temperature stars have the least difference
between absolute visual magnitude and absolute bolometric
magnitude.
● Order of brightness from dimmest to brightest
○ Barnard’s Star
○ Sirius B
○ Sun
○ Canopus
○ Rigel A
● Brightest to Dimmest
○ Antares
○ Polaris
○ Aldebaran A
○ Altair
○ Procyon B
● Hottest to Coolest
○ Aldebaran A
○ Altair
○ Antares
○ Mira
○ Rigel A
d(parsec) = 1/p
Mv is Absolute visual magnitude
Magnitude Distance Formula. 10^(mv-Mv +5)/5 1 Parsec = 3.26 Light Years
1a = Bright Supergiant
1b = SuperGiants
2 = Bright Giants
3 = Giants
4 = Subgiants
5 = Main Sequence
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Win a copy of The Way of the Web Tester: A Beginner's Guide to Automating Tests this week in the Testing forum!
# number formatting
Viggo Oppeg�rd
Greenhorn
Posts: 3
Hey, I'm new to this forum, but you guys seem helpful, so I'll give it a shot and post my question here:
Basically i have a number with two decimals (Currency format), eg 1 000 000.89 dollars that i want to format so that the two decimals are separate from the rest of the number: eg 1 000 000 dollars and 89 cents. How??
Jeff Bosch
Ranch Hand
Posts: 805
You could make a new class with int dollars and int cents as data members.
(Not sure if you're looking for a storage solution or an extraction solution.)
Viggo Oppeg�rd
Greenhorn
Posts: 3
thanks, that's one step ahead, but I still need an extraction method...
Jeff Bosch
Ranch Hand
Posts: 805
If you cast your value to an int, you have the dollars.
If you multiply your value by 100, then take modulus 100, you have your cents.
Good luck!
Viggo Oppeg�rd
Greenhorn
Posts: 3
thanks a heap, that worked great!
Jeff Bosch
Ranch Hand
Posts: 805
Layne Lund
Ranch Hand
Posts: 3061
Originally posted by Jeff Bosch:
If you cast your value to an int, you have the dollars.
If you multiply your value by 100, then take modulus 100, you have your cents.
Good luck!
Or you could subtract the integer dollar value from the first step and then multiply by 100.
Jeff Bosch
Ranch Hand
Posts: 805
Yep. One of the beautiful things about programming is that there is always another valid way to do things!
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Optimized Euler Totient Function for Multiple Evaluations
Euler Totient Function (ETF) Φ(n) for an input n is count of numbers in {1, 2, 3, …, n} that are relatively prime to n, i.e., the numbers whose GCD (Greatest Common Divisor) with n is 1.
Examples:
```Φ(5) = 4
gcd(1, 5) is 1, gcd(2, 5) is 1,
gcd(3, 5) is 1 and gcd(4, 5) is 1
Φ(6) = 2
gcd(1, 6) is 1 and gcd(5, 6) is 1,```
Recommended Practice
We have discussed different methods to compute Euler Totient function that work well for single input. In problems where we have to call Euler’s Totient Function many times like 10^5 times, simple solution will result in TLE(Time limit Exceeded). The idea is to use Sieve of Eratosthenes.
Find all prime numbers upto maximum limit say 10^5 using Sieve of Eratosthenes
To compute Φ(n), we do following.
1. Initialize result as n.
2. Iterate through all primes smaller than or equal to square root of n (This is where it is different from simple methods. Instead of iterating through all numbers less than or equal to square root, we iterate through only primes). Let the current prime number be p. We check if p divides n, if yes, we remove all occurrences of p from n by repeatedly dividing it with n. We also reduce our result by n/p (these many numbers will not have GCD as 1 with n).
3. Finally we return result.
C++
`// C++ program to efficiently compute values` `// of euler totient function for multiple inputs.` `#include ` `using` `namespace` `std;` `#define ll long long` `const` `int` `MAX = 100001;` `// Stores prime numbers upto MAX - 1 values` `vector p;` `// Finds prime numbers upto MAX-1 and` `// stores them in vector p` `void` `sieve()` `{` ` ``ll isPrime[MAX+1];` ` ``for` `(ll i = 2; i<= MAX; i++)` ` ``{` ` ``// if prime[i] is not marked before` ` ``if` `(isPrime[i] == 0)` ` ``{` ` ``// fill vector for every newly` ` ``// encountered prime` ` ``p.push_back(i);` ` ``// run this loop till square root of MAX,` ` ``// mark the index i * j as not prime` ` ``for` `(ll j = 2; i * j<= MAX; j++)` ` ``isPrime[i * j]= 1;` ` ``}` ` ``}` `}` `// function to find totient of n` `ll phi(ll n)` `{` ` ``ll res = n;` ` ``// this loop runs sqrt(n / ln(n)) times` ` ``for` `(ll i=0; p[i]*p[i] <= n; i++)` ` ``{` ` ``if` `(n % p[i]== 0)` ` ``{` ` ``// subtract multiples of p[i] from r` ` ``res -= (res / p[i]);` ` ``// Remove all occurrences of p[i] in n` ` ``while` `(n % p[i]== 0)` ` ``n /= p[i];` ` ``}` ` ``}` ` ``// when n has prime factor greater` ` ``// than sqrt(n)` ` ``if` `(n > 1)` ` ``res -= (res / n);` ` ``return` `res;` `}` `// Driver code` `int` `main()` `{` ` ``// preprocess all prime numbers upto 10 ^ 5` ` ``sieve();` ` ``cout << phi(11) << ``"\n"``;` ` ``cout << phi(21) << ``"\n"``;` ` ``cout << phi(31) << ``"\n"``;` ` ``cout << phi(41) << ``"\n"``;` ` ``cout << phi(51) << ``"\n"``;` ` ``cout << phi(61) << ``"\n"``;` ` ``cout << phi(91) << ``"\n"``;` ` ``cout << phi(101) << ``"\n"``;` ` ``return` `0;` `}`
Java
`// Java program to efficiently compute values ` `// of euler totient function for multiple inputs. ` `import` `java.util.*; ` `class` `GFG{ ` `static` `int` `MAX = ``100001``; ` `// Stores prime numbers upto MAX - 1 values ` `static` `ArrayList p = ``new` `ArrayList();` `// Finds prime numbers upto MAX-1 and ` `// stores them in vector p ` `static` `void` `sieve() ` `{ ` ` ``int``[] isPrime=``new` `int``[MAX+``1``]; ` ` ``for` `(``int` `i = ``2``; i<= MAX; i++) ` ` ``{ ` ` ``// if prime[i] is not marked before ` ` ``if` `(isPrime[i] == ``0``) ` ` ``{ ` ` ``// fill vector for every newly ` ` ``// encountered prime ` ` ``p.add(i); ` ` ``// run this loop till square root of MAX, ` ` ``// mark the index i * j as not prime ` ` ``for` `(``int` `j = ``2``; i * j<= MAX; j++) ` ` ``isPrime[i * j]= ``1``; ` ` ``} ` ` ``} ` `} ` `// function to find totient of n ` `static` `int` `phi(``int` `n) ` `{ ` ` ``int` `res = n; ` ` ``// this loop runs sqrt(n / ln(n)) times ` ` ``for` `(``int` `i=``0``; p.get(i)*p.get(i) <= n; i++) ` ` ``{ ` ` ``if` `(n % p.get(i)== ``0``) ` ` ``{ ` ` ``// subtract multiples of p[i] from r ` ` ``res -= (res / p.get(i)); ` ` ``// Remove all occurrences of p[i] in n ` ` ``while` `(n % p.get(i)== ``0``) ` ` ``n /= p.get(i); ` ` ``} ` ` ``} ` ` ``// when n has prime factor greater ` ` ``// than sqrt(n) ` ` ``if` `(n > ``1``) ` ` ``res -= (res / n); ` ` ``return` `res; ` `} ` `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ``// preprocess all prime numbers upto 10 ^ 5 ` ` ``sieve(); ` ` ``System.out.println(phi(``11``)); ` ` ``System.out.println(phi(``21``)); ` ` ``System.out.println(phi(``31``)); ` ` ``System.out.println(phi(``41``)); ` ` ``System.out.println(phi(``51``)); ` ` ``System.out.println(phi(``61``)); ` ` ``System.out.println(phi(``91``)); ` ` ``System.out.println(phi(``101``)); ` ` ` `} ` `}` `// this code is contributed by mits`
Python3
`# Python3 program to efficiently compute values ` `# of euler totient function for multiple inputs. ` `MAX` `=` `100001``; ` `# Stores prime numbers upto MAX - 1 values ` `p ``=` `[];` `# Finds prime numbers upto MAX-1 and ` `# stores them in vector p ` `def` `sieve(): ` ` ``isPrime ``=` `[``0``] ``*` `(``MAX` `+` `1``); ` ` ``for` `i ``in` `range``(``2``, ``MAX` `+` `1``): ` ` ` ` ``# if prime[i] is not marked before ` ` ``if` `(isPrime[i] ``=``=` `0``): ` ` ` ` ``# fill vector for every newly ` ` ``# encountered prime ` ` ``p.append(i); ` ` ``# run this loop till square root of MAX, ` ` ``# mark the index i * j as not prime` ` ``j ``=` `2``;` ` ``while` `(i ``*` `j <``=` `MAX``): ` ` ``isPrime[i ``*` `j]``=` `1``;` ` ``j ``+``=` `1``;` `# function to find totient of n ` `def` `phi(n):` ` ``res ``=` `n; ` ` ``# this loop runs sqrt(n / ln(n)) times` ` ``i ``=` `0``;` ` ``while` `(p[i] ``*` `p[i] <``=` `n): ` ` ``if` `(n ``%` `p[i]``=``=` `0``): ` ` ` ` ``# subtract multiples of p[i] from r ` ` ``res ``-``=` `int``(res ``/` `p[i]); ` ` ``# Remove all occurrences of p[i] in n ` ` ``while` `(n ``%` `p[i]``=``=` `0``):` ` ``n ``=` `int``(n ``/` `p[i]); ` ` ``i ``+``=` `1``;` ` ``# when n has prime factor greater ` ` ``# than sqrt(n) ` ` ``if` `(n > ``1``):` ` ``res ``-``=` `int``(res ``/` `n); ` ` ``return` `res; ` `# Driver code ` `# preprocess all prime numbers upto 10 ^ 5 ` `sieve(); ` `print``(phi(``11``)); ` `print``(phi(``21``)); ` `print``(phi(``31``)); ` `print``(phi(``41``)); ` `print``(phi(``51``)); ` `print``(phi(``61``)); ` `print``(phi(``91``)); ` `print``(phi(``101``)); ` `# This code is contributed by mits`
C#
`// C# program to efficiently compute values ` `// of euler totient function for multiple inputs. ` `using` `System;` `using` `System.Collections;` `class` `GFG{ ` `static` `int` `MAX = 100001; ` `// Stores prime numbers upto MAX - 1 values ` `static` `ArrayList p = ``new` `ArrayList();` `// Finds prime numbers upto MAX-1 and ` `// stores them in vector p ` `static` `void` `sieve() ` `{ ` ` ``int``[] isPrime=``new` `int``[MAX+1]; ` ` ``for` `(``int` `i = 2; i<= MAX; i++) ` ` ``{ ` ` ``// if prime[i] is not marked before ` ` ``if` `(isPrime[i] == 0) ` ` ``{ ` ` ``// fill vector for every newly ` ` ``// encountered prime ` ` ``p.Add(i); ` ` ``// run this loop till square root of MAX, ` ` ``// mark the index i * j as not prime ` ` ``for` `(``int` `j = 2; i * j<= MAX; j++) ` ` ``isPrime[i * j]= 1; ` ` ``} ` ` ``} ` `} ` `// function to find totient of n ` `static` `int` `phi(``int` `n) ` `{ ` ` ``int` `res = n; ` ` ``// this loop runs sqrt(n / ln(n)) times ` ` ``for` `(``int` `i=0; (``int``)p[i]*(``int``)p[i] <= n; i++) ` ` ``{ ` ` ``if` `(n % (``int``)p[i]== 0) ` ` ``{ ` ` ``// subtract multiples of p[i] from r ` ` ``res -= (res / (``int``)p[i]); ` ` ``// Remove all occurrences of p[i] in n ` ` ``while` `(n % (``int``)p[i]== 0) ` ` ``n /= (``int``)p[i]; ` ` ``} ` ` ``} ` ` ``// when n has prime factor greater ` ` ``// than sqrt(n) ` ` ``if` `(n > 1) ` ` ``res -= (res / n); ` ` ``return` `res; ` `} ` `// Driver code ` `static` `void` `Main() ` `{ ` ` ``// preprocess all prime numbers upto 10 ^ 5 ` ` ``sieve(); ` ` ``Console.WriteLine(phi(11)); ` ` ``Console.WriteLine(phi(21)); ` ` ``Console.WriteLine(phi(31)); ` ` ``Console.WriteLine(phi(41)); ` ` ``Console.WriteLine(phi(51)); ` ` ``Console.WriteLine(phi(61)); ` ` ``Console.WriteLine(phi(91)); ` ` ``Console.WriteLine(phi(101)); ` `} ` `}` `// this code is contributed by mits`
PHP
` 1) ` ` ``\$res` `-= (int)(``\$res` `/ ``\$n``); ` ` ``return` `\$res``; ` `} ` `// Driver code ` ` ` ` ``// preprocess all prime numbers upto 10 ^ 5 ` ` ``sieve(); ` ` ``print``(phi(11).``"\n"``); ` ` ``print``(phi(21).``"\n"``); ` ` ``print``(phi(31).``"\n"``); ` ` ``print``(phi(41).``"\n"``); ` ` ``print``(phi(51).``"\n"``); ` ` ``print``(phi(61).``"\n"``); ` ` ``print``(phi(91).``"\n"``); ` ` ``print``(phi(101).``"\n"``); ` `// this code is contributed by mits` `?>`
Javascript
``
Output:
```10
12
30
40
32
60
72
100```
Time Complexity: O(MAX*log(MAX)+sqrt(n/log(n)))
Auxiliary Space: O(MAX)
This article is contributed by Abhishek Rajput. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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Mathematics
# For an angle ${ x }^{ 2 }$, find :The value of ${ x }^{ 2 }$ it its supplementary angle is three times its complementary angle.
##### SOLUTION
$(180-x^{2})=3(90-x^{2})$
$\Rightarrow 180-x^{2}=270-3x^{2}$
$\Rightarrow 3x^{2}-x^{2}=270-180$
$\Rightarrow 2x^{2}=90$
$\Rightarrow x^{2}=45^{\circ}$
You're just one step away
Subjective Medium Published on 09th 09, 2020
Questions 120418
Subjects 10
Chapters 88
Enrolled Students 87
#### Realted Questions
Q1 Subjective Hard
Find the value of $x$ in each of the following figure, if $I\ ||\ m$.
Asked in: Mathematics - Lines and Angles
1 Verified Answer | Published on 23rd 09, 2020
Q2 Single Correct Medium
Three parallel lines are cut by two transversals as shown in the given figure. If AB$=2$cm, BC$=4$cm and DE$=1.5$cm, then the length of EF is ___________.
• A. $2$cm
• B. $3.5$cm
• C. $4$cm
• D. $3$cm
Asked in: Mathematics - Straight Lines
1 Verified Answer | Published on 17th 08, 2020
Q3 Subjective Medium
Which pair of the dotted line, segments, in the following figure, are parallel. Give reason:
Asked in: Mathematics - Lines and Angles
1 Verified Answer | Published on 23rd 09, 2020
Q4 Subjective Medium
In the given figure, $B \quad AC$ is a straight line.
Find:(i) $x$
(ii)$\angle AOB$
(iii)$\angle BOC$
Asked in: Mathematics - Lines and Angles
1 Verified Answer | Published on 09th 09, 2020
Q5 Subjective Medium
Iron roads $a,\ b,\ c,\ d,\ e$ and $f$ are making a design in a bridge as shown in Fig., in which a $\parallel b,\ c \parallel d,\ e \parallel \ f$. Find the marked angles between $b$ and $c$
Asked in: Mathematics - Lines and Angles
1 Verified Answer | Published on 09th 09, 2020
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# Question Is 221,044 a prime number?
The number 221,044 is NOT a PRIME number.
#### How to check if the number 221,044 is a prime number
A prime number can be divided, without a remainder, only by itself and by 1. For example, 13 can be divided only by 13 and by 1. In this case, the number 221,044 that you looked for, is NOT a PRIME number, so it devides by 1,2, 4, 73, 146, 292, and of course 221,044.
# Question Where is the number 221,044 located in π (PI) decimals?
The number 221,044 is at position 2581159 in π decimals.
Search was acomplished in the first 100 milions decimals of PI.
# Question What is the roman representation of number 221,044?
The roman representation of number 221,044 is CCXXMXLIV.
#### Large numbers to roman numbers
3,999 is the largest number you can write in Roman numerals. There is a convencion that you can represent numbers larger than 3,999 in Roman numerals using an overline. Matematically speaking, this means means you are multiplying that Roman numeral by 1,000. For example if you would like to write 70,000 in Roman numerals you would use the Roman numeral LXX. This moves the limit to write roman numerals to 3,999,999.
# Question How many digits are in the number 221,044?
The number 221,044 has 6 digits.
#### How to get the lenght of the number 221,044
To find out the lenght of 221,044 we simply count the digits inside it.
# Question What is the sum of all digits of the number 221,044?
The sum of all digits of number 221,044 is 13.
#### How to calculate the sum of all digits of number 221,044
To calculate the sum of all digits of number 221,044 you will have to sum them all like fallows:
# Question What is the hash of number 221,044?
There is not one, but many hash function. some of the most popular are md5 and sha-1
#### Here are some of the most common cryptographic hashes for the number 221,044
Criptographic function Hash for number 221,044
md5 777804e05e25409d5f503906765fa9a9
sha1 b7d5c370a6431e423b95ec402c9a614b79cb46a5
# Question How to write number 221,044 in English text?
In English the number 221,044 is writed as two hundred twenty-one thousand and forty-four.
#### How to write numbers in words
While writing short numbers using words makes your writing look clean, writing longer numbers as words isn't as useful. On the other hand writing big numbers it's a good practice while you're learning.
Here are some simple tips about when to wright numbers using letters.
Numbers less than ten should always be written in text. On the other hand numbers that are less then 100 and multiple of 10, should also be written using letters not numbers. Example: Number 221,044 should NOT be writed as two hundred twenty-one thousand and forty-four, in a sentence Big numbers should be written as the numeral followed by the word thousands, million, billions, trillions, etc. If the number is that big it might be a good idea to round up some digits so that your rider remembers it. Example: Number 221,044 could also be writed as 221 thousands, in a sentence, since it is considered to be a big number
#### What numbers are before and after 221,044
Previous number is: 221,043
Next number is: 221,045
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# The Trick of Visual Tracking in Roulette Games
Roulette is considered as one of the easiest casino games. Even novices can place bets at a Roulette table without having to know a lot of details. However, the more seasoned players would like to differ. There are a lot of techniques used in Roulette that are important to understand before you become a pro. One of the methods that the players use to predict the number where the ball will land on the Roulette wheel is called Visual tracking.
Visual tracking as the name suggests means predicting the part of the wheel head where the pill or the ball will stop at the end. This might seem easy, but because the speed of the ball that is rolling on the wheel is really fast, not many people have the patience or the ability to follow the wheel and the ball simultaneously and make predictions of a sure win. So how does this Visual tracking work and how can you accomplish it while playing?
## The Working and Strategy Behind Visual Tracking
The basic aim of Visual tracking as discussed above is to accurately predict where the ball will land on the reels. There are different methods used by different players but most of the experts use zero on the wheel as their reference point. Since the zero is in a different colour, it is easy to follow it when the wheel spins. One eye has to be kept on the zero and the other one has to follow the Roulette ball.
The wheel head and the ball can either spin in the same direction or can also go in opposite directions. If the wheel and the ball go in two different directions, it becomes difficult to follow them initially but the players do get the hang of it with some practice.
The trackers also try to find other reference points in the wheel so that they can easily follow the ball as well as the reference point. The aim of the trackers is to note how many times the Roulette wheel has rotated. A point that is in between the number pocket and the ball track and is also easily visible from the place where the tracker is standing is the best reference point to take.
## What Will You Observe?
While visual tracking you will see that the ball often times moves faster than the wheel. So, for one rotation of the wheel head, the ball might cross your reference point 5-6 times or more. You need to measure the ratio between the speed of the wheel and the ball. This can be only done by seeing how many times the ball crosses your reference point.
As the wheel starts to slow down, the ball also starts to slow down and the number of crossings becomes lesser. The movement of the ball will slow down and the ball will start going lower down towards the pockets. With the slowed-down speed the angle between the reference point and the ball keeps on increasing and at one point the ball will be rotating at half the speed of the wheel. All this is taken into consideration and its exact location or the area where it will land on a pocket can be predicted.
## Keep These Pointers in Mind While Visual Tracking
Visual tracking not only need some skills but you will also need a good position next to the Roulette wheel to accurately see the ball on the wheel. Make sure that you have a clear vision of the ball track. Most of the experts take the last seat of the table to be close to the Roulette Wheel. Also, it is difficult to place the bet when you are already tracking the ball, hence, it is always better to have someone accompany you to the table and help you with betting.
