i could be wrong, was kinda in a rush
PLEASE HELP RIGHT AWAY
1 answer:
Answer:
Approximately 53
Step-by-step explanation:
This is a geometric progression, because each term is found by multiplying the previous one by a constant factor.
In our case, the first term is know (a1 = 3), and the growth factor is also known (q = 1.333333).
What we are tasked to do is to find the 11th (n=11) term of the series, since it asks for AFTER the 10th roll.. That's easy with the base formula:
Let's place the values we have in the equation and solve it:
So, after the 10th roll, you should need 53 dice, approximately.
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i could be wrong, was kinda in a rush
So, check the picture below. It has four sections.
the volume of a pyramid is, 1/3 ( base area)(height).
well, we know the base is a rhombus, and rhombuses diagonals meet at right-angles, therefore the center O has 4 right-angles where they meet.
since we know the angle at A, 50°, and since rhombuses are quadrilaterals with all equal sides, we know all the sides are 6 long, and the angle at C is 50° as well, and the remaining angles are 130°, as shown there.
now, if we tease out one of those triangles from the rhombus, as shown on the 1st section in the picture, we can say use the angle of 65° to get "x" and "y" lengths
now, we'll use those to get the area of the rhombic base.
and then onto the second section
is a right-triangle with a perpendicular line through it, which creates 3 similar triangles, by using the "geometric mean" of "w" there.
anyhow, we used the medium-sized and large-sized triangles there, to draw proportions and get what the "w" distance is, and you can see there what "w" is.
and onto the 3rd section
now that we know how long "w" is, we can use the triangle containing the 40° angle, in order to get the pyramid's height of "h", by using the tangent function, as you see there.
and we also use the same triangle, to get the "z" distance, which is namely the slant-height of the pyramid, however, is also the "altitude" of the triangular faces.
and onto the 4 and last section in the picture
that's a triangular face by herself, it has a base of 6, and an altitude of "z".
now, all that said, let's plug those values to get the volume, and then the surface area, keeping in mind the surface area is just the area of the rhombus plus the area of all four triangles.
 \\\\\\ SA=\left[ \stackrel{\textit{area of the rhombic base}}{4\left( \frac{1}{2}xy \right)} \right]~~+~~\left[ \stackrel{\textit{area of the four triangles}}{4\left[ \frac{1}{2}(6)(z) \right]} \right]](https://tex.z-dn.net/?f=%5Cbf%20V%3D%5Ccfrac%7B1%7D%7B3%7D%5Cleft%5B%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20rhombic%20base%7D%7D%7B4%5Cleft%28%20%5Cfrac%7B1%7D%7B2%7Dxy%20%5Cright%29%7D%20%5Cright%5D%28h%29%0A%5C%5C%5C%5C%5C%5C%0ASA%3D%5Cleft%5B%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20rhombic%20base%7D%7D%7B4%5Cleft%28%20%5Cfrac%7B1%7D%7B2%7Dxy%20%5Cright%29%7D%20%5Cright%5D~~%2B~~%5Cleft%5B%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20four%20triangles%7D%7D%7B4%5Cleft%5B%20%5Cfrac%7B1%7D%7B2%7D%286%29%28z%29%20%5Cright%5D%7D%20%5Cright%5D)
x ≈ 2.54 y ≈ 5.44 w ≈ 2.298 h ≈ 1.93 z = 3
V ≈ 17.7265 SA ≈ 63.58
the volume of a pyramid is, 1/3 ( base area)(height).
well, we know the base is a rhombus, and rhombuses diagonals meet at right-angles, therefore the center O has 4 right-angles where they meet.
since we know the angle at A, 50°, and since rhombuses are quadrilaterals with all equal sides, we know all the sides are 6 long, and the angle at C is 50° as well, and the remaining angles are 130°, as shown there.
now, if we tease out one of those triangles from the rhombus, as shown on the 1st section in the picture, we can say use the angle of 65° to get "x" and "y" lengths
now, we'll use those to get the area of the rhombic base.
and then onto the second section
is a right-triangle with a perpendicular line through it, which creates 3 similar triangles, by using the "geometric mean" of "w" there.
anyhow, we used the medium-sized and large-sized triangles there, to draw proportions and get what the "w" distance is, and you can see there what "w" is.
and onto the 3rd section
now that we know how long "w" is, we can use the triangle containing the 40° angle, in order to get the pyramid's height of "h", by using the tangent function, as you see there.
and we also use the same triangle, to get the "z" distance, which is namely the slant-height of the pyramid, however, is also the "altitude" of the triangular faces.
and onto the 4 and last section in the picture
that's a triangular face by herself, it has a base of 6, and an altitude of "z".
now, all that said, let's plug those values to get the volume, and then the surface area, keeping in mind the surface area is just the area of the rhombus plus the area of all four triangles.
x ≈ 2.54 y ≈ 5.44 w ≈ 2.298 h ≈ 1.93 z = 3
V ≈ 17.7265 SA ≈ 63.58