The ladder, the ground and the wall form a right triangle; the ladder length (L) is the longest side of this triangle. L^2 = h^2 + x^2, where h represents the height of the point on the wall where the ladder touches the wall, and x represents the distance of the base of the ladder from the wall.
We need dh/dt, which will be negative because the top of the ladder is sliding down the wall.
Starting with h^2 + x^2 = L^2, we differentiate (and subst. known values such as x = 5 feet and 4 ft/sec to find dh/dt. Note that since the ladder length does not change, dL/dt = 0. This leaves us with
dh dx
2h ---- + 2x ----- = 0.
dt dt
Since x^2 + h^2 = 15^2 = 225, h^2 = 225 - (5 ft)^2 = 200, or
200 ft^2 = h^2. Then h = + sqrt(200 ft^2)
Substituting this into the differential equation, above:
2[sqrt(200)] (dh/dt) + 2 (5) (4 ft/sec) = 0. Solve this for the desired quantity, dh/dt:
[sqrt(200)] (dh/dt) + (5)(4) = 0, or
dh/dt = -20 / sqrt(200) = (-1.41 ft / sec) (answer)
This result is negative because the top of the ladder is moving downward.
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Answer:
Part 1) The lateral area is 
Part 2) The area of the two bases together is 
Part 3) The surface area is 
Step-by-step explanation:
Part 1) Find the lateral area
we know that
The lateral area of a prism is given by

where
P is the perimeter of the base
h is the height of the prism
so
----> perimeter of rectangle
---> given
substitute

Part 2) The area of the two bases together is?
The area of the rectangular base B is equal to

therefore
The area of the two bases together is

Part 3) The total surface area is
we know that
The surface area is equal to

substitute

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Find slope :
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When the slope is 0, the graph is a horizontal line.
Equation of the graph is y = -4
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Answer: y = -4------------------------------------------------------
Answer:
Triangle A: acute
Triangle B: acute
Triangle C: obtuse
Triangle D: right
Step-by-step explanation:
acute= less than 90 degrees each angle
obtuse=more than 90 degrees for one angle
right=equal exactly 90 degrees for one angle