[ad_1] Dude It’s super easy: def triangle(n): for i in range(1, n +1): print ‘ ‘ * (n – i) + ‘x’ * i Or even: def triangle(n): for i in range(1, n +1): print (‘x’ * i).rjust(n, ‘ ‘) output for triangle(5): x xx xxx xxxx xxxxx Dont just copy this code without comprehending … Read more
[ad_1] Use sine rule:http://media.tiscali.co.uk/images/feeds/hutchinson/ency/c02395.jpg If you have side a and angle A, and you want to work out the length of side b and you have angle b, the formula would be b = sin B * (a/sinA) In Javascript this can be done by var B = 45; //size of angle B var a … Read more
[ad_1] It won’t work because the triangle constructor will attempt to default construct its point members before assigning to them, but point doesn’t have a default constructor. This is because you provide only a point constructor that takes 2 arguments, so the defaulted default constructor is deleted. Instead, you should use a member initialization list … Read more
[ad_1] W.l.o.g. we say n>=m and choose m to be of the form 2^x. (If m is arbitrary <= n as well it only means that we apply the same approach to a second rectangle that is n times m-2^x in size, where x=int_floor(ld(m)) and ld is log to the base of 2 of course.) … Read more
[ad_1] Your logic is vastly more complicated than it needs to be. Simply calculate the difference between the heading and the radial, then normalize that to be between 0 and 360. Then compare that to the three ranges. let radial = … // some radial in the range 0-360 let heading = … // some … Read more
[ad_1] Consider this: public static void printFirstHalf(int m, int n){ if(m>n){ return; } // print asterix for(int i=1; i<=m; i++){ System.out.print(“*”); } System.out.println(); // recurse printFirstHalf(m+1,n); } Do you see where you went wrong with your other method now? If you’re working with recursion for the first time, I understand that it can be difficult … Read more
[ad_1] You rotate single point using rotation matrix. http://en.wikipedia.org/wiki/Rotation_matrix so it end up using matrix multiply eqations. v’ = R * v Where v is point as vector in Cartesian coordinates, R is rotation matrix and v’ is new point. You have to apply this multiplication for each point. 0 [ad_2] solved How can I … Read more
[ad_1] The syntax indeed is incorrect. It should be radial-gradient(circle at right center, #ffbb11 0px, #dd6600 100%). This is also incorrect in the CodePen, and once you fix it, the animation looks different (it has a ‘Click here’ call-to-action when you hover it). It is not the core issue, though. The reason why the animation … Read more
[ad_1] Express that the center of the sphere is equidistant to the three given points and coplanar with them (assuming that the three given points are on a great circle). (X – Xa)² + (Y – Ya)² + (Z – Za)² = R² (X – Xb)² + (Y – Yb)² + (Z – Zb)² = … Read more
[ad_1] In Activity you have to use this to accses to context, cursor..etc but in fragment we use getActivity() instead. 6 [ad_2] solved Check if i’m inside or outside radius of circle
[ad_1] There are of course many ways to set this up. I can remember being confused when starting out, below is an alternative example to give you something else to look at. It is not intended to be fancy but just bare-bones so that you can see a minimal setup of the class with an … Read more
[ad_1] There is lot approaches to check whether point lies inside triangle. The simplest ones, I think: 1) Check whether point is on the same side of all edges (find signs of cross products for edge vectors and vertex-point vectors AB x AP, BC x BP, CA x CP) 2) Represent AP vector in basis … Read more
[ad_1] //Example with mutliple coordinates before creating an arc. AREA DEFINED AS 133830N1450807E TO 132836N1444449E TO 133043N1443814E TO 133515N1443710E THEN CLOCKWISE ON A 15.3 NM ARC CENTERED ON 133416N1445256E TO THE POINT OF ORIGIN //Abbreviation // a // b // m(midangle) (cx,cy,ax,ay,bx,by) // x(lat) // y(long) //Xc=latitude provided in text for center point //Yc=longitude provided … Read more
[ad_1] Yes, it is possible, however, this has nothing to do with programming and is more of a mathematical question. (I would recommend going here https://math.stackexchange.com/) Solving this using basic Algebra, given the slope and y-intercept we can check if a point is on a line by substituting the x and y values. For example, … Read more
[ad_1] If you are using LWJGL you can also use JOML, in which case the following is probably what you might want: import org.joml.*; public class CubePositions { public static void main(String[] args) { /* Cuboid center position */ float px = 10, py = 0, pz = 0; /* Euler angles around x, y … Read more