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| This? Fa = 3*V; Fb = 4*V; Fc = 5*V; Fd = 6*V; a = 1; b = 1; c = 1; d = 1; matrix = Table[ If[Fb - Fa > 5, b = b + a; a = 0]; If[Fc - Fb > 7, c = c + b; b = 0]; If[Fd - Fc > 8, d = c + d; c = 0]; {a, b,... |
| With the wording of your original question I did not understand all the constraints that you have on any possible solution. I suggest you look at all the Latex syntax, determine those terms which appear to not be interpreted by WolframAlpha, write... |
| This {a, b, c, d, e} = {16, 19, 35, 53, 23}; m = 19; Reduce[{a+w m+g==0, b m+g==0}/.Solve[a m+w==b, w, Modulus->90][[1]], g, Modulus->90] Reduce[{a m+w+x m+h==0, b+x*m+h==0}/.Solve[b m+x==c, x, Modulus->90][[1]], h, Modulus->90]... |
| . dd = {Insert Your 795000 Numbers Here}; part = -1; Map[(part++; pos=Ordering[#,1]; Flatten[{#[[pos]], part*5000+pos}])&, Partition[dd,5000]] |
| In V10.1 it displays without the discontinuity. In V11.1 it displays with the discontinuity. Perhaps you can think very carefully, write your own version of TransformedField and see if you can avoid the discontinuity. Perhaps someone can... |
| Perhaps solve 4/3 pi r^3=12.5 kg/19.3 g/cm^3 for r [WolframAlpha link][1] I tried to get it to understand something like weight=12.5 kg, density=19.3g/cm^3, volume=weight/density, solve 3/4 pi r^3=volume for r but had no... |
| I am assuming I can substitute t->u-tt, solve the equation in tt and then reverse the substitution tt->u-t ga[t_] := sd^2/(t + sd^2/ra); gb[t_] := sd^2/(t + sd^2/rb); f = (-2 h[u-t] (gb[t]/sd^2-x (gb[t]-ga[t])/sd^2)+x (x-1)/(2... |
| Try this solut = Flatten[Table[Flatten[{a,b,c,y/.Solve[a*y==3+b/c,y]}],{a,1,3,1},{b,1,2,1},{c,1,2,1}],2] followed by solut[[6]] which gives {2, 1, 2, 7/4} |
| It appears that Mathematica cannot find a closed form solution to your system. EQS = D[S[t], t] == -a S[t] Inf[t]/n - q; EQI = D[Inf[t], t] == a S[t] Inf[t]/n - q - (? + ?)*Inf[t]; DSolve[{EQS, EQI}, {S[t], Inf[t]}, t] |
| When b has not been assigned any value then the value of IntegerDigits [b, 10] will be just IntegerDigits [b, 10], the Length of that will be 2, the value of Part[IntegerDigits [b, 10], 1] will be b and the value of Part[IntegerDigits [b, 10], 2]... |