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I found one of your questions (I’m a blŏŏdy mind reader). It wasn’t yesterday, though. 

Here’s the link. 

(a) Find the total cost function, and the total revenue function.

(b) How many dresses have to be sown and sold per month for her to break even?

(c) Find the profit function, and determine the profit when 90 dresses are sown and sold per month.

(d) How many dresses have to be sown and sold per month to yield a monthly profit of e4,500?

(a)  TC  = 1350 + 34U

       TR = 79U

(b) Breakeven

1350 + 34U  = 79U

1350 = 45U

U = 30

(c) Profit  = TR -TC  

79(90) - [1350  + 34(90)] = 2700

(d)  

4500  = 79U - [ 1350 + 34U]

4500  = 45U - 1350

5850 = 45U

U = 130

Here's a graph of the profit function with e break-even point  and the points in (c) and (d)  plotted  :

https://www.desmos.com/calculator/johprfhcwy

A dressmaker has fixed monthly costs of e1,350. Her variable material costs per dress are e34, and she sells the finished product for e79 each

(b) How many dresses have to be sown and sold per month for her to break even?

(c) Find the profit function, and determine the profit when 90 dresses are sown and sold per month.

(d) How many dresses have to be sown and sold per month to yield a monthly profit of e4,500?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

b) P= 79n - (1350 +34n)

    P= 55n - 1350

    P= 1350/55 = n

breakeven = 24.54 or 25

c) P= 55(90) - 1350 = 3600

d) 55n -1350= 4500

    55n = 5850

    n=5850/55

    n= 106.36 (107)

Just re-post them.......this sometimes happens....I don't know why......!!!!

Do you want us to do your whole assignment for you?  Why would we want to do that?

1/7+2/7^2+1/7^3+2/7^4+...

Note that we can split this into two different sums

[1/7 + 1/7^3 +  1/7^5+......]  +  [ 2/7^2 + 2/7^4 + 2/7^6+....]

The sum of the first series =     [1/7] / [1 - 1/7^2 ]  = 7/48

And the sum of the second series  =  [ 2/7^2] / [ 1 - 1/7^2]  = 1/24

So.....the sum of the whole series =   7/48 + 1/24  =  7/48 + 2/48   =  9/48

When three standard dice are tossed, the numbers a,b,c are obtained. Find the probability that a*b*c=180.

180=18*10=9*2*5*2=2*2*3*3*5

So one of the numbers has to be 5

the other 2 both have to be 6

6*6*5=180

For ease lets assume that the dice are red green and blue.  Any of thes can be the 5 and the other two have to be 6 so that is 3 ways.

How many different combintations are possible.  6*6*6 = 216

So the probabily of getting a product of 180  is    3/216   =  1 / 72

a is correct.

If you show your working for the others I will check it.  They are too hard to do in my head.   :)

The probability is the the coefficient of x^17 in the expansion of the following generating series divided by 6^3.  expand | (x+x^2+x^3+x^4+x^5+x^6)^3

x^18+3 x^17+6 x^16+10 x^15+15 x^14+21 x^13+25 x^12+27 x^11+27 x^10+25 x^9+21 x^8+15 x^7+10 x^6+6 x^5+3 x^4+x^3.

Since the coefficient of x^17 is 3, therefore the exact probability is: 3 / 6^3=3/216 =1/72 =1.39%.

I don't know all those but I do know how to do (d).

\(\dfrac{dy}{dx}\\=\dfrac{d}{dx} \left(\ln \left(x^4\right)+e^x\right)\\=\dfrac{d}{dx}(4\ln x)+\dfrac{d}{dx}(e^x)\)

\(=4\dfrac{d}{dx}(\ln x)+\dfrac{d}{dx} e^x\)

\(=\dfrac{4}x+e^x\)

\(\begin{array}{rl}z_0=e^u&u=2x^2\\\end{array}\\ \dfrac{dz_0}{dx}=\dfrac{dz_0}{du}\times \dfrac{du}{dx}\leftarrow\text{ Chain rule}\\ \;\;\;\;\;\;\!=e^u\times 4x\\ \;\;\;\;\;\;\!=4xe^{2x^2}\)

You can see that I am differentiating z₀ =e^2x^2 using chain rule.

\(\dfrac{dz}{dx}\\ =\dfrac{d}{dx}(z_0)+\dfrac{d}{dx}(6)\\ =4xe^{2x^2}\)

\(\therefore\dfrac{dp}{dx}=\dfrac{dy}{dx}+\dfrac{dz}{dx}=\dfrac{4}{x}+e^x+4xe^{2x^2}\)

1+1-1+1-1+1-1+1+1+1+1-12312312321213123120

= 1+1+1+1+1-12312312321213123120

= 5 - 12312312321213123120

= -12312312321213123115

eight minus negative sixteen = eight plus sixteen = twenty four.

Question: \(f^{-1} \text{ of }f(x)=\frac{7-8x}{3}\)

\(\text{Or }y = \frac{7-8x}{3}\)

\(3y=7-8x\)

\(3y-7=-8x\)

\(x=\dfrac{7-3y}{8}\)

\(f^{-1}(y)=\dfrac{7-3y}{8}\)
\(f^{-1}(x)=\dfrac{7-3x}{8}\)

Finish!!

\(-\sqrt8 + \sqrt 2\)

\(=-2\sqrt2 + \sqrt 2\)

\(= -\sqrt 2\)

Alan answers and explains the details of the solution to this exact question here.

Yeah, but thats like just trial and error. You can't just say 13 because 13.

You start by the easiest fraction: 41/100, then factor by primes: 100 = 2*2*5*5 then check if 41 can be factorized by these numbers (41/5, 41/2) since this doesn't work, the simplest form of 41% is: 41/100.

