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I am srry but there is an answer to this concidering it says, "what is the last interger in her sequence." 

I think your sequence DIVERGES and does not converge !!

1,000,000,  500,000, 2,500,000, 1,250,000, 6,250,000, 3,125,000, 15,625,000.....and blows up!!!

I had a look at the link you provided, thank you kindly...I see it now...have a blessed day..

CPhill,

Thank you for this. I do appreciate...

Hint: Try to write an equation for the context, and use the given f(0)=1 to find the f(10) or find the f(1) to find the f(10). 

Also,,, if they say that the function has shifted 2 units to the right,,, then that would do something to the x axis wouldn't it... 

From what i have understood they are just asking you to translate a line (if its not exponential) 2 units the right and double the y axis... and they give the y intercept as (0,1). Use the information they have given you, and write an equation for it and solve for f(10). 
Now, im going to leave the rest to you. :)


Correct me if i am wrong please.

hope this helps in some way. 

Thank you so so so so sooo so much CPhill! :)

See here, Nirvana  :

 http://clas.sa.ucsb.edu/staff/lee/Secant,%20Tangent,%20and%20Derivatives.htm

it's all good CU! :) 

   1              1            1             1            1

___   +     ___   +    ___  +    ____  +  ____ +  ........     =

1^6           3^6         5^6          7^6         9^6

∞                                           pi^6

∑   1/ [2n +1]^6       =           _____ 

n = 0                                      960

Thanks so much! That was very helpful. Can you please help with the other problems too?

If the discriminant  is < 0,   then we will have non-real roots. In other words, the graph of f(x) will never intersect the x axis.  [ It will always be above the x axis ]

This might help : https://www.desmos.com/calculator/nlbuvyisgg

When p = 8/8 (or, 1)   the graph intersects the x axis twice

When p = 9/8   the graph is tangent to the x axis

When x  = 10/8     the graph is above the x axis

A different approach:

2(x-4)^2 -(x-4) + 7 =

2(x^2-8x+16) -x+4+7 =

2x^2 -17x+ 43

a+b+c = 2-17+43

When the graph of y=2x^2 -x+7 four units to the right, we obtain the graph of y=ax^2+bx+c. Find a+b+c.

2x^2 - x + 7 

The x coordinate of the vertex is     1 / [2*2]  = 1/4

And the y coordinate of the vertex   =   2(1/4)^2 - (1/4) + 7  =  1/8 + 1/4 + 7  =    55/8

When shifted four units to the right the vertex becomes   (17/4, 55/8)

So we have that

2(x -17/4)^2 + 55/8       simplify

2(x^2 - (17/2)x + 289/16)  + 55/8

2x^2 - 17x + 289/8 + 55/8

2x^2 - 17x + 43

So

a + b + c  =   2  - 17   +  43   =    28

there's introduction to number theory by aops! i heard it's really helpful :)

     1      2      1            5  

     2    - 1     2          15  

     3      1   – 1            8  

 The diagonal field of the 1st line is already 1.

 With the 1st row, all other rows in the 1st column are set to 0.

 To the 2nd row the -2-fold of the 1st row is added:

     1      2      1          5  

     0    - 5      0          5  

     3      1    - 1          8  

 The -3-fold of the 1st line is added to the 3rd line:

     1       2       1          5  

     0     - 5      0          5  

     0     - 5    - 4       - 7  

 By dividing the 2nd line by -5 the diagonal element is made to 1:

     1     2     1       5  

     0     1     0     - 1  

     0   - 5  - 4     - 7  

 With the 2nd row, all other rows in the 2nd column are set to 0.

 The 2nd line is added to the 1st line by a factor of -2:

     1     0     1       7  

     0    1      0    - 1  

     0  - 5   - 4    - 7  

 The 5-fold of the 2nd line is added to the 3rd line:

     1     0     1        7  

     0     1     0      - 1  

     0     0   - 4     - 12  

 By dividing the 3rd line by -4 the diagonal element is made to 1:

     1     0     1        7  

     0     1     0      - 1  

     0     0     1        3  

 With the 3rd row, all other rows in the 3rd column are set to 0.

