hscishard
Active Member
It took me ages because I failed to realise that tan 135 is -1 -.- and the calculator was going against me, it was on radians.lol no fair, you've done all this before!
New Question:
Tan B = -1
It took me ages because I failed to realise that tan 135 is -1 -.- and the calculator was going against me, it was on radians.lol no fair, you've done all this before!
New Question:
I really should know the answer to these questions before posting...Tan B = -1
Lol now I got tan B = 5.I really should know the answer to these questions before posting...
I got tanB = -5/2
The sheet I got this from didn't have answers, can someone else check?
Any value of "a", the answer is 3a+2 ......??Hi, I was just wondering if this solution is entirely correct.
Doesn't this imply that for any value for 'a' that the answer is '3a+2'? However doesn't this violate the first condition as a^2=a+2 only has two solutions?
I guess that was like a year 9 paper?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 , hehe .
Very nice.
Wow! Very nice Drongoski. Very creative how you used logs.
I didn't see that; but logs & indices(exponents) are closely linked. What you can do via logs, you should be able to do via corresponding exponential equations. I have not tried that. Doing via logs was already quite a challenge, for me, anyway.Very nice.
I like the change of base part.
But he said you didn't need to use log, didn't he?
Well, that's the beauty of maths . There are many ways of retrieving an answer.Very nice.
I like the change of base part.
But he said you didn't need to use log, didn't he?
Makes sense. Will never be able to think like that in an exam.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 + 1/y = (x+y)/xy
Therefore: z = xy/(x+y) #
That's why past papers are the key to success .Makes sense. Will never be able to think like that in an exam.
So true, a question from a year 9 one I did required Induction, WTF!I guess that was like a year 9 paper?
Those compeitions are soo hard
Haha , it was a senior paper -> ie Years 11 and 12.So true, a question from a year 9 one I did required Induction, WTF!
Wow! Although quite good at indices, I'd not have thought this one out myself. It's so elegant.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 + 1/y = (x+y)/xy
Therefore: z = xy/(x+y) #
Lol, I posted that in the old thread.More Questions !!!
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
For 1:More Questions !!!
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!
Yea..why did you repear so many times?For 1:
Hmmm...On looking at this I don't think it makes sense :S
For Q2.More Questions !!!
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!