Another necessity for visual tracking is the style of a Roulette wheel. The ball track should be of the old style. This makes it easy for the tracker to find a reference point. The ball, on the other hand, should not be too bouncy and should remain at most 5-6 pockets away from the point the dealer drops it into the wheel head. Lastly, the speed at which the wheel is being spun is also important and the croupier should be able to spin the ball at a compatible speed to the wheel head.
## Conclusion
Visual tracking is a great way for players to estimate the area where the pill might end up landing on the reels. However, you do need a lot of practice and patience to do it. Also, it is not always accurate since it is difficult to roughly estimate the speed ratio by just looking at the wheel. Try out Visual tracking during the Roulette game and see if you are able to get the hang of it.
### Blog Summary
Blog Name: The Trick of Visual Tracking in Roulette Games
Posted On: 26/08/2019
Author: Robert Bowron
4 (80%) 1 vote
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and pdfThursday, April 15, 2021 3:24:11 AM0
# Finite And Infinite Series Pdf
File Name: finite and infinite series .zip
Size: 1558Kb
Published: 15.04.2021
The n th partial sum of the series is the triangular number. Because the sequence of partial sums fails to converge to a finite limit , the series does not have a sum. Although the series seems at first sight not to have any meaningful value at all, it can be manipulated to yield a number of mathematically interesting results. For example, many summation methods are used in mathematics to assign numerical values even to a divergent series. These methods have applications in other fields such as complex analysis , quantum field theory , and string theory.
## Infinite Series
You can read a gentle introduction to Sequences in Common Number Patterns. A Sequence is a list of things usually numbers that are in order. When the sequence goes on forever it is called an infinite sequence , otherwise it is a finite sequence. When we say the terms are "in order", we are free to define what order that is!
They could go forwards, backwards Saying " starts at 3 and jumps 2 every time " is fine, but it doesn't help us calculate the:. So, we want a formula with " n " in it where n is any term number. Firstly, we can see the sequence goes up 2 every time, so we can guess that a Rule is something like "2 times n" where "n" is the term number.
Let's test it out:. That nearly worked But mathematics is so powerful we can find more than one Rule that works for any sequence. Can you calculate x 50 the 50th term doing this? In an Arithmetic Sequence the difference between one term and the next is a constant. This sequence has a difference of 3 between each number. In a Geometric Sequence each term is found by multiplying the previous term by a constant. This sequence has a factor of 2 between each number.
The Triangular Number Sequence is generated from a pattern of dots which form a triangle:. By adding another row of dots and counting all the dots we can find the next number of the sequence. Now you know about sequences, the next thing to learn about is how to sum them up. Read our page on Partial Sums.
When we sum up just part of a sequence it is called a Partial Sum. But a sum of an infinite sequence it is called a "Series" it sounds like another name for sequence, but it is actually a sum. See Infinite Series. Hide Ads About Ads. What is a Sequence? Really we could. Example: to mention the "5th term" we write: x 5. Example: 1, 4, 7, 10, 13, 16, 19, 22, 25, Example: 2, 4, 8, 16, 32, 64, , , Note: r should not be 0. This is the Fibonacci Sequence 0, 1, 1, 2, 3, 5, 8, 13, 21, 34 , Sequences also use the same notation as sets: list each element, separated by a comma, and then put curly brackets around the whole thing.
## Service Unavailable in EU region
In the previous section we saw how to relate a series to an improper integral to determine the convergence of a series. Nicely enough for us there is another test that we can use on this series that will be much easier to use. As with the Integral Test that will be important in this section. Since all the terms are positive adding a new term will only make the number larger and so the sequence of partial sums must be an increasing sequence. Therefore, the sequence of partial sums is also a bounded sequence. Then from the second section on sequences we know that a monotonic and bounded sequence is also convergent.
Sequences and Series; Finite and Infinite. Calculus 12, Veritas Prep. We are about to do the coolest theorem in calculus. Not the most important theorem, mind.
## Second Order PDE’s in Finite and Infinite Dimension
In mathematics , the harmonic series is the divergent infinite series.
#### Categories
This list of mathematical series contains formulae for finite and infinite sums. It can be used in conjunction with other tools for evaluating sums. See Faulhaber's formula. See zeta constants. The following is a useful property to calculate low-integer-order polylogarithms recursively in closed form :. Sums of sines and cosines arise in Fourier series. From Wikipedia, the free encyclopedia.
While the English words "sequence" and "series" have similar meanings, in mathematics they are completely different concepts. A sequence is a list of numbers placed in a defined order while a series is the sum of such a list of numbers. There are many kinds of sequences, including those based on infinite lists of numbers. Different sequences and the corresponding series have different properties and can give surprising results. Sequences are lists of numbers placed in a definite order according to given rules. The series corresponding to a sequence is the sum of the numbers in that sequence.
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up. I naturally thought the hyperreal extension of the real numbers would be the next best place to look, but if my resource and my deduction is correct, it isn't.
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# [Haskell-cafe] Harder Question - Chapter 4 - 4.12 - haskell the craft of functional programming - Second Edition
Anton Nikishaev anton.nik at gmail.com
Tue Jul 9 14:06:26 CEST 2013
```Manoel Menezes <manoel.menezes.jr at gmail.com> writes:
> Hi everybody!
>
> I am trying to solve the question for a long time:
>
> [4.12 Harder] Find out the maximum number of pieces we can get by
> making a given number of flat (that is planar) cuts through a solid
> block. It is not the same answer as we calculated for straight-line
> cuts of a flat piece of paper. I find out that this function has
> the following results:
>
> f 0 = 1
> f 1 = 2
> f 2 = 4
> f 3 = 8
>
> That is, from 0 to 3, the flat cuts all the pieces in two other
> pieces, so the number of pieces is doubled.
>
> But, starting from f 4, the flat can not cuts all the pieces, in case
> of f 4, the flat can cut 6 out of the 8 pieces, resulting in 12 pieces
> plus 2 pieces 2 = 14 pieces.
It can cut 7, so it is 2*7+1.
Forget about block borders and suppose you have n-1 planes and add a new
place. It will cut each n-1 planes. Now look at this new plane with
intersection lines (n-1 ones) in it. Each region in plane adds one new
3D piece, so
F(0) = 1;
F(n) = F(n-1) + P(n-1),
where P(n) is max number of pieces slicing plane with n cuts.
Spoiler: cake numbers
--
lelf
```
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https://hub.jmonkeyengine.org/t/changing-the-direction-of-gravity/23404
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# Changing the direction of gravity
I’m not sure where to put code samples so I apologize in advance if this is the wrong forum.
I searched for an example of how to change the direction of gravity since I want my player running around a sphere but didn’t find any way to do that for a charactercontroller. Here is the solution I came up with and hopefully it will help others who are having the same questions.
The steps are
• calculate the vector between the player and the center of the world
• normalize the direction vector
• multiply the normalized vector by my "fallspeed" parameter
Right now I'm calculating the gravity as velocity and not acceleration so the effect isn't "realistic" but its good enough to let me continue my proof of concept.
Also I'm letting the player thrust away from the planet which is simply negating the gravity vector and multiplying that by the thrust parameter.
[java] Vector3f gravityDirection= world.getLocalTranslation().subtract(player.getPhysicsLocation());
player.setWalkDirection(walkDirection);
[/java]
I'm sure this solution is totally obvious to the vets here but like I said I didn't see a solution when I searched.
1 Like
Hi,
I’m trying to achieve something really similar. I would like to be able to set the player’s gravity vector to something other than
[java]CharacterControl player;
//…
player.setUpAxis(0|1|2); // for x, y, z[/java]
I tried various things and the only thing that I could get working was to basically rotate everything else around the player by say 90 degrees, by doing this:
[java]Quaternion rot = new Quaternion();
Vector3f dir = new Vector3f(-9.81f,0f,0f);
rot.fromAngleAxis(FastMath.PI/2, dir);
gameLevel.rotate(rot);
landscape.setPhysicsRotation(rot);[/java]
player.setWalkDirection(Vector3f vector)
only seems to change the view of the player and “walk” him for a tiny bit.
Basically, I want to make a game based on gravity and it would be cool if someone could give me some hints. I did pre-order the “jMonkeyEngine Beginner’s Guide” book to support the project and because I want it on my shelf.
Any ideas much appreciated.
I don’t use the physics, but I believe each physics object has a method to set the gravity for that object?
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https://hackage.haskell.org/package/dsp-0.2.5.1/docs/Polynomial-Basic.html
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Polynomial.Basic
Contents
Description
Simple module for handling polynomials.
Synopsis
# Functions
polyeval :: Num a => [a] -> a -> a Source #
Evaluate a polynomial using Horner's method.
polyadd :: Num a => [a] -> [a] -> [a] Source #
polyAddScalar :: Num a => a -> [a] -> [a] Source #
polysub :: Num a => [a] -> [a] -> [a] Source #
Subtract two polynomials
polyscale :: Num a => a -> [a] -> [a] Source #
Scale a polynomial
polymult :: Num a => [a] -> [a] -> [a] Source #
Multiply two polynomials
polymultAlt :: Num a => [a] -> [a] -> [a] Source #
polydiv :: Fractional a => [a] -> [a] -> [a] Source #
Divide two polynomials
polymod :: Fractional a => [a] -> [a] -> [a] Source #
Modulus of two polynomials (remainder of division)
polypow :: (Num a, Integral b) => [a] -> b -> [a] Source #
Raise a polynomial to a non-negative integer power
polysubst :: Num a => [a] -> [a] -> [a] Source #
Polynomial substitution y(n) = x(w(n))
polysubstAlt :: Num a => [a] -> [a] -> [a] Source #
polyPolySubst :: Num a => [a] -> [[a]] -> [a] Source #
Polynomial substitution y(n) = x(w(n)) where the coefficients of x are also polynomials.
polyderiv :: Num a => [a] -> [a] Source #
Polynomial derivative
polyinteg :: Fractional a => [a] -> a -> [a] Source #
Polynomial integration
roots2poly :: Num a => [a] -> [a] Source #
Convert roots to a polynomial
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https://www.eoas.ubc.ca/~mjelline/EPS104.html
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EPS 104 An Introduction to Mathematical Methods in Geophysics
Instructor: Mark Jellinek (markj@seismo.berkeley.edu, McCone 377)
Meeting times:
Office hours: Wednesday 2-4 pm.
Text: Advanced Engineering Mathematics by Kreysig.
A Course Outline
Weeks 1 – 8. Solution and analysis of ordinary differential equations (ODEs) with applications.
Topics: Techniques for solving 1st order and 2nd order, as well as systems of, linear ODEs. In addition to “standard solution methods” we will discuss Laplace Transform methods and investigate the application of Bessel and Green’s functions in geophysical problems. Other special functions (e.g. Legendre polynomials and Airy functions) may be covered, depending on the interests of students.
Weeks 9 – 14. Derivation, solution and analysis of partial differential equations (PDEs) encountered in geophysical problems.
Topics: Conservation laws (i.e. “balance equations”) in geophysics, Reynolds Transport Theorem, Divergence Theorem, Dimensional Analysis and Scaling, Fourier Series, and solution of linear PDEs by Separation of Variables. Depending on time (and student inclination) we will also explore the application of Laplace Transform, Fourier Transform or similarity solution methods to PDEs encountered in geophysics.
Weeks 15 - 16. Working with geophysical data: an introduction to time series analysis.
Topics: An introduction to spectral analysis.
Course Goals For Students
1) Improve math skills. The student will be able to classify and solve (where possible) a large variety of ODEs. The student will acquire mathematical tools and strategies useful for understanding and (where possible) solving certain partial differential equations encountered in geophysics. The student will be able to apply the basic tools of time series analysis to understand appropriate geophysical data sets.
2) Learn how to learn and understand math. The student will develop a personalized approach for constructing conceptual and technical knowledge of mathematical techniques used in geophysics.
3) “Invert” a mathematical representation of a geological system to determine its physical meaning, significance and limitations. The student will be able to reconstruct and understand existing mathematical representations of geological systems. The student will be able to discern and articulate verbally the strengths and weaknesses of such models.
4) Create a mathematical representation of a geological system. The student will be able to represent a hypothesis mathematically. The student will understand the strengths and limitations of the model system as well as the method of solution.
5) Discuss mathematical models and analysis techniques clearly and intelligently.
Assessment
Notebook 40%
Quizzes 20%
There will be a quiz every other Wednesday (7 in total). Quizzes will only address basic facts and the routine of solving math problems (i.e. pattern recognition). Students can bring 1 sheet (1 side of one page) of review notes to each quiz. Students may drop the lowest grade.
Midterm Paper / Talk (Topics due 4th week of class; Paper due: tba; Talk to be scheduled) 20%
Final Paper / Talk (Paper due on last day of class; Talk to be scheduled) 20%
Notebooks
Your notebooks are not just for class notes in this course. A central goal of this course is that each student develops their own strategy for constructing knowledge “from the ground up” about the conceptual and technical basis for mathematical techniques of interest in geophysics. In particular, understanding the conceptual basis for a given mathematical technique is enormously powerful in determining how it may be used or modified. The notebooks for this course are intended to be a record of your learning and will complement your class notes. They should include the following components:
b) Detailed summaries of the knowledge structure for each math topic we cover. What are the main concepts? How do they fit together? For what sorts of problems would you expect a given technique to be useful? There are no rules for constructing the main links between important concepts underlying a math topic. How you will do this depends on how each of you learn. Flow charts, cartoons, text are just a few ways. Be creative. Learn how you like to learn.
c) Detailed summaries of how each mathematical technique is implemented. How are calculations performed in detail (what is the “recipe”)? Under what conditions may such calculations be performed? Are there important limitations?
Papers and Talks
In detail most problems in the Earth sciences are intractable. Consequently, determining ways to identify and pose simpler “analog problems” comprises a major part of basic research in geophysics. The goals of the midterm and final papers are to give you an opportunity to think deeply about how math is applied to develop precise understandings of geophysical problems. Projects can be (but are not limited to) an analysis of an existing model of a geophysical problem of your choice, an analysis of a dataset resulting from your own research, or you may choose to construct a model of a geophysical process that you find interesting. At the end of the semester I will bind all of your papers into an anthology that will be distributed to each member of the class. The goal of the talks, which will accompany your papers, are for you to learn to communicate mathematical and physical ideas in a precise, clear and simple way.
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Discussion in 'C++' started by storyGerald@gmail.com, Dec 4, 2005.
1. ### Guest
Recently, I'm interested in writing very efficient code.
Problem:
there is a sequence { a(0), a(1), ..., a(n-1) } and a very small
positive integer k, write an algorithm without using multiply to
calculate the following formula:
n-1
_____
\
\ ki
/ 2 * a(i)
/_____
i = 0
inline int sum(int a[], size_t n, size_t k) {
int sum = 0;
while (n--) (sum <<= k) += a[n];
return sum;
}
Gerald
, Dec 4, 2005
2. ### Viktor PrehnalGuest
Hi,
I am not mathematician but I have doutbs your code will work.
I think 2^k can be factorised before the sum, so it should look like
(+ sum is defined (as code array) from 0...n-1)
inline int sum(int a[], size_t n, size_t k) {
int sum = 0;
while (n) sum += a[n--];
return sum << k;
}
Regards, Viktor
Viktor Prehnal, Dec 4, 2005
3. ### Guest
My code is absolutely correct, at least at VC6. Maybe you didn't
understand the problem. The coefficient of 2 is k * i, not k.
Would you please try it again? Thank you!
Regards, Gerald
, Dec 5, 2005
4. ### Kai-Uwe BuxGuest
wrote:
> Recently, I'm interested in writing very efficient code.
>
> Problem:
> there is a sequence { a(0), a(1), ..., a(n-1) } and a very small
> positive integer k, write an algorithm without using multiply to
> calculate the following formula:
> n-1
> _____
> \
> \ ki
> / 2 * a(i)
> /_____
> i = 0
>
>
> inline int sum(int a[], size_t n, size_t k) {
> int sum = 0;
> while (n--) (sum <<= k) += a[n];
> return sum;
> }
>
I think, it is pretty close: you are using the Horner scheme to evaluate the
polynomial, and you realize multiplication by powers of 2 by means of bit
shifting. The only wasted instruction that I see is a redundant initial bit
shift of sum at a point where it is known to be 0. Probably, the compiler
can optimize that away.
Keep in mind, however, that efficiency is hard to argue without reference to
a specific platform. On the other hand, platform specific questions are off
topic in this group.
Huh?
Best
Kai-Uwe Bux
Kai-Uwe Bux, Dec 5, 2005
5. ### Guest
I'm sorry to ask questions like that, because I can't use English
fluently. It's not my mother language.
If this sentence hurt somebody, I apologize ^_^.
That wasn't the meaning I want to express.
Regards
Gerald
, Dec 5, 2005
6. ### Guest
I'm sorry to ask questions like that, because I can't use English
fluently. It's not my mother language.
If this sentence hurt somebody, I apologize ^_^.
That wasn't the meaning I want to express.
Regards
Gerald
, Dec 5, 2005
7. ### Guest
I've improved my code now, Here's the improved code:
inline int sum1(int a[], size_t n, size_t k) {
int sum = a[--n];
while (n) (sum <<= k) += a[--n];
return sum;
}
Thank you for giving me suggestions!
Regards, Gerald
, Dec 5, 2005
8. ### Gernot FrischGuest
<> schrieb im Newsbeitrag
news:...
> I've improved my code now, Here's the improved code:
>
inline int sum1(int a[], size_t n, size_t k)
{
int sum = a[--n];
while (n) (sum <<= k) += a[--n];
return sum;
}
you could improve speed by loop-unrolling for the cases:
n==2,3,4,5,6 - test at which point it gets slower.
....and use a register for "sum".
-Gernot
Gernot Frisch, Dec 5, 2005
9. ### Niklas NorrthonGuest
"" <> writes:
> Recently, I'm interested in writing very efficient code.
>
> Problem:
> there is a sequence { a(0), a(1), ..., a(n-1) } and a very small
> positive integer k, write an algorithm without using multiply to
> calculate the following formula:
> n-1
> _____
> \
> \ ki
> / 2 * a(i)
> /_____
> i = 0
>
>
> inline int sum(int a[], size_t n, size_t k) {
> int sum = 0;
> while (n--) (sum <<= k) += a[n];
Modification of object twice without any sequence points in between.
/Niklas Norrthon
Niklas Norrthon, Dec 5, 2005
10. ### red floydGuest
Niklas Norrthon wrote:
> "" <> writes:
>
>
>>Recently, I'm interested in writing very efficient code.
Hoare's Law (also attributed to Knuth): Premature optimization is the
root of all evil.
>>Problem:
>> there is a sequence { a(0), a(1), ..., a(n-1) } and a very small
>>positive integer k, write an algorithm without using multiply to
>>calculate the following formula:
>> n-1
>> _____
>> \
>> \ ki
>> / 2 * a(i)
>> /_____
>> i = 0
>>
My question is why? This sounds very much like homework. On modern
processors, assuming k and a) are integral, multiplication is very
efficient.
And the loss in readability/maintainability is not worth the few extra
CPU cycles. You can precalculate 2^k
long coeff = 1; // 2^k0 = 1
const long coeff_factor = 2 << k;
long sum = 0;
for (int i = 0 ; i < n; ++i)
{
sum += coeff * a;
coeff *= coeff_factor;
}
This is just off the cuff, but it's a heck of a lot more
intelligent compiler will recognize that you're multiplying coeff by a
power of 2 and optimize accordingly.
Worry about correctness and maintainablilty first, and then worry
about "optimizing" small stuff like this -- and only if an profiler
tells you that the code really is a a bottleneck.
red floyd, Dec 5, 2005
11. ### Guest
wrote:
> Recently, I'm interested in writing very efficient code.
>
> Problem:
> there is a sequence { a(0), a(1), ..., a(n-1) } and a very small
> positive integer k, write an algorithm without using multiply to
> calculate the following formula:
> n-1
> _____
> \
> \ ki
> / 2 * a(i)
> /_____
> i = 0
>
>
> inline int sum(int a[], size_t n, size_t k) {
> int sum = 0;
> while (n--) (sum <<= k) += a[n];
> return sum;
> }
>
You should make it correct first before optimising it.
1. You are modifying the variable "sum" twice between sequence points,
which is not good in C or C++. So the answer may well be wrong,
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# Finding inflection points in R from smoothed data
I have some data that I smooth using `loess`. I’d like to find the inflection points of the smoothed line. Is this possible? I’m sure someone has made a fancy method to solve this…I mean…after all, it’s R!
I’m fine with changing the smoothing function I use. I just used `loess` because that’s what I was have used in the past. But any smoothing function is fine. I do realize that the inflection points will be dependent on the smoothing function I use. I’m okay with that. I’d like to get started by just having any smoothing function that can help spit out the inflection points.
