Step One
Find the base area of the large hexagon as though the smaller one was not removed.
Area = 3*Sqrt(3) * a^2 /2 where a is the length of one side of the hexagon
a = 5
Area = 3*sqrt(3) * 25/2 = 75 sqrt(3) / 2 of the large hexagon without the smaller one removed.
Step Two
Find the area of the smaller hexagon. In this case a = 4
Area2 = 3*sqrt(3)*16/2 = 3*sqrt(3)*8 = 24 sqrt(3)
Step Three
Find the area of the thick hexagonal area left by the removal of the small hexagon.
Area of the remaining piece = area of large hexagon - area of the small hexagon
Area of the remaining piece = 75 *sqrt(3)/2 - 24*sqrt(3)
Step Four
Find the volume of the results of the area from step 3
Volume = Area * h
h = 18
Volume = (75 * sqrt(3)/2 - 24*sqrt(3))* 18
I'm going to leave you with the job of changing all of this to a decimal answer. I get about 420 cm^3
3.14 * 4 = 12.56 ( area of poster)
16 (area of square) - 12.56 = 3.44 (remaining area)
Answer:
Step-by-step explanation:
Use photo math
Answer:
True
True
False
Step-by-step explanation:
TRUE
If the equation Ax = 0 has only the trivial solution, then A is row equivalent to the n × n identity matrix
Here's why
If the equation Ax = 0 has only the trivial solution the determinant of the matrix is NOT 0 and the matrix is invertible therefore it is row equivalent to the nxn identity matrix.
TRUE
If the columns of A span ℝ^n , then the columns are linearly independent
Here's why
Remember that the rank nullity theorem states that
![\text{rank}(A) + \text{Nullity}(A) = \text{Dim}(V)](https://tex.z-dn.net/?f=%5Ctext%7Brank%7D%28A%29%20%2B%20%5Ctext%7BNullity%7D%28A%29%20%3D%20%5Ctext%7BDim%7D%28V%29)
According to the information given we know that
![\text{rank}(A) = n \\dim(V) = n \\](https://tex.z-dn.net/?f=%5Ctext%7Brank%7D%28A%29%20%3D%20n%20%5C%5Cdim%28V%29%20%3D%20n%20%5C%5C)
Therefore you have
![n + \text{Nullity}(A) = n](https://tex.z-dn.net/?f=n%20%2B%20%5Ctext%7BNullity%7D%28A%29%20%3D%20n)
and
![\text{Nullity}(A) = 0](https://tex.z-dn.net/?f=%5Ctext%7BNullity%7D%28A%29%20%3D%200)
Which is equivalent to the problem we just solved.
FALSE
If A is an n × n matrix, then the equation Ax = b has at least one solution for each b in ℝ^n
Here's why
Take b as a non null vector and A=0, then Ax = 0 and Ax=b will have no solution.
Answer:
The answer is the Last Graph
D
Step-by-step explanation: