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Since f(x) is divisible by x^3, f(x) is of the form ax^5 + bx^4 + cx^3.

You then want ax^5 + bx^4 + cx^3 + 2 to be divisible by (x + 1)^3.  Using long division, you get the equations

-10a  + 6b - 3c = 0

4a - 3b + 2c = 0

-a + b - c + 2 = 0

==> a = 6, b = 16, c = 12

So f(x) = 6x^5 + 16x^4 + 12x^3.

1 +  2  + 2^2   +  2^3  + 2^4  + 2^5   + 2^6  + 2^7 + 2^8  + ...... + 2^100   =

( 1 + 2 + 2^2)  + 2^3  ( 1 + 2 + 2^2)  + 2^6 ( 1 + 2 + 2^2)  +  ........ + 2^96 + 2^97 + 2^98+ 2^99 + 2^100    =

(  7)   +  8(7)   +  64 ( 7)  +  ...... +  2^96 ( 1 + 2 + 2^2) + 2^99 + 2^100

7  +  8(7)  +  64 (7)  +.........+   2^96 ( 7)   +  2^99 + 2^100

Every factorization  is divisible  by  7

Looking at the last two terms

Note   that      2^99  + 2^100  =   2^99 ( 1 +2)  =   3 * 2^99  =   3 * (2^3)^33

And

2^3  mod 7  =1

So

(2^3)^33 mod 7   = 1

So

3 *  (2^3)^33  mod 7    =

3 (1)   =  3  =   the remainder

c = λν

3 x 10^8   = 98 x 10^6  (λ)

λ = 3 x 10⁸ / 98x10⁶

   = 3.06 m

No : the answer wa possible areas are 1,2,4,5, and 9 . So sum of all possible areas was 21

YOU THROW A ROCK, BUT ASK WHEN THE BALL REACHES 8 FT?   oops!

8 = -16t^2 + 50t+5 

0 = -16t^2 + 50t -3       Quadratic formula shows t =0.0611985     OR   3.0638 SECONDS   

      (One is for thr rock-ball on the way up...the other on the way down)

really, oh oops...I thought no one answered it, but I'll check it out

Here's what I got  :

1            2          11       1010

10        20        101       1100

100      200      110           

1000    2000    1001   

Using  synthetic division

1    [     a       b       c    ]

                     a     b + a

       ________________________

           a     b + a    a + b + c   =  2

2  [  a      b       c    ]

               2a   2b + 4a

    _______________

      a    b +2a     4a  + 2b + c  = 8

We  have  this system

a + b  +  c  =  2

4a + 2b + c  = 8            

We will  have  infinite solutions....however....if we assume  a  monic poynomial  (a = 1)  then we  have a solution

1 + b + c  = 2   ⇒   b + c   =   1    ⇒  c =  1  - b      (1)

4 + 2b + c  = 8      (2)

Sub (1)  into (2)

4 + 2b + 1  - b   =  8

5 + b  = 8

b  =  3

c  = 1 - 3   =   -2  

a  =1     b  =  3     c    =  -2

(x,y) = (5,-21).

Volume  of first cone =  pi (a)^2  h  /  3

Volume of second cone =  pi (b^2) h  / 3

So

Volume  of   cylinder  =  combined volumes of  cones   =   pi * h ( a^2  + b^2)  / 3

So

pi * r^2 *  h    =   pi * h ( a^2 + b^2)  /  3

r^2  =   (a^2 +b^2)   / 3

r  =   sqrt  [  (a^2 +b^2) / 3  ]   =    radius of  the  cylinder

Third solution:  (moomoo 's was the best !)

10b+a = 27 + 10a  + b

9b-9a=27

b+a = 15   multiply by 9  to get

9b+9a= 135       add red equations to get

18b = 162

b = 9                  then    a + 9 = 15      a = 6

69

Call the digits  a  and  b

So

a +  b  =  15     ⇒  b =  15  - a

So

Call  the original number  10a + (15 - a)      and call  the resulting number  10(15 - a)   + a

So     27 added to the otiginal number  = the resulting number

10a  +  ( 15 - a)   +  27 =   10 ( 15 - a)  + a

9a + 42   =  150 -  9a

18a  =   150  -  42

18a  = 108

a=  108 / 18   =   6

And  b  =15  - 6  =  9

The original number=    69

The resulting  number  =  69+ 27 =  96

Equating times  we  have this system

D/ ( R + 6)   =   D/R -  4

D/ (R - 6) =  D/R + 6                       simplify

D / (R + 6)  =  (D  - 4R)  / R

D/ ( R -6)   =  ( D + 6R) / R

DR  =  (R +6) ( D - 4R)

DR  = (R - 6) ( D + 6R)

