Find the volume of the parallelepiped with one vertex at the origin and adjacent vertices at (1, 4, 0), (-2, -5, 2), and (-2, 2, 1).

17 cubic units.

The volume (V) of a parallelepiped with one vertex at the origin is given by the absolute value of the scalar vector product of the vectors at the adjacent vertices.

V = |(a x b) · c|

In this case,

a = (1,4,0) = i + 4k + 0j

b = (-2,-5,2) = -2i -5j + 2k

c = (-2,2,1) = -2i + 2j + k

First, let's calculate the cross product of vectors a and b. i.e a x b as follows:

(i) Arrange the vectors in a matrix form

a x b = | i        j         k |

          | 1       4        0 |

          | -2    -5        2 |

(ii) Calculate the determinant of the matrix

a x b = i(8 - 0) -j(2-0) + k(-5+8)

a x b = i(8) -j(2) + k(3)

a x b = 8i - 2j + 3k

Secondly, calculate the scalar product of the cross product found above and the vector c as follows;

(a x b) . c = (8i - 2j + 3k) · (-2i + 2j + k)

Multiply like terms

(a x b) . c = (8 * -2)(i.i) + (-2 * 2)(j.j) + (3 * 1)(k.k)             [i.i = j.j = k.k = 1]

(a x b) . c = (-16) + (-4) + (3)

(a x b) . c = -16 - 4 + 3

(a x b) . c = -17

Thirdly, find the absolute value of the result found above. i.e

|(a x b) . c| = |-17|

|(a x b) . c| = 17

Therefore, the volume of the parallelepiped is 17 cubic units.

option a.

step-by-step explanation:

if you look closely at the graph, we can notice that the function is not defined at two different points.

if you look at functions b and c, you will notice that both functions are not defined at just one point. the function b is not defined at x= -5/2 and function c is not defined at x=-5. both options are discarded, given that our function should not be defined at two fiferent points.

now, if you look closely at the graph you will notice that the function sketched is not defined at x=5 and x=-2.

the first one, all those points on the line just make a strait line.