Recent content by lamerali

The base of an isosceles triangle is 20 cm and the altitude is increasing at the rate of 1 cm/min. At what rate is the base angle increasing when the area is 100 cm2? Hmmmm...I have the same question but i came up with a completely different answer...is this anywhere near correct? The...

is the rest of this equation anywhere near correct?? :(

Thank you sooo much! couldn't have done that without you! THANKS for alllll the help! :D

but if i plug in A = (1/2) b x h 100 = (1/2) (20) h h is still equal to 10 cm. Is that the only error you see? because that does not effect the rest of the equation...:S

The base of an isosceles triangle is 20 cm and the altitude is increasing at the rate of 1 cm/min. At what rate is the base angle increasing when the area is 100 cm2? I wasnt really sure where to start on this question so i tried my best at an answer. I'm sure I've gone wrong with this...

okey i believe i figured question 5 out: y = \frac{1}{1+tanx} y1 = (1 + tanx)^{-1} = (-1)(1 + tanx)(sec^{-2} sec ^{2} x = - \frac{(sec^2)x}{(1 + tanx)^2} I am still unsure where i am going with question 4 but here is my zillionth attempt :D y = tan^{2}(cos x) my answer...

also are question five and six okay? i don't see how i can come up with any other solutions.

alright...i'm not sure i got this one but here it goes: y = tan ^{2} (cos x) y1 = 2tan cosx + 2 sec^{2}cosx - sin tan^{2} how does it look? :|

great thank you! as for question 3 y = 2x( \sqrt{x} - cotx) my second attempt at an answer: y1 = 2( \sqrt{x} - cotx) + 2x(\frac{1}{2} x^{- \frac{1}{2}} + csc^{2} x = 2( \sqrt{x} - cotx) + \sqrt{x} + 2x csc^{2} x is this anywhere near correct? thanks i am not sure how to...

so would question 2 be... 4 cos^{2} (\pi x) 12 cos ^{2} (\pi x) (-sin (\pi x)) (\pi) (-sin (\pi x)) 12 \pi cos ^{2} (\pi x)

Hi, I'm working with finding the derivatives of trigonometric functions but I'm not confidant with some of my answers. if someone would go over these derivatives i would appreciate it. thanks in advance! determine \frac{dy}{dx} . do not simplify. question 1 y = sec \sqrt[3]{x} my...

Great! :D thank you tiny tim!

Thannnnnnnk you tiny tim! question 1, 2, and 3 are completely clear, i believe i came up with the correct answer. :D However I'm still unsure of question 4. Even if you plug in x = 0 into \frac{sin cosx}{sec x} where sec x is equal to \frac{1}{cos x} my answer comes to 0.84147, also if you...

Hello, i'm having some trouble with evaluating limits if anyone could help me out a bit i would appreciate it. thanks in advance evaluate the following limits: Question 1: lim x -> 0 \frac{2tan^{2}x}{x^{2}} my answer: u = x^{2} as x -> 0 u-> 0 = lim u->0...

Thank you :D