Melody

Brilliant Heureka :))

You have got my 5 points but you should add your own as well :

In fact, you should for all your answers     

If a manufacturer sells x units of a certain product, his revenue R and cost C (in dollars) are given by the following equations. Use the fact that "profit = revenue - cost" to determine how many units he should sell to enjoy a profit of at least $1750.
R = 22x
C = 5000 + 7x + 0.003x2

P=R-C

P=22x - (5000 + 7x + 0.003x^2)

P=22x - 5000 - 7x - 0.003x^2

P= - 0.003x^2 +15x-5000 

- 0.003x^2 +15x-5000 > 1750

- 0.003x^2 +15x- 6750 > 0

 0.003x^2 -15x+ 6750 < 0

roots

{x=4500, x=500} = {x=500, x=4500}

this is a concave up parabola so it will be below 0 in the middle

500 < x < 4500

f(1)=20 , f(n)=3*f(n-1)-60 find the 10th term

I have been thinking about this question ://      

I would do it the same as Heureka but I was just thinking about why this sequence is so well known.

It is important because it is the same logic that would be used for future value of an ordinary annutiity with an original deposit.

Say you start a  bank account off with P dollars. Then you add M dollars at the end of every month

The interest rate is 100r% per month that is r as a decimal. Each month the amount will grow by a factor of  1+r  which I will call R                              R=1+r

Beginning of month 1        F0  = P

End of month 1                  F1=   PR+M

End of month 2                  F2 =  R*F1 +M

                                          F2 = R(PR+M)+M

                                          F2 = PR^2+MR+M                                           

End of month 3                 F3 = F2 *R +M                                   so    F(n) = R*F(n-1)+M

                                          F3 = (PR^2+MR+M) *R +M

                                          F3 = PR^3 + MR^2 + MR +M

                                          F3 = PR^3 + M(R^2 + R +1)

--

End of nth month              Fn = PR^n + M(R^(n-1) + R^(n-2)...............+R +1)

This is a GP        a=1, r=R, n=n

(R^(n-1) + R^(n-2)...............+R +1) 

\(s_n=\frac{1(R^n-1)}{R-1}\\ s_n=\frac{R^n-1}{R-1}\\\)

\(\mbox{End of the nth month= }PR^n+\frac{M(R^n-1)}{R-1}\)

this is what your money will have accumulated to:

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

Relating this to the given problem - which is not a money problem but it is of the same form as an annuity.

f(1)=20 , f(n)=3*f(n-1)-60 find the 9th term     (It is the 9th because it starts at 1 instead of at 0)

P=20     R=3   M=-60           n=9

\(\mbox{End of the nth period= }T_n= PR^n+\frac{M(R^n-1)}{R-1}\\ \mbox{End of the 9th period= }T_{9} =20*3^{9}+\frac{-60(3^9-1)}{3-1}\\ \mbox{End of the 9th period= }T_{9} =20*3^{9}+\frac{-60(3^9-1)}{3-1}\\ \mbox{End of the 9th period= }T_{9} =20*3^{9}-30(3^9-1)\\ \mbox{End of the 9th period= }T_{9} = -196,800\\\)

Great thinking guest.       I think you have got it!!     I wonder if Riddler agrees ?

Hi Bertie,

Andre Massow was trying to find your access problem, there is a couple in you position but not many.

As far as I know he has not been able to reproduce the problem so I guess he hasn't solved it either. ://

It is nice to say hello to you though :D

The other one was an answer to a diophantine equation.  :)

I suggested it just as a play question. :))

Tues 24/11/15

Puzzle question for everyone.    Thanks Heureka and guest.

http://web2.0calc.com/questions/help-please_14808

i^61       Thanks Melody and CPhill

http://web2.0calc.com/questions/i-61

Proof using vectors.    Thanks Heureka

http://web2.0calc.com/questions/geometry-problem#r3

@@ What is Happening?  [Wrap4]   Mon 23/11/15   Sydney, Australia Time 6:50 pm   ♪ ♫

Good afternoon/morning,

We had some great answers today from CPhill, geno3141, anonnymous4338 and riddler, jc and other guests.

 Thank you

Interest Posts:

If you ask or answer an interesting question, you can private message the address to me (with copy and paste) and I will include it.  Of course only members are able to do this.  I quite likely will not see it if you do not show me.  

1) Riddler's riddles.    Geno got one but the other is unanswered. :)   Thanks Geno and Ridler :)

Let the picture be  4z by 3z     cm 

The frame is 2 cm bigger on all sides so it is 4z+4 and  3z+4  cm

Area of frame

A=(4z+4)(3z+4) = 12z^2 + 16z + 12z + 16 = 12z^2 + 28z + 16 

12z^2 + 28z + 16 =100

12z^2 + 28z -84 =0

6z^2 + 14z -42 =0

3z^2 + 7z -21 =0

\(z = {-7 \pm \sqrt{49+252} \over 6}\\ z = {-7 \pm \sqrt{301} \over 6}\\ z>0 \;\;so\\ z = {-7 + \sqrt{301} \over 6}\\ z = {-7 + \sqrt{301} \over 6}\\ z\approx 1.72489\)

3z=5.1746

4z=6.8996

So, if I have understood the question properly then the pic is  approx  5.17 by 6.90 cm

They are not the same thing.

Try watching this.

https://www.khanacademy.org/math/trigonometry/less-basic-trigonometry/law-sines-cosines/v/law-of-cosines-example

Find the greatest common factor. 8m3, 9m3 Write your answer as a constant times a product of single variables raised to exponents Can some one help me with this?

The greatest common factor of 8 and 9 is 1

and the greatest common factor of m^3 and m^3 is m^3

so

the greatest common factor of    \(8m^3\;\;and\;\;9m^3\;\;is\;\;1m^3\)