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Vsevolod [243]
3 years ago
9

What type of polygon would a slice of a hexahedron at a vertex create? Explain. What type of polygon would a slice of an icosahe

dron at a vertex create? Explain.

Mathematics
1 answer:
harina [27]3 years ago
3 0

Answer:

  • hexahedron: triangle or quadrilateral or pentagon
  • icosahedron: quadrilateral or pentagon

Step-by-step explanation:

<u>Hexahedron</u>

A hexahedron has 6 faces. A <em>regular</em> hexahedron is a cube. 3 square faces meet at each vertex.

If the hexahedron is not regular, depending on how those faces are arranged, a slice near a vertex may intersect 3, 4, or 5 faces. The first attachment shows 3- and 4-edges meeting at a vertex. If those two vertices were merged, then there would be 5 edges meeting at the vertex of the resulting pentagonal pyramid.

A slice near a vertex may create a triangle, quadrilateral, or pentagon.

<u>Icosahedron</u>

An icosahedron has 20 faces. The faces of a <em>regular</em> icosahedron are all equilateral triangles. 5 triangles meet at each vertex.

If the icosahedron is not regular, depending on how the faces are arranged, a slice near the vertex may intersect from 3 to 19 faces.

A slice near a vertex may create a polygon of 3 to 19 sides..

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30 points Someone help im very confused.
Alchen [17]

Answer:

y=cotx

Step-by-step explanation:

5 0
2 years ago
Find the coefficient of the x^3 in the expansion of (2x-9)^5
il63 [147K]

Use the binomial theorem:

\displaystyle (2x-9)^5 = \sum_{k=0}^5 \binom5k (2x)^{5-k}(-9)^k = \sum_{k=0}^5 \frac{5!}{k!(5-k)!} 2^5 \left(-\frac92\right)^k x^{5-k}

The <em>x</em> ³ terms occurs for 5 - <em>k</em> = 3, or <em>k</em> = 2, and its coefficient would be

\dfrac{5!}{2!(5-2)!} 2^5 \left(-\dfrac92\right)^2 = \boxed{6480}

8 0
3 years ago
A ball of radius 15 has a round hole of radius 5 drilled through its center. Find the volume of the resulting solid. Hint: The u
OverLord2011 [107]

Answer:

The volume of the ball with the drilled hole is:

\displaystyle\frac{8000\pi\sqrt{2}}{3}

Step-by-step explanation:

See attached a sketch of the region that is revolved about the y-axis to produce the upper half of the ball. Notice the function y is the equation of a circle centered at the origin with radius 15:

x^2+y^2=15^2\to y=\sqrt{225-x^2}

Then we set the integral for the volume by using shell method:

\displaystyle\int_5^{15}2\pi x\sqrt{225-x^2}dx

That can be solved by substitution:

u=225-x^2\to du=-2xdx

The limits of integration also change:

For x=5: u=225-5^2=200

For x=15: u=225-15^2=0

So the integral becomes:

\displaystyle -\int_{200}^{0}\pi \sqrt{u}du

If we flip the limits we also get rid of the minus in front, and writing the root as an exponent we get:

\displaystyle \int_{0}^{200}\pi u^{1/2}du

Then applying the basic rule we get:

\displaystyle\frac{2\pi}{3}u^{3/2}\Bigg|_0^{200}=\frac{2\pi(200\sqrt{200})}{3}=\frac{400\pi(10)\sqrt{2}}{3}=\frac{4000\pi\sqrt{2}}{3}

Since that is just half of the solid, we multiply by 2 to get the complete volume:

\displaystyle\frac{2\cdot4000\pi\sqrt{2}}{3}

=\displaystyle\frac{8000\pi\sqrt{2}}{3}

5 0
3 years ago
Help my on these two questions plss
Mariana [72]

7. Loan amount = $12000

Monthly payment = $380

Duration of the repayment = 3 years = 3(12) = 36 months.

Total amount Jason repaid = 36 × 380 = 13680

Interest on loan amount = amount repaid - loan amount

= 13680 - 12000

= 1680

Hence, total amount Jason paid in interest on loan = $1680 and the correct option is (D).


8. Loan amount = $35000

Monthly payment = $315

Duration of the repayment = 10 years = 10(12) = 120 months.

Total amount Gerald repaid = 120 × 315 = 37800

Interest on loan amount = amount repaid - loan amount

= 37800 - 35000

= 2800

Hence, total amount Gerald paid in interest on loan = $2800 and the correct option is (B).

3 0
3 years ago
I need help doing this.
Elanso [62]

a[n] = 7n - 16

The fifth term is a[5]

a[5] = 7(5) - 16 = 35 - 16  = 19

Answer: C. 19

3 0
3 years ago
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