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11

The date is different because Wolfram Alpha is wrong about when the Battle of Agincourt took place. Alpha is returning 25 October 1415 in the Gregorian calendar as the date of the battle, but the battle took place on that date in the Julian calendar (the calendar in use at the time). That date corresponds 3 November 1415 in the proleptic Gregorian ...


9

Here's a start.... x = "healed"; y = "pay"; Intersection[ WolframAlpha["synonyms of " <> y, {{"Synonyms:WordData", 1}, "ComputableData"}, InputAssumptions -> {"*MC.%7E-_*WordData-"}], WolframAlpha["rhymes with " <> x, {{"Rhyme:WordData", 1}, "ComputableData"}, PodStates -> {"Rhyme:WordData__More"}, InputAssumptions -> ...


6

data = WolframAlpha["median income in Maryland counties", {{"PropertyRanking:ACSData", 1}, "ComputableData"}]; data // TableForm


6

Ctrl+= gasoline heat of combustion * density results in: Quantity[32.7, "Kilojoules"/"Centimeters"^3] To get energy per gallon: UnitConvert[%, "Joules"/"Gallons"] (* Quantity[1.23782965*^8, "Joules"/"Gallons"] *)


6

As Guesswhoitis already suggested, this is a machine precision issue. So let us do your computation with arbitrary precision numbers. For doing so, all machine numbers have to be replaced with arbitrary precision numbers, otherwise the computation falls back to machine numbers. In the following command I have done this by placing `30 after each machine ...


6

As pointed out by Michael Seifert in a comment GeoGraphics[Polygon[Ctrl+=Italy]] should work (it worked for me), but if you want do it without using Wolfram's server, try GeoGraphics[Polygon[Entity["Country", "Italy"]], GeoBackground -> None]


5

Facebook has deprecated the API that Mathematica relied on to gather this information. Official source.


5

Case closed but the topic should be left here for future visitors who use V9. So: Bug introduced in 9.0 and fixed in 10.0.0


5

airports = {"ABE", "ABI", "ACT"}; AirportData["Properties"] data = AirportData[#, {"IATACode", "Name", "Cities", "Latitude", "Longitude"}] & /@ airports; data // Grid airports = {"ABE", "ABI", "ACT"}; EDIT: A quicker method would be to download a .csv file and filter it. data = Import["http://ourairports.com/data/airports.csv"]; The ...


5

Using Mathematica (v10) Reverse[SortBy[ EntityClass["AdministrativeDivision","USCountiesMaryland"][{"Name","MedianHouseholdIncome"}], Last]] // TableForm


4

WolframAlpha for a quick overview:


4

At a crude level, the evaluator that does the TrigExpand can be emulated by a FixedPoint or ReplaceRepeated. If we want to see the steps behind this particular kind of evaluation, it seems possible that a simple substitute for the built-in TrigExpand can be implemented based on FixedPointList instead of FixedPoint, to keep track of the steps. Here is a ...


4

You're pretty close. The following code should do it: Entity["Airport", #]& /@ airports


4

Note This answer was meant for a bit of fun but instead does a good job of managing expectations for Entity in the WL at present. Image Identify and Google Image Search Wolfram have heavily promoted their ImageIdentify project (http://blog.wolfram.com/2015/05/13/wolfram-language-artificial-intelligence-the-image-identification-project/) which might be a ...


4

This is a documentation oversight. The function FlightData is not available in Mathematica as of this time.


4

This =Solve[64((x-Sqrt[x^2-1])/(x+1))^2-25(1-Sqrt[x-1]/(x+1))^2+9==0, x] and this WolframAlpha["Solve[64((x-Sqrt[x^2-1])/(x+1))^2-25(1-Sqrt[x-1]/(x+1))^2+9==0,x]"‌​] solve immediately. Note: One reason for using WolframAlpha to handle some calculations is that you can sometimes see the steps involved, see additional information, make use of curated ...


4

Your input should be: Series[ (I x + y Exp[I p]) Tanh[Pi (I x + y Exp[I p])] Log[(I x + y Exp[I*p])^2 + a^2] I y Exp[I p], {y, 0, 2}, Assumptions -> {x > 0, y > 0, a > 0, p ∈ Reals} ] Asterisks can be used for multiplication, but a space is more usual. I removed unnecessary parentheses. The more important points are all functions (Exp, ...


