Recent content by DeanBH

so i have y = 1/sqroot(3x+2) find volume when rotated around x, regions are x=2 and x=0 equation needed: V= integral Pi*y^2*dx so. i do intergral pi* (1/sqroot(3x+2))^2 * dx so i get pi integral 1/(3x+2) dx so how do i integrate 1/sqroot(3x+2) ? can someone take me...

i don't think you understand what I'm confused with, i know the answer is 2.34 so i put it into check xN+1 = cuberoot(17.5-2xn) 3.34=cuberoot(17.5-2*2.34) 3.34=2.34 do you understand what i don't understand.

Homework Statement First of all this is revision not homework =) question is Xn+1 = Cuberoot ( 17.5-2xn) answer lies between 2 and 3. i know the answer is 2.34 but what i don't get is why it is xn+1 = equation, because when you put xn=2.34 into the equation you get 2.34 out...

it's more of a dancing peanut butter jelly banana type smile.

in every element energy is used up to link the protons/neutrons together etc. I am looking for a table that shows how the energy required increases as the number of protons/neutrons increases in the nucleus of the elements. i've searched Google quite a while a can't seem to come across...

how much did the lab charge

Is this that schools complaint board?

my supremacy over chess-land came to an abrupt end =(

nvm... its constant

i know that pCosX = 6 and 0.2(pSinX+25)= 6 pSinx = 5 pCosx = 6 how do i solve for X.. I'm not sure how this works

lectures, and hard open questions that gets people involved and thinking IMO.

R = V/I just follow the graph across at 15 volts and find the Watts. then the volts by the watts.

guessed, that the volts in the battery would have to make up for the difference. but why is the amps 0.06 out of 0.032, if the volts were 5.2 instead of 4.2 in the right part of the circuit it wouldn't fit the rule RI=v

no idea

Why do i have to count one of those volt readings as negative, and why inst it 1/r it looks like a a parallel