Recent content by toosm:)ey

Okay, that makes sense and therefore: \lim_{h\rightarrow0^+} \frac{1}{h}} (\tan(h+(\pi/6))}- \frac{1}{\sqrt3}}) equals \lim_{h\rightarrow0^+} \frac{1}{h}} (\tan(h+(\pi/6))}- \tan(\frac{\pi}{6}})) \frac{d}{dx}}\tan(x) = \sec^2(x) \sec^2(\frac{\pi}{6}}) = \frac{4}{3}} and...

\lim_{h\rightarrow0} \frac{1}{h}} (\frac{sin(h+(\pi/6))}{cos(h+(\pi/6))}} - \frac{1}{\sqrt3}}) \lim_{h\rightarrow0} \frac{1}{h}} (\frac{sin(x)cos(\pi/6) + cos(x)sin(\pi/6)}{cos(x)cos(\pi/6) + sin(x)sin(\pi/6)}} - \frac{1}{\sqrt3}}) \lim_{h\rightarrow0} \frac{1}{h}} (\frac{sin(x)(\sqrt3/2)...

Sure, but how do I use the code thing? Or is there an easy way to type out the work I have?

problem number 2

Hello everyone! I have a calculus midterm next week and I need some help solving old midterm questions. I'm a first year calculus student so they're not hard, but I can't figure them out! I am told to solve without using the tan addition formula.

Remember that P=IV You have the total voltage, find the total resistsance. Then find how much voltage is used in R3.

I am going for 23 times. :) It crosses at 12am, but 24 hours later, that is the next day, so it does not count.

Brain teasers?!? These are just cruel!

Thanks again =) 1) Good thinking... Saves a lot of time doing it your way. I never liked the chain rule anyway =) 2) My answer is not changing. Are you getting the same values for x and y, that is, the rectangle is a square? And are you doing the same calculations? Also when calculating...

Hello and thanks again cookie! The Master of Calculus =D I've completed questions 1 and 2, could you check those for errors please? Number 3 I am still stuck on. 1) If P=8x, then dP/dt=8dx/dt. As dp/dt=12, then dx/dt = 3/2 ? and if 2x+2y=p, then 2dx/dt + 2dy/dt = dp/dt...

I have been given an assignment by my teacher, only I can't answer 3 of the questions. 1) A rectangle is expanding so that its length is always three times its width. The perimetre of the rectangle is increasing at a rate of 12cm/minute. Find the rate of increase in area of the rectangle when...

I understand... you're right However, because of the lack of time, I don't have time to do all those calculations... :frown: My website is at http://ca.geocities.com/physicsroxmysox/ , I am sure you can find lots of mistakes in my work, because you're all really intelligent at Physics...

But I do not know how much energy is Elastic energy because I do not know the ropes k value.

I am assuming that a is the acceleration, v is velocity, and t is time. Without the acceleration or time, I can't find the velocity. I am in High School Calculus. We have learned how to differentiate equations, but not those with so many variables. Once the rope begins the elongate...

Using mass = 75kg height, before = 200m, after = 140m Energy before = Energy After Potential + Kinetic = Potential + Kinetic mgh + 0.5mv^2 = mgh + 0.5mv^2 (75)(9.81)(200) +(0.5)(75)(0^2) = (75)(9.81)(140) + (0.5)(75)(v^2) 147150J + 0 = 103005J + 37.5v^2 44145J = 37.5v^2 v^2 =...