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If group  A  can assemble 16 cars in 8 hours, they can build 2 cars per hour

So  each hour in this group....we have 5 labor hours for the men and 6 labor hours for the women

And  group B can assemble  38 cars in 15 hours, so they can build 38/15 cars per hour

And each hour in this group, we have 6 labor hours for the men and 8 labor hours for the women

And  call the portion  of a car built  by a man  in one hour 1/x

And call the  portion of a car built by a women in one hour 1/y

So....we have this system

5 (1/x)  +  6 (1/y)   = 2

6(1/x) + 8(1/y)  = 38/15

Solving this system produces  x  = 5   and y  = 6

So....a man can produce 1/5 of a car per hour and a women produce 1/6 car per hour

So  for group C which consists of 8 men and  3 women.....

The number of cars produced in 10 hours is : 

8 man hrs per hour * (portion of a car produced by a man in one hour) * 10 hours   plus

3 women hrs per hour * (portion of a car produced by a women in one hour) * 10 hours  =

8 (1/5) (10)  + 3(1/6)(10)  =

80/5  +  30/6  =

16  +  5  =

21  cars

Expand    (a+b+c)(d+e+f+g)

=ad + ae + af + ag + bd + be + bf + bg + cd + ce + cf + cg [12 terms]

(x + 7 ) / ( x^2 - x - 2)   =   A  / (x - 2)   +  B / ( x + 1)

We can use partial fractions, here

We can write

( x + 7)  /[( ( x + 1)( x - 2)]  =  A / ( x - 2)  + B / ( x + 1)

Multiply through by  (x + 1) ( x -2)

x + 7  = A ( x + 1)  +  B( x - 2)    simplify

x + 7   = Ax   +  A   +   Bx - 2B

x + 7  =  ( A + B) x  +  ( A - 2B)

Equating coefficients, we have

A + B   = 1

A - 2B  = 7      subtract the first equation from the second 

-3B =  6

B  = -2

And  we can find  A  as  A + B  = 1 ⇒    A - 2  = 1   ⇒  A  = 3

So  (A, B)   = ( 3, - 2)

Thanks Gavin,

Thanks for posting this great examination of Pascal's Triangle. 

I have posed a link to this in our Reference thread (it is in the sticky notes) so that people should be able to find it well into the future. 

The secrets of   PASCAL'S TRIANGLE

GYanggg  has provided us with this great explantaion / examination of Pascal's triangle.

You say:

 "what is the number of partitions of the number 10 where the order matters" should be

"(9 choose 0)+(9 choose 1)+(9 choose 2)+...+(9 choose 9)=29=512"

But you are using combinations where order does NOT matter. It is "permutations" where order or position matters!!. Isn't that a contradiction???

Let's take this by stages

r^3m                        r ^4m                                                        r  ^4m/n  

______   =   _______        raising all this to 1/n  =      ______

r^-m * t⁴ⁿ           t ⁴ⁿ                                                                       t  ⁴

And

r  ^1/n                                                  (   t ^2/m )^m                  t ²

_____  raised to the  -m    =    _______   =        ______ 

  t  ^2/m                                               (    r ^1/n ) ^m               r ^m/n

So....putting all this together, we have

r ^4m/n   *  t²                 r ^3m/n

_________     =    ________

t⁴  *   r ^m/n                    t ²

Thank you for the hint! Your help is appreciated!

Let the other number = N

GCD[N, 315] = 9, solve for N

N = 9

LCM[N, 315] =7!, solve for N

N = 16

So, in order to satisfy both GCD and LCM, then:

N = 9 x 16 = 144, so that:

GCD[144, 315] = 9, and

LCM[144, 315] =7!, So;

N = 144 - the second number.

Max0815, you are very close; all you need to consider is the potential overlap. Look at the diagrams below.
 

The divisors of 40 are

1  2  4  5   8  10   20   and  40

But  x  is a positive integer..  and since x + 3  divides 40......the possible values for x are

1   2   5    7   17    37

So  when

x = 1       LCM  =  4  ( impossible....the LCM is at least 40)

x = 2       LCM  =  10

x = 5       LCM   = 40

So  x  =  5   and  the  other  integer  is  8

Um...Help?!

Yeah, I get it now.

Thanks, a lot CPhil.

The answer sheet I have to solve this question had many flaws, so I guess it was wrong again (it often is...)

Your explanation was great! Really thorough and detailed.

Thanks again. 

OK....glad to help   (if I can )  !!!

thank you so much dude  ill probably back later again lol. since im taking geo online. 

