How do the number of faces, vertices, and edges of a cube compare to the number of faces, vertices, and edges of a tetrahedron? A cube has more faces than a tetrahedron. A cube has more vertices than a tetrahedron. A cube has more edges than a tetrahedron. ANSWERS: 1: 2 2: 4 3: 6

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2

yx = 36

step-by-step explanation:

cos f = opposite side/ hypotenuse

cos f = yw / yx

we know cos f = 4/9 and yw =16

4/9 = 16/yx

multiply each side by yx

4/9 yx = 16/ yx * yx

4/9 yx = 16

multiply each side by 9/4

9/4 * 4/9 yx = 16 *9/4

yx = 36

10+4√2 is the answer. i dont have the time to explain.

the first is 13 the second is 1 the third is 8 and the fourth is 7