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Ah...yes...i see what you mean...I need to re-think this  !!!

aren't you subtracting arrangements like "B-E-G-C-A-D-F" twice?

We are counting all the possible words  (arrangements) = [ 5040 ]    and just subtracting those where "beg"  or "cad" appears  = 2 [120]  =   5040 - 240  = 4800

i'm probably wrong but...

shouldn't the answer be (number of possible words)-(number of possible words with beg)-(number of possible words with cad)+(number of possible words with beg and cad)=4800+(number of possible words with beg and cad)?

3.   f(x)  =  ³√x      ⇒   f(x)  =  -2³√x  -  1

The  "-"  out front reflects the  graph across the x axis

The "2"  "vertically stretches " the graph away from the x axis

The  "-1"  at the end shifts the graph down 1 unit

2.  f(x)  =  2 √x

Note that we cannot put any negative values in the square root  [ this generates a non-real number ]

So  the domain is   [ 0, inf )

And  out of a positive square root, we can never get any  negative values

So.....the range is the same   [0, inf )

Here's a graph : https://www.desmos.com/calculator/hndidezpmc

1.  f(x)  =   ³√x   +  2

We can put in any real number for x.....so the domain  is just  (-inf, inf)

And any real  number can be generated from the function

So.....the range is also   (-inf, inf)

Here's the graph  :  https://www.desmos.com/calculator/sd0menwuyg

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

We'll use a sin graph since at x=0 the buoy is at the y midline....

the period is 8 seconds  (highest point to lowest point and back to highest point)

Frequency is   1/period     1/8

Starts at 20  (sine wave shifted up 20 m)

Amplitude  2   (up 2 meters then down 2 meters below norma ocean depth)    .....( I think this is what you meant in your question.....if it bobs UP 2m then back to normal depth then down 2m)

2sin((2pi x)/8)   + 20

Picture:

CPhill i have no idea about it!? 

#

Can  put these in any one of 5 positions    12, 23, 34, 45, 56

And for each of these, there are 4!  ways to arrange the other letters

#

what do you mean byt he first line?i thought there were 5! ways to arrange them???

can you pls make it a little simple....i really cant see the permutations in them....its too confusing!Thanks!

?huh?

i mean i get the joke but what?

I assume the question is:

$\mathbf{\text{A sequence of positive integers with $a_1=1$ and $a_9+a_{10}=646$ is formed $\\$ so that the first three terms are in geometric progression, $\\$ the second, third, and fourth terms are in arithmetic progression, $\\$ and, in general, for all $n\ge1,$ $\\$ the terms $a_{2n-1}, a_{2n}, a_{2n+1}$ are in geometric progression, $\\$ and the terms $a_{2n}, a_{2n+1},$ and $a_{2n+2}$ are in arithmetic progression. $\\$ Let $a_n$ be the greatest term in this sequence that is less than $1000$. Find $ a_n $.}}$

Let AP = arithmetic progression,

Let GP = geometric progression

$\text{AP Formula:}\\ \begin{array}{|llcll|} \hline \text{ $t =$ term }\\ \text{ $i,j,k = $ indices}\\\\ t_i(j-k) + t_j(k-i) + t_k(i-j) = 0 \\\\ & i = 2n & t_i = a_{2n} \\ & j = 2n+1 & t_j = a_{2n+1} \\ & k = 2n+2 & t_k = a_{2n+2} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline t_i(j-k) + t_j(k-i) + t_k(i-j) &=& 0 \\ a_{2n}(2n+1-(2n+2)) + a_{2n+1}(2n+2-2n) + a_{2n+2}(2n-(2n+1)) &=& 0 \\ \mathbf{-a_{2n} + 2a_{2n+1} - a_{2n+2}} &\mathbf{=}& \mathbf{0} \\ \hline \end{array}$

$\text{GP Formula:}\\ \begin{array}{|llcll|} \hline \text{ $t =$ term }\\ \text{ $i,j,k = $ indices}\\\\ t_i^{j-k} \times t_j^{k-i} \times t_k^{i-j} = 1 \\\\ & i = 2n-1 & t_i = a_{2n-1} \\ & j = 2n & t_j = a_{2n} \\ & k = 2n+1 & t_k = a_{2n+1} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline t_i^{j-k} \times t_j^{k-i} \times t_k^{i-j} = 1 \\\\ a_{2n-1}^{2n-(2n+1)} \times a_{2n}^{2n+1-(2n-1)} \times a_{2n+1}^{2n-1-(2n)} = 1 \\\\ \mathbf{a_{2n-1}^{-1} \times 2a_{2n}^{2} \times a_{2n+1}^{-1} } &\mathbf{=}& \mathbf{1} \\ \hline \end{array}$