Here’s the code I use:
``````x = seq(1,15)
y = c(4,5,6,5,5,6,7,8,7,7,6,6,7,8,9)
plot(x,y,type="l",ylim=c(3,10))
lo <- loess(y~x)
xl <- seq(min(x),max(x), (max(x) - min(x))/1000)
out = predict(lo,xl)
lines(xl, out, col='red', lwd=2)
``````
From the perspective of using R to find the inflections in the smoothed curve, you just need to find those places in the smoothed y values where the change in y switches sign.
``````infl <- c(FALSE, diff(diff(out)>0)!=0)
``````
Then you can add points to the graph where these inflections occur.
``````points(xl[infl ], out[infl ], col="blue")
``````
From the perspective of finding statistically meaningful inflection points, I agree with @nico that you should look into change-point analysis, sometimes also referred to as segmented regression.
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https://github.com/blab/ncov-phylodynamics/commit/8c3a3ac14777ddafcca7ae651d5d5d31869c3572
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# blab / ncov-phylodynamics
Update Markdown
@@ -45,16 +45,17 @@ The Mathematica notebook ncov-phylodynamics.nb contains code to analyze result ### Rate and TMRCA We find substitution rate consistent with previous work of 0.9 × 10-3 (95% CI 0.5-1.4 × 10-3) substitutions per site per year. We find a median TMRCA of 3 Dec, 2019 (95% CI 30 Oct to 17 Dec). We find a median TMRCA of 3 Dec (95% CI 30 Oct to 17 Dec). ### Effective population size and exponential growth rate These phylodynamic approaches can estimate effective size of the virus population by examining rates of coalesce through time. These phylodynamic approaches can estimate effective size of the virus population by examining rates of coalescence through time. Here, we estimated the exponential growth rate as 35.4 (95% CI 9.6-50.0) per year. This translates to a doubling time of 7.2 (95% CI 5.0-12.9) days. This coincides closely with doubling time reported by modeling groups looking at reported cases in China ([Wu et al][Wu et al]). Here, we plot timescale of coalescence {% eqinline N_e \tau %} through time:  {% eqinline N_e \tau %} is what is directly measured by phylodynamic methods and is measured in years. @@ -67,21 +68,22 @@ We assume generation time {% eqinline \tau %} to be 7.5 days following [Li et al Additionally, effective population size {% eqinline N_e %} can be translated into prevalence with knowledge of the variance in offspring distribution. High variance in distribution of secondary cases reduces prevalence relative to {% eqinline N_e %} as described by [Volz et al][Volz et al]. This reduction is equal to {% eq \sigma^2 = \frac{1}{E[R_0]} + \frac{1}{k} + 1 %}, {% eq \sigma^2 = \frac{1}{E[R_0]} + \frac{1}{k} + 1, %} where {% eqinline E[R_0] %} is the mean number of secondary cases and {% eqinline k %} is the dispersion parameter of secondary cases. We assume {% eqinline E[R_0] %} to be between 1.8 and 2.8 following [Wu et al][Wu et al] and others. We assume that variance of secondary cases is at most like SARS with superspreading dynamics with {% eqinline k=0.15 %}, but allow for less variance with {% eqinline k=0.30 %}. Thus, we convert BEAST estimates of {% eqinline N_e \tau %} to point prevalence {% eqinline I %} by following {% eq I = N_e \tau \times \frac{\sigma^2}{\tau} %} Thus, we convert BEAST estimates of {% eqinline N_e \tau %} to point prevalence {% eqinline I %} by following {% eqinline I = N_e \tau \times \sigma^2 / \tau %}. We arrive at the following estimate of prevalence through time:  We estimate a median prevalence on 8 Feb of 28,500 currently infected with a 95% uncertainty interval of between 7500 and 104,300 currently infected. ### Total incidence We estimate incidence in each serial interval and then calculate a cumulative incidence total:  We estimate a median total incidence on 8 Feb of 55,800 total infections since start of epidemic with a 95% uncertainty interval of between 17,500 and 194,400 total infections.
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# toRad() Javascript function throwing error
I'm trying to find the distance between two points (for which I've latitudes & longitudes) using the technique described here at Calculate distance between two latitude-longitude points? (Haversine formula)
The codes are as below Javascript:
```var R = 6371; // Radius of the earth in km
var dLat = (lat2-lat1).toRad(); // Javascript functions in radians
var dLon = (lon2-lon1).toRad();
var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.sin(dLon/2) * Math.sin(dLon/2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c; // Distance in km
```
But when I try to implement it, an error shows up saying Uncaught TypeError: Object 20 has no Method 'toRad'.
Do I need a special library or something to get .toRad() working? because it seems to be screwing up on the second line.
You are missing a function declaration.
In this case toRad() must be defined first as:
```/** Converts numeric degrees to radians */
if (typeof(Number.prototype.toRad) === "undefined") {
Number.prototype.toRad = function() {
return this * Math.PI / 180;
}
}
```
according to the code segment all at the bottom of the page
Or in my case this didn't work. It may because i needed to call toRad() inside jquery. Im not 100% sure, so i did this:
```function CalcDistanceBetween(lat1, lon1, lat2, lon2) {
//Radius of the earth in: 1.609344 miles, 6371 km | var R = (6371 / 1.609344);
var R = 3958.7558657440545; // Radius of earth in Miles
var dLat = toRad(lat2-lat1);
var dLon = toRad(lon2-lon1);
var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.sin(dLon/2) * Math.sin(dLon/2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c;
return d;
}
/** Converts numeric degrees to radians */
return Value * Math.PI / 180;
}
```
I needed to calculate a lot of distances between the points for my project, so I went ahead and tried to optimize the code, I have found here. On average in different browsers my new implementation runs almost 3 times faster than mentioned here.
```function distance(lat1, lon1, lat2, lon2) {
var R = 6371; // Radius of the earth in km
var dLat = (lat2 - lat1) * Math.PI / 180; // deg2rad below
var dLon = (lon2 - lon1) * Math.PI / 180;
var a =
0.5 - Math.cos(dLat)/2 +
Math.cos(lat1 * Math.PI / 180) * Math.cos(lat2 * Math.PI / 180) *
(1 - Math.cos(dLon))/2;
return R * 2 * Math.asin(Math.sqrt(a));
}
```
You can play with my jsPerf (which was vastly improved thanks to Bart) and see the results here.
Why not simplify the above equation and same a few computations?
Math.sin(dLat/2) * Math.sin(dLat/2) = (1.0-Math.cos(dLat))/2.0
Math.sin(dLon/2) * Math.sin(dLon/2) = (1.0-Math.cos(dLon))/2.0
I was having the same problem.. looking at Casper's answer, I just did a quick-fix: Ctrl+H (Find and Replace), replaced all instances of .toRad() with * Math.PI / 180 . That worked for me.
No idea on the browser performance speeds etc, though.. My use case only needs this when the user clicks on a map.
I changed a couple of things:
```if (!Number.prototype.toRad || (typeof(Number.prototype.toRad) === undefined)) {
```
and, I noticed there was no checking for the arguments. You should make sure the args are defined AND probably do a parseInt(arg, 10) / parseFloat on there.
### PHP Transferring Photos From One Oracle Database Table to Another
I am attempting to transfer a set of photos (blobs) from one table to another across databases. I'm nearly there, except for binding the photo parameter. I have the following code:
### How to calculate the inverse of the normal cumulative distribution function in python?
How do I calculate the inverse of the cumulative distribution function (CDF) of the normal distribution in Python?
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Mathematica for High School Students
• To: mathgroup at smc.vnet.net
• Subject: [mg24097] Mathematica for High School Students
• From: "David Park" <djmp at earthlink.net>
• Date: Tue, 27 Jun 2000 00:51:53 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com
```Hi Ted,
Well, I can see that my high school notebooks didn't come off too favorably
with you; but I would like to continue the discussion because it interests
me; because we do have a difference of opinion here; and because I do value
your opinion. I hope you don't mind if I post this to MathGroup because I
tried to get a discussion going on the topic about a month ago. Maybe I can
stimulate more discussion from the larger group.
(There were two notebooks: one on solving simple equations step by step
using pure functions and Map, and one centered around Gauss' feat of adding
the numbers 1 to 100.)
____________________________________________
Ted: The way Gauss added the numbers from 1 to 100 is very easy to explain
without using Mathematica.
______________________
Perhaps, but most mathematical ideas can be explained without the aid of
Mathematica. The vast majority of mathematical books don't even mention
Mathematica. The purpose of the notebooks was to teach something about
Mathematica using an idea that might appeal to students. So, it doesn't seem
to be a reasonable argument to say that we can teach Mathematica without
using it.
____________________________________________
Ted: The way I see one shouldn't learn about Mathematica features that are
not
fully intuitive until their 3rd year of college (except in a Computer
Science class about Mathematica). Even in the 3rd and 4th years of college I
would only encourage gradual introduction of less than intuitive features in
Mathematica.
_____________________
And what features of Mathematica are not "fully intuitive" so that the
student "shouldn't learn" them? Would you hazard to put forward a list of
Mathematica statements to which young students would be restricted? Or a
list that would be banned from their use?
Are the functional programming statements ones that you would call not
"fully intuitive"? Is it easier to understand the For statement, than to
understand the Map statement? I can't remember exactly, but it seems that it
was in junior high school that I learned that an equation remains true if we
do equal operations on each side of the equation (other than dividing by
zero). So if you are going to ban pure functions and Map, how would you
implement that simple concept with Mathematica, which presumably the student
Would you allow the young student to use Solve?
I believe that there is much work to do in learning how to apply tools such
as Mathematica to secondary education. It is not something that I am
directly involved in, and have no expertise in; but I am still interested.
I would like to put forward several arguments for allowing and encouraging
the full use of Mathematica for the INTERESTED student.
First, there is the great advantage to the student of just gathering sheer
mathematical experience. For most students this is not easy to get. Don't
get in his way, but help him. It is much easier to see a new idea, when one
has the mathematical experience that the idea fits on to.
There are mathematical ideas that are far easier to teach and learn with
Mathematica than out of a text book, such as some of the functional
programming ideas. You can actually see them in action. It is not that the
ideas are difficult or abstruse; they just don't fit into the rigid
curriculum. There is no reason why a bright student shouldn't be exposed to
these ideas.
I don't think that any young student interested in Mathematica should be
shackled in his use of it, or even worse, admonished for using it. I have a
sister fifteen years younger than me. One day when I was home from college,
and reading a book, she said: "I wish I had a book to read." "You don't have
any books?" "They gave us one book, but I've read it." "Why don't you take
out some books from the library." "O No! We're not allowed to do that."
Later I related this experience to a primary school teacher, expressing my
disapproval of such restrictions. Did I ever get a piece of her mind! There
was a very "scientific" method of teaching students how to read. If they
jumped ahead and read something out of order, the whole plan was ruined!
How do we know that the young student isn't another Gauss? We shouldn't
erect any road blocks. We should help as far as we can and see how far he
goes. Richard Feynman's high school teacher put him on to doing variational
calculations. Would you say that he shouldn't have been permitted such
things until his third or fourth year in college? Don't worry, there is
little chance that a student will charge ahead of his or her capabilities; a
far greater danger is that he will fall short of his capability.
David Park
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## We'll teach you how.
“An expense ratio is the amount companies charge investors to manage a mutual fund or exchange-traded fund (ETF). The expense ratio represents all of the management fees and operating costs of the fund. The expense ratio is calculated by dividing a mutual fund’s operating expenses by the average total dollar value for all the assets within the fund.” Investopedia
About 8 years ago, I was put on the first retirement plan committee at my day job. Until that point, my employer had a pretty generous 401(k) with a match. But there was nobody regularly monitoring the 401(k), its funds, or the fees charged by the broker, the plan administrator, and the individual funds in the plan.
There were two of us on the committee–the controller and myself–that became obsessed with lowering the expense ratio. For those of you who aren’t familiar with expense ratios, they are the fees you pay to maintain your 401(k), expressed in terms of a percentage. When the committee took over management of the 401(k), the average expense ratio of the funds in the plan was 3.5%.
That meant that for every \$100 that someone had in the 401(k), they had to spend \$3.50 per year. \$1,000.00 meant that they had to spend \$35.00 a year. \$100,000 meant \$3,500. You may be looking at those numbers and thinking “What’s the big deal? \$3.50 per year is nothing. Seems like you’re chasing pennies.”
Here’s the deal, though. An average return in a well-balanced 401(k) might be 6% to 8% per year. As a matter of fact, that’s the percentage you’ll often see as the assumed return in a retirement calculator. And over the life of a retirement plan, for every year that you have a 20% return (“hello, 2012”), most years are much lower (“hello, 2008”). Thus, planners don’t like to count on more than 6% or 8% (game theory regression to the mean?).
Let’s do a little bit more math. That means in an average year, \$100 in your 401(k) will earn you between \$6.00 and \$8.00. That sounds pretty good…..until you subtract the expense ratio. If you put \$100 in the plan and it earns 6%, at the end of the year you have \$106. But then you have to subtract the expense ratio (\$106*.035) and you are left with only \$102.29. Now let’s suppose inflation is 3%, which is what the Federal Reserve targets as ideal inflation. That means the \$100 you invested is actually worth less than when you invested it. What a total rip-off. More than half of your investment gain went to money managers and brokers.
There are entire blogs dedicated to lowering expense ratios. Financial independence retire early (FIRE) folks lose sleep over them. No one wants to give away the money they earn on their investments. For five years, the retirement committee at my day job negotiated and re-negotiated the fees our 401(k) participants were paying. Every time we spoke to our plan broker, we asked for lower rates. Eventually, through negotiation, we were able to get the average expense ratio on the 401(k) down to between 1.25% and 1.5%, which put our plan in the 99th percentile for plans its size.
So right now, you are probably asking yourself why I just wrote 500 words on 401(k) expense ratios in a blog about buying and selling houses. Well, the answer is easy. For most people, their home is their largest investment. And while I don’t personally think you should plan on retiring based on your home’s equity, when you start to prepare to sell your home you should think of it like an investment.
Historically, home prices have been tied directly to inflation. While in some locations during some time periods, home prices outpace inflation (like Boise over the last couple of years), and others lag behind (say maybe Detroit), eventually, given a long enough time, house prices will be pretty even with inflation. That means if your home was worth \$200,000 when you bought it, after 10 years at the average inflation of 3% it would be worth \$268,780.00. An extra \$68,780 in equity sounds pretty good, doesn’t it?
Here’s the thing though: your home may have appreciated 68,780, but that appreciation wasn’t cost-free. If you’re like most Americans, you needed a mortgage to buy your house. No shame in that; not very many people have \$200,000 in cash lying around. For this real estate thought experiment, let’s assume that you have a standard 30 year mortgage, and that you put 20% down. Rates have been historically low for the last decade, so let’s give this mortgage a relatively modest 4.0% interest rate. That means in 10 years, you will have paid \$58,138.25 in interest. You have only \$10,641.75 in gain left after your interest payments. Ouch!
Now, here’s where this math stuff starts to really get scary if you aren’t paying attention. If you pay a realtor or real estate agent the standard 6% commission to sell your house, you will be paying \$16,126.80. That means if you sell your house, you’ll actually be getting \$5,485.05 less for your house than you paid for it! After 10 years! Now you still will get a check when you close, because you’ve also been paying down your mortgage during that time. But, your house, as an investment, has been a loser. It has a negative rate of return.
How do you make money on the investment that is your house? Well, on the transaction side of things you do exactly what the retirement committee did. You have to lower your expense ratio. You can do that in two ways. First, you can lower the amount of interest you pay on your house. In a few weeks we’ll interview a lender to give you some tips on exactly how to do that. But at most we are talking about ½% to 1% you might be able to save, and that is if your credit is bad or rates go down. And one thing to note about interest rates is that they are often pegged to other economic indicators, so there isn’t much room to negotiate. However, saving one percent is still worth it and you should totally make that effort.
But there is another way to save a whole lot of cash on a real estate transaction. Just two paragraphs ago we talked about the two major expenses in a real estate transaction: the interest rate and the real estate commissions. You can save 6% of the value of your home by selling it yourself. In the example above, if you didn’t use a realtor on your \$200,000 house, it swings the transaction from being a \$5,500 loss to almost an \$11,000 gain. Hey, a gain is better than a loss any day of the week. What’s more, you’ll have a 5.32% return on investment. Not quite 6%, but pretty dang close, and a reasonable gain for an investment.
If you are worried about selling your house yourself, of maybe have never really thought about it, we are here to teach you how. Revostate’s tools and tutorials can help you turn your biggest investment into an investment that actually gives you a return instead of a loss.
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# hw1css - CSE 105 Introduction to the Theory of Comptuation...
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CSE 105: Introduction to the Theory of Comptuation Fall 2010 Problem Set 1 Instructor: Daniele Micciancio Due on: Wed. Oct 6, 2010 Guidelines: solutions to the homework should be submitted electronically, following the instructions on the class website. As part of your solutions to this assignment you should submit: 1. A file with the text of your solutions to all 4 problems, in pdf format. 2. 4 jflap files containing the automata for problems 2 and 3. The main file should be in pdf format (no doc files, etc.) and be clearly divided into 4 sections (and respective subsections) corresponding to the 4 problems. There is no jflap file for problems 1 and 4. All your files (writeup.pdf,2a.jff,2b.jff,3a.jff,3b.jff) should be named and zipped together according to the instructions. Homeworks should be submitted on the due date before class time. 1 Simple DFA questions Let M be the DFA with states Q = { q 1 , q 2 , q 3 , q 4 , q 5 }, input alphabet Σ = { u, d } , transition function δ , start state q 2 and accept states F = { q 2 } where δ is given by the following transition table. δ q 1 q 2 q 3 q 4 q 5 u q 5 q 1 q 2 q 3 q 4 d q 2 q 3 q 4 q 5 q 1 1. Give 3 example strings that are accepted by the DFA 2. Give 3 example strings that are not accepted by the DFA 3. Give a high level English description of M using complete sentences and proper grammar and punctu- ation. This description should be no more than two reasonably short sentences
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# Description
Physics is a physical world simulator and playground -- you can add squares, circles, triangles, or draw your own shapes, and see them come to life with forces (think gravity, Newton!), friction (scrrrrape), and inertia (ahh, slow down!).
## Screenshots
A motor drives the centre cross shape, while the outer edges are pinned to stop the blender breaking. One motor is used to drive a circle connected via a belt to another pinned wheel that bounces a weight to make a puppet dance. More puppet dancing Running dog using two motors Using a motor on a circle to drive a piston that slowly releases balls one by one Motor driven earthquake simulator Example illustrating transverse wave propagation. Example illustrating longitudinal wave propagation. Example illustrating the pen tool attached to the end of a pendulum. Tracing multiple objects A simple gear and motor generated in Turtle Art rendered in Physics a chain generated in Turtle Art
## Ideas to try
• Build a catapult to simulate Angry Birds
• Build a machine that sorts different sized balls in to two buckets, large and small, with no ball jams.
• A cyclic mechanism for lifting balls from the bottom of the screen to the top, again and again.
• Try dropping 2 different mass objects at the same time.
• Experiment with pendulums of different lengths and masses.
• Ping pong, can you make a device that hits a ball back and forth across the screen?
• Try building a mechanical binary clock.
• Convert rotation into parallel motion using pistons.
• Experiment with touching one motorised circle against one pinned circle of various sizes.
• Try building a rag-doll puppet and make it dance in a convincing way.
• Use just links and circles to make a structurally sound Eiffel Tower.
• Try building a ratchet mechanism.
• Try building an analogue clock face where the min hand goes around 60 times for each hour.
• Modify the program itself: Modifying Physics
## Tools
Activity Toolbar Button
Opens the activity toolbar (See Below)
Tools Toolbar Button
Opens the tools toolbar (See below)
Play/Pause Button
allows you to stop time and start it again, allowing you to build constructions without them falling to the ground while you work. You can now use the Grab tool while time is stopped to easily re-arrange the position of objects. <Ctrl+space>
Erase Pen Button
Erase all the points drawn by the pen tool
Erase-all Button
Erase all objects in current simulation
Activity Title
Change the name of your project here
Activity Description
Collaboration
Open the collaboration palette to share your project
Save pen data
Save pen data to the Journal as a JSON-encoded stream
Save pen data (CSV)
Save pen data to the Journal as a comma-separated values database
Draw Tool Button
click and hold to draw any shape you'd like! <Ctrl+D>
Circle Tool Button
circles of any radius. Click (center) drag and release (outer edge). <Ctrl+C>
Triangle Tool Button
triangles (equilateral) of any size and initial rotation. Click (center of base) drag and release. <Ctrl+T>
Box Tool Button
rectangles of any dimension. Click (corner) drag + release (opposite corner). <Ctrl+B>
Polygon Tool Button
as many sides as you would like. Draw your own N-gon. Click, drag and release for each point, to end return to the start point. <Ctrl+P>
Grab Tool Button
drag existing objects around with the mouse. <Ctrl+G>
Motor Tool Button
click an object to pin and drive it with clockwise rotation. It is easiest to stop the simulation, place your object, add its motor, and resume the simulation again. <Ctrl+M>
Pin Tool Button
pin a shape to the screen, it can rotate about the pin. If you want to lock a shape in place, use two or more pins to stop it rotating. <Ctrl+O>
Joint Tool Button
connect two objects together with a rod. Click on any object, drag to another object and release to create the joint (each end of joint allows rotation). <Ctrl+J>
Chain Tool Button
add a flexible chain to connect two objects together. Click and hold to draw any shape. If you start or stop on an object, it becomes connected to the end of the chain (like the joint tool)
Pen Tool Button
add a pen to any object to trace the motion of that object
Erase Tool Button
click on an object to erase it, or click and hold to draw a line of destruction--erasing everything in its path! <Ctrl+E>
Mass
feather, wood, and anvil represent light to dense
Bounciness
basketball, baseball, and bowling ball represent very bouncy to inelastic
Friction
ice skate, shoe, and sneaker represent low to high coefficient of friction
The motor has a palette to set speed (fast and slow) and direction (counter-clockwise and clockwise)
The chain has a palette to set the coarseness and weight of the chain elements
Hints and tips: Single clicking (no drag) with the circle, triangle or box tool, will add a default sized shape. Once you have used a shape tool, it remembers the last shape you made with it, a single click will add a clone of that last shape.