We  need to take the  cube root of the  ratio  of  the  volume of the larger solid to the smaller to  find the scale factor

So  we have   ∛ [ 1125 / 576  ]  = ∛  [ 125/64 ] =   5/4

So.....the correseponding side of the larger solid  =

Side length of the smaller solid  *  ( scale factor of larger solid to the smaller )

12  *  (5/4)  =

60  / 4  =

15  inches

Asd an adddendum.....when we are talking about volume comparisons of similar solids....we are either taking a cube root to find a scale factor....or we are cubing the scale factor to find a voliume

For surface area......we   are either taking a square root to find a scale factor, or we are squaring the scale factor to find a surface area

The surface area  of  Y  to X  is   (  891/ 539)  = 81/49

For any dimension from Y to X.....we can take the square root of 81/49  = 9/7

So....this is the scale factor  for  each dimension from Y to X

So....the height of Y  will be  =

Height of X  * ( scale factor of Y to X )   =

28  * (9/7) =

28 / 7  *  9   =

4  * 9  =

36 cm

The  scale factor  of the larger pyramid to the smaller is 12/9  =  4/3

The volume of the larger will be equal to

Volume of the smaller pyramid  *  ( scale factor ) ^3

So we have

162  * (4/ 3)^3  =

162  *  (64  / 27)  =

384  in^3

The scale factor from B  to A  is  (12/18)  = (2/3)

So...the surface area  of  B  =   the surface area  of A  *  the scale factor of B to A squared

So we have

1836  *  (2/3)^2  =

1836 *  (4/9)  =

816  cm^2

OK

When we cut all the washers out of the sheet   (  and before we cut any holes in the washers )  the area occupied by each washer  is just  pi  ( outside radius)^2  = pi (1 cm)^2  = pi cm^2

[Note that if the diameter of each washer is 2 cm....then the radius of each is just 1 cm  ]

So....the area occupied by each washer is that of a circle with a radius of 1 cm  =

pi (radius)^2  = pi  (1 cm)^2   =  pi  cm^2

So.....the total are occupied by these washers is  just  7500* pi  cm^2

And the area of the sheet  is  just  (300 cm * 100 cm)  = 30000 cm^2

So....the amount of excess metal surrounding the washers   is just the total area  of the sheet  less the total area occupied by all the washers   = [ 30000  - 7500 pi ] cm^2    (1)

When we take all the washers off the original sheet  [ 7500 of them ] we will cut the centers out  of each of them.....the amount of metal in the center  (the area of metal that will form the hole in the washer )  is just  pi * ( inside radius)^2  = pi  (1/2 cm)^2  = pi/4 cm^2

Since we have 7500 of these....the total area of excess metal from this part is just

7500 *[ pi /4 cm^2]  =   1875 pi  cm^2     (2)

So  the total amout of metal  that we have  to cut washers from is now just  the sum of (1)  and (2)  =

[30000 cm^2  - 7500 pi cm^2]  + 1875 pi cm^2  =  30000cm^2  - 5625 pi cm ^2  =

[ 30000  - 5625 pi ] cm^2   = area of the new sheet 

And each washer  cut from this sheet will  occupy the same area as before  = pi cm^2

So....the total number   of washers we can cut from the new sheet  is

[ Area  of new sheet  ] / [ Area of each washer ] =  [ 30000 - 5625pi] cm^2  /  pi  cm^2   =

[ 30000  - 5625 pi ]  / pi   =

about   3924  washers

Does that make sense  ??

Thanks CPhill 

So sorry, but could you please explain the final part of your explanation again? I'm trying to fully grasp the concept.

This is (part of) what you said: 

"So....the excess metal  surrounding all the washers is  just

Area of the sheet  - area  occupied by the washers =

(30000  - 7500pi) cm^2   (2)

So....the total amount of excess metal available  for forming new washers  is just  the sum of   (1)  and (2)  =  [  1875 pi  +  (30000 - 7500pi) ] =  [30000 - 5625 pi ]  cm^2

So....the total number of   possible washers   that can be made with a new sheet   is just

[  Area  of metal available  ]  / [ Area  occupied by each washer ]   =

[ 30000 - 5625pi ] cm^2  / [ pi] cm^2  ≈  3924 additional washers"

I just need help on the bolded part. Especially [pi] cm^2.

Thanks again.

I enjoyed it and I enjoyed it well.

Yup!

I totally agree with you, that's why I put together this list of all the patterns hidden inside the Pascal's  triange. 

I hope you enjoyed.