 To the 1st line the -1x of the 3rd line is added:

     1     0     0       4  

     0     1     0     - 1  

     0     0     1       3  

 The last column contains the solutions.

https://prnt.sc/pc1zxj

\((2-i)z+(2+i)\bar{z} = 20\\ z = x + i y\\ (2-i)(x+iy) + (2+i)(x-i y) = 20\\ (2x+y) - i(x-2y)+(2x+y)+i(x-2y)=20\\ 2x+y=10\\ y=-2x+10\\ y=0 \Rightarrow x=5\\ \text{Intersection is at $z=5$}\)

  I love the image of the sniper morphing into Nikita Khruschev and pounding his shoe... funny, and prophetic too.  Wonderful idea.  As long as we're devising alternatives, why not simply omit the sentence?  Don't have the sniper throw down the revolver at all.  Of course, that deprives us of the opportunity to second guess the author for his error about the firearm discharging.  

.

yes sir thank you !

To calculate the cost of Polly's ingredients

[ 2 2 5 4 ]  [ 3]

                   5              =   [ (2(3) + 2(5) + 5(2) + 4(6) ] =  [ 6 + 10 + 10 + 24] =  [ 50]

                   2

                  [6]

Notice that this is the last entry in  the matrix  [ 37  ]

                                                                           92

                                                                         [  50 ]

To get the cost of  Patricia's ingredients  we would have

[ 6 6 4 6 ] [ 3 ]

                  5       

                  2

                [ 6 ]            The result of this should be 92   [the second entry]

And finally, the cost of Pauline's ingredients is

[ 3 4 1 1 ] [ 3]

                  5

                  2

                [6 ]        The result of this should be 37  [the first entry ]

Does this make sense ?????

This is lengthy.......be careful......even one mistake is deadly!!!

2   1   -3       -19

1   2   1          -4

1  -1   5         21                multiply R1 by (-1/2)   add to R2

2     1      -3     -19

0   3/2    5/2     11/2           multiply R2  by  2

1   -1        5       21   

2    1     - 3         -19

0    3      5            11

1    -1     5            21              multiply R1 by (-1/2)   add to R3

2    1       - 3      -19

0    3        5         11

0   -3/2   13/2     61/2         multiply  R3  by 2

2     1    -3         -19

0     3     5           11

0    -3    13         61               multiply R2 by (-1/3)  add to R1

2      0    -14/3    -68/3

0      3      5         11

0      -3     13       61              add R2 to R3   put result in R3.....multiply R1  by 3

6   0      -14           -68    

0    3        5            11

0    0       18           72             divide R1   by 2 , R3  by 18

3     0    - 7           -34

0    3      5             11

0    0      1              4            multiply R3 by 7.....add to R1

3     0     0             -6

0     3     5             11

0     0     1              4           multiply R3 by -5....add to R2

3  0   0       -6

0  3   0      - 9

0   0 1         4            divide R1, R2  by 3

1  0   0     - 2

0   1   0    - 3

0   0   1      4

Read across  and the solutions are  (x, y , z}  =   {-2, -3, 4 }

is the result for the 3 * 1 matrix the answer for #2?

If she took out as much money  each day from day 8 to day 12  as she put in from day 0 to day 4, she should have $200 in her account [ her beginning balance ] on Day 12

So....she must have taken more out each day since she has $0  on day 12

Yep....good job!!!

Now....just perform the multiplication of the two matrices

 [ You should end up with the last answer in the resulting 3 x 1 matrix  ]

On day 0....the balance was ≈ $200

On day 4, the balance was $400

So.....we can  form a linear equation from day 0  to day 4

The slope  is  [ 400 - 200] / [4 - 0 ]  =  200/4  =  50

This indicates that the balance is increasing by $50  from day 0 to day 4

So

Day 0  = $200

Day 1 = $250  = B(1)

Day 2  = $300 = B(2)

Day 3 = $350  = B(3)

Day 4 = $400