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Current Year 12 Thread Mark I (2012-2013)

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Reply 7080

L'Evil Fish

Original post by Robbie242

is there a more simple way? that sounds very confusing for me personally

Hmmm... Find gradient the point before and point after...?

My Username is really Poisson Distribution :mmm:

Reply 7081

Robbie242

Original post by L'Evil Fish

Hmmm... Find gradient the point before and point after...?

My Username is really Poisson Distribution :mmm:

break down the whole question using the latex thing then I can do the other 15 questions :P

Reply 7082

usycool1

Original post by L'Evil Fish

Hmmm... Find gradient the point before and point after...?

My Username is really Poisson Distribution :mmm:

There's no need to - the question doesn't ask for the max and min values. :tongue:

Reply 7083

justanotherindigo

Original post by mynameisntbobk

It's alright :smile:

I know what you mean. They beat around the bush all the time

Yeah thats true, and my teacher didn't bother explaining it ): And something else:
when the heart is contracting powerfully its called systolic blood pressure, so this means that diastolic blood pressure is the opposite. Am I right since that contraction is systole and relaxation is diastole? (teacher didnt explain this either) Thank you! :biggrin:

Reply 7084

Robbie242

Original post by usycool1

There's no need to - the question doesn't ask for the max and min values. :tongue:

Usy help me I missed the whole lesson he set homework I understand the factorising and the turning point but what about finding whether the line starts at the top or the bottom! thanks

Reply 7085

L'Evil Fish

Original post by Robbie242

break down the whole question using the latex thing then I can do the other 15 questions :P

I have no idea how to use latex properly... Imagine you differentiate... You can fund the gradient at any point.

So if you find the gradient at three consecutive points eg:

0 1 and 2

And fins them out... Lets say the gradients are:

4 0 -2

Then it goes up turns back down...

Which is a maximum :smile:

My Username is really Poisson Distribution :mmm:

Reply 7086

Abject Testament

Original post by Robbie242

Sup guys I missed triple maths because of the interview anyway can somebody help me.

Now I have the equation y=3x+2x^2 -x^3
factorised to make y=x(3-x) (x+1)
BUT heres the problem I get x=0 x=3 and x=-1
I'm finding it hard since my teacher literally didn't see me at all at the start of curve sketching.
My question is does anybody know how to see which way the curve goes I know theres something about positive negative but how do I get there and whats a nice method? Thanks

This is a 3rd order polynomial. Since the coefficient of the highest power is negative, the curve itself will be negative.

Reply 7087

L'Evil Fish

Original post by usycool1

There's no need to - the question doesn't ask for the max and min values. :tongue:

I thought he wants turning points? :colondollar:

My Username is really Poisson Distribution :mmm:

Reply 7088

mynameisntbobk

Original post by L'Evil Fish

For you... It'll be C2 :smile:

The one I did only works in the form:

(1+x)^n

X can be anything! So (1+4x) woukd work but in place of x you put 4x...

Try this:

(1+2x)^3

My Username is really Poisson Distribution :mmm:

Okay umm

1 + 8x + (3)(2)/2! 8x^2 + (3)(2)(1)/3! 8x^3 + (3)(2)(1)(0)/4

So um 1 + 8x^2 + 3 + 8x^2 + 2 + 8x^3
So 8x^3 + 16x^2 + 6
I did something wrong :s-smilie:

Reply 7089

Robbie242

Original post by Abject Testament

This is a 3rd order polynomial. Since the coefficient of the highest power is negative, the curve itself will be negative.

but what if I have one like x^3 + 5x^2 + 4x how do I do this since it turns out negative whats a met hod you can use to double check or just generally use for most solutions?

Reply 7090

L'Evil Fish

Original post by mynameisntbobk

Okay umm

1 + 8x + (3)(2)/2! 8x^2 + (3)(2)(1)/3! 8x^3 + (3)(2)(1)(0)/4

So um 1 + 8x^2 + 3 + 8x^2 + 2 + 8x^3
So 8x^3 + 16x^2 + 6
I did something wrong :s-smilie:

(1+2x)^3

= 1 + 3x + (3)(2)/2! (2x)^2 + (3)(2)(1)/3! (2x)^3

= 1 + 3x + 12x^2 + 8x^3.

Where did the 8s come from?

My Username is really Poisson Distribution :mmm:

Reply 7091

mynameisntbobk

Original post by L'Evil Fish

(1+2x)^3

= 1 + 3x + (3)(2)/2! (2x)^2 + (3)(2)(1)/3! (2x)^3

= 1 + 3x + 12x^2 + 8x^3.

Where did the 8s come from?

My Username is really Poisson Distribution :mmm:

Okay, I don't understand this then. The 8s came from 2x^3.

Reply 7092

usycool1

Original post by Robbie242

Usy help me I missed the whole lesson he set homework I understand the factorising and the turning point but what about finding whether the line starts at the top or the bottom! thanks

If it's

-x^3

(minus), then the line will start from the top (from the left). If it's

x^3

(positive), then the line will start from the bottom (from the left). :smile:

Reply 7093

Abject Testament

Original post by Robbie242

but what if I have one like x^3 + 5x^2 + 4x how do I do this since it turns out negative whats a met hod you can use to double check or just generally use for most solutions?

You just check whether the coefficient of the highest power is negative or positive. If it's negative, then you have a negative curve. Like this question, the highest power is

x^3

, the coefficient is positive, therfore the graph of

x^3 + 5x^2 + 4x

starts and ends in the positive quadrants.

Reply 7094

TheSnail

ANY PHYSICS PEEPS ABOUT

I took up physics AS this week (dropped eng. lit) and i'm stuck on something

When a cell is connected directly across a high resistance voltmeter the reading is 1.50V. When the cell is shorted though a low resistance ammeter the current is 1.5A.
What is the emf and internal resistance of the cell?

other questions on this worksheet have involved manipulating them into simultaneous equations and solving it that way...not sure whether this one is the same.

I believe you use the equation

V = emf - Ir or write it as IR = emf - Ir

I think you also use R=V/I

please help!

(edited 11 years ago)

Reply 7095

silentlife

Original post by g.k.galloway

I think so...

Reply 7096

L'Evil Fish

Original post by mynameisntbobk

Okay, I don't understand this then. The 8s came from 2x^3.

That's the last one :smile:

(2x)^3 = 8x^3

My Username is really Poisson Distribution :mmm:

Reply 7097

TheSnail

Original post by silentlife

I think so...

i've used that to calculate the resistance of the voltmeter (if that's even right) but dont know what to do now!!!!!

Reply 7098

mynameisntbobk

Original post by L'Evil Fish

That's the last one :smile:

(2x)^3 = 8x^3

My Username is really Poisson Distribution :mmm:

Yeah, I'll look this up tomorrow after school

Reply 7099

silentlife

Original post by g.k.galloway

i've used that to calculate the resistance of the voltmeter (if that's even right) but dont know what to do now!!!!!

The voltage is 1.5V - you've stated it... Unless I've read this completely wrong... Don't over think AS physics, especially the electricity section - it's really that easy...

Edit: what's with the simultaneous equations - you're begin to make me question myself...