Answer:
Please find attached the image of the quadrilateral TRAM after a rotation of -90 degrees, created with MS Excel
Step-by-step explanation:
The given coordinates of the vertices of the quadrilateral TRAM are;
T(-5, 1), R(-7, 7), A(-1, 7), M(-5, 4)
By a rotation of -90 degrees = Rotation of 90 degrees clockwise, we get;
The coordinates of the preimage before rotation = (x, y)
The coordinates of the image after rotation = (y, -x)
Therefore, we get for the the quadrilateral T'R'A'M', by rotating TRAM -90 degrees as follows;
T(-5, 1) → T'(1, 5)
R(-7, 7) → R'(7, 7)
A(-1, 7) → A'(7, 1)
M(-5, 4) → M'(4, 5)
The image of TRAM after -90 degrees rotation is created by plotting the derived points of the quadrilateral T'R'A'M' on MS Excel and joining the corresponding points as presented in the attached diagram.
Answer:
v ≈ (3.28512, 20.74146)
Step-by-step explanation:
v = 21(cos(81°), sin(81°))
v ≈ (3.28512, 20.74146) . . . . (x, y) components
The notation varies among authors. The vector can be written as ...
(r, θ) = (21, 81°)
r∠θ = 21∠81°
r cis θ = 21 cis 81°
v = 21·e^(i·9π/20)
(x, y) ≈ (3.28512, 20.74146)
v = 3.28512i +20.74146j . . . . perhaps this is the vector notation you want (i and j are unit vectors in the x- and y-directions, respectively)
I would go with the first one
Draw a diagram to illustrate the problem as shown in the figure below.
Euclid is placed at the origin at (0,0).
Apollonius is 12 m north and 9 m east of Euclid, so its coordinate is (9,12).
Pythagoras is at the arbitrary position (x,y) so that is is at distance d from Euclid and 2d from Apollonius.
From the distance formula, obtain
d² = x² + y² (1)
(2d)² = (x-9)² + (y-12)²
or
4d² = (x-9)² + (y-12)² (2)
Substitute (1) into (2).
4(x² + y²) = x² - 18x + 81 + y² - 24y + 144
3x² + 3y² + 18x + 24y = 225
Divide by 3.
x² + 6x + y² + 8y = 75
Create perfect squares.
(x+3)² - 9 + (y+4)² - 16 = 75
(x+3)² + (y+4)² = 10²
Answer:
The path of Pythagoras is a circle of radius 10 m, centered at (-3, -4).