## Did you know?
There is a Plugin for Turtle Blocks that lets you create projects for the Physics Activity:
Use these blocks to create objects that are added to the bodylist and jointlist of a Box2d database used by the Physics Activity. Objects are positioned by the turtle x,y and colored based on the current pen attributes. From left to right:
• density: set the object density (0 is light; 1 is heavy)
• friction: set the object friction (0 is slippery; 1 is sticky)
• bounciness: set the object restitution (0 is rigid; 1 is bouncy)
• start polygon: use the current turtle x,y position to specify the first vertex of a polygon
• add point: use the current turtle x,y position to add a vertex to a polygon
• end polygon: use the current turtle x,y position to specify the last vertex of a polygon
• end filled polygon: use the current turtle x,y position to specify the last vertex of a filled polygon
Note: The polygon must be 'normal', e.g., no crossed lines; no holes. Since Box2d does not support concave polygons, all polygons are converted to triangles (triangulation). (In Turtle Blocks, these triangles are shown by slight variations in color.)
• triangle: add a triangle object at the current turtle x,y position
• circle: add a circle object at the current turtle x,y position
• rectangle: add a rectangle object at the current turtle x,y position
• gear: add a gear object at the current turtle x,y position with the number of teeth as the argument
• gear radius: returns the radius of a gear with the number of teeth as the argument (useful for positioning gears)
• motor: add a motor at the current turtle x,y position (attached to the object at this position)
• pin: add a pin at the current turtle x,y position (attached to the object at this position) to prevent movement
• pen: add a pen at the current turtle x,y position (attached to the object at this position) to trace motion
• joint: add a joint between the object at the current turtle x,y position and the object at x, y
• save as Physics project: save the current bodylist and jointlist as a Physics project in the Journal
Note: The current model is cleared whenever the Erase button is pressed or a Clean block is run.
##### Examples
For Turtle Blocks versions 107 - 183 File:TAgear.ta File:Gear.physics
For Turtle Blocks verisons 184 - [1]
Chain Turtle Blocks project file: [2]
File:Physics-plugin.tar.gz Physics plugin
# Development
There are quite a few code layers to contend with. A regular Python Activity acts as a Sugar wrapper to Physics which is written using SugarGames which itself wraps Pygame, Physics then uses Elements as a wrapper for Box2D.
## Video
Please do post videos with feedback, talking while you're trying something for the first time is particularly insightful as it can highlight those initial expectations for user interface behaviour. Bonus points to Dennis for being brave enough to be part of the first wave!
UI feedback Video unfortunately there's a seg fault at the end, ticket from Dennis is #1194.
Sugar at the Boston Museum, for more information see Bill Kerr's blog.
## Release Notes
### 17
ENHANCEMENT:
• Chain tool behaves like magic pen: you can make loops and closed paths
### 16
ENHANCEMENT:
• Add sharing to magic pen and polygon
• Clean up of some internals
### 15
ENHANCEMENT:
BUG FIX:
• Fix problem with pause/play button
### 14
ENHANCEMENTS:
• New translations
• GTK3 conversion by Ignacio Rodriguez
### 13
ENHANCEMENTS:
• Added palettes to set density, bouciness, and friction (w/Sai Vineet)
• Added palette to set motor speed, rotation (w/Sai Vineet)
• Added chain tool (w/Sai Vineet)
BUG FIX:
• Updated Box2d version to eliminate some memory leaks (Sai Vineet)
### 12
ENHANCEMENTS:
• Added option to joints to set collideConnected = False
• Added erase traces (Sai Vineet)
• Added save/restore of pens and traces (Sai Vineet)
BUG FIXES:
• Removed cjson dependency for elements
• pep8 cleanup (Sai Vineet)
### 11.9
• Removed cjson dependency for elements
• Added option to joints to set collideConnected = False
### v5
• Added support for new toolbar design for Sugar 0.86 (old toolbars still work for Sugar 0.84 and back)
• Objects can now be moved with the Grab tool while the simulation is paused!
• Tool code cleanup and refactoring
• Includes latest translations
• Many thanks go to Peter Froehlich, Akash Gangil, Craig Macomber, and the translation team.
### v4
• Include latest translations
### v3
• Journal state saving now supported!
• MIME type support added so Physics Journal entries can be sent to others (application/x-physics-activity)
• Fixed Activity title text input so you can name your work correctly (olpcgames glitch)
• New Grab toolbar icon (hand)
• New Polygon toolbar icon (irregular polygon shape)
• Cleaned up toolbar order
• Single click behaviour so that Circle, Triangle, and Box tool add default sized shapes
• Single click behaviour for all tools, so that a subsequent single click creates a clone of the last shape made with that tool.
• Erase tool now erases (one by one) pins/joints/motors from a shape, before finally removing the shape itself.
• Using <Ctrl>+key for all keyboard shortcuts (was causing PyGame input focus issues when typing a title)
• Using the Sugar standard arrow cursor for the PyGame canvas (well, a fake one)
• Cleaned up the Activity icon
• Upgraded to new version of Elements 0.13
• Upgraded to new version of Box2D
• Picked up Pootle (July 3rd) translations
#### user notes
I really struggle to get any of the motor actions to work. Can you post some videos of your own successes? It's hard to build curriculum when you don't know how something is supposed to work.
### v2
• Migrated to Sugar Labs
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# ch05 - Joh_Ch05.qxd 2:39 PM Page 2 CHAPTER 5 Dosage...
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Dosage Calculations After completing this chapter, you should be able to: • Solve one-step pharmaceutical dosage calculations. • Set up a series of ratios and proportions to solve a single dosage calculation. • Determine what information you will need to solve for, in addition to any given information, to properly calculate dosages. • Convert pediatric weights from pounds to kilograms. • Accurately determine dosages based on mg/kg/day. • Define common sig (signa) codes used on prescriptions. INTRODUCTION Proper dosing of medications is important to ensure patient safety. Calculating dosages, dosage regimens, and compounding formulas involves the use of simple math principles. You can solve many of these problems by setting up ratios and proportions using the information given in the question and keeping like units consistent. CHAPTER 5 2 L earning Objectives
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Chapter Five Dosage Calculations 3 Sig Refresher The sig portion of the prescription order, meaning signa , is where the instruc- tions for the patient are written. Pharmacy technicians enter the information from the prescription order into the computer. The sig is an important value to remember in order to properly determine pediatric dosages. The following are some of the more common sigs you will find on prescriptions: qd every day qod every other day d daily bid twice a day tid three times a day qid four times a day q4h every 4 hr q6h every 6 hr q8h every 8 hr q4–6h every 4–6 hr prn as needed Depending on the workplace, you may also see sigs such as the following: q3d every three days qmwf every Monday, Wednesday, and Friday qw every week Dosage Calculations Dosage calculations include calculating the number of doses, dispensing quantities, and ingredient quantities; these calculations are performed in the pharmacy on a daily basis. The pharmacy technician must have a full work- ing knowledge of how to perform these calculations. To perform dosage calculations, you will utilize the information and principles introduced in the previous chapters of this book. You can solve these calculations by setting up ratios and proportions, keeping like units consistent, and cross-multiplying. CALCULATING THE NUMBER OF DOSES To calculate the number of doses, you should first determine which informa- tion presented is actually applicable to the question. Too often mistakes are made on dosage calculations because we overcomplicate them. EXAMPLE 5.1 How many 1-tsp doses are in a 4 oz bottle of Prozac ® Liquid Solution 20 mg/5 mL? Rx Prozac ® Solution tsp. po qd Disp. # 4 oz .. TT = = = = = = = = = = = = = =
4 Chapter Five Dosage Calculations Let’s look at the information that has been provided: tsp po—the dose qd—the frequency 4 oz—the quantity dispensed Prozac ® Solution 20 mg/5 mL—the drug name and strength 120 mL—the quantity of the stock bottle The question is simply asking how many doses make up the total amount being dispensed. The strength of the drug, frequency of dosage and quantity
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Next: WILSON-BURG SPECTRAL FACTORIZATION Up: CAUSALITY AND SPECTAL FACTORIZATION Previous: Toeplitz methods
## Kolmogoroff spectral factorization
With Fourier analysis we find a method of spectral factorization that is as fast as Fourier transformation, namely for a matrix of size N. This is very appealing. An earlier version of this book included such an algorithm. Pedagogically, I didn't like it in this book because it requires lengthy off-topic discussions of Fourier analysis which are already found in both my first book FGDP and my third book PVI.
The Kolmogoroff calculation is based on the logarithm of the spectrum. The logarithm of zero is minus infinity -- an indicator that perhaps we cannot factorize a spectrum which becomes zero at any frequency. Actually, the logarithmic infinity is the gentlest kind. The logarithm of the smallest nonzero value in single precision arithmetic is about -36 which might not ruin your average calculation. Mathematicians have shown that the integral of the logarithm of the spectrum must be bounded so that some isolated zero values of the spectrum are not a disaster. In other words, we can factor the (negative) second derivative to get the first derivative. This suggests we will never find a causal bandpass filter. It is a contradiction to desire both causality and a spectral band of zero gain.
The weakness of the Kolmogoroff method is related to its strength. Fourier methods strictly require the matrix to be a band matrix. A problem many people would like to solve is how to handle a matrix that is almost'' a band matrix -- a matrix where any band changes slowly with location. Blind deconvolution A area of applications that leads directly to spectral factorization is blind deconvolution.'' Here we begin with a signal. We form its spectrum and factor it. We could simply inspect the filter and interpret it, or we might deconvolve it out from the original data. This topic deserves a fuller exposition, say for example as defined in some of my earlier books. Here we inspect a novel example that incorporates the helix.
solar
Figure 9
Raw seismic data on the sun (left). Impulse response of the sun (right) derived by Helix-Kolmogoroff spectral factorization.
Solar physicists have learned how to measure the seismic field of the sun surface. If you created an impulsive explosion on the surface of the sun, what would the response be? James Rickett and I applied the helix idea along with Kolmogoroff spectral factorization to find the impulse response of the sun. Figure shows a raw data cube and the derived impulse response. The sun is huge so the distance scale is in megameters (Mm). The United States is 5 Mm wide. Vertical motion of the sun is measured with a video-camera like device that measures vertical motion by doppler shift. From an acoustic/seismic point of view, the surface of the sun is a very noisy place. The figure shows time in kiloseconds (Ks). We see about 15 cycles in 5 Ks which is 1 cycle in about 333 sec. Thus the sun seems to oscillate vertically with about a 5 minute period. The top plane of the raw data in Figure (left panel) happens to have a sun spot in the center. The data analysis here is not affected by the sun spot so please ignore it.
The first step of the data processing is to transform the raw data to its spectrum. With the helix assumption, computing the spectrum is virtually the same thing in 1-D space as in 3-D space. The resulting spectrum was passed to Kolmogoroff spectral factorization code. The resulting impulse response is on the right side of Figure . The plane we see on the right top is not lag time ;it is lag time Ks. It shows circular rings, as ripples on a pond. Later lag times (not shown) would be the larger circles of expanding waves. The front and side planes show tent-like shapes. The slope of the tent gives the (inverse) velocity of the wave (as seen on the surface of the sun). The horizontal velocity we see on the sun surface turns out (by Snell's law) to be the same as that at the bottom of the ray. On the front face at early times we see the low velocity (steep) wavefronts and at later times we see the faster waves. This is because the later arrivals reach more deeply into the sun. Look carefully, and you can see two (or even three!) tents inside one another. These inside tents'' are the waves that have bounced once (or more!) from the surface of the sun. When a ray goes down and back up to the sun surface, it reflects and takes off again with the same ray shape. The result is that a given slope on the traveltime curve can be found again at twice the distance at twice the time.
Next: WILSON-BURG SPECTRAL FACTORIZATION Up: CAUSALITY AND SPECTAL FACTORIZATION Previous: Toeplitz methods
Stanford Exploration Project
4/27/2004
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# How many MBs are there in a GB, TB, KB or Byte? ( Conversion Tool )
2018-05-24T15:27:51+00:00May 24th, 2018|Web Hosting|
Web hosting disk space and monthly transfer limits, bandwidth, RAM etc. can sometimes be confusing. Particularly with conversions between Kilobytes, Megabytes, Gigabytes and Terabytes.
As a web hosting company, we hear questions like these a lot;
• How many MBs are there in a GB?
• Is 1 GB equal to 1000 or 1024 MB?
• How many MBs are there in a Terabyte? etc.
With help of this simple MB to GB converter and the short article explaining all the units, I hope to make things simpler for you.
## Convert Bytes to KB, MB, GB and TB
You can use the following converter to easily convert any measure of disk space or bandwidth to the other easily.
Just add a value for either Bytes, Kilobytes, Megabytes, Gigabytes or Terabytes and the tool will calculate the rest for you, on the fly.
### What is a Byte (B)?
In a computer, information is stored as bits and bytes, at the smallest scale. A bit is too small to be of much use. 8 bits grouped together make one Byte.
A Byte therefore, is one of the two smallest units of measurement used to measure data.
### What is a Kilobyte (KB)?
A Kilobyte is a unit of measure for digital storage or data, which is equal to 1024 Bytes.
Some decades ago, a Kilobyte used to the most popular unit of measurement for data, but since the volume of information has increased significantly, it has been replaced by Gigabyte as the most popular one.
### What is a Megabyte (MB)?
A Megabyte (MB) is a unit of data measure equal to 1024 KBs or 1,048,576 Bytes.
### What is a Gigabyte (GB)?
The most popular unit of measure – particularly in the web hosting industry – these days, the Gigabyte (GB) is equal to 1024 MBs or 1,048,576 KB.
### What is a Terabyte (TB)?
A Terabyte is a unit of data measure which is equal to 1024 GBs or 1,048,576 MBs.
### Why Does my Hard Drive Have Less Space Than that Advertised?
To a hard disk manufacturer, one KB is 1000 bytes, one MB is 1000 KB, and one GB is 1000 MB.
Which is technically an accurate definition given the prefixes; Kilo meaning 1,000, etc.
Given that definition, if a hard disk is advertised as 1TB, it contains 1 * 1000 * 1000 * 1000 * 1000 = 1,000,000,000,000 bytes of space. The hard disk manufacturer therefore, advertises it as a 1TB hard drive.
When actually, the 1 TB hard drive would be able to store only 931 GB or 0.909 TB of actual data.
Manufacturers of RAM however, do not manufacture their products in even groups of 1,000. When you’re buying memory, a KB is 1024 bytes, an MB is 1024 KB, and a GB is 1024 MB.
Start Making Money Today with the Inspedium Affiliate Program!
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# How to calculate steel quantity from drawing for Beam?
Hello, guys here we will learn how to calculate steel quantity in the beam which is given in the drawing.
Suppose we are given the length of beam 10′(Feet) and Depth of beam 18″(Inches). The arrangement of steel bars in the beam is given below.
Let the hook be 9D. Where D is the diameter of the steel bars.
Spacing between the stirrups be 6.3″ at the corners and 12″ at the middles.
Clear Cover be 1″.
Bent up bars are making angle of 45 degree with horizontal.
Hence, this is the drawing at the site. It is given by the designer, not by the surveyor. Surveyors have to calculate the quantity of steel from the given drawing below.
Now, let us move ahead and see “how to calculate steel quantity from drawing” for beam step by step.
### Step-2 Calculate Quantity for Bent Up Steel Bars
After Calculating length of bent up bars, calculate the Total Weight.
Now, We can Calculate the weight of steel bars by the following formula. Remember the length of steel bars should be in feet and the diameter should be in mm to use this formula.
Here, we calculate the total length of stirrups, and then we can calculate the weight by using the following formula given below.
Total Length of Stirrups = (Length of one stirrups * No. of stirrups)
Hence the total quantity of steel bars required for different grade of steel bars is given below.
For #6 grade we need 48 Kg Steel.
For #3 grade we need 11 Kg Steel.
## Video
Hence, in this way we can calculate the quantity of steel bars in a beam Step by Step.
I hope this article on “How to calculate steel quantity from drawing for Beam?” remains helpful for you.
Happy Learning – Civil Concept
Contributed by,
Civil Engineer – Ranjeet Sahani
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# Homework Help: Frictionless physics problem
1. Feb 13, 2008
### psychfan29
1. A 1.5 kg rock is being twirled in a circle on a frictionless surface using a horizontal rope. The radius of the circle is 2.00m and the rope makes 100 revolutions in 1.00 minutes. What is the tension in the rope?
This problem has totally confused me. I don't even know what equation to use that will relate tension to the other givens in the problem!
2. What is the apparent weight of a 75.0 kg person traveling at 100 km/hour:
a) over the peak of a hill with a radius of curvature = to 500m
and
b) at the bottom of a hollow of the same radius
I'm totally lost!!! HELP!!!!!!!!!!!!!
2. Feb 13, 2008
### greeniguana00
1) Here are the equations you need:
Speed = Distance / Time
Centripetal Force = Mass * (Speed^2) / Radius
So, if you know the Centripetal Force, what do you think the tension on the rope would be?
2) You should be able to use the equation for centripetal force I gave you here to find the force needed to keep the person on his path. It will be the same magnitude for both at the bottom of the hill and the top, just opposite in direction. When at the top of the hill, some of the force of gravity will be that force keeping the person on the path, and the remaining part will be the apparent weight. At the bottom of the hill, the force of gravity is pushing the person off the path, meaning the normal force pushing the person back on the path (the person's apparent weight) is greater.
Force of Gravity = Mass * 9.8m/s^2 down
3. Feb 13, 2008
### psychfan29
I still don't know what equation to use to relate tension to the other values.....
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Author: Oscar Cronquist Article last updated on December 28, 2018
The image above shows an array formula in cell B3 that calculates numbers based on the numerical ranges in cell range E3:F6.
Array formula in B3:
=SMALL(IF(COUNTIFS(\$E\$3:\$E\$6, "<="&ROW(\$1:\$21),\$F\$3:\$F\$6, ">="&ROW(\$1:\$21)), ROW(\$1:\$21)), ROW(A1))
To enter an array formula, type the formula in a cell then press and hold CTRL + SHIFT simultaneously, now press Enter once. Release all keys.
The formula bar now shows the formula with a beginning and ending curly bracket telling you that you entered the formula successfully. Don't enter the curly brackets yourself.
### Explaining formula in cell B3
#### Step 1 - Create a sequence
The ROW function returns a row number based on a cell reference, if the cell reference has multiple rows then the row function returns an array of numbers.
ROW(\$1:\$21)
returns
{1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15; 16; 17; 18; 19; 20; 21}
#### Step 2 - Check if number is in sequence
The COUNTIFS function checks if a number is larger or equal to the start value and smaller or equal to the end value. If both conditions are met the COUNTIFS function returns 1.
COUNTIFS(\$E\$3:\$E\$6, "<="&ROW(\$1:\$21),\$F\$3:\$F\$6, ">="&ROW(\$1:\$21))
becomes
COUNTIFS(\$E\$3:\$E\$6, "<="&{1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15; 16; 17; 18; 19; 20; 21},\$F\$3:\$F\$6, ">="&{1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15; 16; 17; 18; 19; 20; 21})
becomes
COUNTIFS({1;5;11;19}, "<="&{1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15; 16; 17; 18; 19; 20; 21},{3;6;13;21}, ">="&{1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15; 16; 17; 18; 19; 20; 21})
and returns
{1; 1; 1; 0; 1; 1; 0; 0; 0; 0; 1; 1; 1; 0; 0; 0; 0; 0; 1; 1; 1}
Tip! Use an Excel defined Table to create dynamic cell references that you don't have to adjust if more ranges are added or deleted.
#### Step 3 - Return number if in range
The IF function has three arguments, the first one must be a logical expression. If the expression evaluates to TRUE then one thing happens (argument 2) and if FALSE another thing happens (argument 3).
IF(COUNTIFS(\$E\$3:\$E\$6, "<="&ROW(\$1:\$21),\$F\$3:\$F\$6, ">="&ROW(\$1:\$21)), ROW(\$1:\$21))
becomes
IF({1; 1; 1; 0; 1; 1; 0; 0; 0; 0; 1; 1; 1; 0; 0; 0; 0; 0; 1; 1; 1}, ROW(\$1:\$21))
becomes
IF({1; 1; 1; 0; 1; 1; 0; 0; 0; 0; 1; 1; 1; 0; 0; 0; 0; 0; 1; 1; 1}, {1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15; 16; 17; 18; 19; 20; 21})
and returns
{1; 2; 3; FALSE; 5; 6; FALSE; FALSE; FALSE; FALSE; 11; 12; 13; FALSE; FALSE; FALSE; FALSE; FALSE; 19; 20; 21}.
