An octagonal pyramid has 9 vertices. The base of this shape is an octagon, which will give it 8 vertices when the triangles that form the sides are considered. Those triangles will lead up to the apex (top) of the pyramid, and that will be the 9th vertex.
Area of rectangle is lenth by breath.So mm by mm will give area of mm square.
Many ways: Join the opposite vertices; Join the midpoint of each side to the midpoint of the opposite side; Draw three lines parallel to one of the sides at distances of a quarter, half and three-quarters of the width. Then there are ways that give equal but non-congruent parts.
32* * * * *An octahedron is a generic term for a polyhedron with eight faces. There are a number of different configurations and these will give rise to a different number of vertices. Some examples:Heptagon based pyramid: 8 vertices, 28 angles.Hexagon based prism: 12 vertices, 36 anglesQuadrilateral based dipyramid: 6 vertices, 24 angles.
The answer depends on what and where the points are. Since you have provided no information on that it is not possible to give a sensible answer.
The first step to finding a triangle's center of gravity is to calculate the average of the x-coordinates and y-coordinates of the triangle's vertices. This will give you the coordinates of the centroid, which is the point where the center of gravity lies.
Well this is my thought depending on where the point of dilation is the coordinates of the give plane is determined. The point of dilation not only is main factor that positions the coordinates, but the scale factor has a huge impact on the placement of the coordinates.
-- If they give you one set of 'x' and 'y' coordinates, then you have the location ofone point on the line. One point doesn't have a slope.-- If they give you two sets of 'x' and 'y' coordinates, then you have the locations oftwo points on the line. The slope of the straight line between two points is(the difference between the 'y' values) divided by (the difference between the 'x' values)
rectangle or paralellogram
You can follow the following steps. * First, you determine the slope between the two points. Just calculate delta-y / delta-x (that is, difference in y-coordinates, divided by the difference in x-coordinates, between the two points). * Next, you use the point-slope formula, to get an equation for the line. You can use any of the two points for this; each of the points will give you an equation that looks different, but the two equations are equivalent, if you do everything correctly. * Finally, solve the resulting equation for "y"; that will give you the equation in slope-intercept form.
The four vertices of the rectangle are (x, 0), (x, 3-x2), (-x, 3-x2) and (-x,0) so that the sides are of length 2x and 3-x2 Therefore, the perimeter, P, is 4x + 6 - 2x2 dP/dx = 4 - 4x which is 0 when x = 1 when x = 1, P = 4 + 6 - 2 = 8 units. d2P/dx2 = -4 < 0 which confirms that x = 1 does give a maximum.
Make a pentagon, and draw lines from each point to each other point. That would give you a five-pointed star within a pentagon. Plant the cabbages on the 5 vertices of the pentagon, and on the five points where the internal lines (the diagonals) cross.Make a pentagon, and draw lines from each point to each other point. That would give you a five-pointed star within a pentagon. Plant the cabbages on the 5 vertices of the pentagon, and on the five points where the internal lines (the diagonals) cross.Make a pentagon, and draw lines from each point to each other point. That would give you a five-pointed star within a pentagon. Plant the cabbages on the 5 vertices of the pentagon, and on the five points where the internal lines (the diagonals) cross.Make a pentagon, and draw lines from each point to each other point. That would give you a five-pointed star within a pentagon. Plant the cabbages on the 5 vertices of the pentagon, and on the five points where the internal lines (the diagonals) cross.
The coordinates of a point are in reference to the origin, the point with coordinates (0,0). The existence (or otherwise) of an angle are irrelevant.
you are correct!
It is other users who give trust points. They give you these points for seeing useful answers you have submitted.
No, a rectangle cannot ever be a rhumbus - nor a rhombus (to give its correct selling).
the area of a rectangle = length x widthwe can rearrange this to give uslength of a rectangle = area/ width