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18

apple = Interpreter["Company"]["Apple"]["Image"] Interpreter["Company"]["GE"]["Image"] Also works for the continent and respects colours: Interpreter["Company"]["Siemens"]["Image"] Update Interpreter["Company"]["Wolfram"]["Image"] For Apple addicts: ImageFilter[Max[Flatten[#]] - Min[Flatten[#]] &, ...


13

Another variation: SemanticInterpretation["AAPL Logo"] This method is nice because you can do at once: logos = SemanticInterpretation["AAPL, TSLA, GE and MSFT Logos"]; Column[logos, Frame -> All, FrameStyle -> Directive[Red, Thick]]


12

Use Interpreter without federal state or country. Works even for small german towns :) Interpreter["City"]["Memmingen"] GeoPosition[%] GeoGraphics[%] EDIT: Also you could ask for the airport: town = Interpreter["City"]["Memmingen"]; airport = Interpreter["Airport"]["Memmingen"]; GeoPosition[{town, airport}] or just use the nearest airport ...


11

You could use the official ICAO abbreviation: Entity["Airport", "KLAX"] which makes sense because you will only be able to use officially named airports most of the time anyway. You can always get those (or the city details) via a W|A query and work your way from there (no googling involved): or like this (although it misses Van Nuys):


9

QuantityMagnitude[WolframAlpha[#, {{"Result", 1}, "ComputableData"}]] & /@ {"Apple company revenue", "Apple company interest expense", "Apple company employees"} (* {1.828*10^11, 3.84*10^8, 97000} *) Update: Exchange rate conversion WolframAlpha[#, {{"Result", 1}, "ComputableData"}] & /@ {"Sony company revenue", "Sony company interest ...


8

The following may be useful as starters: WolframAlpha["femme from french to german",{{"Translation:TranslationData", 1}, "ComputableData"}] (* {" Frau | Ehefrau", " Frau"} *)


8

The problem seems to be that many companies, even some of the biggest, well known firms are not known in Entity["Company","name"] form to Mathematica. This holds for companies like Apple, Microsoft, General Electrics etc. I believe that CompanyData needs entities to be of this form. If you want to discover a company's entity representation using W|A ...


8

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

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

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


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]


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 ...


5

Something like this? WolframAlpha["apple logo", "PodImages", IncludePods -> "Image:FinancialData"]


5

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


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 ...


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 =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 ...


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"] *)


3

Thanks to your nice answers and comments I found a solution to my problem: FlightPath[cities_List, pro_: "AzimuthalEquidistant"] := Module[{lc = Length@cities, color, dest, dist, entity, legend, pair, path}, color = {Red, Darker@Green, Blue, Black, Orange, Purple}; entity = Interpreter["City"][#] & /@ cities; dest = GeoPosition@First@entity; ...


3

First, you should understand that WolframAlpha does not simply execute Mathematica code as typed. Rather, there is a parsing stage. And the two parses are simply different in this case. We can see this by taking a closer look in Mathematica: WolframAlpha["x = 10; N[ 2 Pi (x -11/8)/LambertW[(x -11/8)/E]]", {{"Result", 1}, "ComputableData"}] (* Out: ...


3

I am not sure why you do not want to use DSolve but it works perfectly in your case. f[t_] := Piecewise[ {{t^2 , 0 <= t < 1}, { 1, t > 1}}] DSolve[ {y''[t] - y[t] == f[t] , y[0] == 0, y'[0] == 0 }, y[t], t] gives correct solution of your equation. If you insist using Laplace transform in this problem, this can be done as follows, lap = ...


2

Select[Range[23], GCD[24, #] == 1 &] {1, 5, 7, 11, 13, 17, 19, 23} This also works: Select[Range[23], CoprimeQ[24, #] &]


2

Using exact numbers or higher precision numbers gives the result that matches google's output: WolframAlpha["convert 1/4 miles to inches", {{"Result", 1}, "ComputableData"}] (* 15840 in *) WolframAlpha["convert .250 miles to inches", {{"Result", 1}, "ComputableData"}] (* 15840 in *) Another example: WolframAlpha["convert 1/5 miles to inches", ...


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

Wolfram|Alpha is more willing to change the input because the input is generally natural language. Alpha has interpreted hw has two single-letter symbols, h and w. Mathematica has interpreted hw as a single two-letter symbol. Because you asked Mathematica to solve for h and h does not appear in the expression, there is no solution. Adding a space tells ...



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