Preliminary mathematics marathon (1 Viewer)

[hscishard]

lol no fair, you've done all this before!

New Question:

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.

Tan B = -1

 

[ohexploitable]

Tan B = -1

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]

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?

Lol now I got tan B = 5.

Nvm you're right.
I used TanA+TanB/1+TanATanB

How gay

 

[Gussy Booo]

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?

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]

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 .

I guess that was like a year 9 paper?
Those compeitions are soo hard

 

[Drongoski]

 

[hscishard]

Very nice.
I like the change of base part.

But he said you didn't need to use log, didn't he?

 

[Gussy Booo]

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) #

 

[Drongoski]

Very nice.
I like the change of base part.

But he said you didn't need to use log, didn't he?

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]

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.
Look above for an alternative solution.

 

[hscishard]

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) #

Makes sense. Will never be able to think like that in an exam.

 

[Gussy Booo]

Makes sense. Will never be able to think like that in an exam.

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%.

 

[AAEldar]

I guess that was like a year 9 paper?
Those compeitions are soo hard

So true, a question from a year 9 one I did required Induction, WTF!

 

[Gussy Booo]

So true, a question from a year 9 one I did required Induction, WTF!

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

 

[Gussy Booo]

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!

 

[Drongoski]

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) #

Wow! Although quite good at indices, I'd not have thought this one out myself. It's so elegant.

 

[hscishard]

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

Lol, I posted that in the old thread.

 

[AAEldar]

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!

For 1:



Hmmm...On looking at this I don't think it makes sense :S

 

[hscishard]

For 1:



Hmmm...On looking at this I don't think it makes sense :S

Yea..why did you repear so many times?

It is right however

 

[hscishard]

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!

For Q2.
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