#### Step 4 - Extract k-th smallest number
The ROWS function keeps track of the numbers based on an expanding cell reference. It will expand as the formula is copied to the cells below.
SMALL(IF(COUNTIFS(\$E\$3:\$E\$6, "<="&ROW(\$1:\$21),\$F\$3:\$F\$6, ">="&ROW(\$1:\$21)), ROW(\$1:\$21)), ROWS(\$A\$1:A1))
becomes
SMALL({1; 2; 3; FALSE; 5; 6; FALSE; FALSE; FALSE; FALSE; 11; 12; 13; FALSE; FALSE; FALSE; FALSE; FALSE; 19; 20; 21}, ROWS(\$A\$1:A1))
becomes
SMALL({1; 2; 3; FALSE; 5; 6; FALSE; FALSE; FALSE; FALSE; 11; 12; 13; FALSE; FALSE; FALSE; FALSE; FALSE; 19; 20; 21}, 1)
and returns 1 in cell B3.
The array formula in cell B3 lists numbers not in ranges specified in cell range E3:F6.
Array formula in B3:
=SMALL(IF(FREQUENCY(IF((COLUMN(\$B2:\$U2)>\$F\$3:\$F\$6)+ (COLUMN(\$B2:\$U2)<\$E\$3:\$E\$6), COLUMN(\$B2:\$U2), ""), COLUMN(\$B2:\$U2))=ROWS(\$E\$3:\$E\$6), ROW(\$2:\$22), ""),ROW(A1))
### Get the Excel file
numbers-inside-range.xlsx
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Frequent Visitor
## Total Average Measure based on parameters from another measure
The Measure is for each individual month is working correctly, but the total is calculating the average for every month.
I need to calculate the total average based on a criteria set by another measure. Only average the months that have a value for the referenced measure. I have used filters to exlude those values, I have use summarize and I still cannot get the correct total average.
WOEX =
//REFERENCE VALUE
var ttL = CALCULATE(SUMX(oge,OGE[Value]), FILTER(OGE, OGE[Cat] ="xloe" && OGE[Category] = "8/8THS TTL LEASE OPERATING EXP"))
// CALCULATE ONLY MONTHS THAT HAVE A VALUE FOR CORREPONDING REFERENCE VALUE
var surf = CALCULATE(AVERAGE(OGE[Value]), FILTER(oge, OGE[Cat] = "woe" && OGE[Category] = "SURFACE REPAIRS AND MAIN"&& ttl <> 0))
Return
CALCULATE(AVERAGEX(SUMMARIZE(OGE, Account_id[account_id]),surf), FILTER(OGE,ttl<>0))
Result Table:
268.77 is the average for all months. The correct answer I'm looking for is 358.36.
1 ACCEPTED SOLUTION
Frequent Visitor
I solved it.
For what ever reason the two measure where not seeing the realtionship to one another, so instead I created a calculated column for TTL and filtered the new measure to where TTL <> 0 and it corrected it.
2 REPLIES 2
Frequent Visitor
I solved it.
For what ever reason the two measure where not seeing the realtionship to one another, so instead I created a calculated column for TTL and filtered the new measure to where TTL <> 0 and it corrected it.
Super User II
WOEX =AVERAGEX(VALUES([Month]),
//REFERENCE VALUE
var ttL = CALCULATE(SUMX(oge,OGE[Value]), FILTER(OGE, OGE[Cat] ="xloe" && OGE[Category] = "8/8THS TTL LEASE OPERATING EXP"))
// CALCULATE ONLY MONTHS THAT HAVE A VALUE FOR CORREPONDING REFERENCE VALUE
var surf = CALCULATE(AVERAGE(OGE[Value]), FILTER(oge, OGE[Cat] = "woe" && OGE[Category] = "SURFACE REPAIRS AND MAIN"&& ttl <> 0))
Return
CALCULATE(AVERAGEX(SUMMARIZE(OGE, Account_id[account_id]),surf), FILTER(OGE,ttl<>0)))
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# Why correlation length diverges at critical point?
It is not the correlation length of the system that you should look at, but the correlation of the fluctuations. If T>>Tc the spins are randomly oriented and the lenghtscale of fluctuations is very small. As you get closer to Tc, the fluctuations become more correlated, and lenghtscale increases toward infinity. Similarly for the ferromagnet at temperatures much less than Tc, all spins are aligned. The fluctuations at 0 < T << Tc have short correlation lengths. As you heat the system, it is still mostly ordered, but the number of spins pointing in the opposite direction increases, and so does the correlation length of these fluctuations
I think your trouble is that a correlation length $\xi$ is not to be interpreted as correlation in the sense of statistics, e.g.
$\frac{<(s(x)-\lt s(x) \gt)(s(y)-<s(y)>)>}{\sqrt{<\left( s(x)-\lt s(x)>\right)^2 \lt(s(y)-\lt s(y)>)^2 \gt)}}$,
but rather defined via $<(s(x)-< s(x)>)(s(y)-<s(y)>)>=e^{-|x-y|/\xi}$
( see for example (https://physics.stackexchange.com/q/59690).
Assume that, at zero temperature, all spins are "frozen" and perfectly aligned and hence perfectly correlated (in the statistical sense). However, since $s(x)=<s(x)>$ and $s(y)=<s(y)>$ in this case, it follows that $\xi=0$ .
As far as the second part of the question is concerned: Critical points are phase transitions that correspond to fixed points in the renormalization group flow. What this means is that the process of consecutively dividing the spin lattice into blocks, integrating them out and constructing a new Hamiltonian between those blocks has reached a fixed point: The form of the Hamiltonian does not change any longer, only its parameters (couplings) get re-adjusted with any further block-spin operation. This in turn means that the system has lost its scale and has become scale free. So if I were to take two pictures of the material, one of size one inch and the other one of size one micro-inch, you could not tell me which one is which. The only way to describe this mathematically is by assuming a power-law which yields $\xi(T) \sim (T-T_c)^{-\nu}$ where $T_c$ is the critical temperature and $\nu$ is the scaling dimension that is not necessarily integer.Hence the correlation length diverges at the critical point.
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# Write Which of the Following Statement Are True? Justify Your Answer. the Sets P = {A} and B = {{A}} Are Equal. - Mathematics
The sets P = {a} and B = {{a}} are equal.
#### Solution
False
= {a}
= {{a}} = {P}
P$\neq${P
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 11 Mathematics Textbook
Chapter 1 Sets
Exercise 1.4 | Q 4.5 | Page 16
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### Home > PC3 > Chapter 13 > Lesson 13.3.2 > Problem13-114
13-114.
Use the definition of instantaneous rate of change to show that the equation of the tangent line to the curve $y=12\sqrt{x}$ at the point $\left(4, 24\right)$ is $y = 3x + 12$.
You are given a point, so now you need the slope of the tangent line.
Use the point - slope form for a linear function.
$m= \lim \limits_{ h \to 0}\frac{12\sqrt{4+h}-12\sqrt{4}}{h}$
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# solution
1. (c) The above series is actually the prime no. ones. Here each prime no. starting from 11 would get multiplied by 2. 11 × 2 = 22 13 × 2 = 26 17 × 2 = 34 ………… 41 × 2 = 82 43 × 2 = 86 Hence, option (c) is correct.
2. (d) The above series follows a multiplication pattern 1 × 2 = 2 3 × 4 = 12 5 × 6 = 30 …………… 11 × 12 = 132 13 × 14 = 182 Hence, option (d) is correct
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3.b
4. (b) The above series follows the pattern given below :- 1 × 1 = 1 2 × 2 = 4 4 × 4 = 16 8 × 8 = 64 …………… 64 × 64 = 4096 Hence, option (b) is correct.
5. (e) In the above series, the numbers follows this pattern:-
9 + 5 = 14; 14 – 2 =12
12 + 5 = 17; 17 – 2 = 15
15 + 5 = 20; 20 –2 = 18
18 + 5 = 23; 23 – 2 = 21
Hence, Option (e) is the answer. Here actual answer is 23, 21
6. a
7.d
8.(c) In the above series, the next term is derived from previous one in certain pattern: –
=> 4 = 1 × 1 + 3
14 = 4 × 2 + 6
51 = 14 × 3 + 9 and so on.
Here, consider term 218
=> Now according to above pattern,
=> 51 × 4 + 12Þ
=>216 should come, but 218 is being wrongly inserted
Hence, option (c) is correct
9.e
10. (c); In the above mentioned series, the next term is derived from previous one by dividing a no. following a certain pattern.
=>630 3.5 = 180
=>180 3 = 60
=>60 2.5 = 24
Here instead of 26, 24 should come in above series.
Hence, option (c) is correct.
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CGS3269.HW1.Fall09
# CGS3269.HW1.Fall09 - representation(15 points a(18 10(13 10...
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UCF School of Electrical Engineering & Computer Science CGS 3269 Computer Architecture Fall 2009 DUE 09/23/09 1.- Using 4 bit numbers, for example (5) 10 = ( 0101) 2 Write all positive numbers and all negative numbers that can be represented with four bits in sign-magnitude, one’s complement, and two’s complement. (20 points) 2.- Using binary numbers of 8 bits(i.e. (28) 10 = ( 0001 1100) 2 ). Write the numbers from (1) 10 to (20) 10 in binary, octal, and hexadecimal.(15 points) 3.- Convert: (10 points) a) from (18) 10 to (?) 2 b) from (10011100011) 2 to (?) 16 c) from (10011100011) 2 to (?) 8 d) from (10011100011) 2 to (?) 10 4.- Convert to binary and solve the following arithmetic operations using one’s complement representation : (15 points) a) (18) 10 + (13) 10 b) (18) 10 - (13) 10 c) -(18) 10 - (13) 10 5.- Convert to binary and solve the following arithmetic operations using two’s complement
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Unformatted text preview: representation : (15 points) a) (18) 10 + (13) 10 b) (18) 10- (13) 10 c)-(18) 10- (13) 10 6.- Using the MARIE computer assembly language, write a program that computes the following expression: z a * b * c. The computer will read in the input values a, b, and c from the keyboard, and the final result (z) have to be displayed. In addition, every time an input value is read in, it must be displayed on the screen. Remember that the instruction set does not have an instruction to execute multiplication. The program must be tested in the MARIE simulator . (25 points) Submit through Webcourses. 1.- A word document with the answers to question 1 to 5. 2.- A separate file with the MARIE assembly program....
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# MATLAB Basemap Data (R2017b)
Jiro‘s pick this week is MATLAB Basemap Data by MathWorks Mapping Team.
In the new R2017b release, there is a new interactive geographic function called geobubble which displays bubbles at geographic locations, with the size and color representing other parameters.
tsunamis = readtable('tsunamis.xlsx');
tsunamis.Cause = categorical(tsunamis.Cause);
gb = geobubble(tsunamis,'Latitude','Longitude', ...
'SizeVariable','MaxHeight','ColorVariable','Cause')
gb =
GeographicBubbleChart with properties:
Basemap: 'darkwater'
MapLayout: 'normal'
SourceTable: [162×20 table]
LatitudeVariable: 'Latitude'
LongitudeVariable: 'Longitude'
SizeVariable: 'MaxHeight'
ColorVariable: 'Cause'
Use GET to show all properties
We can focus on a particular area, for example North America, with geolimits.
geolimits([10 65],[-180 -60])
The background image you see is the basemap. By default, the basemap you’re seeing here is called “darkwater” and is included with the product. You can change the Basemap property to five other maps: grayland, bluegreen, colorterrain, grayterrain, landcover. These maps are automatically accessed over the Internet. For those with no or low Internet connection, they can download an offline version of the maps.
grayland
gb.Basemap = 'grayland';
pause(2)
bluegreen
gb.Basemap = 'bluegreen';
pause(2)
colorterrain
gb.Basemap = 'colorterrain';
pause(2)
grayterrain
gb.Basemap = 'grayterrain';
pause(2)
landcover
gb.Basemap = 'landcover';
pause(2)
EDIT (to illustrate the exploratory nature of geobubble, as mentioned by Rob):
One of the use cases for geobubble is interactive exploration. For example, I can zoom in with my mouse to see a particular location in greater detail.
R2017b
There are plenty of new, great features in the new R2017b. Take a look at what’s new.
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Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :
# Class 8 Maths Chapter 8 Comparing Quantities MCQs
Class 8 Maths Chapter 8 Comparing Quantities MCQs (Questions and Answers) are provided online for free. These multiple-choice questions are prepared by our subject experts according to the CBSE syllabus (2022-2023) and NCERT guidelines. The objective questions are also available chapter-wise at BYJU’S to make students learn each concept thoroughly and help them to score better marks in exams. Also, learn important questions for class 8 Maths here.
Practice more and test your skills on Class 8 Maths Chapter 8 Comparing Quantities MCQs with the given PDF here.
## MCQs Questions on Class 8 Chapter 8 Comparing Quantities
Multiple choice questions (MCQs) are available for Class 8 Chapter 8 Comparing Quantities chapter. All the problems have four multiple options, in which one is the right answer. Students have to solve each question and choose the correct answer.
1. The ratio of speed of cycle 12 km per hour to the speed of scooter 36 km per hour is
A. 1:2
B. 1:3
C. 1:4
D. None of the above
Explanation: Speed of cycle/Speed of scooter = 12/36 = ⅓
2. The ratio of 10m to 10 km is:
A. 1/10
B. 1/100
C. 1/1000
D. 1000
Explanation: 10m/10km = 10m/10000m = 1/1000
3. The percentage of 1:4 is:
A. 75%
B. 50%
C. 25%
D. 100%
Explanation: 1:4 = 1/4
(1×100)/(4×100) = 1/4 x 100% = 0.25 x 100% = 25%
4. The percentage of 2:5
A. 20%
B. 50%
C. 60%
D. 40%
Explanation: 2:5 = ⅖ = ⅖ x 100% = 0.4 x 100% = 40%
5. If 50% of students are good at science out of 20 students. Then the number of students good at science is:
A. 10
B. 15
C. 5
D. 11
Explanation: 50% of students out of 20 students = 50% of 20
= (50/100) x 20
= ½ x 20
= 10
6. The price of a motorcycle was Rs. 34,000 last year. It has increased by 20% this year. The price of motorcycle now is:
A. Rs. 36,000
B. Rs. 38,800
C. Rs. 40,800
D. Rs. 32,000
Explanation: 20% of Rs.34,000 = 20/100 x 34,000 = Rs.6800
New price = Rs. 34,000+Rs.6800
= Rs. 40,800
7. An item marked at Rs. 840 is sold for Rs. 714. The discount % is:
A. 10%
B. 15%
C. 20%
D. 25%
Explanation: Discount = Marked Price – Sale Price
= 840 – 714
= Rs. 126
Discount % = (126/840) x 100% = 15%
8. A person got a 10% increase in his salary. If his salary was Rs. 50000, then the new salary is:
A. Rs. 55000
B. Rs. 60000
C. Rs. 45000
D. Rs. 65000
Explanation: Previous salary = Rs. 50000
10% of Rs.50000 = (10/100) x 50000 = Rs. 5000
New salary = Rs. 50000 + Rs. 5000
= Rs. 55000/-
9. The cost of the article was Rs. 15500 and Rs. 500 was spent on its repairing. If it is sold for a profit of 15%. The selling price of the article is:
A. Rs.16400
B. Rs.17400
C. Rs.18400
D. Rs.19400
Explanation: Total cost = 15500+500 = 16000
Profit % = (Profit/Cost price) x 100
Profit = (Profit% x Cost price)/100
P = (15×16000)/100 = 2400
Selling Price = Profit + cost price = 2400 + 16000 = Rs.18400
10. Waheeda bought an air cooler for Rs. 3300 including a tax of 10%. The price of the air cooler before VAT was added is:
A. Rs. 2000
B. Rs. 3000
C. Rs. 2500
D. Rs. 2800
Explanation:10% VAT on Rs.100 will make it Rs.110
So, for price including VAT Rs.110, the original price is Rs.100
Then, Price including VAT Rs. 3300, the original price = Rs. (100/110) x 3300 = Rs. 3000.
11. The ratio of 50 paise to Rs. 1 is:
A. 1 : 2
B. 2 : 1
C. 1 : 1
D. 1 : 5
Rs. 1 = 100 paise
50 paise : 100 paise
50 : 100
1 : 2
12. The Ratio of 10m to 1 Km is Equal To:
A. 1/10
B. 1/100
C. 1/1000
D. 1000
Explanation: 10m/1km = 10m/1000m = 1/100
13. The ratio of 1m and 100 cm is:
A. 1:100
B. 1:10
C. 10:1
D. 1:1
Explanation: One meter is equal to hundred centimeters.
1m = 100 cm
Thus, their ratio will be:
⇒ 1m:100cm
⇒ 100cm:100cm
⇒ 1:1
14. The percentage of ratio 2:3 is:
A. 44.44%
B. 55.55%
C. 66.66%
D. None of the above
Explanation: 2:3 = ⅔
Converting into percentage, we have to multiply the ration by 100.
⅔ x 100% = 200/3 = 66.66..%
15. The ratio 4 : 25 converted to percentage is:
A. 8%
B. 4%
C. 16%
D. 25%
4 : 25 =4/25 x 100% = 16%
16. Convert the fraction ⅛ into a percentage.
A. 12.5 %
B. 25%
C. 8%
D. 16%
Explanation: Multiply the given fraction by 100 to get the percentage value.
⅛ x 100 = 12.5%
17. The ratio of speed of a motorbike 50 km per hour and speed of cycle 20 km per hour is:
A. 2:5
B. 5:1
C. 5:2
D. 1:2
Explanation: Speed of motorbike is 50km/hr
Speed of cycle is 20km/hr
Ratio of their speeds are = 50/20 = 5/2 = 5:2
18. If 40% of students passed in the Mathematics test, out of 40 students in the class. What is the number of students who have passed?
A. 10
B. 16
C. 5
D. 11
Explanation: 40% of students out of 40 students will be:
= (40/100) x 40
= 2/5 x 40
= 16
19. 25% of 60 students of a class likes to play football. How many students does not like to play football?
A. 45
B. 15
C. 25
D. 50
Explanation: Number of students = 60
25% of 60 students = 25/100 x 60 = 15
Number of students who does not play football = 60 – 15 = 45
20. Out of 40 students, 25% passed in a class. How many students did actually pass?
A. 10
B. 20
C. 30
D. 40
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# Surjection from a Noetherian ring induces open map on spectra?
Let $A$ be a Noetherian ring, $f: A\rightarrow B$ a surjective ring map, then should the induced map on spectra $f^*: Spec(B)\rightarrow Spec(A)$ be an open map? In Atiyah and Macdonald, Chapter 1, Exercise 21, $f^*$ is already a closed map and is a homeomorphism from $Spec(B)$ onto the closed subset $V(\ker f)$ of $Spec(A)$.
Since $f: A\rightarrow B$ trivially satisfies the "going-down property" defined in Chapter 5, Exercise 10 of Atiyah and Macdonald, and if the conclusion of Chapter 7, Exercise 24 is true, then $f^*$ must be an open map, and by Chapter 1, Exercise 21, we only need to show the image of $f^*$, i.e. $V(\ker f)$, is open in $Spec(A)$. Is it true?
So the problem is reduced to
If $A$ is a Noetherian ring, $f: A\rightarrow B$ a surjective ring map, then should the closed subset $V(\ker f)$ of $Spec(A)$ is also open in $Spec(A)$?
Since any closed subset of $Spec(A)$ is of the form $V(\ker f)$ for some surjective ring map $f: A\rightarrow B$, we have also reduced the problem to
In the space $Spec(A)$ (having the Zariski topology), where $A$ is a Noetherian ring, does the collection of open sets and of closed sets coincide?
• Dear Lao-tzu, As Martin points out in his answer below, you mean surjection, not epimorphism. E.g. $\mathbb Z \to \mathbb Q$ is an epimorphism in the cat. of rings, and the induces map on Specs is neither open nor closed. Regards, Feb 13, 2014 at 3:06
You ask if every closed subset of a noetherian affine scheme is open, which is - of course - absolutely wrong. Always look at examples first! What about $\mathbb{A}^1$?
• If the answer of my question is negative, then this mean the conclusion of Chapter 7, Exercise 24 is wrong, even in the particular case when the map is surjection, or equivalently, the $A$-algebra $B$ is generated by a single element $1\in B$. Feb 13, 2014 at 3:16
For your highlighted questions: a topological space has a nonempty proper subset which is both open and closed iff the space is disconnected. Thus in a (Noetherian) ring with no idempotents $\ne 0, 1$ (e.g. a local ring or a domain), no proper nonempty closed set is open.