3

Based on the documentation, I would say that this is a bug. Note, however, that you can perhaps get at the data you want via the call EntityValue[Entity["Country", "UnitedStates"], EntityProperty["Country", "ElectricityTotal", "Date" -> 2007]] though the relationship between "ElectricityTotal" and "ElectricityProduction" is less than clear. I would ...


3

Update According to @xzczd in my comment: Simply removing the outermost braces will work, i.e. {0,0},{1,2},{5,3},{6,5}. Original Answer (0,0) (1,2) (5,3) (6,5) worked. To efficiently convert your lists to this form: convertlist[expr_] := StringReplace["(0,0) " <> ToString[expr], {"{{" -> "(", "}}" -> ")", "}, {" -> ") ("}] Comparing ...


3

There is an easier way using WolframAlpha:


3

In Mathematica (version 10): weight=UnitConvert[Quantity[4 10^6, "USDollars"]/(Entity["Element", "Gold"]["Price"]), "Pounds"] (* Quantity[228.315, "Pounds"] *) UnitConvert[ weight/(Entity["Element", "Gold"]["Density"]), "Liters"] (* Quantity[5.36442, "Liters"] *)


2

Mathematica (version < 10) FinancialData["XAU/USD"] returns the price of one Troy Ounce, but the equivalence to kilograms isn't available in Mathematica v9 as far as I see. So: i = Quiet@Import["https://www.google.com/search?&q=troy+ounce+to+kilograms", "HTML"]; troyToKg = ToExpression@First@StringCases[i, "One troy ounce is" ~~ x__ ~~ " grams" ...


2

I just tried it, it works on Alpha as is. It just did not like the Exclusions -> None but it still took it. Most of the times, Wolfram Mathematica language will also work in Wolfram Alpha as well. I like to think of it as the Wolfram Alpha language being an almost superset of the Wolfram Mathematica language. Plot[Piecewise[{{10*x - 0.5*5*x^2, 0 < x ...


1

This answer suggested by @Hubble07 works. subsetQ[list1_, list2_] := WolframAlpha[ "is " <> ToString[list1] <> "a subset of " <> ToString[list2], {{"Result", 1}, "ComputableData"}]


1

I found that wolfram|Alpha wasn't good enough, by RhymeZone is a great site for this, and they even support near rhymes. Here's an example: Rhymes[r_] := Rhymes[r] = Module[{url, links, rhy}, url=StringTemplate["http://www.rhymezone.com/r/rhyme.cgi?Word=`1`&typeofrhyme=perfect&org1=syl&org2=l&org3=y"][r]; ...


1

Perhaps this unsophisticated attempt is a starting point. whales = {"have hair", "live birth", "breathe air", "live in water", "have fins", "can swim"}; fish = {"lay eggs", "have scales", "breathe water", "live in water", "have fins", "can swim"}; tint = Intersection[whales, fish]; w = Complement[whales, tint]; f = Complement[fish, tint]; r1 = ...


1

You can use Rationalize to convert numbers to exact numbers gammaex = 0.2506 // Rationalize[#, 0] &; omega[t_] = 2.43163218375*10^7*Exp[1700*(1/298.15 - 1/(273.15 + t))] // Rationalize[#, 0] &; w[t_] = (3.414105049212413*10^12)/(omega[t]) // Rationalize[#, 0] &; v[t_] = Sqrt[661.6469313477045*(t + 273.15)] // Rationalize[#, 0] &; ...


1

A plot like that in the reference provided in the Question can be obtained as follows. Solve the given equation (with x normalized to a for convenience and without loss of generality): Reduce[x == Exp[1/x]/(5 - 5 Exp[1/x]), x] (* Element[C[1], Integers] && -1 + E^x^(-1) != 0 && -5 + ProductLog[C[1], 5*E^5] != 0 && x == (-5 + ...


1

Okay, I'm slowly going through the questions involving elliptic integral evaluations. Yet again, none of the software mentioned in this thread have managed to produce a "clean" expression. For the benefit of future readers, here's a tidier closed form for your perusal: N[2 (7 Sqrt[5] + 2 (10 EllipticE[2 π/3, 11/12] + EllipticF[2 π/3, 11/12])/Sqrt[3])/9, 20] ...


1

Try WolframAlpha["Forest Gump", "FullOutput"] runs about 1 sec. on my computer. Hint: Make sure you have fast Internet connection. Here an extract of screenshot of result



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