So we have:

$\begin{array}{|lrcll|} \hline (1) & \mathbf{a_{2n+2}} &\mathbf{=}& \mathbf{2a_{2n+1} -a_{2n} } \\ (2) & \mathbf{a_{2n+1} } &\mathbf{=}& \mathbf{\dfrac{ a_{2n}^{2} }{a_{2n-1} } } \\ \hline \end{array}$

$\mathbf{a_1=1 }\\ \begin{array}{|rcll|} \hline a_3 &=& \dfrac{a_2^2}{a_1} \quad a_1 = 1 & | \quad \text{ Formula $(2)$} \\ \mathbf{a_3} & \mathbf{=} & \mathbf{ a_2^2 } \\ \hline a_4 &=& 2a_3-a_2 \quad a_3 = a_2^2 & | \quad \text{ Formula $(1)$} \\ &=& 2a_2^2-a_2 \\ \mathbf{a_4} & \mathbf{=} & \mathbf{ a_2(2a_2-1) } \\ \hline a_5 &=& \dfrac{a_4^2}{a_3} & | \quad \text{ Formula $(2)$} \\ a_5 &=& \dfrac{a_2^2(2a_2-1)^2}{a_2^2} \\ \mathbf{a_5} & \mathbf{=} & \mathbf{ (2a_2-1)^2 } \\ \hline a_6 &=& 2a_5-a_4 & | \quad \text{ Formula $(1)$} \\ &=& 2(2a_2-1)^2-a_2(2a_2-1) \\ \mathbf{a_6} & \mathbf{=} & \mathbf{ (2a_2-1)(3a_2-2) } \\ \hline a_7 &=& \dfrac{a_6^2}{a_5} & | \quad \text{ Formula $(2)$} \\ a_7 &=& \dfrac{(2a_2-1)^2(3a_2-2)^2}{(2a_2-1)^2} \\ \mathbf{a_7} & \mathbf{=} & \mathbf{ (3a_2-2)^2 } \\ \hline a_8 &=& 2a_7-a_6 & | \quad \text{ Formula $(1)$} \\ &=& 2(3a_2-2)^2-(2a_2-1)(3a_2-2) \\ \mathbf{a_8} & \mathbf{=} & \mathbf{ (3a_2-2)(4a_2-3) } \\ \hline a_9 &=& \dfrac{a_8^2}{a_7} & | \quad \text{ Formula $(2)$} \\ a_9 &=& \dfrac{(3a_2-2)^2(4a_2-3)^2}{(3a_2-2)^2} \\ \mathbf{a_9} & \mathbf{=} & \mathbf{ (4a_2-3)^2 } \\ \hline a_{10} &=& 2a_9-a_8 & | \quad \text{ Formula $(1)$} \\ &=& 2(4a_2-3)^2-(3a_2-2)(4a_2-3) \\ \mathbf{a_{10}} & \mathbf{=} & \mathbf{ (4a_2-3)(5a_2-4) } \\ \hline \end{array}$

$\mathbf{a_9+a_{10}=646}\\ \begin{array}{|rcll|} \hline \mathbf{a_9+a_{10}} & \mathbf{=}& \mathbf{646} \\ (4a_2-3)^2+(4a_2-3)(5a_2-4) &=& 646 \\ (4a_2-3) (4a_2-3+5a_2-4) &=& 646 \\ (4a_2-3) (9a_2-7) &=& 646 \\ \ldots \\ 36a_2^2-55a_2-625 &=& 0 \\\\ a_2 &=& \dfrac{55\pm\sqrt{55^2-4\cdot36\cdot(-625)} }{2\cdot 36} \\\\ &=& \dfrac{55\pm\sqrt{93025} }{72} \\\\ &=& \dfrac{55\pm305 }{72} \\\\ a_2 &=& \dfrac{55+305 }{72} \\\\ \mathbf{a_2 } & \mathbf{=} & \mathbf{5} \\\\ && \text{or} \\\\ a_2 &=& \dfrac{55-305 }{72} \\\\ a_2 & = & -3.47222222222 \\ && \text{no solution, because $a_n$ is a sequence}\\ && \text{of positive integers!}\\ \hline \end{array}$

