Further Maths AQA GCSE exams

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Neither...its come up in multiple past papers that I've done. It's basically Pythagoras and trig in 3 dimensions, so it might be under that in the spec?

Oh I see. I thought you meant intersection between say the line

\displaystyle \frac{x-2}{3} = y+5=\frac{z-1}{2}

and the plane

\displaystyle x-2y+z=2

.
I know what you're talking about now. :wink:

Reply 21

Wolfram Alpha

Original post by B_9710

You said factorise

\displaystyle x^{-4}-25y^2

. Expand yours out and see what it comes out as. You posted in the question a power of x^-2.
Unless that was a mistake in your question?

Ah, I've just realised it is indeed an error in my question. The power of is supposed to be positive. Sorry about that.

Reply 22

B_9710

Original post by Wolfram Alpha

Ah, I've just realised it is indeed an error in my question. The power of is supposed to be positive. Sorry about that.

The answer you posted is correct then.

Reply 23

Fox Corner

Original post by zara_ruby

Just though I would start a thread about the further maths exams and how everyone is revising.

Thanks for making this thread OP! I moved this into the Maths Exams forum and also put it on the GCSE Exam Directory Thread (which is here) so other students can find it easily :h:

Reply 24

Wolfram Alpha

Hello everyone. I have a question.
1/2x2a^2xsin30 = 18.
If I wanted to get rid of the 1/2, I multiply both sides by two. Does this mean the new equation is as follows:
2a^2xsin30 = 18.
I wasn't very sure since in the first equation, not everything is being halved. Thanks in advance.

Reply 25

niamhdoesmaths

Original post by Wolfram Alpha

No, as you haven't multiplied the rhs by 2. It should be (I think):
2a²×sin30 = 36

Reply 26

samb1234

Original post by Wolfram Alpha

Assuming it's 2(a^2) and not (2a)^2 the easiest way is just to go straight to a^2 sin30=18

Reply 27

B_9710

Original post by Wolfram Alpha

Is the question

\displaystyle \frac{1}{2}\times 2a^2 \sin 30^{\circ} = 18 ?

If so the 1/2 and the 2 on the left hand side of the equation cancel right away.

(edited 7 years ago)

Reply 28

Wolfram Alpha

Original post by B_9710

Is the question

\displaystyle \frac{1}{2}\times 2a^2 \sin 30^{\circ} = 18 ?

If so the 1/2 and the 2 on the left hand side of the equation cancel right away.

Ah, I see. Thanks very much. Since I post quite a few maths questions, may I just ask how I can post the proper...equation so it looks like yours do. It's much less confusing.

Reply 29

samb1234

Original post by Wolfram Alpha

Ah, I see. Thanks very much. Since I post quite a few maths questions, may I just ask how I can post the proper...equation so it looks like yours do. It's much less confusing.

http://www.thestudentroom.co.uk/wiki/LaTex

Reply 30

Wolfram Alpha

Original post by NiamhM1801

No, as you haven't multiplied the rhs by 2. It should be (I think):
2a²×sin30 = 36

Oh yes, I just realised. Thanks for your help Niamh.

Reply 31

Wolfram Alpha

Original post by samb1234

http://www.thestudentroom.co.uk/wiki/LaTex

Thanks very much.

Reply 32

samb1234

Original post by Wolfram Alpha

Thanks very much.

No problem, just to warn you it isn't the most intuitive thing to use

Reply 33

niamhdoesmaths

Original post by Wolfram Alpha

Oh yes, I just realised. Thanks for your help Niamh.

No problem, although the other persons method of just cancelling the ½ and the 2 is easier (I didn't see it at first)

Reply 34

Ano123

Hi, I need help on a particularly awkward question.
The question is :

Simplify fully the expression

\displaystyle \frac{3x^3}{x^2} \times \sqrt{\frac{9}{x^2}}

.
I get an answer but it's just cant be right.
Could someone quickly just post an answer so I can see if I'm right?
Thank you!

Reply 35

niamhdoesmaths

Original post by Ano123

Hi, I need help on a particularly awkward question.
The question is :

Simplify fully the expression

\displaystyle \frac{3x^3}{x^2} \times \sqrt{\frac{9}{x^2}}

.
I get an answer but it's just cant be right.
Could someone quickly just post an answer so I can see if I'm right?
Thank you!

I got 9*

Reply 36

Ano123

Original post by NiamhM1801

I got 9*

So you're saying for any x value, the expression will always be equal to 9. That's interesting.
Try x=-1 into the original expression and see what answer you get. Post the answer you get when x=-1.

Reply 37

niamhdoesmaths

Original post by Ano123

So you're saying for any x value, the expression will always be equal to 9. That's interesting.
Try x=-1 into the original expression and see what answer you get. Post the answer you get when x=-1.

Ah, now I think you should have 2 solutions? 9 and -9? As √9 has 2 answers?

Reply 38

Ano123

Original post by NiamhM1801

Ah, now I think you should have 2 solutions? 9 and -9? As √9 has 2 answers?

Not quite.

\sqrt 9 = 3

\sqrt 9 \neq -3

.
What's actually happening here is you made a mistake when square rooting x^2 . If you put any negative number as x into the expression, say -1 you get

\sqrt {(-1)^2} = \sqrt 1 = 1

. So as you can see if x is a negative number then it returns a positive number.
So actually

\sqrt {x^2} = | x |

.
The vertical lines mean the absolute value. Meaning we just take the magnitude of the number. So |-2| = 2.
|-100|=100 etc.

Reply 39

niamhdoesmaths

Original post by Ano123

Not quite.

\sqrt 9 = 3

\sqrt 9 \neq -3

.
What's actually happening here is you made a mistake when square rooting

x^2 . If you put any negative number as x into the expression, say -1 you get

\sqrt {(-1)^2} = \sqrt 1 = 1

. So as you can see if x is a negative number then it returns a positive number.
So actually

\sqrt {x^2} = | x |

.
The vertical lines mean the absolute value. Meaning we just take the magnitude of the number. So |-2| = 2.
|-100|=100 etc.

Woah there's no way this could be in further maths right? And I thought that √9 could be 3 or -3? As (-3)² is 9?
What did you get as your answer then? This has really confused me!