$ v. $$ v. $$$ Forum

NoLieAbility · Post by **NoLieAbility** » Sun Jun 01, 2014 11:52 pm

What does each term denote? Is it a measured by 1-9 = $, 10-99 = $$, and 100+ = $$$? Is there a set format that is used?

This feels like something that I should have been able to find for myself, but I'll be damned if I could find it.

malleus discentium · Post by **malleus discentium** » Mon Jun 02, 2014 12:07 am

$ vs $$ vs $$$ are relative rather than absolute markers. Some people assign quartiles to them ($ is 25% tuition, $$ half etc.) but that's not standard and inferring any absolute rather than relative value from $ vs $$ vs $$$ isn't going to be useful.

Clyde Frog · Post by **Clyde Frog** » Mon Jun 02, 2014 2:40 pm

ManoftheHour · Post by **ManoftheHour** » Mon Jun 02, 2014 2:47 pm

Clyde Frog wrote:$$$$

Nucky · Post by **Nucky** » Mon Jun 02, 2014 3:20 pm

If you have to ask, you can't afford it.

John Everyman · Post by **John Everyman** » Mon Jun 02, 2014 4:05 pm

Asked this question a little while ago and was nearly unanimously told quartile percentages. Not sure why anyone would be against establishing a standard here.

transferror · Post by **transferror** » Mon Jun 02, 2014 4:12 pm

Nucky wrote:If you have to ask, you can't afford it.

TheSpanishMain · Post by **TheSpanishMain** » Mon Jun 02, 2014 4:29 pm

$$$$ > $.

Clyde Frog · Post by **Clyde Frog** » Tue Jun 10, 2014 6:03 pm

$$+$$=$$$$

ManoftheHour · Post by **ManoftheHour** » Thu Jun 12, 2014 2:50 pm

$$$$-$$=$$

Sinatra · Post by **Sinatra** » Fri Jun 13, 2014 1:11 am

$x=?, if x=4?

ScottRiqui · Post by **ScottRiqui** » Fri Jun 13, 2014 1:21 am

S + I = $

Nova · Post by **Nova** » Fri Jun 13, 2014 1:46 am

Sinatra wrote:$x=?, if x=4?

$$$$

ManoftheHour · Post by **ManoftheHour** » Fri Jun 13, 2014 1:51 am

Sinatra wrote:$x=?, if x=4?

x=?/$, 4=?/$

Clyde Frog · Post by **Clyde Frog** » Sat Jun 14, 2014 3:12 am

Clyde Frog wrote:$$$$

ManoftheHour · Post by **ManoftheHour** » Sat Jun 14, 2014 3:21 am

Clyde Frog wrote:
Clyde Frog wrote:$$$$

TheSpanishMain · Post by **TheSpanishMain** » Sat Jun 14, 2014 3:07 pm

££££

Attax · Post by **Attax** » Mon Jun 16, 2014 10:46 am

$$^2=$$$$

ScottRiqui · Post by **ScottRiqui** » Mon Jun 16, 2014 11:02 am

Attax wrote:$$^2=$$$$

Shame on you - should be $$$.

John Everyman · Post by **John Everyman** » Mon Jun 16, 2014 11:03 am

√$$$$ = $$

Attax · Post by **Attax** » Mon Jun 16, 2014 11:19 am

ScottRiqui wrote:
Attax wrote:$$^2=$$$$
Shame on you - should be $$$.

Why's that? $$^2 = $$*$$=$$$$ and $$$$^(1/2)=$$

But what'd d/d$ f($$)?

ScottRiqui · Post by **ScottRiqui** » Mon Jun 16, 2014 11:27 am

Attax wrote:
ScottRiqui wrote:
Attax wrote:$$^2=$$$$
Shame on you - should be $$$.
Why's that? $$^2 = $$*$$=$$$$ and $$$$^(1/2)=$$

But what'd d/d$ f($$)?

Exponentiation before multiplication, remember? Only the second '$' gets squared, then you multiply the result by the first '$' to give you $$$.

Attax · Post by **Attax** » Mon Jun 16, 2014 11:37 am

ScottRiqui wrote:
Attax wrote:
ScottRiqui wrote:
Attax wrote:$$^2=$$$$
Shame on you - should be $$$.
Why's that? $$^2 = $$*$$=$$$$ and $$$$^(1/2)=$$

But what'd d/d$ f($$)?
Exponentiation before multiplication, remember? Only the second '$' gets squared, then you multiply the result by the first '$' to give you $$$.

Oh, I meant $$ as like a representation of 2 total. But in that case, since only the second $ gets squared and it is a singular unit it would be $*$^2 where $^2=$ thus $$^2=$$

ScottRiqui · Post by **ScottRiqui** » Mon Jun 16, 2014 11:40 am

Attax wrote: Oh, I meant $$ as like a representation of 2 total. But in that case, since only the second $ gets squared and it is a singular unit it would be $*$^2 where $^2=$ thus $$^2=$$

Just do use parentheses, bro:

($$)^2 = $$$$

Attax · Post by **Attax** » Mon Jun 16, 2014 11:43 am

ScottRiqui wrote:
Attax wrote: Oh, I meant $$ as like a representation of 2 total. But in that case, since only the second $ gets squared and it is a singular unit it would be $*$^2 where $^2=$ thus $$^2=$$
Just do use parentheses, bro:

($$)^2 = $$$$

What do you think I am? A STEM person?!?!?!

So then d/d$ of f($$) where f($$)=($$)^2 would = 2($$)

The world is pure again!

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