The outside of a picture frame measures 14 in and 20 in 196in^2 of the picture shows find the thickness of the frame
1 answer:
You might be interested in
Refer to the diagram shown below.
When x = 30 ft, the cable is at 15 ft, therefore y(30) = 15.
That is,
a(30 - h)² + k = 15 (1)
Also, because the distance between the supports is 90 ft, therefore
y(0) = 6 ft, and y(90) = 6 ft
That is,
a(-h)² + k = 6 (2)
a(90 - h)² + k = 6 (3)
From (2) and (3), obtain
a(90 - h)² = ah²
90² - 180h + h² = h²
180h = 90²
h = 45 ft.
From (1) and (2), obtain
225a + k = 15
2025a + k = 6
Therefore
1800a = -9
a = - 0.005
k = 15 - 225(-0.005) = 16.125 ft
Answer:
The equation for the cable is
y = - 0.005(x - 45)² + 16.125
A graph of the solution verifies that the solution is correct.
When x = 30 ft, the cable is at 15 ft, therefore y(30) = 15.
That is,
a(30 - h)² + k = 15 (1)
Also, because the distance between the supports is 90 ft, therefore
y(0) = 6 ft, and y(90) = 6 ft
That is,
a(-h)² + k = 6 (2)
a(90 - h)² + k = 6 (3)
From (2) and (3), obtain
a(90 - h)² = ah²
90² - 180h + h² = h²
180h = 90²
h = 45 ft.
From (1) and (2), obtain
225a + k = 15
2025a + k = 6
Therefore
1800a = -9
a = - 0.005
k = 15 - 225(-0.005) = 16.125 ft
Answer:
The equation for the cable is
y = - 0.005(x - 45)² + 16.125
A graph of the solution verifies that the solution is correct.