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Can someone please do this step by step. Giving explanation as well. Thanks in advanced.

Hi there (:. Well part (a) should be pretty straight forward. Its a rectangular hyperbola, in quadrants 1 and 3, due to it being positive. Now, the question asks for the hyperbola to be drawn from 1-->2. Therefore, we are only taking the first quadrant into consideration. Also, we should draw...

Someone please do this.

But they have to satisfy both equations. Also, an angle of 0, is not an acute angle ;)

More Questions :D !!! 1) If 2^(2p).3^(p) = 12^(x), find p in terms of x and hence find p such that 2^(2p).3^(p)=144 2) If sinA = 2sinB and tanA = 3tanB, find A and B to the nearest degree, given that A and B are acute angles Have fun!

Haha :), it was a senior paper -> ie Years 11 and 12.

That's why past papers are the key to success :D. You've seen the question now. If you see a similar question in an exam, you've increased your chances to solve it by at least 60%.

Well, that's the beauty of maths (: . There are many ways of retrieving an answer. Look above for an alternative solution.

Wow! Very nice Drongoski. Very creative how you used logs. But here is an alternative solution: Let 2^x = 5^y = 10^z = A Then: 2 = A^(1/x) , 5 = A^(1/y) , 10 = A(1/z) Now, 10 = 2 x 5. .'. A(1/z) = A^(1/x) x A^(1/y) = A^(1/x + 1/y) Now, all the "front values" are the same. So: 1/z = 1/x +...

Bahahah! Funny picture. However, I think Stephen Hawking does believe in a GOD, but not a PERSONAL GOD. I have his book right here next to me: "God Created The Integers - Stephen Hawking"-.

Any value of "a", the answer is 3a+2 ......?? a^3 = 3a+2 my friend, not a. I got this question from The Australian Maths Competition :D, hehe (: .

Point taken.

HAHAHAH! OMGGGGGG! thats soo funny. i remember looking at the screen and saying...wait..why are my sin and cos the other way around. Then i just switched em around. Good Q. >.< I would so lose a mark there.

Do my question: If 2^x = 5^y = 10^z, show that z = xy/x+y

Looks to easy =/... Um....If we express it in mod arg form we get --> 1cis(pi/3). Now..according to De Moivre [1cis([i/3)]^8 = 1^8cis(8pi/3). But this doesn't lie number the principle argument. So if we play around..it becomes --> 1^8cis(5pi/3). This furthermore gives: cos5pi/3+isin5pi/3. We...

Yes. Something like that (:. If it even saves me 1 minute. I'll take it.

=sin(-)/1+cos(-) + 1+cos(-)/sin(-) =sin(-)^2 + (1+2cos(-)+cos(-)^2)/sin(-)(1+cos(-)) =sin(-)^2+1+2cos(-)+(1-sin^2(-))/sin(-)(1+cos(-)) =2+2cos(-)/sin(-)(1+cos(-)) =2(1+cos(-))/sin(-)(1+cos(-)) =2/sin(-) =2cosec(-)

A clue for the second question. NOTE: Does not require logs. It's just a harder application of your normal algebra. If 2^x = 5^y = 10^z, show that z = xy/x+y 2x5 = 10. Hopefully that gets you somewhere.

Well done. :) . Question was testing more than just knowledge. 3a+2 is in linear form. The highest degree of a, is one.