Always.
Yes, it is possible to calculate the chromaticity coordinates using absorbance values. The best way to calculate the chromaticity coordinates using absorbance values is by using the formula x = x/x+y+z.
Select any one of the vertices and draw all the diagonals from that vertex. This will divide the polygon (with n sides) into n-2 triangles. Use the coordinates of the vertices of each triangle to calculate its area, and then add the areas of these triangles together.
A dodecagon is a regular polygon that can be drawn using rotations. These are normally drawn in a Geometer's Sketchpad.
You cannot really compare whether the area or perimeter is bigger, since they have different units. If you want to compare which one is numerically bigger, that depends primarily on what units you are using, as well as the magnitude of the lengths.
i drew a picture using a regular polygon for a dogs head
Yes, it is possible to calculate the chromaticity coordinates using absorbance values. The best way to calculate the chromaticity coordinates using absorbance values is by using the formula x = x/x+y+z.
Once you know the coordinates, you can use the distance formula to find the lengths of the sides, then using that, you can find the area.
Coordinates: R is (-6, 2) and T is (1, 2) Length of side RT is 7 units using the distance formula
Select any one of the vertices and draw all the diagonals from that vertex. This will divide the polygon (with n sides) into n-2 triangles. Use the coordinates of the vertices of each triangle to calculate its area, and then add the areas of these triangles together.
you have all the coordinates wrong
try with file_open command it will work. if not possible asume that as a image and then using coordinates you can get a value bye chanti
#include<iostream> #include<vector> #include<assert.h> // Typdefs to hide unnecessary implementation detail and remove verbosity. using Coordinates = std::pair<int,int>; using Polygon = std::vector<Coordinates>; // Calculate the area of a given polygon. int area (const Polygon& polygon) { // The given polygon must list all vertices in the correct sequence // either clockwise or anti-clockwise. It does not matter which // vertex begins the sequence, but the last vertex is assumed to // join the first. // Initialise an accumulator. int accumulator = 0; // A polygon with less than 3 vertices is no polygon! if (polygon.size() < 3) return accumulator; // The last vertex is the previous vertex to the first vertex. // We'll deal with this specific pair of vertices first. size_t previous = polygon.size() - 1; // Iterate through all vertices in sequence. for (size_t current=0; current<polygon.size(); ++current) { // The previous and current vertices form an edge. We need to calculate // the area of the imaginary rectangle extending horizontally from this // edge until it meets the Y-axis. This edge may not be vertical so we // also extend in the opposite direction by the same amount. That is, // for vertices {x1, y1} and {x2, y2}, the imaginary rectangle's opposing // corners will be at imaginary vertices {0, y1}, {x1+x2, y2}. The area // of this imaginary rectangle is therefore (x1+x2)*(y1-y2). // Note: the imaginary rectangle's area may be negative but that's OK. // It'll simply be subtracted from the accumulator and that's exactly // what we want. accumulator += (polygon[previous].first + polygon[current].first) * (polygon[previous].second - polygon[current].second); // The current vertex now becomes the previous vertex // in readiness for the next iteration. previous = current; } // Remove the sign (make absolute). accumulator *= (accumulator<0) ? -1 : 1; // At this point the accumulated total is guaranteed to be an even // number (as we'll see). But let's play safe and assert that fact. assert (accumulator % 2 == 0); // Since each imaginary rectangle was exactly double what we needed // (because we extended in both directions), divide the accumulated // total by 2. It's more efficient to do that here, once, rather // than for each individual rectangle. We don't have to worry about // fractions since the accumulator is guaranteed to be even. return accumulator / 2; } // Driver to test the function. int main() { Polygon square; square.push_back (Coordinates (0, 0)); square.push_back (Coordinates (0, 10)); square.push_back (Coordinates (10, 10)); square.push_back (Coordinates (10, 0)); assert (area (square) == 100); std::cout << "Square (10 x 10) has area " << area (square) << std::endl; Polygon triangle; triangle.push_back (Coordinates (0, 0)); triangle.push_back (Coordinates (0, 6)); triangle.push_back (Coordinates (8, 0)); assert (area (triangle) == 24); std::cout << "Right-angled triangle (width 8, height 6) has area " << area (triangle) << std::endl; Polygon irregular; irregular.push_back (Coordinates (0, 0)); irregular.push_back (Coordinates (0, 14)); irregular.push_back (Coordinates (8, 14)); irregular.push_back (Coordinates (8, 4)); irregular.push_back (Coordinates (12, 4)); irregular.push_back (Coordinates (12, 0)); assert (area (irregular) == 128); std::cout << "6-sided irregular polygon has area " << area (irregular) << std::endl; Polygon line; line.push_back (Coordinates (0, 0)); line.push_back (Coordinates (0, 4)); line.push_back (Coordinates (0, 8)); assert (area (line) == 0); std::cout << "3 points on a line of length 8 has area " << area (line) << std::endl; }
No. Because tessellation is about using lost (infinitely many) copies of a polygon to cover a surface, One polygon does not comprise a tessellation.
A dodecagon is a regular polygon that can be drawn using rotations. These are normally drawn in a Geometer's Sketchpad.
Using systematic polygon naming, a twenty-two sided polygon would be called an icosakaidigon.
You cannot really compare whether the area or perimeter is bigger, since they have different units. If you want to compare which one is numerically bigger, that depends primarily on what units you are using, as well as the magnitude of the lengths.
i drew a picture using a regular polygon for a dogs head