# User Portlet Valeriu Ungureanu
Discussions
There are a lot of solutions. In your code there is only one error. You must remove **i** from **AppendTo[]**: In:= x = {1, 2, 3, 4, 5}; y = {1, 2, 3, 4, 5}; result = {} For[i = 1, i
Dear Alex, I have found the source of the error in the Mathematica 10. **Table[0, data[[Length[data], 1]]]** must have an iterator form **Table[0, {i, data[[Length[data], 1]]}]**. So, the following code works perfectly in Mma 10: data =...
The formula is true for some constraints on the values of a and b. First, a>0. Second, a+b>0 or a>-b. Third, 1+b/a>0 follows from the precedent two constraints. Considering the truth of these inequalities, it may be elementary proved...
Sorry Sander for the duplication. Your comment appeared only after I posted mine.
Sure, there are other approaches, too. E.g.: sol1 = Flatten@Solve[t^4 + t^3 - 2 t^2 - t + 1 == 0, {t}, Reals]; w1 = sol1[[3, 2]] w2 = sol1[[4, 2]] or w1 = t /. sol1[] w2 = t /. sol1[]
I have fixed some variables and used Minimize[] for the rest of variables... Unfortunately, this is all I can do for the moment!
Does the following code satisfy your needs? Plot3D[2 x + y, {x, 0, 1}, {y, 0, 1}, RegionFunction -> (#2 >= 0 && #1 \[Minus] #2 >= \[Minus]1 && #1 + #2 Automatic, PlotRange -> All] ![enter image description here] ...
You may use also the function Reduce[]: In:= Reduce[3^(2 x - 1) == 27, x, Integers] Reduce[3^(2 x - 1) == 27, x, Reals] Out= x == 2 Out= x == 2
Try Manipulate: a = 1; Manipulate[ PolarPlot[a (1 + 2 Sin[t/2]), {t, 0, j}, Ticks -> False, PlotRange -> {{-3, 1.5}, {-2.5, 2.5}}], {j, 4 Pi/50, 4 Pi, 4 Pi/50}] Sure, you can export animation as a gif file: ![enter image...
The minimum value $0$ of the function $z=Abs[4x+6y-18]$ is achieved on the straight line defined by the equation $4x+6y-18=0$. So, you want to solve the optimization problem: $x+y \rightarrow min,$ s.t. $4x+6y-18=0,$ $x \geq 0, y\geq 0.$ ...