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Refer to here:

I think admin has to choose you for that. You have to prove that you will be a good moderator and that you'll contribute to the forum I guess.

Let q  be the number of quarters and d  be the number  of dimes

Then we have this system :

q  +  d    =  600  →   d = 600  - q    (1)

.25q + .10d  = 123.75      (2)

Sub (1)  into (2)  for d and we have that

.25q + .10 ( 600 - q)   = 123.75    simplify

.25q  + 60  -.10q  =  123.75

.15q   + 60 = 123.75      subtract 60 from both sides

.15q  =  63.75      divide both sides by  .15

q = 425  → she has  425 quarters

thanks

0 / 0   shows  an "error"  because  this is an undetermined quantity

Remember that  if b is not equal to 0 and  b divides a, then  a / b   =  c   

So...for instance

6/2 = 3      and note that    3 * 2   =  6

But

0/ 0    could equal anything     because   anything * the denominator  =   0

sqrt169 = 13

13*5 = 65

Thanks for the A2A. 

There is no gain or loss in the deal as 800 rupees is given and you recieve 800 rupees as well.

Cheers!

In the figure below, the horizontal lines are parallel and the bisectors of the two interior angles intersect at T.

Which is m?

Given A(6, -17), B(-6, 17), C(4, 9), D(-4, 4), the midpoint of which segment is at the origin?

a and b here are not rational

I guess I shouldn't have put the emphasis on the word "irrational" in my earlier reply.  However, I think the terms with the surds should come in  complementary pairs.  For example, the cubic can be rewritten as follows:

All solutions are (I suspect) irrational, but two are in the form a + b√3 and a - b√3.

Stage 1: the two captains are called  -->  total number = 2

Stage 2: four new players are called  -->  total number = 6

Stage 3: eight new players are called  -->  total number = 14

Stage 4: sixteen new players are called  -->  total number = 30

Stage 5: twenty new players are called  -->  total number = 50

Thanks for the A2A.

90 - 90(.70) = 27 dollars

Cheers!

thanks for the note:) 

(127)^4−x     =    9^2x−1

should be written as

127^(4−x)     =    9^(2x−1)

Looking for real solutions.

\(127^{4−x}     =    9^{2x−1}\\ log(127^{4−x} )    =    log(9^{2x−1})\\ (4-x)log(127 )    =    (2x-1)log(9)\\ 4log127+log9 = 2xlog9+xlog127\\ 4log127+log9 = x(2log9+log127)\\ x=\frac{4log127+log9}{2log9+log127} \\ \)

x is approximately 2.335

ok thanks Alan.

Hi guest,

I looked here....... but how do you know for sure that they are all irrational ?

True!  Perhaps I just got lucky!

I just had fun drawing this curve

\(y=a(x^4-8x^3+8x^2+52x-77)\)

In the case of this picture 'a' is negative

It is known that the number of Blue is seven times the number of Green: \(B=7\cdot G\) 
Also that, \(B=63\). Then calculate green: 

\(B=63=7\cdot G \Rightarrow G=63/7=9\).

The pieces altogether is \(B+G+R=63+9+23=95\)

Assuming there are 23 pieces of red kuti kuti.

hoped i helped

hi thanks for the response! I get it except for the - 4/5 in the second question. Why is it not y=(2/5)x - 2?

(oh wow i'm really losing my mind from sleep deprivement aren't i :-) 2/5 * 2 = 4/5)

Alan i don't think thats right. There are 3rd degree polynomials with rational coefficients that have only irrational roots (for example x³-3x+1)

Ok, got it.  Thanks Alan  

x^-3 = 1/27

x^6 = 300

15 + 3 * t^04 = 21

12x^2/3 = 48

Thank you very much!

Exponentiation takes priority over negation, so -2^4 is -(2^4).  If you want the negation to take priority you would need to emphasise this by writing (-2)^4.

In the graph a line intersects the points A(0,1) and B(5,3). 

a. Give an equation for line l

b. Give an equation for line m that is parallel to line l and goes through the point of intersection C(2,0)