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Rina8888 [55]
3 years ago
12

6. (09.02 MC)

Mathematics
1 answer:
Llana [10]3 years ago
5 0

Answer:

Hello,

Step-by-step explanation:

v_x(t)=ln(t-2)\ \Longrightarrow\ a_x(t)=\dfrac{1}{t-2} \\\\v_y(t)=e^{2t-t^2}\ \Longrightarrow\ a_y(t)=e^{2t-t^2}*(2-2t)\\\\t=3\ i\ suppose\ \\\\a_x(3)=\dfrac{1}{3-2}=1\\\\a_y(t)=e^{6-9}*(2-6)=-\dfrac{4}{e^3}

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An object dropped from a height of 600 feet has a height, h(t), in feet after t seconds have elapsed, such that h(t)=600 - 16t^2
gulaghasi [49]

Answer:

t as a function of height h is  t = √600 - h/16

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Step-by-step explanation:

Function for height is h(t) = 600 - 16t²

where t = time lapsed in seconds after an object is dropped from height of 600 feet

t  as a function of height h

replacing the function with variable h

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substituting h = 50 in the equation

t = √600 - 50/16

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