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4 years ago

Name the edges of an hexagonal pyramid

Mathematics

1 answer:

aleksandrvk [35]4 years ago

4 0

Given:

Hexagonal pyramid

To find:

The edges of an hexagonal pyramid.

Solution:

Edges means lines which connecting to vertices.

Edges in the base of the pyramid:

AB, BC, CD, DE, EF, FA

Edges in the triangular shape of the pyramid:

AG, BG, CG, DG, EG, FG

Therefore edges of an hexagonal pyramid are:

AB, BC, CD, DE, EF, FA, AG, BG, CG, DG, EG, FG

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The position function of a particle in rectilinear motion is given by s(t) = 2t^3 - 21t^2 + 60t + 3 for t ≥ 0. Find the position

avanturin [10]

Your first and second derivatives which are the velocity and the acceleration respectively are correct.
The particle has reverses direction when v(t) ≤ 0 so at interval 2 < t < 5 sec.
You can substitute now t = 2 for s(2) and a(2) for first changing direction. Also t = 5 for s(5) and a(5) for second changing direction.

6 0

4 years ago

Compute the determinant using a cofactor expansion across the first row. Also compute the determinant by a cofactor expansion do

V125BC [204]

Answer:

Step-by-step explanation:

It is given that

$\Delta=\begin{vmatrix}3&0&3\\2 &3&3\\0 &4&-2\end{vmatrix}$

By cofactor expansion across the first row, we get

$\Delta=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}$

$\Delta=3\left[(-1)^{1+1}\begin{vmatrix}3&3\\4&-2\end{vmatrix}\right]+0\left[(-1)^{1+2}\begin{vmatrix}2&3\\0&-2\end{vmatrix}\right]+3\left[(-1)^{1+3}\begin{vmatrix}2&3\\0&4\end{vmatrix}\right]$

$\Delta=3\left[-18\right]+0\left[(-1)(-4)\right]+3\left[8\right]$

$\Delta=-54+0+24$

$\Delta=-30$

Therefore, the value of determinant is -30.

By cofactor expansion across the second column, we get

$\Delta=a_{12}C_{12}+a_{22}C_{22}+a_{32}C_{32}$

$\Delta=0\left[(-1)^{2+1}\begin{vmatrix}2&3\\0&-2\end{vmatrix}\right]+3\left[(-1)^{2+2}\begin{vmatrix}3&3\\0&-2\end{vmatrix}\right]+4\left[(-1)^{3+2}\begin{vmatrix}3&3\\2&3\end{vmatrix}\right]$

$\Delta=0\left[(-1)(-4)\right]+3\left[(-6)\right]+4\left[(-1)3\right]$

$\Delta=-18-12$

$\Delta=-30$

Therefore, the value of determinant is -30.

7 0

4 years ago

Daredevil Danny was cliff jumping. He jumped off a cliff that was 1 4 . 2 14.214, point, 2 meters above the surface of the water

iragen [17]

Question:

Daredevil Danny was cliff jumping. He jumped off a cliff that was 14.2 meters above the surface of the water. At the end of his jump, Danny was 2.3 meters below the surface of the water.

The following equation describes this situation 14.2 + (-16,5) = -2.3 .

What does -16.5 tell us?

Answer:

The displacement between his current position and his initial position

Step-by-step explanation:

Given

$Initial\ Position = 14.2m$