Recent content by Thylacine

I can see how I got an answer by using the wrong method (1/cos(x))^2 . Thanks for the help. I should have looked at a graph of the curve before I started. MIF has wasted so much of my time with wrong answers.

Find the volume of the solid formed if the curve y = sec^-1(x) is rotated about the x-axis from x = 0, 0.5, using the trapezoidal rule with 5 subintervals. MIF challenge exercise 7 Q10 (a Thanks for any help. I get a different answer to BOB.

Thanks for the feedback. The question comes from Maths in Focus, HSC course, Book 2, 2003 by Margaret Grove p 475, Challenge exercise 12. Q10 (c) I guess the answer in the back of the book is wrong. Grrrr It's so frustrating when you spend ages on a question, because the editors of a textbook...

A squad of 8 is chosen at random from 3 baseball teams with 10 players in each team. Find the probability that Joe from the B team and Dan from the A team will be chosen. My answer: (28C6 . 2C2)/ 30C8 = 28/435...

Thanks just have to work out theta.

A triangle ABC is right-angled at C. Angle BAC = theta. D is the point on AB such that DC is perpendicular to AB. i) You are given that 2AD + 7BC = 5AB Simplify: 2cos theta + 7tan theta ii) Hence find theta

rumblroar your brilliant, I substituted the wrong values. I've just put the "y's" in the right place and it fell into place. A bit of dancing round the room a few whoops. Thanks

Expand tan(x+2y) I get this far (tanx+2tanx/(1-〖tan〗^2 x))/(1-tanx 2tany/(1-〖tan〗^2 y)) but then I can't get to the answer in BOB. Please help!

Can anyone help me with this. The flow rate of water out of a hole in a tank is known to be proportional to the squareroot of the height of water above the hole. That is, dV/dt = sq root(h) Suppose the tank has constant cross-sectional area A, show that the height of the water in the...

Can someone help me with this. The flow rate of water out of a hole in a tank is known to be proportional to the squareroot of the height of water above the hole. That is, dV/dt = sq root(h) Suppose the tank has constant cross-sectional area A, show that the height of water in the tank is...

Yep that makes sense. Thanks

I'm not comparing the bases ie (-2 and 2). I'm looking at the exponents ie x and 4 then 1 and 0

l og3 81 = x then 3x = 81 3x = 34 therefore x = 4 11= 1 10 = 1 11 = 10 1 = 0 Why is this wrong / right ????

The velocity of a particle is given by v=100 + 280e^-kt ms^-1 . If the velocity decreases by 20% after 50 seconds , find the % decrease in velocity after 3 minutes.

Thankyou both for your help. I realise now I was treating the x - x^2 as though it was in brackets. ie loge(x-x^2). This caused major problems when trying to find the second derivative. Thanks again.