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
Preliminary mathematics marathon (1 Viewer)
- Thread starter ohexploitable
- Start date
Tan B = -1
ohexploitable
hey
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?
hscishard
Active Member
Gussy Booo
Mathematics <3
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 .hscishard
Active Member
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
Those compeitions are soo hard
hscishard
Active Member
Very nice.}\ \ =\ \ \frac{y}{log_{5}5\ +\ log_{5}2}\ \ =\ \ \frac{y}{1\ +\ \frac{y}{x}}\ \ \ =\ \ \frac{xy}{x+y})
I like the change of base part.
But he said you didn't need to use log, didn't he?
Gussy Booo
Mathematics <3
Wow! Very nice Drongoski. Very creative how you used logs.}\ \ =\ \ \frac{y}{log_{5}5\ +\ log_{5}2}\ \ =\ \ \frac{y}{1\ +\ \frac{y}{x}}\ \ \ =\ \ \frac{xy}{x+y})
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) #
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.
Gussy Booo
Mathematics <3
Well, that's the beauty of maths
. There are many ways of retrieving an answer.Look above for an alternative solution.
hscishard
Active Member
Makes sense. Will never be able to think like that in an exam.
Gussy Booo
Mathematics <3
That's why past papers are the key to success
.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%.
So true, a question from a year 9 one I did required Induction, WTF!
Gussy Booo
Mathematics <3
Haha
, it was a senior paper -> ie Years 11 and 12.
Gussy Booo
Mathematics <3
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!
!!!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!
Wow! Although quite good at indices, I'd not have thought this one out myself. It's so elegant.
hscishard
Active Member
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!
Hmmm...On looking at this I don't think it makes sense :S
Last edited:
hscishard
Active Member
Yea..why did you repear so many times?For 1:
Hmmm...On looking at this I don't think it makes sense :S
It is right however
hscishard
Active Member
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!
It's funny.
For the first part A and B could be 0
But for the Tan part, A can be 60 and B can be 30. Or 0