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 X(1-0.4)/80000=X(80000)(1-0.4)/60000

I am not sure what you are trying to do but the only solution for this is X=0

LHS=X(1-0.4)/80000=48000X

RHS=X(80000)(1-0.4)/60000=0.8X

The only way that LHS=RHS is if X=0

-----------------------

LHS stands for left hand side

RHS stands for right hand side

$$\begin{array}{rcr}
\textcolor[rgb]{1,0,0}{1}\times5\times7=1*35&=&35\\
\textcolor[rgb]{1,0,0}{2}\times5\times7=2*35&=&70\\
\textcolor[rgb]{1,0,0}{3}\times5\times7=3*35&=&105\\
\textcolor[rgb]{1,0,0}{4}\times5\times7=4*35&=&140\\
\textcolor[rgb]{1,0,0}{5}\times5\times7=5*35&=&175\\
...&=&...\\
\textcolor[rgb]{1,0,0}{n}\times5\times7=n*35&\approx&2009\\
\end{array}$$

$$n\approx\dfrac{2009}{5\times7}=\dfrac{2009}{35}=57.4$$

 Between 1 and 2009, how many numbers are multiples of 5 and 7:

$$\boxed{n=57}$$

Is it bad that he has one just like that?

Thanks jboy314.   ♬     

Multiply both sides by -4/7 and you get

v= -49*-4/7

v= -7*-4

v= 28

check

-7/4*28=-49  

Okay Sorry and thanks for the explanation.  

My apologies Melody.

I'm the second anonymous, not the anonymous that posed the original question. I was simply answering your (implied) question as to what GCSE stood for.

Well sorry I still have no idea.  Remember this is an international forum, there is no point using acronyms etc that only have meaning in your country or in your state, province or district because most people will not have any idea what you are talking about.  

It stands for General Certificate of Secondary Education, and it represents one huge c**k-up (if I'm allowed to use that phrase, if not, one huge mistake,) in the educational system in England.

Let me see.

I'm going to use my own letters.

R is the yearly repayment.

A is the original amount of the loan

n = number of years

i = yearly interest rate as a decimal.  For example if the rate is 6% per annum then i=6/100=0.06

$$A=R\times \frac{1-(1+i)^{-n}}{i}$$

Now the aim is to make n the subject

$$\begin{array}{rll}
\frac{Ai}{R}&=& 1-(1+i)^{-n}\\\\
(1+i)^{-n}&=& 1-\frac{Ai}{R}\\\\
log\left((1+i)^{-n}\right)&=& log\left(1-\frac{Ai}{R}\right)\\\\
-n\times log(1+i)&=& log\left(1-\frac{Ai}{R}\right)\\\\
n &=&\:\dfrac{-log\left(1-\frac{Ai}{R}\right)}{log(1+i)}\\\\
\end{array}$$

It is interesting.

We all agree that when you divide by zero the answer is undefined.  But we all have different reasoning.

Some of our reasons appear to be contradictory.  

I have done it via limits but anonymous appears to be correct when s/he says that it is not a limit question.

 It has just got me thinking a little.  That is all.

$$\.. \hspace{80pt} \ \lim\limits_{x \to +0}{\frac{c}{x}} = \infty\ where\ c = positive\: const \\\\

\.. \hspace{80pt} \ \lim\limits_{x \to -0}{\frac{c}{x}} = -\infty\ where\ c = positive\: const$$

This is why it is undefined.  If you approach zero from above it is +infinity.  If you approach from below it is negative infinity.

Here is a graph (in this one c=1 but it could be any pos constant.)

$$V_1=\dfrac{4}{3}\pi r_1^3$$

$$V_2=\dfrac{4}{3}\pi r_2^3 \quad | \quad r_2=2r_1$$

$$V_2=\dfrac{4}{3}\pi (2r_1)^3= \underbrace{\dfrac{4}{3}\pi r_1^3}_{V_1}*\underbrace{2^3}_8$$

$$V_2=8V_1=8*71\;cm^3=568\;cm^3$$

What does GCSE stand for?  I have no idea?

Look here

tangent condition: $$7.75^2+4^2=8.75^2$$

$$7.75^2+4^2\stackrel{?}{=}8.75^2$$

$$(\frac{31}{4})^2
+(4)^2\stackrel{?}{=}(\frac{35}{4})^2
\quad | \quad *4^2$$

$$31^2+(4^2)^2
\stackrel{?}{=}35^2$$

$$31^2+16^2
\stackrel{?}{=}35^2$$

$$961+256 \stackrel{?}{=}1225$$

$$1217\neq1225$$   not a tangent!

Wish him well from us

We all pitched in a bought him a gift.  We hope he likes it.

Sorry Ornstein but you are not on the right track.

You have to make the denominators the same so that you can compare them.

Erin 1/3=4/12

Lauren 2/8=1/4=3/12

4/12 is more than 3/12 so Erin is the most successful.

$$\dfrac{L}{\sqrt{L}}*\sqrt{(\sqrt{L}+\sqrt{K})^2-x^2}$$

You are right. I'm sorry, I made a mistake... the first equation should be (a "+" not a "-" there):

Then that means, that he only brought the "L - (the big fraction)^2 * X^2" to the same denominator resulting in:

and then he simplified the "L^2" and the "(Sqrt(L)+Sqrt(K))^2" inside of the big sqare root with the fraction outside of the square root thus resulting in:

Does this make sense to you?

Thank you!

$$L-K\neq\left(
\sqrt{L}+\sqrt{K}
\right)^2$$

$$3d\equiv 1 \bmod 1288 \Rightarrow 3d =1288m+1 \Rightarrow 1288m+1 \mbox{ must be divisible by 3 without rest!}$$

$$\Rightarrow 1288m \equiv 2 \bmod 3$$

$$\text{here } \quad 1288\equiv 1 \bmod 3 \quad \mbox{ and } \quad \textcolor[rgb]{1,0,0}{2}*1288\equiv \textcolor[rgb]{1,0,0}{2} \bmod 3$$

$$\mbox{so }\quad m=2 \quad \mbox{ and } \quad d=\frac{1288*2+1}{3}=\frac{2577}{3}=859$$

$$\boxed{3*\textcolor[rgb]{1,0,0}{859}\equiv 1 \bmod 1288}$$

-7 (2a -3b)

=-7*2a- - 7*3b

= -14a+21b

This is modular arithmetic (clock arithmetic).  

d = (1+1288*n)/3  where n is a positive integer (not all positive integers!).

In fact the smallest positive integer for n that gives a positive integer for d is n=2, for which d = 859.

$${\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{859}} = {\mathtt{2\,577}}$$

$${\mathtt{2\,577}} \,{mod}\, {\mathtt{1\,288}} = {\mathtt{1}}$$

Generally, n = 2+3k, where k is 0, 1, 2, ... etc. will result in a value of d that satisfies mod(3d,1288)=1