 Ps. since 41 is a prime number you can just immediately tell that 41/100 is as far as you can go.

-6(0.1x+0.5)=0.6

0.1x+0.5=0.1

0.1x=-0.4

x=-4

Tc=5000+20n

Tr=40n

Pr=40n-(5000+20n)=20n-5000

P(500)=20×500-5000=5000

Break even

Pr=0.    So.    N=250unit

Pr =10000

N=7500 units

a number that does something without thinking 

Yes, you're correct!

You could change this:   log_x(0,01)  =  -2  into exponential form   --->   x^-2  =  0,01

                                                                                                      --->   x^-2  =  1/100

                                                                                                     --->   x²  =  100

                                                                                                      --->   x  =  10

thank you!!

(a) They save e 750 per month for 8 years at an annual rate of interest of 2%, credited monthly. How much do they have at the end of 8 years?

(b) They then take out a 50 year mortgage to buy the house, and can afford e 700 per month. If the annual rate of interest on the mortgage is 2%, debited monthly, how much can they borrow?

(c) (i) How much can they afford to pay for their house, including the deposit saved?

(ii) How much net interest will they pay over the 50 year period?

ANSWERS:

a) The couple will have saved a total of =78,009.56 Euros in 8 years.

b) The couple can borrow up to =265,361.97 Euros.

c) - (i) Since they can borrow the amount in b), they can afford a house worth: 265,361.97 + 78,009.56=343,371.53 Euros.

     (ii) The couple will pay a total of =154,638.03 Euros in interest over the 50-year period.

To find the probability that Doug survives, I'm going to find the probability that he walks the plank and subtract that number from 1.

For Doug to die in the first round, the other three must live and he must roll either an 8 or a 9.

First, what's the probability that anyone rolls either an 8 or a 9?

An eight can be rolled in 5 ways; either 6&2, 5&3, 4&4, 3&5, or 2&6.

A nine can be rolled in 4 ways; either 6&3, 5&4, 4&5, 3&6.

So, there are 9 different ways that anyone can lose out of 36 possible combinations: 9/36  =  1/4.

This also means that here is a 3/4 probability that the person wins.

For Doug to die on the first round, the other three must live and he must die:  

     (3/4)(3/4)(3/4)(1/4)  =  (3/4)³(1/4)  =  27/256

For Doug to die on the second round, all must live on the first round ( (3/4)⁴ ), the other three must live on the second round and he must die on the second round:  (3/4)⁴ · (3/4)³(1/4)  =  (3/4)⁷(1/4)

For Doug to die on the third round, all must live on the first two rounds ( (3/4)⁸ ), the other three must live on the third round and he must die on the third round:  (3/4)⁸ · (3/4)³(1/4)  =  (3/4)¹¹(1/4)

This can go on forever; what we must do is to add all these fractions together:

     (3/4)³(1/4)  +  (3/4)⁷(1/4)  +   (3/4)¹¹(1/4)  +  ...

This is an infinite geometric sequence whose first terms is  (3/4)³(1/4)  and whose common ratio is  (3/4)⁴.

The formula for the sum of an ininite geometric sequence with a ratio in the range:  -1 < r < 1 is:

     Sum  =  a / ( 1 - r)          where a is the first term and r is the common ratio.

     Sum  =  [ (3/4)³(1/4) ] / [ 1 -  (3/4)⁴ ]  =  27/175

But, this is the probability that he loses; the probability that he lives is  1 - 27/175  =  148/175.

x₀ = a

x₁ = b

x_{n + 2} = ( 1 + x_{n+ 1} ) / x_n

x₂₀₁₂ = ?

Let's write out a few terms:

x₀  =  a

x₁  =  b

x₂ = ( 1 + b ) / a  =  ( b + 1 ) / a

x₃  =  [ 1 + ( b + 1 ) / a ] / b          

Multiply this term by  a / a  and simplify:

x₃  =  [ 1 + ( b + 1 ) / a ] / b   ·  a / a     --->   x₃  =  [ a + ( b + 1 ) ] / ( ab )  

x₃  =  ( a + b + 1 ) / ( ab )

x₄  =  [ 1 + ( a + b + 1 ) / ( ab ) ] / ( b + 1 ) / a

Multiply this term by  (ab) / (ab)  and simplify:

x₄  =  [ 1 + ( a + b + 1 ) / ( ab ) ] / ( b + 1 ) / a   ·  ( ab ) / ( ab )   --->   [ ab + a + b + 1 ] / [ b( b + 1) ]

Factor the numerator:  ab + a + b + 1  =  a(b + 1) + (b + 1)  =  (a + 1)(b + 1)

x₄  =  [ (a + 1)(b + 1) ] / [ b( b + 1) ]  

x₄  =  (a + 1) / b

x₅  =  [ 1 + (a + 1)/b ] / [ (a + b + 1)/(ab)

Multiply this term by  (ab)/(ab)  and simplify:

x₅  =  [ ab + a(a + 1) ] / [ (a + b + 1) ]  =  [ a(b + a + 1 ] / (a + b + 1) ]

x₅  =  a          Which is just x₀!

x₆  =  [ 1 + a ] / [ (a + 1) / b ]

Multiply this term by  b/b  and simplify:

x₆  =  [ b(1 + a) ] [ (a + 1) ]   

x₆  =  b           Which is just x₁!

This shows that it repeats every 5 terms.

To find out the value of term #2012, divide 2012 by 5 and note the remainder.

2012 / 5  =  402  remainder 2.

Thus term #2012 equals terms #2     --->     x₂₀₁₂  =  x₂  =  (b + 1) / a