Check the picture below, notice those three quadrilaterals, bearing in mind that a square and rhombus both have equal sides all around.
How do I do question 11? Please help me thank you
1 answer:
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All the points that are 6 units from (-1, 1) are those on the circle
(x+1)^2 +(y-1)^2 = 36
For y=0, the two points of interest satisfy
(x+1)^2 +1 = 36
(x+1)^2 = 35 . . . . . . subtract 1
x+1 = ±√35
x = -1±√35
The points you seek are (-1-√35, 0) and (-1+√35, 0), about (-6.916, 0) and (4.916, 0).
(x+1)^2 +(y-1)^2 = 36
For y=0, the two points of interest satisfy
(x+1)^2 +1 = 36
(x+1)^2 = 35 . . . . . . subtract 1
x+1 = ±√35
x = -1±√35
The points you seek are (-1-√35, 0) and (-1+√35, 0), about (-6.916, 0) and (4.916, 0).
Answer:
23
Step-by-step explanation:
Here is the complete question
Find the volume of the parallelepiped with one vertex at the origin and adjacent vertices at (1, 0, -3), (1, 2, 4), and (5, 1, 0).
Solution
We find the volume of the parallelepiped by making a 3 × 3 column matrix whose columns are the corresponding coordinates of the vertices of the parallelepiped.
So, (1, 0, -3), (1, 2, 4) and (5, 1, 0)
The determinant of A is the volume of the parallelepiped. So,
detA = 1(2 × 0 - 4 × 1) - 1(0 × 0 - (-3) × 1) + 5(0 × 4 - (-3) × 2)
= 1(0 - 4) - 1(0 + 3) + 5(0 + 6)
= 1(-4) - 1(3) + 5(6)
= -4 - 3 + 30
= 23
So the volume of the parallelepiped is 23