$\text{Let $a_n$ be the greatest term in this sequence that is less than $1000$. Find $ a_n $}$

$\begin{array}{|lclcl|} \hline a_1 && &=& 1 \\ a_2 && &=& 5 \\ a_3 &=& 5^2 &=& 25 \\ a_4 &=& 5\cdot 9 &=& 45 \\ a_5 &=& 9^2 &=& 81 \\ a_6 &=& 9\cdot 13 &=& 117 \\ a_7 &=& 13^2 &=& 169 \\ a_8 &=& 13\cdot 17 &=& 221 \\ a_9 &=& 17^2 &=& 289 \\ a_{10} &=& 17\cdot 21 &=& 357 \\ a_{11} &=& 21^2 &=& 441 \\ a_{12} &=& 21\cdot 25 &=& 525 \\ a_{13} &=& 25^2 &=& 625 \\ a_{14} &=& 25\cdot 29 &=& 725 \\ a_{15} &=& 29^2 &=& 841 \\ \color{red}a_{16} &=& 29\cdot 33 &\color{red}=& \color{red}957 \qquad a_n < 1000\\ a_{17} &=& 33^2 &=& 1089 \qquad a_n > 1000 \\ \hline \end{array}$

$\mathbf{\text{$a_n$ is $ 957$}}$

$\text{GP:}\\ \begin{array}{|r|r|rrrl|} \hline n & \text{ratio} & a_{2n-1} & a_{2n} & a_{2n+1} \\ \hline 1 &5 &1 & 5 & 25 \\ 2&1.8 &25 & 45 & 81 \\ 3&1.4\ldots &81 & 117 & 169 \\ 4&1.30769230769 &169 & 221 & 289 \\ 5&1.23529411765 &289 & 357 & 441 \\ 6&1.19047619048 &441 & 525 & 625 \\ 7&1.16 &625 & 725 & 841 \\ 8&1.13793103448 &841 & 957 & 1089 \\ \ldots & \ldots& \ldots& \ldots& \ldots \\ \hline \end{array}$

$\text{AP:}\\ \begin{array}{|r|r|rrrl|} \hline n & \text{common difference} & a_{2n} & a_{2n+1} & a_{2n+2} \\ \hline 1 &20 & 5 & 25 & 45 \\ 2 &36 &45 & 81 & 117 \\ 3 &52 &117 & 169 & 221 \\ 4 &68 & 221 & 289 & 357 \\ 5 &84 &357 & 441 & 525 \\ 6 &100& 525 & 625 & 725 \\ 7 &116& 725 & 841 & 957 \\ \ldots & \ldots& \ldots& \ldots& \ldots \\ \hline \end{array}$

(-1) - (-9)

Think of negative numbers as COLD numbers (the freezer has a negative temp)

an positive numbers are HOT numbers.

Here you have  a temp of -1  you want to take away some cold (because -9 is a cold number)

If you take cold out of a room then the room gets warmer!

So take away a negative can replaced by adding.

Taking  away some cold air has the same effect as adding some warm air.

 Both times the room gets warmer.   Do you get it?

      - - = + + = +

so

(-1)-(-9) = (-1) + (+9) = -1+9 = 8

or

(-1)-(-9) = -1 - -9 = -1+9 = 8

Thank you Alan that is ever so kind of you. 

I will continue asking you to check my answers for now but hopefully at some point I will learn Python and be able to use it to do my own checks. I really like the idea of programming again and I'd like to be able to check my own probablility solutions too.

Since you have said it is alright, I will keep asking you to check my answers. At least I will for the moment.

Thank you, I really appreciate your input and help :)

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

Melody,

(1) You don’t need any of the MATLAB add-ons to run Monte Carlo simulations.

(2)  Python is an increasingly popular programming language these days. I haven’t used it myself, so can’t say much about it, but I suspect it’s fairly easy to use (and can certainly be used for MC simulations), especially if you have previous programming experience in Fortran. 

(3) MC in Excel is certainly clumsy!

(4) I’m also happy to continue running MC checks!

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  ??