CPhill

2.  SQUARE

a) Vowels are always together

Note that  the vowels can appear in any one of these positions  [ V = vowel ]

V V V _ _ _

_ V V V _ _

_ _ V V V _

_ _ _ V V V

There are 4 positions where the vowels can appear together.....and for each of these the vowels can be arranged in 3! ways  =  6 ways    and the other letters can be arranged in 3!  = 6 ways

So......the total possible "words" that can be made where the vowels appear together is

4 *  6 * 6    =   144 "words"  =  144 arangements

b) Vowels are never together 

The vowels can appear in these positions

V _ V _ V _

_ V _ V _ V

So there are 2 possibilities here.....and for each of these, the vowels can be arranged in 3! ways = 6 ways  and the other 3 letters can be arranged in 3! ways  = 6 ways

So.....the total possible arrangements where the vowels never appear together is  just :

2 * 6 * 6    =    72  arrangements

c)  UAE  never appear together

First  note that the total possible  arrangements is just  6!  = 720

Let's count the number of arrangements where UAE  do appear together

U A E _ _ _

_ U A E _ _

_ _ U A E _

_ _ _ U A E

There are 4 of these....and for each, the other letters can be arranged in 3!  = 6 ways

So....the total arrangements where UAE appear together  is just :

4 * 3!  =  4 * 6   =  24

So....the number of arrangements where UAE never appear together  is just

Total arrangements  - Arrangements where UAE appear together  

720  - 24   =  696 arrangements

There will  be   12 x 12   = 144 possible outcomes

Outcome     Frequency      

2                    1                

3                    2

4                    3

5                    4

6                    5

7                    6

8                    7

9                    8

10                  9

11                10

12                11

                  ____ 

                    66

So....... the probaility of a number  >  12  =  

[ 144  - 66 ]   / 144  =

78 / 144  =  

39 / 72   =

13 / 24

Thanks for spotting my omission, X²

To find the domain restrictions, we just need to find the values that make the denominator = 0

So

x^2 + 2x  - 3   =   0       factor

(x + 3) ( x - 1)  = 0

Setting  each fact or to  0   and solving for x we have that

x +  3  = 0                             x  - 1  = 0

x  = -3                                    x  = 1

So x cannot   equal  -3   or  1   

Original ratio of experienced players to total players  =  4/10

When we add x experienced players, we get the new ratio of experienced players to total players =

[ x + 4 ] / [ x + 10]

The final number of players is just   x + 10

1 /  [ x - 1 ]   +  2 / x    =  x / [ x - 1 ]

Subtract the first fraction from  both sides

2 / x  =     x /  [ x - 1 ]  -   1  /  [  x - 1 ]

2 / x  =      [ x - 1] / [ x - 1]         cross-multiply

2[ x - 1 ]  =   x [ x - 1 ]       simplify

2x  - 2  = x^2  - x            subtract 2x from both sides....add 2 to both sides

x^2  - 2x  - x  + 2  = 0     simplify

x^2 - 3x +  2  =  0        factor

(x - 2) ( x - 1)  =  0

Setting each factor to 0  and soving for x produces  x = 2  or x  = 1

The second solution  makes two denominators inthe original problem  = 0

So....x = 2  is the only solution

The third answer is correct

√[ 2x  - 1 ]  - x + 2  = 0    rearrange as

√ [2x - 1 ]  =  x  - 2         square both sides

2x - 1  =  ( x-2)^2

2x  - 1  =  ( x - 2) ( x - 2)

2x - 1  =  x^2  -  4x +  4        subtract  2x from both sides, add 1 to both sides

x^2 - 4x - 2x + 4 + 1  = 0

x^2 - 6x + 5  =  0     factor

(x - 5)  ( x  - 1)  =  0

Setting each factor to  0 and solving for x  we get the possible solutions of

x = 5    or  x  = 1

Note that  x  = 5 solves the original equation   but  x  = 1 doesn't

So ... the last answer is correct 

1.     Rectangle 1 has length x and width y. Rectangle 2 is made by multiplying each dimension of Rectangle 1 by a factor of k, where k > 0.

(a)   Are Rectangle 1 and Rectangle 2 similar? Why or why not?

(b)   Write a paragraph proof to show that the perimeter of Rectangle 2 is k times the perimeter of Rectangle 1.

(c)   Write a paragraph proof to show that the area of Rectangle 2 is  k^2 times the area of Rectangle 1.

Not as bad as it looks

a)    Note that the   dimensions of Rectangle 2 are just   2x  and 2y

They are similar  because the ratio of their corresponding dimensions is the same

To prove this :

 Length of Rec 2     =  2x                   Width of Rec 2   =    2y

_____________         __   =  2           ___________          __     =  2

Length of Rec 1           x                     Width of Rec 1          y 

b)    You can write your paragraphs based on these  :

Perimeter  of Rec 1  =  2(W + L )    =  2 (x + y)

Perimeter of  Rec 2   =  2 (W + L )  =  2 (2x + 2y)  = 2 (2 (x + y))  =  4(x + y)   =

2 * [ 2(x + y) ] =   2 times the perimeter of Rec 1

c)

Area of Rec 1   =  xy

Area of Rec 2  =  2x * 2y  =   4 [xy]   =  4 times the area of Rec 1

Note that the scale factor here is   36/30   = 48/40  = 6/5

Since the prisms are the same height....the volume of the larger one will  be  the one with the base of the  triangle on the right

Since the prisims are the same height...the volume of the larger prism will be :

(scale factor)^2  as  much as the  smaller  =

(6/5) *  (6/5)      =  (6/5)^2  =    36/25   as much

Proof

Volume of smaller prism  =  base * height  =  (1/2)(40)(30) * height  =  600 * height

Volume of the larger prism  = base * height  = (1/2) (48)(36) * height  = 864 * height

So

Ratio  of  Larger Volume to Smaller    =    864 / 600   [  the height "cancels" ] 

And note that    600 * (6/5)^2  = 600 *  36/25  =   864

Does that help  ??