Want this question answered?
No because they have 7 vertices
It is true.
if it is a true cube then yes, all 6 faces are congruent
If you mean -4+6=4, then it is not true.
Yes, 6 is a factor of 30. It evenly divides into 30. 6 x 5 = 30
No because they have 7 vertices
Square pyramid: 5 Cube: 6
They can have faces which are triangles, quadrilaterals, pentagons, hexagons or heptagons. It has 8 faces and can have 6 to 12 vertices.
hex = 6, hept = 7, oct = 8, non or enne = 9.
They are all even numbers.
It is true.
They all have 6 sides and each side are the same length.
They all have 6 valence electrons
All the elements in group 6 are called the transition metals. This is also true for groups 3-12.
A heptagon has seven sides, a hexagon six, a nonagon nine, and an octagon eight. So in 5 heptagons, 2 hexagons, 4 nonagons, and 3 octagons there are (5*7)+(2*6)+(4*9)+(3*8) = 35 + 12 + 36 + 24 = 107 sides altogether.
if it is a true cube then yes, all 6 faces are congruent
Equilateral triangles can tile a plane, but regular heptagons cannot; nor can they tile the plan together. Where vertices meet (at a point on the plane) there is a complete turn of 360°. Each vertex of an equilateral triangle is 60°; 360° ÷ 60° = 6, a whole number of times, so a whole number of equilateral triangles can meet at a vertex of the tiling. Each vertex of a regular heptagon is 128 4/7°; 360° ÷ 128 4/7° = 2 4/5 which is not a whole number, so a whole number of regular heptagons cannot meet at a vertex of the tiling, so there will be gaps. With one regular heptagon there are 360° - 128 4/7° = 232 3/7°, but this cannot be divided by 60° a whole number of times, so one regular heptagon and some equilateral triangles cannot meet at a vertex of the tiling without gaps. With two regular heptagons there are 360° - 2 x 128 4/7° = 102 6/7°, but this cannot be divided by 60° a whole number of times, so two regular heptagons and some equilateral triangles cannot meet at a vertex of the tiling without gaps. With three or more regular heptagons, they will overlap when trying to place them on a plane around a point - leaving no space for any equilateral triangles.