Melody

Oh you lot prattle a lot, and zegroes, if you are going to do it on the calc anyway why not do it all at once?

$${\frac{{\mathtt{0.3}}{\mathtt{\,\times\,}}{\mathtt{52}}}{{\mathtt{65}}}} = {\frac{{\mathtt{6}}}{{\mathtt{25}}}} = {\mathtt{0.24}}$$

Thumbs up from me too!

Yes and then they will be mine

$$5^2+2^{3x+2}\\

=25+2^{3x}2^2\\

=25+8^x\times4\\$$

think about this 

$${\mathtt{\,-\,}}{\frac{{\mathtt{9}}}{{\mathtt{0.1}}}} = -{\mathtt{90}}$$

$${\mathtt{\,-\,}}{\frac{{\mathtt{9}}}{{\mathtt{0.01}}}} = -{\mathtt{900}}$$

$${\mathtt{\,-\,}}{\frac{{\mathtt{9}}}{{\mathtt{0.001}}}} = -{\mathtt{9\,000}}$$

$$\frac{-9}{n}\quad \mbox{ approaches} -\infty \mbox{ as n }\rightarrow\: ^+0$$

BUT n can approach 0 from above or below.

$${\mathtt{\,-\,}}{\frac{{\mathtt{9}}}{-{\mathtt{0.1}}}} = {\mathtt{90}}$$

$${\mathtt{\,-\,}}{\frac{{\mathtt{9}}}{-{\mathtt{0.01}}}} = {\mathtt{900}}$$

$${\mathtt{\,-\,}}{\frac{{\mathtt{9}}}{-{\mathtt{0.001}}}} = {\mathtt{9\,000}}$$

This time 

$$\frac{-9}{n}\quad \mbox{ approaches} +\infty \mbox{ as n }\rightarrow\: ^-0$$

n cannot approach -infinity and +infinity at the same time and end up at the same place so -9/0 must be undefined.  I think indeterminant is the more common term these days but to the best of my understanding both terms mean the same thing.

Does that make sense to you?

think about it 

2,4,8,16,32,64,128    I have just kept multiplying by 2

So 2^7=128

like Alan said, x=7

thank you NinjaDevo,

It is a really good idea to refer people to a similar question like you have done here.    

Hi NinjaDevo,

You give great detailed answers along with showing how to check, that is really great.

However think about this step

 6x²+5x-4=0

(3x + 4)(2x - 1) = 0

I think that for many people this is the step that causes the greatest amount of problems and you have not described how you did it.

Here is a you tube clip to show you how to do it.

A good answer Sally, thank you

Just make sure the scores are sorted smallest to largest - or vice versa first.

$$y=ax^2+bx+c\qquad a\ne0$$     

this is a parabola.

If a>0 then it is concave up, If a<0 then it is concave down.

The y intercept is c

$$\mbox{The axis of symmetry is }\qquad x=\dfrac{-b}{2a}\\

\mbox{Substitute in this value of x to get the y co-ordinate of the vertex}$$

The roots, x intercepts are 

$$x=\dfrac{-b}{2a}\pm \dfrac{\sqrt{b^2-4ac}}{2a}$$

That is usually enough.