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Business maths (unlike finance) are not too difficult once you learn the terms and how to apply the related maths.

a) A 3% discount applies because it is paid within 30 days.

The Net amount due = net price - partial payment credit

Net amount = 28900 - 650 (Returned goods)= $28,250 (5000/0.97) reflects 3% discount

Net amount = 28250 - 5154.64

Net amount due = $23,745

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b) n/60 is the net date. It is 60 days from April 25 and that is June 24.

To conveniently count the days, you can use excel spread sheet or this link

There are a few factoring forms worth knowing, three of which are:

--- Difference of squares:  A² - B²  =  (A + B)(A - B)

--- Difference of cubes:     A³ - B³  =  (A - B)(A² + AB + B²)

--- Sum of cubes:             A³ + B³  =  (A + B)(A² - AB + B²)

In  x^6 - y^6,  first use difference of squares:

  =  (x³)² - (y³)²  to get:  (x³ + y³)(x³ - y³)    

Now use both sum of cubes and difference of cubes:

  =  [ (x - y)(x² + xy + y²) ][ (x + y)(x² - xy + y²) ]

  =  (x - y)(x + y)(x² + xy + y²)(x² - xy + y²)

What is x^3 + 1/8 when completely factored

$$x^3+\frac{1}{8}\\ \\
=x^3+\frac{1}{2^3}\\ \\
=x^3+ ( \frac{1}{2} ) ^3 \quad | \quad \textcolor[rgb]{1,0,0}{a^3+b^3 = (a+b)(a^2-ab+b^2) } \\\\
a = x \text{ and } b = \frac{1}{2} \\\\
x^3+ ( \frac{1}{2} ) ^3 = (x+\frac{1}{2})(x^2-x*\frac{1}{2}+(\frac{1}{2} )^2 )$$

I had a post agreeing with you chris but it has disappeared and I have changed my mind.  

I think I still like my answer.  There are only 5 places those those 3 can sit, that is true, but you need to take to acount where everyone else is sitting as well.  Anyway I am too tired to do more now.  See you tomorrow. :))

Ok....we know an angle, an adjacent side, and we're trying to find the opposite side.....so....the tangent function seems to be what we need...so we have....

tan (24) = opp/ 16.9     multiply both sides by 16.9

16.9 tan (24) = about 7.5 cm

i see :) how i think i kinda get it now. What if those 3 guys cannot sit with each other in any way, what would be the solution if each of the 3 dont want to be sitted to each other? what would you subtract to the total?

$${\mathtt{8}}{\mathtt{\,\times\,}}\left({\mathtt{\,-\,}}{\frac{{\mathtt{3}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}\right) = {\mathtt{\,-\,}}\left({\mathtt{9}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{8}}\right)\right) \Rightarrow \left\{ \begin{array}{l}\end{array} \right\}$$

You never need to appologise for admitting that you do not understand.  we want you to learn.  We do not to pretend to learn.

there are 144 ways those people can sit together. BUT the question asks you how many ways can they be seated so that they are NOT together.  That is why I subtracted from the total number of seating arrangements possible which is 7!

Oh my way any 2 of those guys can sit together, just not all 3 of them. :)

Here's my contribution to this ....

The total arrangements for any 7 people sitting in a row is 7!  = 5040

We can seat the 3 people together in five different ways.....(seats 1 - 5)

And for each of these there are 3! ways they can be arranged

So, the total number of ways they can sit together is:

(5 * 3!) = 30

So....the probability that they don't sit together is

1 - ( the probability they do sit together) =

1 - 30/5040  = 0.994047619047619    = about 99%

I don't know if this is correct......if anuone spots a flaw in my logic.....I'll be glad to know!!!

if there are 4!*3!= 144 total ways those guys can sit why subtract it to 7! ? when only those 3! guys dont want to sit with other? im just not at this sorry for asking things like this?

thank you very much CPHILL. and how to you solve it?

Verymuch appreciated the effort :) im having problems in trying to analyze probability problems and btw thankyou again :) to follow up on a question the 3 guys cannot sit with each other right?

I honestly don't get this......

What is the probability that 7 people can seat with a certain 3 people not in a consecutive order?

Lets so those 3 people are glued together.

Then it would be like seating 4 people.  

Sorry I can't count it would be 4 people and the cojoined triplets - that is 5 'bodies'.  I will correct it from here.

there are 5! ways of sitting 5 people in a  row = 120 ways

now thos 3 people can be joined together in any order so ther would be 3! ways = 6 ways.

So ther are 120*6= 720 ways those 3 people can be seated together.

now there are 7! ways of seating 7 people in a row.  That is 5040 ways

5040-720=4320  so there are 4320 ways to seat everyone without those 3 individuals sitting together.

So the prob that they will not be sitting together is  4320/5040

$${\frac{{\mathtt{4\,320}}}{{\mathtt{5\,040}}}} = {\frac{{\mathtt{6}}}{{\mathtt{7}}}} = {\mathtt{0.857\: \!142\: \!857\: \!142\: \!857\: \!1}}$$

Now it is a running joke around here - none of us are really probablity experts. So don't trust me to much :))

CPhill answer and mine now agree.  I believe this answer is correct    

OK....that's a horse of a different color.....!!!

So we have

cos^-1 (((3*sqrt(3)+4) /(5*sqrt(10)))   =  54.43573387782 degrees

And there you go.....!!!

I sure will!!!!

in a row i guess

Just subtract 3 from the "x value" of each coordinate and add 4 to each "y" value.....

Are they sitting in a row or around a table?

Can 2 of those people sit together?

It's an old joke between us, happy7......Melody and I are known to give some very "creative" answers to probability questions........some of them even turn out to be correct.....!!!!

Well 5*5=25

1*25=25

Woah....wait....what???

Yep......out of our minds !!!! LOL!!!

Not real high.  We must be sick.  This is a real anomoly.  LOL 

So x=2?

$${\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{4}}}{\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,-\,}}{\mathtt{3}} = {\mathtt{27}}$$