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14 мая 2023 09:21
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y(0) = 1/2 cos 0 - 3 = 


y(π/6) = 1/2 cos π/6 - 3 =


y(π/3) = 1/2 cos π/3 - 3 =

 

y(π/2) = 1/2 cos π/2 - 3 =


y(2π/3) = 1/2 cos 2π/3 - 3 =

 

y(5π/6) = 1/2 cos 5π/6 - 3 =


y(π) = 1/2 cos π - 3 = 


y(7π/6) = 1/2 cos 7π/6 - 3 =


y(4π/3) = 1/2 cos 4π/3 - 3 = 


y(3π/2) = 1/2 cos 3π/2 - 3 =


y(5π/3) = 1/2 cos 5π/3 - 3 = 


y(11π/6) = 1/2 cos 11π/6 - 3 =


y(2π) = 1/2 cos 2π - 3 = 

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It looks like you have provided a set of initial conditions for a function y(x). Each initial condition specifies the value of y at a particular value of x. To fully determine the function y(x), we would need more information, such as an equation or formula that relates y and x. Without this additional information, we cannot say much more about the function y(x).
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14 мая 2023 09:27
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