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A tractor is moving at a rate of 12 ft/s away from a building which is 90 ft high. How fast is the distance between the tractor and the top of the building increasing when the tractor is 120 ft away from the base of the building

Respuesta :

elliottou

Answer:

53.66 ft/s

Step-by-step explanation:

c = [tex]\sqrt{a^2+b^2}[/tex]

c' = [tex]\sqrt{2aa' + 2bb'}[/tex]

c' = [tex]\sqrt{2(12)(120)+2(90)(0)}[/tex]

the height of the building does not change, so therefore b' is 0

c' = 53.66 ft/s

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