19. Tru Green charges an estimated fee of
1 answer:
You might be interested in
This problem will be solved both analytically and graphically.
The epicenter lies on a circle with radius=5 from X(3,3). Therefore
(x-3)² + (y-3)² = 25 (1)
Y(-13,-5) has the epicenter on a circle with radius=13, therefore
(x+13)² + (y+5)² = 169 (2)
Z (-5,3) has the epicenter on a circle with radius = 5, therefore
(x+5)² + (y-3)² = 25 (3)
Subtract (1) from (3).
(x+5)² - (x-3)² = 0
x² + 10x + 25 - x² + 6x - 9 = 0
16x = -16
x = -1
From (1), obtain
16 + (y-3)² = 25
(y-3)² = 9
y = 0 or y = 6
Check answers with (2).
When x=-1, y=0. obtain
(x+13)² + (y+5)² = 169 (Correct, Accept)
When x=-1, y=6, obtain
(x+13)² + (y+5)² = 265 (Incorrect, Reject)
The epicenter is at (-1,0).
Graphical solution (see the figure below).
From X (3,3), draw a circle with radius =5.
From Z (-5, 3), draw a circle with radius = 5.
The only point where all three circles intersect is (-1, 0) approximately.
Answer: The epicenter is at (-1,0)
The epicenter lies on a circle with radius=5 from X(3,3). Therefore
(x-3)² + (y-3)² = 25 (1)
Y(-13,-5) has the epicenter on a circle with radius=13, therefore
(x+13)² + (y+5)² = 169 (2)
Z (-5,3) has the epicenter on a circle with radius = 5, therefore
(x+5)² + (y-3)² = 25 (3)
Subtract (1) from (3).
(x+5)² - (x-3)² = 0
x² + 10x + 25 - x² + 6x - 9 = 0
16x = -16
x = -1
From (1), obtain
16 + (y-3)² = 25
(y-3)² = 9
y = 0 or y = 6
Check answers with (2).
When x=-1, y=0. obtain
(x+13)² + (y+5)² = 169 (Correct, Accept)
When x=-1, y=6, obtain
(x+13)² + (y+5)² = 265 (Incorrect, Reject)
The epicenter is at (-1,0).
Graphical solution (see the figure below).
From X (3,3), draw a circle with radius =5.
From Z (-5, 3), draw a circle with radius = 5.
The only point where all three circles intersect is (-1, 0) approximately.
Answer: The epicenter is at (-1,0)