Note that f'(x) can be rewritten algebraically a f'(x)= x -(1/x). Now, take the integral (anti-drivitive) of f'(x) with respect to 'x' to obtain f(x). So, Int(x-(1/x),x)=((x^(2))/2)-ln(abs(x))+C To understand this you must first look at the rules of integra...
You are almost right but since u=-t^(2) the limits are [0,-25] then put them into e^(u) and you get, e^(-25)-e^(0)= e^(-25)-1 Then divide that answer by 5 thus the answer is, (e^(-25)-1)/5
sqrt(-16)=sqrt(-1)sqrt(16)=i sqrt(16)=4i sqrt(-25)=sqrt(-1)sqrt(25)=i sqrt(25)=5i I think you can see it now. Just remember i=sqrt(-1). And don't forget your properties of sq-roots.
Plug in dp/dt =-14 kpa/min because it is decreasing at a rate. So solving for dv/dt= (1-(V^(1/4)dP/dt))/((1/4)PV^(-3/4)). Your answer so be dv/dt= (1-((340)^(1/4)(-14))/((1/4)(87)(340)^(-3/4)).
(square root 5) times (square root 12) x > 0, (square root x) times (square root 3x)
what impact did the Reformation have on Western society
I need a fable to copy.