DR  = DR  +  6D - 4R^2  - 24R

DR  = DR - 6D + 6R^2   -36R

0  =  -4R^2 + 6D  -  24R

0  =   6R^2  - 6D  -36R             add  these

0 = 2R^2  + 60R

2R ( R - 30)  = 0

Setting the second factor to  0   and  solving for R   produces  the  normal  rate of     30 Km/ hr

And

-4 (30)^2  +  6D  - 24 (30)   = 0

-3600 +  6D  - 720   = 0

6D  =  3600 +  720

6D  =  4320

D = 4360  /  6  =   720  km =   distance of the journey

Possible numbers

87

78 

96

69

The number has to be smaller than its opposite so

78

69

87-78=9

96-69=27

So the number is 69

Don't know where ya went....

   derivative of the velocity function is the acceleration function....you got that !

      derivative of the velocity function also shows the slope of the velocity function at each point

             where the slope goes from positive to negative    or   from  negative to positive  

                  the acceleation is too....these are the points where velocity will be a maximum (maybe negative or  maybe positive value)

                      these occur where the acceleration function = 0   (slope = 0    as it goes from pos to neg or  neg to pos)

the derivative of the given velocity funtion is given as    -10(s^2-4s-4) / (s+4)^2

                                                                                       -10(s^2-4s-4)/(s+4)^2 = 0

                                                                                         -10s^2 +40s+40 = 0                 Use Quadratic Formula

                                                                                             s = 2+- 2 sqrt2               these are the points where accel = 0

Plug these values into the orignal Velocity equation to calculate the values......pick the one that is maximum (here I do not know if they want most positive or greatest magnitude)

Use s =3    in the derived ACCELERATION equation to calculate the acceleration at s = 3 

   The graph above should help to verify your answers......  

First one

g(x) =  (x + 1)^2 =  x^2  + 2x + 1

f(x)  = 4x  - 7

f(x) + g(x)   =  x^2 + 6x   -   6  = s(x)

s(3)  =  (3)^2  + 6(3)  - 6    =   21

Second one

f(x)   -  g(x)     =  ( 2x^2 + 3x - 9)  - (  5x + 11) =  2x^2  - 2x  - 20

f(x) - g(x) + h(x)    =    (2x^2  - 2x  - 20)  +  ( -3x^2 + 1)  =  -x^2 -2x -19

Last one

f(g)   means  that   we are  putting   g  into  f....so we have

3 ( x^2 -5x   -1 )  +   2  =

3x^2 - 15x  - 3  + 2

3x^2  - 15x   - 1

Didn't Max Wong answer this a few days ago?

im an idiot... i probably should have known LNVOLUTE wasn't a word...

3 out of 4 would be 4 combinations

2 out of 4 would be 6 combinations

4*6=24

24*120=2880

To find the inradius

Calculate the semi-perimeter

semi-perimeter, S,     (4 + 5 +6)   /  2  =    15/2 =  7.5

Use  Heron's  Formula  to find the Area, A

A = sqrt  [ S  ( S -A)(S- B) (S - C) ]  =   sqrt [ 7.5 * 1.5 * 2.5 * 3.5 ]

A^2  =    (7.5 * 1.5 * 2.5 * 3.5) =  98.4375

Area =   rS     

A/S =  r   =  A  /7.5

Circumradius =      product of the  sides  /  (4  * area)  =      4*5*6   / (4 * area) =  120 / (4A)   = 30 / A =   R

So

R /r   =       (30 / A)

               _________ =       (30 / A)   ( 7.5 / A)   =  225 / A^2 = 225/ 98.4375     ≈ 2.29

                  A / 7.5    

Let   sqrt  ( 4x   - 3) =    a

So we have

a  +   10 / a  = 7                multiply through by  a

a^2  + 10   = 7a                    rearrange as

a^2  - 7a  + 10  =   0           factor as

(a - 5)  ( a - 2)  =   0         set each factor to  0 and solve for a

a  - 5  =  0                        a   -  2    =  0

a  =  5                              a   =   2

So

sqrt ( 4x  - 3)  =   5               or                   sqrt (4x - 3)   = 2

                                    square both  sides

4x   - 3   =  25                       4x  - 3   =  4

4x   =  28                              4x  =     7

x  = 7                                      x =  7/4                are  the two solutions

Here is a graph that may help you verify the answers you find:

This question a) and b) are not dependent....

maximum velocity ...and acceleration at s = 3    are not the same

    take derivative of velocity , set = 0    to find s value where velocity is maximum

          use this value of 's' in the given velocity equation to calculate the VALUE of the velocity at the 's' you found....

im talking about question a. the way these problems are supposed to work is a is seperate from b, but b relys on a