Respuesta :
The light must be installed 4 feet away from the west wall using slope-intercept form.
Assume the distance between you and the west wall is x. Let
y be equal to the ceiling height of this distance from the west wall.
y equals 8 when x equals 0 and when y equals 10 when x equals 3.
Equation across the slope of a straight line: y = mx + b.
m - slope, b - intersection with the y-axis.
Divide the change in y by the corresponding change in x to get m.
As a result, m equals (10.5-8) ÷ (3-0) = 2.5 ÷ 3 = 0.83
When x equals 0, b is the value of y.
As a result, b equals 8.
The equation of the line expressing the ceiling height is y = 2.5 ÷ 3 × x + 8.
when x = 0, y = 8
when x = 3
y = 2.5 ÷ 3 × 3 + 8
= 2.5 + 8
= 10.5
The equation accurately models the tilt of the ceiling as it runs from west to east.
The maximum height is 12 feet.
y = 2.5 ÷ 3 × x + 8
12 = 2.5 ÷ 3 × x + 8
take 8 off both sides,
4 = 2.5 ÷ 3 × x
(4 × 3) ÷ 2.5 = x
x = 4.8
The ceiling is 12 feet high, and the distance between the west wall and the ceiling is 4.8 feet.
That's the furthest you can reach from the west wall, given the maximum height of 12 feet.
x is obtained by dividing both sides by 2.5
= 10 ÷ 2.5
= 4 feet
Read more about slope intercept form at
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The question is -
A cathedral ceiling is 8 feet high at the west wall of a room. As you go from the west wall toward the east wall, the ceiling slants upward. Three feet from the west wall, the ceiling is 10.5 feet high.
You want to install a light in the ceiling as far away from the west wall as possible. You intend to change the bulb, when required, by standing near the top of your small stepladder. If you stand on the highest safe step of your stepladder, you can reach 12 feet high. How far from the west wall should you install the light?
