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2 answers:
Answer:
Option A. His solution is accurate.
Step-by-step explanation:
Charles factors the expression
Let's factorize the given expression
We will take out common from the given expression
After analyzing Charles process of factorizing we can conclude that his solution is accurate.
Option A. is the solution.
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Answer:
The boat is approaching the dock at a speed of 3.20 ft/s when it is 4 feet from the dock.
Step-by-step explanation:
The diagram of the situation described is shown in the attached image.
The distance of the boat to the dock along the water level at any time is x
The distance from the person on the dock to the boat at any time is y
The height of the dock is 5 ft.
These 3 dimensions form a right angle triangle at any time with y being the hypotenuse side.
According to Pythagoras' theorem
y² = x² + 5²
y² = x² + 25
(d/dt) y² = (d/dt) (x² + 5²)
2y (dy/dt) = 2x (dx/dt) + 0
2y (dy/dt) = 2x (dx/dt)
When the boat is 4 ft from dock, that is x = 4 ft,
The boat is being pulled at a speed of 2 ft/s, that is, (dy/dt) = 2 ft/s
The speed with which the boat is approaching the dock = (dx/dt)
Since we are asked to find the speed with which the boat is approaching the dock when the boat is 4 ft from the dock
When the boat is 4 ft from the dock, x = 4 ft.
And we can obtain y at that point.
y² = x² + 5²
y² = 4² + 5² = 16 + 25 = 41
y = 6.40 ft.
So, to the differential equation relation
2y (dy/dt) = 2x (dx/dt)
when x = 4 ft,
y = 6.40 ft
(dy/dt) = 2 ft/s
(dx/dt) = ?
2 × 6.40 × 2 = 2 × 4 × (dx/dt)
25.6 = 8 (dx/dt)
(dx/dt) = (25.6/8) = 3.20 ft/s.
Hope this Helps!!!