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1
AIEEE 2006
+4
-1
Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature $${T_0},$$ while Box contains one mole of helium at temperature $$\left( {{7 \over 3}} \right){T_0}.$$ The boxes are then put into thermal contact with each other, and heat flows between them until the gases reach a common final temperature (ignore the heat capacity of boxes). Then, the final temperature of the gases, $${T_f}$$ in terms of $${T_0}$$ is
A
$${T_f} = {3 \over 7}{T_0}$$
B
$${T_f} = {7 \over 3}{T_0}$$
C
$${T_f} = {3 \over 2}{T_0}$$
D
$${T_f} = {5 \over 2}{T_0}$$
2
AIEEE 2005
+4
-1
A gaseous mixture consists of $$16$$ $$g$$ of helium and $$16$$ $$g$$ of oxygen. The ratio $${{Cp} \over {{C_v}}}$$ of the mixture is
A
$$1.62$$
B
$$1.59$$
C
$$1.54$$
D
$$1.4$$
3
AIEEE 2005
+4
-1
A system goes from $$A$$ to $$B$$ via two processes $$I$$ and $$II$$ as shown in figure. If $$\Delta {U_1}$$ and $$\Delta {U_2}$$ are the changes in internal energies in the processes $$I$$ and $$II$$ respectively, then
A
relation between $$\Delta {U_1}$$ and $$\Delta {U_2}$$ can not be determined
B
$$\Delta {U_1} = \Delta {U_2}$$
C
$$\Delta {U_2} < \Delta {U_1}$$
D
$$\Delta {U_2} > \Delta {U_1}$$
4
AIEEE 2005
+4
-1
The figure shows a system of two concentric spheres of radii $${r_1}$$ and $${r_2}$$ are kept at temperatures $${T_1}$$ and $${T_2}$$, respectively. The radial rate of flow of heat in a substance between the two concentric spheres is proportional to
A
$$In\left( {{{{r_2}} \over {{r_1}}}} \right)$$
B
$${{\left( {{r_2} - {r_1}} \right)} \over {\left( {{r_1}{r_2}} \right)}}$$
C
$${\left( {{r_2} - {r_1}} \right)}$$
D
$${{{r_1}{r_2}} \over {\left( {{r_2} - {r_1}} \right)}}$$
EXAM MAP
Medical
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# Opposite Gameelementary P.e. Games
## 1) Stuff the Turkey
Skill Concepts: Shooting, Chasing/Fleeing, Teamwork, Cardiovascular Endurance
Critical thinking covers many aspects of thought including planning, reasoning, logic and reflection. Physical education class, or P.E., can incorporate all the aspects of critical thinking in many. Adjectives online game, with question and answer activities, grammar exercises, and vocabulary building sections to learn or practice using adjectives modifying nouns. These activities focus on adjectives modifying nouns through 4 different areas. There are vocabulary and spelling exercises students can use to learn or practice the opposite pairs.
Equipment: 2 MultiGoals; 10 balls; 6 hula hoops; 20 polyspots; 4 cones; pinnies
Games-for-All-Reasons suggests allowing a player who misses a shot to take a second crack at it, but if the player misses his second shot, he must go back to the beginning. Line Basketball. Teaching Ideas suggests playing the basketball-variation Line Basketball. This game mainly utilizes passing abilities and has a slight focus on shooting skills. The PE Game Ideas section provides you with Physical Education resources which will help you to plan PE Warm Up Games, PE Tag Games, PE Thinking Games, and PE Coordination Games. Within each section you will find a whole range of different games which will excite and challenge your students.
Set Up:
• Place the MultiGoals on opposite ends of the activity area. If using a basketball court, place the goals near the end lines.
• If there is not a line dividing the playing area in half, you will need to make a center line using additional polyspots or cones.
• Put 10 polyspots in a circle around each MultiGoal, spacing them approximately 3 yards (steps) away from the goal, to mark the goal area. This will be referred to as the “No Guarding Zone.”
• Make a square using 4 cones in opposite corners of the gym spacing them about 3 yards (steps) apart. These squares will be “Jail.”
• Scatter 2 hula hoops on each half. The hoops are called “Safety Circles.”
• Take the remaining two hula hoops and put one on each side of the gym in the opposite corners with respect to where the jails are. Place 5 balls in each of these 2 hoops.
• Divide the group into two teams and use pinnies to distinguish the teams.
How to Play:
• The game is called “Stuff the Turkey,” and is the opposite of Capture the Flag.
• The object of the game is to successfully get the 5 balls (“stuffing”) from the hula hoop on your side of the gym, to your goal on the opposite side of the gym. The ball must make it into the center of the MultiGoal (the “turkey”).
• Explain the rules of the game:
• Once you cross the center line, anyone on the other team can tag you.
• If you get tagged, you must go to jail.
• If you are holding a ball, you are NOT allowed to be a tagger. (Note: A player is not allowed to drop the ball, tag someone, and then pick it back up. Rather, that player must return the ball to their hula hoop or get it to their goal before they can help tag.)
• If you are carrying a ball when you get tagged, you must return it to your hula hoop before going to jail.
• If you throw or shoot a ball attempting to make a shot and you miss, then you are in jail (return the ball to your side).
• The circles around the goals are No Guarding Zones. The defense is NOT allowed inside this area. The offense can enter this area to make a goal, but the offense is not safe inside this area… if the defense tags you, go to jail.
• The 2 hula hoops on each side of the gym are Safety Circles. The defense cannot tag a player standing inside a hula hoop (limit 1 player per hoop).
• In order to be freed from jail, someone from that team must touch one of the 4 cones without being tagged. Once freed from jail, every person (including the player that freed jail) gets, and must take, a free walk back (all the way back to their side). They identify themselves by walking with their hands on their heads.
• Once a team successfully deposits all 5 of their balls into their goal, the game ends. Reset the game by having players return to their sides, put the balls back in the starting hoops, and you are ready to play again!
Check This Out
• If players become really good at this game and start to play too much defense, try limiting the number of taggers per side. Give foam noodles that are cut to an 18 inch length to 3-5 players on each side. These are the only players that can tag an opponent that crosses onto their side. After each round, switch taggers.
• Try this variation! Allow a player to throw a ball from their side of the field to someone in their jail in order to set them free. If the ball is caught everyone gets a free walk back, but if the ball is missed the player who threw it must return the ball to the hula hoop and then join the player(s) in jail.
## Opposite Game Elementary P.e. Games To Play
Skill Concepts: Throwing, Teamwork, Muscular Endurance
Equipment: Omnikin Balls; hula hoops; 20 cones; foam balls (at least 1 per player)
Set Up:
• Place a hula hoop in center of the playing area with the Omnikin ball inside (the Omnikin ball represents the Mayflower).
• Place ten polyspots/cones on both sides of the center line going parallel about 5 yards away. The area in the middle is the ocean.
• Divide group into two teams and have each team line up on opposite ends of the playing area.
• Select one player from each team to be the captain. Remind the group that only the bravest of players should volunteer for this job because the ocean can be rough at times (i.e. the balls will be flying around).
• Evenly divide the balls between the two teams.
How to Play:
• On your signal to start, players try to throw the foam balls at the Omnikin Ball.
• All players must remain behind the line of spots/cones. Only the 2 players selected should be in the ocean.
• The job of the captain is to collect the balls in the ocean and roll them back to their teammates. They are not allowed to touch the Omnikin Balls for any reason and they are not allowed to interfere with the balls that are being thrown/shot.
• A team wins when the Omnikin ball (“Mayflower”) touches land (the line of spots/cones).
Check This Out
• Add more than one Omnikin ball. These can be sisters of the Mayflower. This creates multiple targets for teams to throw toward.
## 3) Turkey Tag
Equipment: 3 pool noodles (cut to about 12 inches), small cones
Set Up:
• Create a playing area large enough for all players.
• Mark a space using the cones approximately 10×10 paces. This space can be anywhere the teacher finds most convenient. This will serve as the fence for the farm.
• Select 3 players to be “It” (they get the noodles). They are the farmers.
How to Play:
• Explain to students that wild turkeys can fly, but farm turkeys cannot. The reason is that wild turkeys get lots of exercise and eat a balance diet, where farm turkeys do not.
• When tagged (by a farmer), students must squat down and waddle like a turkey to the farm (squared off space of cones).
• In order to escape the farm (and get back into the game), the turkeys in the farm must exercise in order to build up enough strength to fly over the fence. Let students pick any exercise and do it 5-10 times (depending on age and fitness level).
## 4) Thanksgiving Stations
Make any stations you want and gear them toward the season. Here are some ideas.
“THANKS”
Focus: Overhand throw
Set up large pieces of paper with 1 letter on each piece to spell “THANKS”. Place polyspots on the floor for throwers to stay behind.
Object is to throw the fleece ball at the “T”. Once students have hit the “T” they move down to the “H” and so on to hit every letter and spell out the word THANKS.
“STUFF THE TURKEY”
Focus: Underhand toss
Cut out a picture of a turkey and tape it to an empty bucket. Use bean bags as the stuffing.
Object is to toss the stuffing (bean bags) into the turkey (bucket) to stuff him for Thanksgiving Day.
“MASH POTATOES”
Focus: jumping, landing
Scatter the stepping stones (or poly spots can serve as a substitute) in a small area of the floor.
The object is to have the students jump from one potato to the next to mash them.
“CAPTURE THE TURKEY”
Focus: Striking
Decorate white balloons with a magic marker with a face and colored tail feathers. Give students a small paddle.
Object is to see how long students can keep the turkey (balloons) flying (in the air) using the paddles.
## 5) Turkey Trot Obstacle Course
Focus: Fitness
Set up a course with various obstacles or tasks along the way.
Examples: hula hoops jump through, mats to climb over, exercises to perform, tunnels to crawl through, targets to throw at, etc. Use whatever equipment you have to make any stations that will be fun for your students!
## What’s up Phys Ed Friends!
Today I wanted to share a fun update to an old game I’ve been playing forever to work on soccer skills. Specifically this game helps students focus on dribbling with feet using control and moving to open spaces. I hope it’s a great addition to the next soccer unit you’re planning in your Physical Education Yearly Plan.
Check out the video below to see me explain it to my kids:
• For those interested links to my Wireless Mic or TV Setup
• Here’s a super poor quality video of the original post on PE Universe (2010)
• This is basically a soccer style version of Fire and Ice
If you don’t have time for the video, check out the summary below.
## Soccer Noodle Tag
Have students come and sit in the middle of the playing area so you can explain the game to them.
In this game there will be people with a noodle trying to tag you from the shoulders down. If tagged, just like Noodle Dance Tag, you will do a fist pump and move side to side (or another type of dance or movement if you want to switch it up). Review on renault can clip china cloneauto diagnostic tool.
There will also be a few people dribbling a ball with their feet (if you have possession of a ball you are invincible and can’t get tagged).
When the players with the ball see that you are frozen, they will call your name and give you a pass using the inside of the foot. IF you can trap the pass, you are unfrozen and can take the ball and try to pass to someone else. The person who passed the ball to you will now run around and try to avoid being tagged.
Equipment Needed: 4-5 noodles and 5-6 gatorskin balls or soccer balls for every 25-30 students
3 Ways to get frozen:
• Falling Down, Getting Tagged, Going outside the boundary lines
• If you have a ball you are invincible and cannot be frozen, if you get tagged while in possession of the ball you should continue playing the game
Setup:
• Give out about 4-5 noodles for every 25-30 kids
• Give out about 5-6 balls for every 25-30 kids (more if they are super low skill level, less to make it harder) I like to use soccer balls for older students and gator skin balls for younger students.
After distributing equipment, tell students to find open space and wait for the music to start. Once the music starts the game begins.
After a few minutes of play stop students and rotate the noodles to students who are raising their hand to signal that they have not been a tagger yet. Play again until all students have had a chance to be a tagger.
Easier Variation 2:
## Opposite Game Elementary P.e. Games Youtube
For K-1st Grades you can make it easier by allowing them to simply tag the frozen person by touching them with the ball (dribbling it into their feet) in order to unfreeze them, this will be easier than trying to trap the ball from a pass. Once the ball touches the frozen player, they are unfrozen and will try to dribble to ball to another student who is frozen
Refinements:
• Make sure you are dribbling with control, using soft touches
• Taggers – make sure you are tagging softly below the shoulders
• Call out a name before making a pass – Communication is important!
Variation:
• Have people spread their legs and make a bridge when frozen
• Instead of trapping it have the player with the ball dribble the ball between the frozen student’s legs to unfreeze them (PE Universe Video Example Here)
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https://www.physicsforums.com/threads/splash-mechanics-on-mars.952576/
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# Splash mechanics on Mars
• Yoni
#### Yoni
They found an underground lake on Mars last week!
Given that the lake exists and is actually liquid (and has a surface), two obvious questions come to mind:
1. Will I be able to beet my earthly current record in bouncing a pebble there?
2. If I release a pebble from a given height here and there, which splash will be higher?
Cheers to space exploration!
They found an underground lake on Mars last week!
Given that the lake exists and is actually liquid (and has a surface), two obvious questions come to mind:
1. Will I be able to beet my earthly current record in bouncing a pebble there?
Do you mean "skipping" a pebble? What record do you mean here -- number of skips or distance travelled?
The gravity on Mars is only 38% of the gravity here on Earth (http://coolcosmos.ipac.caltech.edu/ask/73-How-strong-is-the-gravity-on-Mars-), so a rock would travel farther after each skip. As far as the number of skips, I don't know how the lower gravity would affect this.
As on Earth, you would not want to use a pebble, but would want to choose a rock with a flat surface. As English is probably not your first language, a pebble is a small stone, typically roundish. A good skipping stone would tend to be larger than a pebble.
Yoni said:
2. If I release a pebble from a given height here and there, which splash will be higher?
Seems pretty obvious to me that with lower gravity on Mars, the splash from a pebble would be higher there.
Will I be able to beet my earthly current record in bouncing a pebble there?
I assume you mean "beat" and "skip", not "beet" and "bounce". Given that Mars has about 1/3 the gravity of Earth, I don't see how you could not beat the record but there are some complications. You'd have to be wearing a spacesuit so your throwing ability would be limited. THEN of course, there is the real killer which is that it is about a mile underground which is REALLY going to complicate rock skipping
I did mean "beat" and "skip", and I guess our pebbles here are flatter than yours... :)
In any case, while it is trivial that the distance would be longer, My first guess was that the mean number of skips would not depend on gravity (if all other factors are similar).
Regarding the height of a splash: Let's take it to the extreme. In very low gravity, surface tension completely governs the process. The rock would meet the water with very low velocity and the surface tension would pull it into the water, and I would presume make no splash at all.
My first guess was that the mean number of skips would not depend on gravity (if all other factors are similar).
That's a good point. I had not looked at it that way, I was just thinking of the distance.
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# Answer to Question #17813 in C++ for Asma
Question #17813
1- Write a program that prompts the user to input a positive integer and then saves the number with the digits reversed into anther integer variable called Rev.
2- A number is known as palindrome number if the reverse of the digits is equal to the number itself.
Example:
Enter the Number
12621
The Reversed is: 12621
The number: 12621 is Palindrome
1
2012-11-02T13:05:22-0400
#include <iostream>
#include <string>
using namespace std;
int main(){
cout<<"Enter number\n";
string a, Rev;
cin>>a;
Rev = a;
Rev.reserve();
cout<<"The reversed is: "<< Rev<<endl;
bool ok = true;
for (int i = 0 ; i < Rev.size() / 2 ; i++){
& if (Rev[i] != Rev[Rev.size()-1-i]){
& cout<<"The number "<<a<< " isn't Palindrome\n";
& ok = !ok;
& break;
& }
}
if (ok) cout<<"The number "<<a<< " is Palindrome\n";
return 0;
}
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Submit order
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https://socratic.org/questions/how-do-you-find-the-exact-value-of-arc-tan-sqrt3
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# How do you find the exact value of arc tan(-sqrt3)?
##### 1 Answer
May 14, 2015
$\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$ and $\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$
Since $\sin \left(- \theta\right) = - \sin \theta$ and $\cos \left(- \theta\right) = \cos \theta$ for any $\theta$, we have
$\tan \left(- \frac{\pi}{3}\right) = \sin \frac{- \frac{\pi}{3}}{\cos} \left(- \frac{\pi}{3}\right) = \frac{- \sin \left(\frac{\pi}{3}\right)}{\cos} \left(\frac{\pi}{3}\right)$
$= - \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = - \frac{\sqrt{3}}{1} = - \sqrt{3}$
So $\arctan \left(- \sqrt{3}\right) = - \frac{\pi}{3}$
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Solves a numerical or symbolic system of ordinary differential equations.
ode(
f,
var,
times,
timevar = NULL,
params = list(),
method = "rk4",
drop = FALSE
)
Arguments
f
vector of characters, or a function returning a numeric vector, giving the values of the derivatives in the ODE system at time timevar. See examples.
var
vector giving the initial conditions. See examples.
times
discretization sequence, the first value represents the initial time.
timevar
the time variable used by f, if any.
params
list of additional parameters passed to f.
method
the solver to use. One of "rk4" (Runge-Kutta) or "euler" (Euler).
drop
if TRUE, return only the final solution instead of the whole trajectory.
Value
Vector of final solutions if drop=TRUE, otherwise a matrix with as many rows as elements in times and as many columns as elements in var.
References
Guidotti E (2022). "calculus: High-Dimensional Numerical and Symbolic Calculus in R." Journal of Statistical Software, 104(5), 1-37. doi:10.18637/jss.v104.i05
Other integrals: integral()
Examples
## ==================================================
## Example: symbolic system
## System: dx = x dt
## Initial: x0 = 1
## ==================================================
f <- "x"
var <- c(x=1)
times <- seq(0, 2*pi, by=0.001)
x <- ode(f, var, times)
plot(times, x, type = "l")
## ==================================================
## Example: time dependent system
## System: dx = cos(t) dt
## Initial: x0 = 0
## ==================================================
f <- "cos(t)"
var <- c(x=0)
times <- seq(0, 2*pi, by=0.001)
x <- ode(f, var, times, timevar = "t")
plot(times, x, type = "l")
## ==================================================
## Example: multivariate time dependent system
## System: dx = x dt
## dy = x*(1+cos(10*t)) dt
## Initial: x0 = 1
## y0 = 1
## ==================================================
f <- c("x", "x*(1+cos(10*t))")
var <- c(x=1, y=1)
times <- seq(0, 2*pi, by=0.001)
x <- ode(f, var, times, timevar = "t")
matplot(times, x, type = "l", lty = 1, col = 1:2)
## ==================================================
## Example: numerical system
## System: dx = x dt
## dy = y dt
## Initial: x0 = 1
## y0 = 2
## ==================================================
f <- function(x, y) c(x, y)
var <- c(x=1, y=2)
times <- seq(0, 2*pi, by=0.001)
x <- ode(f, var, times)
matplot(times, x, type = "l", lty = 1, col = 1:2)
## ==================================================
## Example: vectorized interface
## System: dx = x dt
## dy = y dt
## dz = y*(1+cos(10*t)) dt
## Initial: x0 = 1
## y0 = 2
## z0 = 2
## ==================================================
f <- function(x, t) c(x[1], x[2], x[2]*(1+cos(10*t)))
var <- c(1,2,2)
times <- seq(0, 2*pi, by=0.001)
x <- ode(f, var, times, timevar = "t")
matplot(times, x, type = "l", lty = 1, col = 1:3)
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## RolyPoly 2 years ago (again)Solve by Laplace transform: $\frac{d^2y}{dx^2} + a \frac{dy}{dx} - 2a^2y=0$ y(0)=6, y'(0) = 0
1. RolyPoly
$\frac{d^2y}{dx^2} + a \frac{dy}{dx} - 2a^2y=0$$sY'-0 +a(sY-6)-2a^2Y=0$$s(sY-6) +a(sY-6)-2a^2Y=0$$(s^2+as-2a^2)Y=6a+6s$$Y=\frac{6a+6s}{(s+2a)(s-a)}$So far so good? Or...??
2. experimentX
|dw:1355844666804:dw|
3. RolyPoly
y(0)=6, y'(0) = 0
4. experimentX
sorry ... i didn't see, looks like the transform of d^2y/dx^2 is incorrect.
5. experimentX
6. RolyPoly
I guess so.. $L(y'') = s^2Y - s(6) -0=s^2Y-6s$What's wrong?
7. experimentX
$sY'-0 +a(sY-6)-2a^2Y=0$ this should be like $s^2 Y - 6s + a(sY - 0) -2a^2 Y = 0$
8. RolyPoly
$L(\frac{dy}{dx}) = sY - f(0)$Right?
9. experimentX
yep!!
10. experimentX
woops!! sorry I misread ... i put opposite!!
11. experimentX
$s^2 Y - 0 - 6 + a(sY - 6) - 2a^2 Y = 0$
12. RolyPoly
It's okay! But what's wrong with my workings?
13. RolyPoly
$L(y'') = s^2Y - s(6) -0=s^2Y-6s$
14. experimentX
the second step!! lol ... i again put it wrong!!
15. RolyPoly
The second step is right. You just get the third immediately.. $L(y'') = sL(y')-f'(0) = s^2 Y-sf(0) - f'(0)$ Anyway.. I still get the third step..?!
16. experimentX
Y' ... well i thought you differentiated it.
