That's right and thanks for answering my question.
Math - 4th grade
It could be a line because a line segment.
Runner A is initially 6.0 mi west of a flagpole and is running with a constant velocity of 4.0 mi/h due east. Runner B is initially 4.0 mi east of the flagpole and is running with a constant velocity of 5.0 mi/h due west. How far are the runners from the flagpole when they meet?
Circle O is centered at the origin and has a radius of 5. Circle O' is the image of circle O after a translation of T1,2. What is the center of circle O'? What is the radius of circle O'?
What single rotation is equivalent to Rotate 40 degrees circle, Rotate 60 degrees circle, rotate -10 degrees
Use the chain rule d/dx[f(x)^(n)]= (n(f(x))^(n-1))*(f'(x)).
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.
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