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#91417 Quadratic question

Posted by John on 30 November 2006 - 10:03 PM in Mathematics

Use the discriminant firstly to see if the roots are real and distinct (2 roots), equal(1 root) or imaginary(no roots).

and after that try using to 2 nearest factors of 15 both positive and negative factors.



#91408 Quadratic question

Posted by John on 30 November 2006 - 09:06 PM in Mathematics

In this case i would use calculus first to find the turning point.

Then solve

3x power2.gif - 15x + 11= a(x+b) power2.gif + c

for a.



#93023 Scottish schools to show climate change film

Posted by John on 18 January 2007 - 09:08 PM in Geography

I think my mate is doing a voice over that that video.



#90163 Trig help

Posted by John on 13 October 2006 - 06:53 PM in Mathematics

I've been asked this question in my uni homework and not sure how to expand it.

Expand sin(A + B + C) in terms of sin A, sin B, sin C, cos A, cos B, cos C.

How would i tackle this?

They also ask the same for cos(A + B + C)


Thankies =D



#100389 Oh Wow Look Who's Stuck ~

Posted by John on 07 October 2007 - 12:15 PM in Physics

This is rather simple Vectors using Pythagoras.

And I don't want to insult you by showing you how to do Pythagoras smile.gif



#100155 What chances would you give me?

Posted by John on 22 August 2007 - 09:44 AM in After School

You'll get in, easily!



#92119 [Quadratic Theory] h/w

Posted by John on 17 December 2006 - 12:27 PM in Problem Questions

A product is the answer from multiplying two or more numbers together, not the summation of two or more numbers.

For question seven use b2 - 4ac = a negative number, to find the smallest possible integer value for K.



#92189 Oxford

Posted by John on 19 December 2006 - 06:20 PM in After School

This should be in the after school section

EDIT: I see it has been moved now.



#91011 Need help on dropping a subject

Posted by John on 15 November 2006 - 11:31 PM in New Sixth Years

If you drop it, they can not force you to keep going to the class time AFTER school.

So you are in the clear.



#94851 Trig Question

Posted by John on 14 March 2007 - 10:41 PM in Mathematics

cos(x+15) = 2/3

take the cos over to the other side to get

x+15 = cos^-1(2/3)
x+15 = 48.1897

then solve for x

x = 48.1897 - 15
x = 33.1897

x is in degrees btw



#93634 Can someone go over a past paper with me, as im struggling and have an exam o...

Posted by John on 03 February 2007 - 08:59 PM in Mathematics

Add me, I'll do my best to help.

MSN address is in my profile



#90287 Buoyancy Force

Posted by John on 19 October 2006 - 06:40 PM in Physics

The Bouyancy Force is created by the force below the object being greater than the force above the object, and if you take the force above the object from the force below the object you get the Bouyancy force IIRC.



#92710 unit 2 - outcome 3 - trigonometry

Posted by John on 11 January 2007 - 08:06 PM in Problem Questions

convert the radians to degrees, do the equation, and then convert the answer back to radians.



#99184 Need a little Help.

Posted by John on 04 June 2007 - 06:41 PM in Mathematics

Reinstall your PDF viewer



#94187 Help with a past paper question

Posted by John on 21 February 2007 - 07:45 PM in Mathematics

Use the chain rule on the trig part, and standard differentiation on the latter



#91043 Another quadratics question.

Posted by John on 17 November 2006 - 03:04 PM in Problem Questions

You are correct.

Now what you need to do is subsitute x and y for the co-ordinate (0,18).

And then solve for a.

Tell me what you get for your answer and I'll let you know if it is correct.



#91463 Integration

Posted by John on 02 December 2006 - 07:14 PM in Mathematics

For the first one:

Break the fraction up and use the rules opf indices to simplify and then integrate.

For the second one:

Multiply out the bracket and do the same from there as you did with the first one.

I know there is a simpler way, but i cant remember how to do it.



#94938 circles

Posted by John on 17 March 2007 - 06:40 PM in Mathematics

start by finding the equation of the radius

from there you find the line perpendicular to to the radius

I think you can go on from there.