Drawing this square and then drawing in the four radii from the center of the cirble to each of the vertices of the square results in the construction of four triangular areas whose hypotenuse is 3 sqrt(2). Draw this to verify this statement. Note that the height of each such triangular area is (3 sqrt(2))/2.
So now we have the base and height of one of the triangular sections.
The area of a triangle is A = (1/2) (base) (height). Subst. the values discussed above, A = (1/2) (3 sqrt(2) ) (3/2) sqrt(2). Show that this boils down to A = 9/2.
You could also use the fact that the area of a square is (length of one side)^2, and then take (1/4) of this area to obtain the area of ONE triangular section. Doing the problem this way, we get (1/4) (3 sqrt(2) )^2. Thus,
A = (1/4) (9 * 2) = (9/2). Same answer as before.
We can factor 120 into
2 * 2 * 2 * 3 * 5
I think there are nine different possibilities
1) 8 * 3 * 5
2) 4 * 2 * 15
3) 4 * 6 * 5
4) 4 * 10 * 3
5) 2 * 2 * 30
6) 2 * 3 * 20
7) 2 * 5 * 12
8) 2 * 6 * 10
9) 2 * 15 * 4
Answer:
the answer is 6x.
Step-by-step explanation:
Answer: SECOND OPTION.
Step-by-step explanation:
It is important to remember that, by definition, the "Resultant vector " is the vector obtained by adding two or more vectors.
For this exercise you need to use the formula for calculate the magnitude of the resultant vector obtained by adding the vectors "a" and "b" shown in the picture attached:
Where "c" is the magnitude of the resultant vector asked in the exercise.
You need to analyze the graph given in the exercise.
You can identify that:
Then the next step, knowing which are "a" and "b", is to substitute values into the formula given at the beginning of this explanation, as following:
Finally, evaluating, you get that "c" is:
Octahedron has 8 faces, icosahedron has 20 faces. Therefore the answer is dodecahedron, it has 12 faces.
