Melody

$${\frac{\left({\mathtt{4.72}}{\mathtt{\,\times\,}}{\mathtt{105}}\right){\mathtt{\,\times\,}}\left({\mathtt{4.72}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{5}}}\right)}{\left({\mathtt{7.6}}{\mathtt{\,\times\,}}{\mathtt{1\,012}}\right)}}{\mathtt{\,\times\,}}\left({\mathtt{7.6}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{12}}}\right) = {\mathtt{231\,149\,407\,114\,624\,505.928\: \!853\: \!754\: \!940\: \!711\: \!5}}$$

Alan,  could you comment on my answer please?

The limit as x approaches 1 from above exists   It is 1 I think

The limit as x approaches 1 from below exists   It is p(x) I think

I don't think that the limit as x approaches 2 exists at all.

Is this right?   

So it is a "Floor" function I think.

I think the answer is "none of these"

The function is continuous for every x that is NOT an integer.  That's what I think anyway.

f(x)=1.81x - 3.7

okay, thanks Alan.

FV=P(1+i)^n

i=0.046/12=0.003833333333

n=20*12 = 240

FV=24000

$${\frac{{\mathtt{24\,000}}}{\left({\left({\mathtt{1.003\: \!833\: \!333\: \!333\: \!333\: \!3}}\right)}^{{\mathtt{240}}}\right)}} = {\mathtt{9\,581.294\: \!137\: \!691\: \!544\: \!508\: \!7}}$$

$9581.30 must be invested

$$\\\frac{x^4}{x^9}=\frac{xxxx}{xxxxxxxxx}=\frac{1}{x^5}\\\\
$all the other x'es cancel out$\\\\
0r \quad \frac{x^4}{x^9}=x^{4-9}=x^{-5}\\\\
\frac{1}{x^5}=x^{-5}$$

Alan, that is a really good explanation.  But most students would not know how to factorise this quadratic.

I have a method that I will go through but I am wondering how you do it?

$$\\15a^2+4a-3=0\\\\
$You need to find two numbers that multiply to 15*-3=-45$\\
$and add to +4$\\
$Since they multiply to a negative one is $ + $ and the other is $ - \\
$I can see straight off that they are +9 and -5$\\
$Now you split 4a up into -5a+9a $\\\\
15a^2+-5a+9a-3=0\\
15a^2-5a \quad +9a-3=0\\
$Now factorise the pairs$\\
5a(3a-1)+3(3a-1)=0\\
$this factorises to$\\
(5a+3)(3a-1)=0$$

Here is Minnie - Dragon wanted me to introduce her.

Have you seen notation like this before or is it just an educated guess?