Asked by Chris
The rate of performance is given by dN/dt=1/(2sqroot(t+1)) where N is the number of unity completed t hours after beginning a new tak.
If 2 units are completed afer 3 hours, how many units are completed after 8 hours?
N= INT 1/2 (t+1)^(-1/2)
I will be happy to critique your thinking. Solve for the constant of integration, and you have it.
I really don't understand the problem for some reason, but my thinking thus far is INT(t+1)^(-1/2)
1/2 INTu^(-1/2)du
1/2((u^(1/2)/(1/2))du
u^(1/2)du
(t+1)^(1/2)
Anything you can do to help would be great!
That is it. YOu can check it by taking the derivative. DO NOT forget the constant of integration, I think you need it.
If 2 units are completed afer 3 hours, how many units are completed after 8 hours?
N= INT 1/2 (t+1)^(-1/2)
I will be happy to critique your thinking. Solve for the constant of integration, and you have it.
I really don't understand the problem for some reason, but my thinking thus far is INT(t+1)^(-1/2)
1/2 INTu^(-1/2)du
1/2((u^(1/2)/(1/2))du
u^(1/2)du
(t+1)^(1/2)
Anything you can do to help would be great!
That is it. YOu can check it by taking the derivative. DO NOT forget the constant of integration, I think you need it.
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