Melody

$$If f(x) =4x-7\\
then\\
$f(9)=4*9-7=36-7=29$\\
$f(a)=4a-7$\\
$f(-3)=4*-3-7=-12-7=-19$\\
$f(0)=4*0-7=0-7=-7$\\\\
See, each time the x that was in the function bracket to start with is \\replaced with the new thing in the function bracket.\\
Does that help?$$

I don't see any question here!  

for 1800$ you can buy 3 computers and 1 printer. if the price of a computer will go down by 500$, then for 1300$ you can buy 2 computers and 2 printers. what is the cost of a computer?

3C+1P=1800  (eqn 1a)  $$\rightarrow$$   P=1800-3C   (eqn 1b)

2(C-500)+2P=1300  (eqn 2a)

2C-1000+2P=1300   (eqn 2b)

2C+2P = 2300    (eqn 2c)

C+P=1150  (eqn 2d)

sub P=1800-3C into equation (eqn 2d)

C+1800-3C=1150

-2C=-650

C=$325

The original cost of the computer is $325

This doesn look right because it means that the price fell ot a negative number which is pretty silly.

But I will keep thinking about it anyway.

Check

1P=1800-3*625 = $825

3c+1P=3*325+825=$1800 correct

If the price of the computers fell by 500 then they fell to 325-500=-$175

Now 2 computers and 2 printers cost -175*2+2*825 = $1300  correct

Okay, the answer is correct.  The only way this works is if the computers have become so obsolete that the company will PAY $175 to have them taken away.

hi anonymous,

Why don't you join up.  It is very easy to do.  Much easier than posting anonymously every time.  

I liked you proof.  I just wrote it in LaTex as I worked through it.  

$$\begin{array}{rll}
e^{\pi i}&=&cos\pi+isin\pi\\
e^{\pi i}&=&-1+0\\
e^{\pi i}&=&-1\qquad \\
ln(e^{\pi i})&=&ln(-1)\qquad \\
ln(-1)&=&\pi i\qquad \\
now&&\\
x&=&i^i\\
lnx&=&lni^i\\
lnx&=&ilni\\
lnx&=&iln(-1)^{1/2}\\
lnx&=&\frac{i}{2}ln(-1)\\
lnx&=&\frac{i}{2}\times \pi i\\
lnx&=&\frac{\pi\times i^2}{2}\\
lnx&=&\frac{\pi\times -1}{2}\\
lnx&=&\frac{-\pi}{2}\\
e^{lnx}&=&e^{\frac{-\pi}{2}}\\
x&=&e^{\frac{-\pi}{2}}\\
x&\in&Z\\
therefore&&\\
i^i&\in&Z\\


\end{array}$$

$$\begin{array}{rll}
\frac{3}{10}&=&\frac{?}{20}\\\\
\frac{3*2}{10*2}&=&\frac{?}{20}\\\\
\frac{6}{20}&=&\frac{?}{20}\\\\
6&=&?\\\\
?&=&6\\\\

\end{array}$$

$$\begin{array}{rll}
5(m+1)&=&7(m-1)\\
5m+5&=&7m-7\\
5m+5-5&=&7m-7-5\\
5m&=&7m-12\\
5m-7m&=&7m-12-7m\\
-2m&=&-12\\
\frac{-2m}{-2}&=&\frac{-12}{-2}\\
m&=&6\\
\end{array}$$

Do you mean on the web2 calculator?

inverse trig is the same as arc trig

so if you want inverse tan just use atan

What rate do you have and I will show you how to work out the effective  annual rate. 

Why can't you do these yourself Natasza?

To find a mean you just add up all the numbers and divide by how many there are.  It is another word for average.

The others can be done either by formula or on your calculator.  I expect there are stat functions on the web2 calc but I will admit that I have not looked.

Now, what in particular would you like help with?

This is included in the sticky threads