Feb 7 answers

Q1

3 - 6x - x^2

= -x^2 – 6x + 3

= -[x^2 + 6x – 3]

= -[(x^2 + 6x) - 3]

= -[(x + 3)^2 – 3 – 9]

= -(x + 3)^2 + 12

Q4

b^2 – 4ac = 0

(-2p – 6)^2 – (4 * (p + 1) * 3p) = 0

(4p^2 + 24p + 36) – (12p^2 + 12p) = 0

-8p^2 + 12p + 36 = 0

2p^2 – 3p – 9 = 0

p = -2/3 and p = 3

**1**

# HSN Maths Quiz

Started by Dave, Jan 13 2005 05:29 PM

23 replies to this topic

### #21

Posted 11 January 2006 - 09:17 AM

### #22

Posted 11 January 2006 - 04:23 PM

QUOTE(duncad @ Jan 11 2006, 06:29 AM)

The maths needs to be input in 'latex' code.

This needs to go between the [

*tex*] [

*/tex*] tags, if you have maths in a sentence, or [

*display*] [

*/display*] if you want to do a few lines of maths.

Here's an example:

The code for this is:

CODE

[display]f(x) = f(0) + \frac{1}{0!}\int_0^x{f'(t) \, dt}[/display]

I'll just demonstrate a few common commands:

x^n

x_n

\frac{a}{b}

\int_{a}^{b} {f(x) \, dx}

\pi

To line up equations at the = sign, use &= like this:

CODE

[display]

y &= (x-1)(x+3) \\

&= x^2 + 3x - x - 3 \\

&= x^2 +2x -3

[/display]

y &= (x-1)(x+3) \\

&= x^2 + 3x - x - 3 \\

&= x^2 +2x -3

[/display]

Gives:

Also, notice that \\ takes a new line in [

*display*] mode.

For more details, there are lots of good tutorials on the web, like this one

### #23

Posted 04 May 2006 - 04:43 PM

Q1 - should it not be :

f'(x)= 1/2(8x-5) ^-1/2 times 8

= 4(8x-5)^-1/2

= 4/the sqaure root of (8x - 5)

If not where am i going wrong?

f'(x)= 1/2(8x-5) ^-1/2 times 8

= 4(8x-5)^-1/2

= 4/the sqaure root of (8x - 5)

If not where am i going wrong?

### #24

Posted 04 May 2006 - 05:23 PM

yes it is

If i am not here i am somewhere else

#### 1 user(s) are reading this topic

0 members, 1 guests, 0 anonymous users