17. experimentX
the answer should be according to mathematica $\left\{\left\{y[x]\to 2 e^{-2 a x} \left(1+2 e^{3 a x}\right)\right\}\right\}$
18. RolyPoly
Sorry :(
19. RolyPoly
Yes... But I can't get it :(
20. experimentX
the final answer is right ... you get $F(s) = \frac{6(a+s)}{s^2 + as - 2a^2}$ looks lkie you factorized it.
21. experimentX
now just take the inverse transform.
22. RolyPoly
Factorize it -> partial fraction -> inverse Right?
23. experimentX
yeah!! that's the way
24. RolyPoly
$F(s) = \frac{6(a+s)}{s^2 + as - 2a^2}=\frac{6(a+s)}{(a+2a)(s-a)}$I suppose..
25. RolyPoly
s+2a typo
26. RolyPoly
$Y = \frac{6a+6s}{(s+2a)(s-a)}=\frac{A}{s+2a}+\frac{B}{s-a}$ A(s-a) + B(s+2a) =6a+6s A+B = 6 2B-A = 6 => B=4 , A=2 $Y =\frac{2}{s+2a}+\frac{4}{s-a}$$y=2e^{-2at}+4e^{at}$Why couldn't I get this answer... :\ Thanks!!
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> > > > Problem 3BSC
# Standard Normal Distribution Identify the requirements
## Problem 3BSC Chapter 6.2
Elementary Statistics | 12th Edition
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Problem 3BSC
Standard Normal Distribution Identify the requirements necessary for a normal distribution to be a standard normal distribution.
Step-by-Step Solution:
Step 1 of 1</p>
In order for a normal probability...
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##### ISBN: 9780321836960
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Users Online Now: 2,542 (Who's On?) Visitors Today: 383,493 Pageviews Today: 1,136,197 Threads Today: 491 Posts Today: 11,293 03:53 PM
Absolute BS Crap Reasonable Nice Amazing
# Swastika in magic squares!
Anonymous Coward
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06/15/2011 09:31 AM
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Swastika in magic squares!
Anonymous Coward (OP)
User ID: 1426730
Finland
06/15/2011 10:00 AM
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Re: Swastika in magic squares!
3 22 21 16 19 15 12 9 10 3
10 4 11 6 17 13 23 20 4 22
9 20 2 14 25 8 18 2 11 21
12 23 18 5 24 7 5 14 6 16
15 13 8 7 1 1 24 25 17 19
19 17 25 24 1 1 7 8 13 15
16 6 14 5 7 24 5 18 23 12
21 11 2 18 8 25 14 2 20 9
22 4 20 23 13 17 6 11 4 10
3 10 9 12 15 19 16 21 22 3
39 54 64 64 39
64 39 64 39 54
64 64 4 64 64
54 39 64 39 64
39 64 64 54 39
Give each individual value an individual color to see it.
It's 5x5 extroverted magic square (I call it quadrasquare) starting from the middle point (1x1x1x1 which gives the total cell value 4)
Bigger ones such as 25x25 quadrasquare gives neat pictures aswell. Somewhat familiar intuition...
Anonymous Coward (OP)
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06/15/2011 10:37 AM
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Re: Swastika in magic squares!
And yes, it's based on lo shu.No coments yet? wtf? :D
Prisoner of Technology
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United States
06/15/2011 10:39 AM
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Re: Swastika in magic squares!
No one has ever seen a perfect circle, nor a perfectly straight line, yet everyone knows what a circle and a straight line are.
Perceived circles or lines are not exactly circular or straight, and true circles and lines could never be detected since by definition they are sets of infinitely small points.
Anonymous Coward
User ID: 1276837
France
06/15/2011 10:41 AM
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Re: Swastika in magic squares!
so what? it's a wheel
Prisoner of Technology
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06/15/2011 10:43 AM
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Re: Swastika in magic squares!
Last Edited by Technological Supremacy on 06/15/2011 11:12 AM
No one has ever seen a perfect circle, nor a perfectly straight line, yet everyone knows what a circle and a straight line are.
Perceived circles or lines are not exactly circular or straight, and true circles and lines could never be detected since by definition they are sets of infinitely small points.
Anonymous Coward (OP)
User ID: 1426730
Finland
06/15/2011 10:45 AM
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Re: Swastika in magic squares!
:swastistar:
Quoting: Prisoner of Technology
Exactly!
The swastistar actually represents an circle :)
Ahh.. 4/sqrt(fibonacci361/fibonacci360) = 4/sqrt(phi) We haz the pi 3.1446055110296931442782343433718357180924882313508929506596
Prisoner of Technology
User ID: 792783
United States
06/15/2011 11:11 AM
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Re: Swastika in magic squares!
Quoting: Prisoner of Technology
Exactly!
The swastistar actually represents an circle :)
Ahh.. 4/sqrt(fibonacci361/fibonacci360) = 4/sqrt(phi) We haz the pi 3.1446055110296931442782343433718357180924882313508929506596
Quoting: Anonymous Coward 1426730
No one has ever seen a perfect circle, nor a perfectly straight line, yet everyone knows what a circle and a straight line are.
Perceived circles or lines are not exactly circular or straight, and true circles and lines could never be detected since by definition they are sets of infinitely small points.
Anonymous Coward (OP)
User ID: 1426730
Finland
06/15/2011 11:37 AM
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Re: Swastika in magic squares!
5 76 75 66 64 56 62 46 51 45 40 35 29 25 27 15 16 5
16 6 26 17 36 19 37 30 47 41 57 52 67 63 77 65 6 76
15 65 4 55 74 54 72 44 61 34 50 24 39 14 28 4 26 75
27 77 28 7 38 18 48 20 58 31 68 42 78 53 7 55 17 66
25 63 14 53 3 43 73 33 71 23 60 13 49 3 38 74 36 64
29 67 39 78 49 8 59 10 69 21 79 32 8 43 18 54 19 56
35 52 24 42 13 32 2 22 81 12 70 2 59 73 48 72 37 62
40 57 50 68 60 79 70 9 80 11 9 22 10 33 20 44 30 46
45 41 34 31 23 21 12 11 1 1 80 81 69 71 58 61 47 51
51 47 61 58 71 69 81 80 1 1 11 12 21 23 31 34 41 45
46 30 44 20 33 10 22 9 11 80 9 70 79 60 68 50 57 40
62 37 72 48 73 59 2 70 12 81 22 2 32 13 42 24 52 35
56 19 54 18 43 8 32 79 21 69 10 59 8 49 78 39 67 29
64 36 74 38 3 49 13 60 23 71 33 73 43 3 53 14 63 25
66 17 55 7 53 78 42 68 31 58 20 48 18 38 7 28 77 27
75 26 4 28 14 39 24 50 34 61 44 72 54 74 55 4 65 15
76 6 65 77 63 67 52 57 41 47 30 37 19 36 17 26 6 16
5 16 15 27 25 29 35 40 45 51 46 62 56 64 66 75 76 5
103 184 175 175 184 184 184 184 103
184 94 184 184 184 184 184 94 184
184 184 103 175 184 184 103 184 175
184 184 184 103 184 103 175 184 175
184 184 184 184 4 184 184 184 184
175 184 175 103 184 103 184 184 184
175 184 103 184 184 175 103 184 184
184 94 184 184 184 184 184 94 184
103 184 184 184 184 175 175 184 103
Anonymous Coward (OP)
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06/15/2011 11:42 AM
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Re: Swastika in magic squares!
Same for 3x3
2 7 9 6 4 2
4 3 8 5 3 7
6 5 1 1 8 9
9 8 1 1 5 6
7 3 5 8 3 4
2 4 6 9 7 2
16 28 16
28 4 28
16 28 16
Anonymous Coward
User ID: 1427599
United States
06/15/2011 12:16 PM
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Re: Swastika in magic squares!
actually, the swastika stands for the sun, when its magnetic field lines become too compressed as the outer layers turn at a different rate than the inner layers, causing them to bunch up like the spring in a wind-up toy, until boom! It quickly unwinds, sending out a massive coronal ejection in all directions. It is warning sign. For looking straight down on the sun while the fields are wound tight, it makes the shape of a swastika... More proof that the ancients had superior or contemporary technology at one time.
aether
User ID: 1412926
United Kingdom
06/15/2011 12:29 PM
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Re: Swastika in magic squares!
actually, the swastika stands for the sun, when its magnetic field lines become too compressed as the outer layers turn at a different rate than the inner layers, causing them to bunch up like the spring in a wind-up toy, until boom! It quickly unwinds, sending out a massive coronal ejection in all directions. It is warning sign. For looking straight down on the sun while the fields are wound tight, it makes the shape of a swastika... More proof that the ancients had superior or contemporary technology at one time.
Quoting: Anonymous Coward 1427599
the cross the Swastika, rotating magnetic fields.
Anonymous Coward
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06/16/2011 06:11 AM
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Re: Swastika in magic squares!
Kimchi
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06/16/2011 02:20 PM
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Re: Swastika in magic squares!
卐
Prisoner of Technology
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United States
06/16/2011 02:35 PM
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Re: Swastika in magic squares!
No one has ever seen a perfect circle, nor a perfectly straight line, yet everyone knows what a circle and a straight line are.
Perceived circles or lines are not exactly circular or straight, and true circles and lines could never be detected since by definition they are sets of infinitely small points.
Anonymous Coward
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Finland
06/19/2011 02:06 PM
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Re: Swastika in magic squares!
Nut to what actual application can the magic square be applied to to improve oneselfs life?
Anonymous Coward
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06/24/2011 07:07 AM
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Anonymous Coward
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06/29/2011 09:36 PM
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Anonymous Coward
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07/08/2011 06:41 PM
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Re: Swastika in magic squares!
Anonymous Coward
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07/08/2011 06:42 PM
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Re: Swastika in magic squares!
Anonymous Coward
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Finland
07/13/2011 08:43 PM
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Re: Swastika in magic squares!
:swastistar:
Quoting: Prisoner of Technology
Exactly!
The swastistar actually represents an circle :)
Ahh.. 4/sqrt(fibonacci361/fibonacci360) = 4/sqrt(phi) We haz the pi 3.1446055110296931442782343433718357180924882313508929506596
Quoting: Anonymous Coward 1426730
Anonymous Coward
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07/14/2011 04:49 PM
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Re: Swastika in magic squares!
(4/sqrt(lim_(n->infinity) F_(n+1)/F_n))
murat
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United States
07/14/2011 04:55 PM
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Re: Swastika in magic squares!
Well played! I enjoy this type of magic.
murat
User ID: 1451949
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07/14/2011 05:05 PM
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Re: Swastika in magic squares!
My spirit guide is very powerful. She has a name and it is only spoken of in the land of dreams. In your video you said her name and her number. I am very impressed. I see good things in your future.
Anonymous Coward
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07/24/2011 08:10 AM
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Re: Swastika in magic squares!
v0rtex666
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07/24/2011 08:18 AM
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Re: Swastika in magic squares!
Cool. Tripping out on magic squares looking for patterns is a productive activity. Squares sometimes are associated with entities or astral realms. What is the source of these squares?
Another day in eternity, basking in the reflected vortex of the chaotic light and the homogeneous metallic quantum foam.
Anonymous Coward
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United States
07/24/2011 10:21 AM
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Re: Swastika in magic squares!
actually, the swastika stands for the sun, when its magnetic field lines become too compressed as the outer layers turn at a different rate than the inner layers, causing them to bunch up like the spring in a wind-up toy, until boom! It quickly unwinds, sending out a massive coronal ejection in all directions. It is warning sign. For looking straight down on the sun while the fields are wound tight, it makes the shape of a swastika... More proof that the ancients had superior or contemporary technology at one time.
Quoting: Anonymous Coward 1427599
Poor souls, the swastika represents harmonious balance in human society by way of sanatana-dharma or the four varnas(as represented by the foundational cross ) and the four asramas(represented by the left facing flags)giving a clock wise motion in balance with nature. That means the natural(God given rules) system of life must balance for the humans to live free and happy(satisfied).
Always remember "It is not nice to fool with mother nature"
Follow the rules and regulation of human life and that is what kind of society you will have. Act like dogs, then it is a dog eat dog world.
Brahmana, ksatriya, vaisya, sudra first, work is foundational. Then the home life of brahmacari, grhasta, vanaprastha, sannyasa.
As is is in heaven, it is on earth.
"According to the three modes of material nature and the work ascribed to them, the four divisions of human society were created by Me. And, although I am the creator of this system, you should know that I am yet the non-doer, being unchangeable."
raja
Anonymous Coward
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07/25/2011 08:35 AM
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Re: Swastika in magic squares!
Cool. Tripping out on magic squares looking for patterns is a productive activity. Squares sometimes are associated with entities or astral realms. What is the source of these squares?
Quoting: v0rtex666
are you sarcastic or not? :D
Anonymous Coward
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07/27/2011 12:44 PM
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Re: Swastika in magic squares!
Anonymous Coward
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07/27/2011 07:43 PM
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# Homework Help: Bijective function
1. Oct 30, 2008
### CarmineCortez
1. The problem statement, all variables and given/known data
Is this function bijective ?
f: [0,1] --> [0,1]
f(x) = x if x E [0,1] intersection Q
f(x) = 1-x if x E [0,1]\Q
2. Relevant equations
3. The attempt at a solution
it is bijective for the rational numbers not sure about the irrationals.
2. Oct 30, 2008
### morphism
Just manually check whether it's 1-1 (consider the different meanings of "f(x)=f(y)") and onto (draw a graph; is there anything in [0,1] that f misses?).
3. Oct 30, 2008
### Dick
Why not? If x is irrational and in [0,1] then 1-x is irrational and in [0,1]. The function is pretty easy to invert.
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C : One-third of previous value in each position of an array
# C Exercises: Array fill, replace each subsequent position of the array by one-third value of the previous
## C Basic Declarations and Expressions: Exercise-129 with Solution
Write a C program that reads an array (length 10), and replaces the first element of the array by a given number and replaces each subsequent position of the array by one-third the value of the previous.
Sample Solution:
C Code:
``````#include <stdio.h>
int main ()
{
int i;
double array_nums[10];
double n;
// Prompt user for input
printf("Input a number:\n");
scanf("%lf", &n);
// Initialize the first element of the array with the input value
array_nums[0] = n;
// Loop to generate and store subsequent array elements
for (i = 1; i < 10; i++)
{
n = n / 3; // Divide the current value by 3
array_nums[i] = n; // Store the result in the array
}
// Print the array elements
printf("\nArray elements:\n");
for (i = 0; i < 10; i++)
{
printf("array_nums[%d] = %.4lf\n", i, array_nums[i]);
}
return 0; // End of program
}
``````
Sample Output:
```Input a number:
35
Array elements:
array_nums[0] = 35.0000
array_nums[1] = 11.6667
array_nums[2] = 3.8889
array_nums[3] = 1.2963
array_nums[4] = 0.4321
array_nums[5] = 0.1440
array_nums[6] = 0.0480
array_nums[7] = 0.0160
array_nums[8] = 0.0053
array_nums[9] = 0.0018
```
Flowchart:
C programming Code Editor:
What is the difficulty level of this exercise?
Test your Programming skills with w3resource's quiz.
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Workshop 8 - Problem 15c Solution
# Workshop 8 - Problem 15c Solution - Current Value Upper...
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Sheet1 Page 1 LINEAR PROGRAMMING PROBLEM MIN 11.2X1+5X2+14X3+9X4 S.T. 1) 1X1+1X3=100 2) 1X2+1X4=150 3) 4X1+3X2<600 4) 6X1+8X2<1080 OPTIMAL SOLUTION Objective Function Value = 2210.000 Variable Value Reduced Costs -------------- --------------- ------------------ X1 0.000 0.200 X2 135.000 0.000 X3 100.000 0.000 X4 15.000 0.000 Constraint Slack/Surplus Dual Prices -------------- --------------- ------------------ 1 0.000 -14.000 2 0.000 -9.000 3 195.000 0.000 4 0.000 0.500 OBJECTIVE COEFFICIENT RANGES Variable Lower Limit
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Unformatted text preview: Current Value Upper Limit------------ --------------- --------------- ---------------X1 11.000 11.200 No Upper Limit X2 No Lower Limit 5.000 5.267 X3 No Lower Limit 14.000 14.200 X4 8.733 9.000 No Upper Limit RIGHT HAND SIDE RANGES Constraint Lower Limit Current Value Upper Limit------------ --------------- --------------- ---------------1 0.000 100.000 No Upper Limit 2 135.000 150.000 No Upper Limit 3 405.000 600.000 No Upper Limit 4 0.000 1080.000 1200.000...
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## This note was uploaded on 09/13/2011 for the course MGMT 304 taught by Professor Kenny during the Spring '11 term at Buena Vista.
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https://curatedcourses.org/math/la/e/linsys/3x3/soln/row_reduce/i
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# Example of solving a 3-by-3 system of linear equations by row-reducing the augmented matrix, in the case of infinitely many solutions
math.la.e.linsys.3x3.soln.row_reduce.i
# Matrix equations with zero or infinitely many solutions
A 3x3 matrix equation Ax=b is solved for two different values of b. In one case there is no solution, and in another there are infinitely many solutions. These examples illustrate a theorem about linear combinations of the columns of the matrix A.
##### math.la.d.mat.eqn
Created On
February 15th, 2017
7 years ago
Views
3
Type
Video
Timeframe
Pre-class
Perspective
Example
Language
English
Content Type
text/html; charset=utf-8
# Introductory worksheet on systems of linear equations and augmented matricesAwaiting Moderation
Focus on computations skills. Convert from linear system to augmented matrix and back. Perform row operations. Solutions included. Uncomment \printanswers line before compiling to show solutions.
##### math.la.e.linsys.3x3.soln.row_reduce.i
CC-BY-NC-SA-4.0
Created On
August 7th, 2018
6 years ago
Views
2
Type
Handout
Timeframe
In-class
Perspective
Example
Language
English
Content Type
text/html; charset=utf-8
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https://www.volcanodiscovery.com/de/erdbeben/erdbeben-info/6972413/mag4quake-Aug-17-2022-Indonesia-BANDA-SEA-antipode3.html
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Kontakt | RSS | EN | DE | EL | ES | FR | IT | RU
# Leichtes Erdbeben der Stärke 4.4 - Banda Sea, 93 km südwestlich von Ambon, Maluku, Indonesien, am Mittwoch, 17. Aug 2022 um 09:44 Lokalzeit - writeAge(1660697087)
Quakes around its antipode during the following 3 Tage
Check different time intervals: 24 hours | 2 days
Datum & Zeit: 17 Aug 2022 00:44:47 UTC -
Ortszeit am Epizentrum: Mittwoch, 17. Aug. 2022 09:44 (GMT +9)
Magnitude: 4.4
Tiefe: 250 km
Epizentrum Breite / Länge: 4.1786°S / 127.5033°E (Banda Sea, Indonesien)
Antipode Breite / Länge: 4.179°N / 52.497°W (Französisch-Guayana)
Primäre Datenquelle: USGS (United States Geological Survey)
Interaktive Karte anzeigen
[kleiner] [größer]
## Quakes within 30 degrees from seismic antipode up to 3 Tage after the quake
The area around the antipode is marked with the big circle on the map.
The total number of quakes that occurred within 3 Tage around the antipode of the quake was 0. This is compared to the average number of 0 quakes during preceding intervals of 3 Tage with a standard deviation (σ) of 0, calculated by analzing all 100 preceding intervals of 3 Tage for the same region.
Interpretation:
Note: At a magnitude of 4.4, it is unlikely that this quake could affect areas around its antipode. If an increase or decrease of quake is detected, it is likely caused by other factors.
There is no significant variation in the number of quakes that occurred around the antipode within 3 Tage.
Mag class Total # Average σ Observed deviation 6+ 0 0 0 n/a 5+ 0 0 0 n/a 4+ 0 0 0 n/a 3+ 0 0 0 n/a Total 0 0 0 n/a
Explanation:
BLACK = Number of quakes within normal statistical range
RED = Significantly lower than normal number of quakes
GREEN = Significantly higher than normal number of quakes
## Unterstützen Sie unsere Arbeit!
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Sources: VolcanoDiscovery / VolcanoAdventures and other sources as noted.
Use of material: Most text and images on our websites are owned by us. Re-use is generally not permitted without authorization. Contact us for licensing rights.
Volcanoes & Earthquakes
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https://jessechaulk.com/posts/125-valid-palindrome/
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### Problem Statement#
A phrase is a palindrome if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.
Given a string `s`, return `true` if it is a palindrome, or `false` otherwise.
#### Example 1:#
``````Input: s = "A man, a plan, a canal: Panama"
Output: true
Explanation: "amanaplanacanalpanama" is a palindrome.
``````
#### Example 2:#
``````Input: s = "race a car"
Output: false
Explanation: "raceacar" is not a palindrome.
``````
#### Example 3:#
``````Input: s = " "
Output: true
Explanation: s is an empty string "" after removing non-alphanumeric characters.
Since an empty string reads the same forward and backward, it is a palindrome.
