This discussion is now closed.

Check out other Related discussions

AQA Core 4 - Monday 10th June 2013 (AM) - Official Thread

Original post by beautywithbrains

Can somebody please help me out on this question? It's jan08 qu3c)
Don't understand what to with the 8 on the bottom?
ImageUploadedByStudent Room1370795658.408430.jpg

ImageUploadedByStudent Room1370795658.408430.jpg

Posted from TSR Mobile

You pull out 1/2, and just multiply the expansion of 1+3/2x^1/2 by 1/2

Reply 341

coolstorybrother

8

Original post by Enginerd.

I may not be around for the remainder of the day, so good luck to everyone sitting the exam tomorrow! And I hope you get the grade you need/want. :smile:

thanks
for the vectors theres some sweet questions earlier on in the thread post 50 ish, what basic geometry are you guys gonna go over? (i.e rhombus , parallelogram etc what else? )

Reply 342

Bord3r

10

Original post by beautywithbrains

Can somebody please help me out on this question? It's jan08 qu3c)
Don't understand what to with the 8 on the bottom?
ImageUploadedByStudent Room1370795658.408430.jpg

Posted from TSR Mobile

Well from the question previously you get the expansion for

Unparseable latex formula:

\sqrt{(1+\frac{3}{2})}[br]\[br]The numerator of part c) is \sqrt{2} \sqrt{(1+\frac{3x}{2})}[br]\[br]So for the overall equation you get { \sqrt{2} \sqrt{(1+\frac{3x}{2})} \over\sqrt{8}}[br] \Rightarrow { \sqrt{2} \sqrt{(1+\frac{3x}{2})} \over 2\sqrt{2}}[br]\[br]\Rightarrow \frac{1}{2} \ multiplied \ by \ your \ expansion \ from \ b)[br]

Reply 343

Barcelona'99

2

Original post by krane

Can anyone help me with the questions on trig, where you get like cosx= 2/3 and have to work out sinx? I get this, but what do you do when it's an obtuse angle?

This probably doesn't make much sense, an example is Jan 2012 Q2 a ii)
:smile:

treat it in the same way. Find the value of sine as you normally would and then refer to the CAST diagram - an obtuse angle is between 90 and 180 degrees (sine is positive). If you were finding tan or cos of an obtuse angle, it would be negative. The values are going to be the same, it's the sign (+ or -) that you need to consider.
Hope that helps

Reply 344

Barcelona'99

2

Original post by krane

Can anyone help me with the questions on trig, where you get like cosx= 2/3 and have to work out sinx? I get this, but what do you do when it's an obtuse angle?

This probably doesn't make much sense, an example is Jan 2012 Q2 a ii)
:smile:

So, in the case of the Jan 2012 question. Alpha is acute and Beta is obtuse.
Tan is positive when the angle is acute, so the value will = 4/3.

But Tan is negative when the angle is obtuse, so the value will = -1/root3

Reply 345

Tobias2

Could some please explain Q 2a)ii) from Jan 2012?

http://filestore.aqa.org.uk/subjects/AQA-MPC4-QP-JAN12.PDF

Unsure how the fact Sin(B) is obtuse effects the question.

Thanks in advance for the help!

Reply 346

iloveitachi

can someone help me with this question please? It's jan 2010 paper last question part bii thanks in advance!

(edited 10 years ago)

Reply 347

beautywithbrains

Original post by Bord3r

Well from the question previously you get the expansion for

Unparseable latex formula:

\sqrt{(1+\frac{3}{2})}[br]\[br]The numerator of part c) is \sqrt{2} \sqrt{(1+\frac{3x}{2})}[br]\[br]So for the overall equation you get { \sqrt{2} \sqrt{(1+\frac{3x}{2})} \over\sqrt{8}}[br] \Rightarrow { \sqrt{2} \sqrt{(1+\frac{3x}{2})} \over 2\sqrt{2}}[br]\[br]\Rightarrow \frac{1}{2} \ multiplied \ by \ your \ expansion \ from \ b)[br]

Omg thank you! Finally get it!!

Posted from TSR Mobile

Reply 348

Dirtybit

20

Original post by The H

That's true, the Jan 13 vectors Q was fairly easy compared to others, so... get ready for a hard one this time :mad:

Yeah most likely will be :rolleyes:

Reply 349

Hunarench95

13

Original post by iloveitachi

can someone help me with this question please? It's jan 2010 paper last question part bii thanks in advance!

Work out your dh/dt from the original expression for h (as you would normally). Then rearrange the original expression for h, make e^(t/4) the subject, then substitute back into your dh/dt. Should fall out :smile:

Reply 350

Dhaden

13

How you guys learning the vectors stuff except from doing past papers?

Reply 351

JTD77

Hey, could some help me with question (3b) from the January 2010 paper?

Reply 352

Enginerd.

21

Original post by coolstorybrother

thanks
for the vectors theres some sweet questions earlier on in the thread post 50 ish, what basic geometry are you guys gonna go over? (i.e rhombus , parallelogram etc what else? )

Yep, those kinds of questions trip me up! Make sure you know that isosceles triangle has 2 equal sides.

Original post by iloveitachi

can someone help me with this question please? It's jan 2010 paper last question part bii thanks in advance!

dh/dt means height per hour, it is growing at a rate of 13 mm per hour, so replace the dh/dt by 13 and solve.

Reply 353

iloveitachi

Original post by Hunarench95

Work out your dh/dt from the original expression for h (as you would normally). Then rearrange the original expression for h, make e^(t/4) the subject, then substitute back into your dh/dt. Should fall out :smile:

Could you please show me? Cause I integrated it but I'd like to understand the method used in the ms and thanks a lot for answering! :smile:

Reply 354

iloveitachi

Original post by Enginerd.

Yep, those kinds of questions trip me up! Make sure you know that isosceles triangle has 2 equal sides.

dh/dt means height per hour, it is growing at a rate of 13 mm per hour, so replace the dh/dt by 13 and solve.