Recent content by ve3rfd

This is the question A 25 ft long ladder is resting against a vertical wall and the bottom of the ladder is moving out at a fixed rate of 2.00 ft per second. Graph x and y as functions of time. and I did x which is easy and the y = mx + b was y = 1/2x + 0 I just need some help with y?

I am having trouble getting the slope? Because the graph is a curve and the y value decreases as time increases.

I am having trouble figureing out the y = mx + b of this if the table of values look like Distance Time (s) 25.000 0.000 24.920 1.000 24.678 2.000 24.269 3.000 23.685 4.000 22.913 5.000 21.932 6.000 20.712 7.000 19.209 8.000 17.349 9.000 15.000 10.000 11.874 11.000 7.000 12.000...

Hi, I had a question that wanted me to find the instantaneous rate of change of V with respect to h for h = 0.60cm. The equation was V = 1/6*pi*h^3 + 2.00*pi*h and I retrieved an answer of 6.85 cm/s. Could anyone tell me if that is correct? Thanks.

Okay, Thanks a lot i didn't get the d^3.

So would the new equation be 1/3 4 - d * 4 - d (pi r^2)?

Thats the thing i am having trouble with, starting the equation? I know I need the Volume for a cylinder equation but how do I use it to figure it out?

I am having trouble with this problem. Any help how to start it would be appreciated? http://img297.imageshack.us/img297/1382/untitledsk9.jpg