Vertex: 0,4.92
focus: 0,4.8575
axis of symmetry: x = 0
directrix: y = 4.9825
Answer:
16 ?
Step-by-step explanation:
Answer:
146.41
Step-by-step explanation:
third order determinant = determinant of 3×3 matrix A
given ∣A∣=11
det (cofactor matrix of A) =set (transpare of cofactor amtrix of A) (transpare does not change the det)
=det(adjacent of A)
{det (cofactor matrix of A)} ^2 = {det (adjacent of A)}
^2
(Using for an n×n det (cofactor matrix of A)=det (A)^n−1
)
we get
det (cofactor matrix of A)^2 = {det(A) ^3−1
}^2
=(11)^2×2 = 11^4
=146.41
Given the figure of a regular pyramid
The base of the pyramid is a hexagon with a side length = 6
The lateral area is 6 times the area of one of the side triangles
So, the side triangle has a base = 6
The height will be:
![\begin{gathered} h^2=6^2+(\frac{\sqrt[]{3}}{2}\cdot6)^2=36+27=63 \\ h=\sqrt[]{63} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20h%5E2%3D6%5E2%2B%28%5Cfrac%7B%5Csqrt%5B%5D%7B3%7D%7D%7B2%7D%5Ccdot6%29%5E2%3D36%2B27%3D63%20%5C%5C%20h%3D%5Csqrt%5B%5D%7B63%7D%20%5Cend%7Bgathered%7D)
so, the lateral area =
![6\cdot\frac{1}{2}\cdot6\cdot\sqrt[]{63}=18\sqrt[]{63}](https://tex.z-dn.net/?f=6%5Ccdot%5Cfrac%7B1%7D%7B2%7D%5Ccdot6%5Ccdot%5Csqrt%5B%5D%7B63%7D%3D18%5Csqrt%5B%5D%7B63%7D)
