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#101 Re: Help Me ! » Find two numbers from a single number » 2006-01-23 05:56:31
Maybe I don't fully understand the question but:
1+2=3
#102 Re: Help Me ! » Factorising Quadratics » 2006-01-23 05:54:29
solved
(-2x + 3)(x + 5) or (2x - 3)(-x - 5)
#103 Re: Help Me ! » Factorising Quadratics » 2006-01-23 02:58:13
(ax + b) (cx + d)
so
-2x^2 - 7x + 15
-[2x^2 + 7x - 15]
-[(2x - 3)(x + 5)]
my problem is removing the square set of brackets here.
#104 Re: Help Me ! » Factorising Quadratics » 2006-01-23 02:53:55
I am stuck on the question y=-2x^2 - 7x + 15 where I need to sketch the curve.
x = 0 => y = 15
-2x^2 - 7x + 15
-[2x^2 + 7x - 15]
Here, I can see that the two numbers I need are +10 and -3 since:
10 * -3 = 2 * -15
and
10 - 3 = 7
so would the answer be...
#105 Re: Help Me ! » diffrentiation » 2006-01-17 12:47:16
∫dy/dx(x^4 + 6x^3 + 4x^2 + 21x + 4)dx = ∫4x^3 + 21x^2 + 8x + 21
#106 Re: Help Me ! » diffrentiation » 2006-01-17 12:45:04
The question is about differentiation so I assume you need to change ((x + 1)(x + 2))² into
((x + 1)(x + 2))*((x + 1)(x + 2)) = (x² + 3x + 2)*(x² + 3x + 2) = x^4 + 6x^3 + 4x^2 + 21x + 4
∫(x^4 + 6x^3 + 4x^2 + 21x + 4)dx = ...
or somthing like that?
#107 Re: Help Me ! » A really obvious but interesting question » 2006-01-16 03:47:18
I think the problem was, anya was flipping one fraction but still dividing them. I was emphasising that you need to multiply it once flipped. Like you say, it's easier to multiply it that way 
#108 Re: Help Me ! » A really obvious but interesting question » 2006-01-15 21:30:22
Easy Method!!!
Take the 2 fractions you wish to divide. Flip one of them over. Then multiply the 2 new fractions together (multiply the top numbers, then multiply the bottom numbers). That will give you the correct answer.
4 1
_ Divided by _
9 2
becomes
4 2
_ Multiplied by _ <----- This part has been flipped over
9 1
and thus
4 * 2 = 8
9 * 1 = 9
So your answer would be 8/9
8
_
9
#109 Re: Help Me ! » help help » 2006-01-13 02:03:26
y? cos
TEHE
#110 Re: Help Me ! » Integrating fractions » 2006-01-13 01:46:41
Since I did my GCSE mathematics at a state school about 3 years ago (I skieved off or was stoned all the time), I remember nothing of it. Starting at Advanced GCSE level now means there's a lot that I'm expected to know that I don't 
Do you know any good resources for learning algebra? I'm sure if I revised for a week I'd stop making (as many) silly mistakes.
#111 Re: Help Me ! » Integrating fractions » 2006-01-13 01:30:19
3/(4x^5) = 12x^-5
∫(12x^-5)dx = (12x^-4)/-4 + c = -12/4x^4 + c = -3x^4 + c
Is this a case of bad algebra or am I taking a wrong step in my integration?
#112 Re: Introductions » Aloha » 2006-01-12 23:53:19
I found out about the site from my Cybernetics buddy at Reading University. That's where I hope to go once I get my A-level.
#113 Re: Help Me ! » Integrating fractions » 2006-01-12 23:46:48
same answer
sorry that my formatting is pretty c r a p p y
I think it should be
∫(2x^2 - x/2)dx = ...
#114 Re: Help Me ! » Integrating fractions » 2006-01-12 22:55:07
∫dy/dx = ∫2x^2 - x/2
y = (2x^3)/3 - (x^2)/2(2) + C = (2x^3)/3 - (x^2)/4 + C
#115 Re: Help Me ! » Integrating fractions » 2006-01-12 22:28:49
I'll make a somewhat wild guess
∫dy/dx = ∫2x^2 - x/2
y = (2x^3)/3 - ((x^2)/2) / 2x + C
*edit - forgot the + c
#116 Help Me ! » Integrating fractions » 2006-01-12 22:20:13
- RickyOswaldIOW
- Replies: 7
The question is:
integrate the function 2x^2 - x/2
The first part would bt (2x^3)/3 but I am unsure of how to integrate the fraction.
*notices the bottom of the screen*
cool update
#117 Re: Help Me ! » help help » 2006-01-12 22:18:08
There are a good few people on this board who will help you, but you must have a little patience. They are probably in bed 
#118 Re: Help Me ! » A really obvious but interesting question » 2006-01-11 19:17:55
I think you are right about flipping one of the fractions, only you need to then multiply each side (not divide)
(4/9) / (1/2) = (4/9) * (2/1)
4 * 2 = 8
-----------
9 * 1 = 9
But then again, I'm pretty terrible with algebra
#120 Help Me ! » Tangent to Normal. » 2006-01-10 14:49:07
- RickyOswaldIOW
- Replies: 3
I've been working out the gradient (and whole equation) of the tangent of certain points on a curve but this new set of questions wants the equation of the normal. I figure I'd just work as I had been and get the equation of the tangent, then simply flip over the gradient and reverse the sign for the normal i.e.
tangent; y = 7x - 13
normal; y = -1/7x - 13
The full sum is as follows:
y = 2x^2 - 5x + 3 at (2, 1)
dy/dx = 6x - 5
12 - 5 = 7
(y - 1)/(x - 2) = 7
y - 1 = 7(x - 2) = 7x - 14
y = 7x - 13
and thus the normal; y = -1/7x - 13.
The book claims x + 3y = 5
#121 Re: Help Me ! » Differentiation » 2006-01-08 19:22:31
Why does 1/(4x) = 4^-1 * x^-1? I was taught to recirpocal i.e. simply bring the bottom of the fraction to the top and change the sign of the power. Why have you split it into 4^-1 * x^-1?
P.S. thanks for explaining the fractions, it makes a lot of sense.
#122 Re: Help Me ! » Differentiation » 2006-01-08 16:41:00
in addition, could you check this (I think the book has put the answers in the wrong order)
y = 4 + 1/4x at x = -1
= 4 + 4x^-1
dy/dx = -4x^-2 = -4/x^2 = -4/1 = -4. i.e. the Gradient of the Tangent on the curve at the stated point (x = -1) is -4 and thus the Gradient of the Normal is 1/4. My book claims it is the other way round and with opposite signs (tang is -1/4 and norm is 4)!!
#123 Re: Help Me ! » Differentiation » 2006-01-08 16:14:51
Could you explain how to work out a fraction over a fraction?
-1 / (1 / 4) = -4
I've never actually been taught this
#124 Re: Help Me ! » Conundrum » 2006-01-07 16:40:36
I'm not saying it can't be solved...
#125 Re: Help Me ! » Simplify » 2006-01-07 04:31:12
No offence intended. I play online FPS games and my language in that would get me banned in a flash here. You **...