``````
### Solution#
#### Python#
``````class Solution:
def isPalindrome(self, s: str) -> bool:
"""
T: O(N)
S: O(N)
"""
# Create a new list that only has letters and numbers
values = [ch.lower() for ch in s if s.isalnum()]
# Initialize two pointers
a, b = 0, len(values) - 1
# Walk from the front and back to try to find a
# pair that is not equal. Return False if such a
# is found.
while a < b:
if values[a] != values[b]:
return False
a += 1
b -= 1
# The string is a valid palindrome. Return True
return True
``````
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# Interactive Real Analysis
Next | Previous | Glossary | Map
## 6.5. Differentiable Functions
### Theorem 6.5.15: l Hospital Rules
If f and g are differentiable in a neighborhood of x = c, and f(c) = g(c) = 0, then
provided the limit on the right exists. The same result holds for one-sided limits.
If f and g are differentiable and f(x) = g(x) = - then
provided the last limit exists.
### Proof:
The first part can be proved easily, if the right hand limit equals f'(c) / g'(c): Since f(c) = g(c) = 0 we have
Taking the limit as x approaches c we get the first result. However, the actual result is somewhat more general, and we have to be slightly more careful. We will use a version of the Mean Value theorem:
Take any sequence {xn} converging to c from above. All assumptions of the generalized Mean Value theorem are satisfied (check !) on [c, xn]. Therefore, for each n there exists a number cn in the interval (c, xn) such that
Taking the limit as n approaches infinity will give the desired result for right-handed limits. The proof is similar for left handed limits and therefore for 'full' limits.
The proof of the last part of this theorem is left as an exercise.
Next | Previous | Glossary | Map
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# Steps in Constructing a c-Chart
Attribute control charts arise when items are compared with some standard and then are classified as to whether they meet the standard or not. The control chart is used to determine if the rate of nonconforming product is stable and detect when a deviation from stability has occurred. The argument can be made that a LCL should not exist, since rates of nonconforming product outside the LCL is in fact a good thing; we WANT low rates of nonconforming product. However, if we treat these LCL violations as simply another search for an assignable cause, we may learn for the drop in nonconformities rate and be able to permanently improve the process.
The c Chart measures the number of nonconformities per “unit” and is denoted by c. This “unit” is commonly referred to as an inspection unit and may be “per day” or “per square foot” of some other predetermined sensible rate.
Steps in Constructing a c Chart
1. Determine cbar.
2. There are k inspection units and c(i) is the number of nonconformities in the ith sample.
3. Since the mean and variance of the underlying Poisson distribution are equal,
4. Thus,
and the UCL and LCL are:
5. Plot the centerline cbar, the LCL and UCL, and the process measurements c(i).
6. Interpret the control chart.
Example:
```Farnum Example:
data is from Farnum (1994):
Modern Statistical Quality Control and Improvement, p. 248
Non-conforming
Day Errors/1000 lines
1 6
2 7
3 7
4 6
5 8
6 6
7 5
8 8
9 1
10 6
11 2
12 5
13 5
14 4
15 3
16 3
17 2
18 0
19 0
20 1
21 2
22 5
23 1
24 7
25 7
26 1
27 5
28 5
29 8
30 8
Calculations:
CBAR = 4.4667
UCL = cbar + 3*sqrt(cbar) = 10.80701366
LCL = cbar - 3*sqrt(cbar) = -1.873680327 = 0
(when LCL < 0, set LCL = 0)
Day CL UCL LCL NonConforming
1 4.4667 10.80701366 0 6
2 4.4667 10.80701366 0 7
3 4.4667 10.80701366 0 7
4 4.4667 10.80701366 0 6
5 4.4667 10.80701366 0 8
6 4.4667 10.80701366 0 6
7 4.4667 10.80701366 0 5
8 4.4667 10.80701366 0 8
9 4.4667 10.80701366 0 1
10 4.4667 10.80701366 0 6
11 4.4667 10.80701366 0 2
12 4.4667 10.80701366 0 5
13 4.4667 10.80701366 0 5
14 4.4667 10.80701366 0 4
15 4.4667 10.80701366 0 3
16 4.4667 10.80701366 0 3
17 4.4667 10.80701366 0 2
18 4.4667 10.80701366 0 0
19 4.4667 10.80701366 0 0
20 4.4667 10.80701366 0 1
21 4.4667 10.80701366 0 2
22 4.4667 10.80701366 0 5
23 4.4667 10.80701366 0 1
24 4.4667 10.80701366 0 7
25 4.4667 10.80701366 0 7
26 4.4667 10.80701366 0 1
27 4.4667 10.80701366 0 5
28 4.4667 10.80701366 0 5
29 4.4667 10.80701366 0 8
30 4.4667 10.80701366 0 8
```
c – Chart:
Handpicked Content: Steps in Constructing a Median And Range Control Chart
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## THREE-DIMENSIONAL TELEVISION SYSTEM (Aug, 1953)
Why use those annoying glasses when you could stare through slits cut in a pipe?
>>|
Next >>
2 of 2
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2 of 2
THREE-DIMENSIONAL TELEVISION SYSTEM
By Paul A. O’Neal
YOUR FIRST LOOK at 3-D TV will be just as startling and realistic as when you first viewed the new 3-D movies at your local motion-picture theater.
Three-dimensional vision is actually easy to accomplish on television. Whereas in cinematography there are many problems in producing 3-D in large auditoriums, TV can be utilized in a small room and need provide for only a few viewers at any one time. There is no need for using two films and keeping them matched, and no wide-angle screen or throw-away Polaroid glasses are required.
Actually, in TV we have to provide only a simple system whereby basically the images come alternately from the left and the right lens at the transmitting station. This may be accomplished by using any efficient system of optics which will put an image from one lens on the TV-camera plate immediately followed by the alternate lens image, or it may be accomplished with two television cameras switched electronically.
In any event, the two lenses should be approximately as far apart as the human eyes. It is desirable to make the image not only appear to have three dimensions in the television receiver, but also to make it protrude out of the screen so the announcer seems to hold his hand out into the room. This startling illusion is illustrated in Fig. 3. You have the desire to shake hands with him, or to reach out and grasp a box of candy that he may offer. This illusion may be exaggerated by having the pickup lenses farther apart than the eyes of the viewer. In the days to come, the expert stereo-TV cameraman will have to know his angles and protrusions.
The principles recently demonstrated by U. A. Sanabria at American Television Inc., involve the basic system illustrated in the block diagram, Fig. 1. A square-wave generator alternately cuts off electronically first one camera, then the other, producing 15 images from each camera. Then these are transmitted in the usual fashion using the same impulses, and nothing need be done to upset the transmitting standards. This studio arrangement with the TV-picture cameras and the 3-D periscope is being inspected and checked by Mr. Sanabria in photos C and D.
At the receiving end, photos A and B, the television receiver does not have to be touched. The picture-viewing device is tubular in form, and the experimental model employs an ordinary magnetic synchronous motor which is used to twirl a little piece of metal called a butterfly. When this flat piece of metal, with a cutout in each alternate corner flops over, it opens the right eye first, then the left one in step with the electric switch at the camera end. Since all TV stations are local, the ordinary 60-cycle power system will be adequate for general use.
Mr. Sanabria states that it is possible to utilize many tricks to eliminate the mechanical “lorgnette,” and suggests a checkerboard system of Polaroid, Fig. 2, having a right and left alternate series of checkers, and then using interdots at the transmitter to correspond with every other image. Then, with ordinary Polaroid spectacles, no moving parts have to be used. * * *
1 comment
1. TechEBlog » 3D Television - 1953 says: June 24, 20069:41 pm
[…] [via ModernMechanix] […]
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# Excel Spreadsheets
## Best Concrete Mix Design Quantity Calculation Excel Sheet | How to Calculate Cement Sand Aggregate Quantity in Concrete Excel Spreadsheet
Calculate Cement Sand Aggregate How to calculate cement sand and aggregate quantity in concrete for any grade of concrete such as cement sand aggregate calculation for M20, how to calculate cement sand and aggregate quantity in M25 grade of concrete is a major problem for Civil Engineer. This problem is solved after reading the article ‘How to Calculate Cement Sand And …
## Calculate Quantity of Cement Mortar in Brickwork Excel Easy Way
Civil Experience very well knows that when you do the task to Calculate Quantity of Cement Mortar in Brickwork for your client, its required perfect quantity without any error. If you go wrong you can be embarrassed but don’t take pressure you will get a solution in Civil Experience so join us fast for perfect learning. In this …
## Civil Software Training by Dynamic-Struct Engineering Consultant and Software Training Institute
Er. Mohammad Sohel Abdul Hafeez is M.Tech structure engineer and also takes online classes for Civil engineering students. If you are a civil engineer then you should have good knowledge about structural engineering. If you want to become a structural engineer, then you should take this free demo lecture which will 100% prove to be …
## Best Isolated Footing Design EXCEL Sheet | How to Design Isolated Footing | Isolated Footing Design Excel Sheet As Per IS 456
In today’s civil experience article Design Isolated Footing problem is solved and your isolated RCC footing time decrease (In 2 minutes), accuracy also increases so let’s go for it. Isolated Footing Design Excel Sheet Today, we come with the new and important part of the estimation series is an isolated footing design as per IS …
## [SECRET] How to TWO WAY SLAB Design Excel Sheet As per IS 456 | How Professionals Create Two Way Slab Design
One Way Slab and Two way slab design calculation is the most difficult task in structural designing but today Civil Experience solve bar bending schedule for two way slab excel as per IS Code excel sheet (as per IS 456) all problems are in one excel file Design of Two Way Slab According to an expert civil engineer, Two way slab design excel sheet …
## Haunch Bar Cutting Length Calculation Excel Sheet
Haunch Bar Calculation for estimation of reinforcements bar needs in the work is the most essential part because without estimation you have no idea about how many quantities are required for the construction project. Civil Experience going to provide an excel sheet for Haunch Bar Cutting Length Calculation free in this article but before you …
## [Advance] Concrete Mix Design as per IS CODE by Excel Sheet | Concrete mix design as per is 10262:2019
What is the secret behind creating a mix design of concrete as per IS 10262 fluently in just seconds with high accuracy? and how I make concrete Mix Design as an engineer? Are you a civil engineer but face a problem in the mixed design of concrete creation for a construction site? or Spend a lot of time behind …
## Collection Of Civil Engineering Design Spreadsheet | Top Civil Engineering Excel Sheets | Best Civil Excel Sheet
Civil Engineering Excel Sheets – Excel is provided by Microsoft and is very powerful which is used largely in the structural designing, estimation of materials, and thcivil engineering calculation excel sheete calculation of the quantity of materials in the civil engineering field. Why Expert Civil Engineer Use Excel Sheet? Most top engineers and structural designers/consultants use …
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# Two Charged Masses Suspended From Strings
• Geromy
In summary, the problem involves two similar charged masses hanging from silk threads and forming an angle θ. Using Coulomb's Law and trigonometry, it can be shown that the distance between the two masses at equilibrium is x = ((q2L)/(2∏ε0mg))1/3. The forces acting on the masses are the force due to charges and the force of gravity, which form a right triangle with the force of tension as the hypotenuse. By equating the forces and using trigonometric ratios, the final equation can be derived.
Geromy
## Homework Statement
Two similar tiny balls of mass m are hung from silk threads of length L and carry equal charges q. The angle formed by the two strings is bisected by an imaginary line, forming angle θ. Assume that θ is so small that tan θ can be replaced by its approximate equal, sin θ. The distance between the two charged masses, at equilibrium, is x. Show that, for equilibrium,
x = ((q2L)/(2∏ε0mg))1/3
## Homework Equations
Clearly, Coulomb's Law is a vital piece of this problem, and other sources that I've checked suggest such things as the constancy of the ratio of the sides of a triangle divided by the sine of that side's opposite angle and combining the electric force and gravitational force into the components of the force of tension.
## The Attempt at a Solution
Despite having a couple of ideas about how to start, I haven't made much progress. I noticed that there is an identical problem here (https://www.physicsforums.com/showthread.php?t=305517) that suggested the constant sine ratio rule, but I can't for the life of me get to the right answer. Any help would be appreciated.
Haha, so, I figured it out! I thought I would post here in case anyone wants an explanation.
Consider the forces acting on a single one of the charged masses: the force due to the charges is kq2/x2 - x is the radius between the two charges, since it's the distance they're separated by.
The force due to gravity is also pretty straightforward: it's simply mg.
Since these two forces are perpendicular, they form two sides of a right triangle. The hypotenuse of this triangle is equal to the final force acting on the charged mass, the force of tension. This triangle contains angle θ, since the force of tension is in the direction of the silk thread.
Since now we've reintroduced θ, we can start using trigonometry. Tan θ = Opp/Adj, which in this case is our force due to electric charge (F) over the force due to gravity (W), or F/W.
However, Tan θ = Sin θ, which equals x/2L
when you combine equations, you get 2kq2L = mgx3, and from here it's a simple matter of algebra to get to the final equation.
Well done!
Welcome to PF!
ehild
## 1. How do the charges in the masses affect each other?
The charges in the masses can either attract or repel each other, depending on their polarity. Like charges (positive-positive, negative-negative) will repel each other, while opposite charges (positive-negative) will attract each other.
## 2. What factors determine the strength of the force between the charged masses?
The strength of the force between the charged masses is determined by the magnitude of the charges and the distance between them. The larger the charges and the closer they are, the stronger the force will be.
## 3. Can the masses be suspended at any distance from each other?
No, the masses must be suspended at a specific distance from each other in order for the forces to be balanced. If the distance between the masses is too small, the forces will be too strong and the system will become unstable.
## 4. How does the length of the strings affect the motion of the masses?
The length of the strings can affect the motion of the masses by changing the tension in the strings. If the strings are shorter, the tension will be greater, causing the masses to move faster. If the strings are longer, the tension will be lower, causing the masses to move slower.
## 5. What happens if the charges in the masses are equal?
If the charges in the masses are equal, the forces between them will be balanced and the masses will remain stationary. This is known as electrostatic equilibrium.
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'Believe to Achieve'
# Maths
Week 6- Multiplication and Division
This week we have been consolidating our understanding of the grid method and have been multiplying 2-digit numbers by single-digit numbers. Some of us have also looked at multiplying 2-digit numbers by 2-digit numbers.
Week 5- Multiplication and Division
This week we have been looking at working out multiplication questions using the grid method. We have used our understanding of arrays to help us.
Week 4- Multiplication and Division
This week we have looked at arrays to help us work out multiplication problems. We will be starting to look at putting these into the grid method next week.
Week 3 - Multiplication and Division
Over the past couple of weeks the children have been learning how to multiply and divide by 3, 4 and 8. We used arrays and the bar model to multiply and sharing and grouping to solve division problems. At the end of the week we played a division board game to consolidate our understanding of the 8 times table.
This week we have started our new topic; multiplication and division. We have started by looking at the 3 times table and how this helps us understand division. We are about to move onto the 4 times table!
Maths Challenge
Can you recite the 3 times table to your family? Can you write down the division facts to match?
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# Create acceleration with direction
I want to move my object with an acceleration in the proper direction. For example if I have these values:
``````friction = 0.3;
direction = 180;
speed = 5;
``````
For making it clear, I need a formula to create a linear acceleration.
-
You can calculate the invidual components by using sin and cos:
``````xSpeed = sin(direction)*speed;
ySpeed = cos(direction)*speed;
[...]
xPos += xSpeed;
yPos += ySpeed;
``````
Accleration is simply increasing the speed.
``````xAccel = sin(accelDirection)*accleration;
yAccel = cos(accelDirection)*accleration;
// [...]
xSpeed += xAccel;
ySpeed += yAccel;
``````
-
Actually before posting this question I wrote this solution. Your answer will be completed by the Xavier Arias Botargues's answer. Thanks. – MahanGM Jul 22 '12 at 9:58
Acceleration is the rate at which velocity changes. If it is zero, then you will keep moving at your start velocity. If it is larger than zero, and in the same direction as the current velocity, velocity will increase.
``````velocity = old_velocity + acceleration;
position = old_position + velocity
``````
If acceleration = 2, then your velocity will increase by 2 for each timestep.
-
This is similar to other 2 answers. Thanks. – MahanGM Jul 22 '12 at 10:01
To do any movement or physics calculations you should always use the time spent since the last frame (dt or delta); this depends on the main loop implementation, but is generaly passed to a method called update as parameter. (If this is not available assume dt = 1)
To update speed with acceleration and time and position with speed and time you can:
``````speed += acceleration * dt;
position += speed * dt;
``````
Note that acceleration, speed and position are vectors, so they have the operations:
``````speed -= speed * friction
Yeah I didn't mention the time passing in my question. I'm using `timeGetTime()` to calculate previous frame. Thanks. – MahanGM Jul 22 '12 at 9:59
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# Better Programming with Snap! 6.1
After today's first day of SnapCon 2020 I couldn't help myself to have a closer look at the map()-block and hyperblocks. You will find a description of the differences in a new blog post at http://gigers.com/blog/?p=375.
The blog post itself is written in German, but you can just enjoy the pictures.
As several attendees of the conference have already realised, will lead to adjustments in the curriculum. Basically they fulfill a prophesy Stanislaw Lem has made way back in his "Summa technologiae". Only when certain aspects of technology become simple to handle, will humans be able to understand more sophisticated concepts well. Hyperblocks definitely fall into that category.
HI, welcome to the forum!
The version with MAP can be made a little simpler even pre-hyperblocks:
Wow, that was super quick of you. Thank you for the nice, very well-written blog post!
Thanks a lot for letting me know about this version. It is much more elegant. I have included it into my blog post.
I use this method to modify r in RGB. In teaching, it is found that after RGB is set to 0,0,0,0, similar matting effect can be achieved.
I'm now trying to use it in my 12-year-old (seventh grade) teaching and it's understandable to kids. Now I'm thinking about the distortion of sound, the filter of pictures.
(English is not good, from the translator, please forgive me.)
Maybe it is a good idea to hide some complexity:
And then do all the calculations for the individual pixels separately (just a very simple example which switches the RGB values around):
This would be a little less intimidating if you made it
but there's no need for rgba_local at all; just REPORT (LIST ...)
(You could get rid of the others, too, but I agree that there's an increase in self-documentingness with them.)
Thanks again. It says it right in the block.
So here is an example which actually does some calculations.
Could be fun to give students an original picture and an alternative version and let them try to replicate it by adjusting the pixelfilter function.
(I'm assuming that you didn't really mean for all of the ITEM blocks to have ITEM 1 OF, and that you want some blue value even if the given value is less than 50.)
My version has the computation of blue right in the LIST block. I'm not sure if that helps or hurts, compared to making a B variable.
Why would you report LIST?He didn't do something wrong.
(joke)
Would you rather have a number be in the report block? Perhaps some arbitrary big number, maybe 18001767679 or something like that.
You are correct on both accounts.
For students I will keep the RGB-channels separately, as I intend to use the block as a blueprint for other filter experiments.
Thanks again.
Or 1e+10000000000000000000
This version while slightly less flexible than the one above works with hyperblocks and should be therefore much faster.
I did some testing and on a fast computer, you can adjust the parameters with a slider nearly in realtime.
instead of doing it would be better if you use the if then else reporter instead
There is always more to learn. With the reporter it looks much better.
it's what it was meant to do, report
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A251809 Decimal expansion of 3*sqrt(2)*Pi^3/128. 4
1, 0, 2, 7, 7, 2, 2, 5, 8, 5, 9, 3, 6, 8, 5, 8, 5, 6, 7, 8, 7, 9, 2, 5, 6, 6, 1, 8, 0, 0, 2, 2, 5, 5, 7, 6, 7, 2, 1, 0, 1, 0, 0, 3, 1, 8, 5, 3, 6, 9, 9, 7, 4, 6, 5, 3, 3, 1, 0, 8, 4, 7, 5, 5, 1, 8, 5, 2, 5, 7, 7, 7, 2, 4, 6, 8, 5, 8, 4, 9, 6, 8, 0, 3, 5, 1 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET 1,3 REFERENCES L. B. W. Jolley, Summation of series, Dover Publications Inc. (New York), 1961, p. 64 (formula 340). LINKS G. C. Greubel, Table of n, a(n) for n = 1..10000 FORMULA Equals Sum_{i >= 0} (-1)^floor(i/2)/(2i+1)^3 = +1 +1/3^3 -1/5^3 -1/7^3 +1/9^3 +1/11^3 - ... EXAMPLE 1.027722585936858567879256618002255767210100318536997465331084755185... MATHEMATICA RealDigits[3 Sqrt[2] Pi^3/128, 10, 90][[1]] PROG (PARI) 3*sqrt(2)*Pi^3/128 \\ G. C. Greubel, Jul 27 2018 (MAGMA) R:= RealField(); 3*Sqrt(2)*Pi(R)^3/128; // G. C. Greubel, Jul 27 2018 CROSSREFS Cf. A153071: Sum_{i >= 0} (-1)^i/(2i+1)^3. Cf. A233091: Sum_{i >= 0} 1/(2i+1)^3. Sequence in context: A011052 A153738 A159790 * A016639 A138341 A153520 Adjacent sequences: A251806 A251807 A251808 * A251810 A251811 A251812 KEYWORD nonn,cons AUTHOR Bruno Berselli, Dec 10 2014 STATUS approved
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