They mark the first one. Let's hope you did well in question 5

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## Philip's Content

There have been 13 items by Philip (Search limited from 19-January 21)

### #102262 Quick question

Posted by Philip on 31 May 2009 - 09:45 AM in Geography

They mark the first one. Let's hope you did well in question 5

### #102234 Wrote in pencil

Posted by Philip on 26 May 2009 - 06:00 PM in Physics

They might be lenient on you

But, you never know cos a friend of mine failed his Product Design prelim cos he wrote in pencil!

Let's just hope their fair...

### #102219 One Quick Question...

Posted by Philip on 23 May 2009 - 07:18 PM in Mathematics

well thats what iv always been told =) good luck!

Thanks, i took your advice. Loads of my friends at school got a right bollocking from the chief invigilator cos they didn't have a pencil for the objective test!

### #102170 One Quick Question...

Posted by Philip on 20 May 2009 - 08:12 PM in Mathematics

I know it sounds stupid but i can't remember which one your allowed to use

### #102158 Paper 2 2005

Posted by Philip on 20 May 2009 - 03:15 PM in Mathematics

log4(5-x) - log4(3-x) = 2 x<3

Haha i just did this one 5 minutes ago

Right, you already know log_{4}(5-x)-log_{4}(3-x)=2

So you re-arrange to log_{4}(5-x)-log_{4}(3-x)=log_{4}4^2

**Remember**: log_{4}4=1

Then you can cancel out the logs:

(5-x)/(3-x)=4^2

16(3-x)=(5-x)

48-16x=5-x

5-15x=48

-15x=43

x=-43/15

Hope this helped! Any questions just reply back.

### #102154 2006 Paper 2 Question 12 b

Posted by Philip on 20 May 2009 - 01:26 PM in Mathematics

Thanks!

Handily you can find the HSN worked solution at Google Books but here's how I would do it (not too different really)...

Remember that extremes (that is the mins and maxes) can occur at stationary points or

__the start and end of the given range__(underlined as that's probably what you've done wrong )

So, A = 80 - 12

*x*- 48^-1

dA/d

*x*= -12 + 48

*x*^-2 = 0 (for stat. point)

-12 + 48/

*x*^2 = 0

48/

*x*^2 = 12

12

*x*^2 = 48

*x*^2 = 4

*x*= 2 or -2

Thus

*x*= 2 as

*x*> 0.

So find A for the

__values of x...__

**three**When

*x*= 1, A = 80 - 12x1 - 48/1^2 = 20

When

*x*= 2, A = 80 - 12x2 - 48/2^2 = 32

When

*x*= 4, A = 80 - 12x4 - 48/4^2 = 20

Therefore: Max of A is 32 when

*x*= 2 and Min of A is 20 when

*x*= 1 or 4

For the three values at the end: did you just use the x=1 and x=4 from the graph to find the mins and maxs?

What i did wrong was that i found the minimum x=2, but didn't know you just subbed in the other two values.

Thanks for the help

### #102147 2006 Paper 2 Question 12 b

Posted by Philip on 20 May 2009 - 10:43 AM in Mathematics

Thanks!

### #102146 2007 Past Paper Question

Posted by Philip on 20 May 2009 - 10:38 AM in Mathematics

1) Work out point B to be (7,8)

2) Radius of the large circle = square root of ((g squared) +(f squared) - c)

i.e,square root of ((7squared)+(8squared)-77) = square root of 36 = 6

therefore radius of large circle = 6

3) Between the point B and the outer of the circle there are three sets of radii this means that the radii of the smaller

circles is the radius of the large circle divided by three, i.e 6 divided by 3=2, this is the length of each of the small radii

because the question tells us they are congruent.

4) The y-coordinate is the same for each of the centres cause they are parallel to the x-axis. So for D y=8

5) To get the x-coordinate for D use the x-coordinate from B. So it would be x-coordinate + radius of large circle + radius

of small circle, i.e 7+6+2=15.

6) The point D has coordinates (15,8)

7) Put these numbers into the equation of a circle ((x-a)squared + (y-b)squared = radius squared)to give us:

(x-15)squared + (y-8)squared = 2squared

Hope this helps, if you need anymore explanation, just ask!

Thanks!

### #102111 2007 Past Paper Question

Posted by Philip on 18 May 2009 - 12:09 PM in Mathematics

2007 Paper 1 Question 5

2007 Paper 1 Question 5

I get the centre of the circle. But i don't know what to do from there!

Would appreciate some help, thanks

### #101880 Recurrence Relations

Posted by Philip on 08 April 2009 - 01:30 PM in Mathematics

The monks prepare 8000litres of spirit estimating that this will produce at least 4000litres of whisky in 12 years time. Are they right?

Hope i get this right...

Ok, so we know that they prepare 8000 litres of spirit. And each year it loses 3.5% of volume; or you could look at it as having 96.5% of its original volume (which is probably easier here). And the time is 12 years.

This is just a depreciation question.

Therefore the numbers we have taken out of the question can be put like this:

(Don't know why but the 2 of '12' doesn't want to be powered, just imagine it is )

Firstly, we have the number being inputted - 8000. Then we just multiply it by the 96.5% or 0.965 - as 1.0 would be 100%; 0.5 would be 50%. This number is the 'powered' by the number of years in question - in this case it's 12.

So all that is left to do is put the number into the equation, to get:

Once you have written that, you write a conclusion saying: "After 12 years there is 5217 litres left. As 5217>4000, the monks are correct."

Hope this helped!

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