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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.

omg omg tell me how. lol. i wanted to see working out, so i could kno how to do it. explain plzzzzzzzzzzzzzzz. + how u got the centre. they're both right btw.

Lets see your working out? [=

The values you have given me, EXACTLY MATCH: y=10sec(theta/2) So, you are looking at the wrong question (:. However, make sure you understand the concepts that I explained in my previous posts (y)