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Here are parts (1) and (2)

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When the polynomial P(x) = 6x3 + kx2 + x − 2 is divided by x + 2, the remainder is 0. Which of the following is also a factor of P(x)?

A:3x-1

B:3x+1

C:2x-1

D:x-1

If x+2 is a factor then f(-2)=0

so

\(P(2) = 6*(-2)^3 + k*(-2)^2 + (-2)− 2=0\\ 6*-8 + 4k-4=0\\ 4k=52\\ k=13\\ so\\ p(x)=6x^3+13x^2+x-2\)

If 3x-1 is a factor then x=1/3  will be a zero I can check this with the calcuator

6*(1/3)^3+13*(1/3)^2+(1/3)-2 = 0    So  (3x-1) IS a factor

If 3x+1 is a facor then  f(-1/3) = 0

6*(-1/3)^3+13*(-1/3)^2+(-1/3)-2 = -1.1111111111111111    No 3x+1 is not a factor

If 2x-1 is a factor then f(1/2)=0

6*(1/2)^3+13*(1/2)^2+(1/2)-2 = 2.5    No 2x-1 is not a factor

If x-1 is a factor then f(1)=0

6*(1)^3+13*(1)^2+(1)-2 = 18     No that isn't a factor either     

I have determined this without any algebraic division.   

having trouble solving this, If anyone could even give me a tip to get started.

What I don't understand is,

Q1 how do you get the v^2 into an ode? how do I get rid of the square? If i derive it, I'll loose the drag coefficient d and thus can't solve Q2. If I was to then get v on it's own into an ODE. I'm not sure how to find d as i'll still have m as an unknown.

Q5 How do I find the spring stiffness k as I have the other unknowns of c and m.

Any help will be very greatly appreciated!

When the polynomial P(x) = 6x³ + kx² + x − 2 is divided by x + 2, the remainder is 0.

Which of the following is also a factor of P(x)?

A:3x-1

B:3x+1

C:2x-1

D:x-1

polynomial long division:

\(6x^3 + kx^2 + x - 2 : x + 2 = 6x^2+kx - 12x -2k+25 + \underbrace{\dfrac{4k-52}{x+2}}_{=0} \\ 6x^3 + kx^2 + x - 2 : x + 2 = 6x^2+kx - 12x -2k+25 + 0 \)

remainder is 0:

\(\begin{array}{|rcll|} \hline \dfrac{4k-52}{x+2} &=& 0 \\ 4k-52 &=& 0 \\ 4k &=& 52 \\ \mathbf{ k } & \mathbf{=} & \mathbf{13} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 6x^3 + kx^2 + x - 2 : x + 2 &=& 6x^2+kx - 12x -2k+25 + 0 \quad & | \quad k = 13 \\ &=& 6x^2+13x - 12x -26+25 \\ &=& 6x^2+x -1 \\ &=& (3x-1)(2x+1) \\\\ \mathbf{ 6x^3 + kx^2 + x - 2 } & \mathbf{=} & \mathbf{(x+2)(3x-1)(2x+1)} \\ \hline \end{array} \)

A:3x-1

We are not here to do all your homework for you.

Please post just one and try and answer others from the answer you are given.

These are all log questions.

Here are some  log rules.

Thank you! You are correct!

Thank you! You are correct!

Thank you! You are correct!

Here is a graph:

First, realize that the drivative of e^x    IS   e^x

   We can find the slope at point a = ln3  by substituting this in to

        e^x     e^(ln3) = 3   = slope

AT point  ln 3 ,  e^(ln3)       or    ln 3 , 3

so   using   y = mx + b form

      3 = (3) (ln3) + b      solve for b

3-3ln3 = b

equation of the line then becomes

y = 3 x  + 3- 3ln3

Wow!   NOTHING ELSE seems to be troublesome for ya !!!  

Thanks, EP.....multiplication always caused me problems....LOL!!!

y = x^2 + 8x + 18       complete the square on x

y = x^2 + 8x + 16 + 18 - 16

y = (x + 4)^2 + 2

(y - 2) = (x + 4)^2    (1)

In the form 

4p (y - 2)  = ( x + 4)^2   .....it's clear from (1) that  p =1/4

This parbola turns upward

The vertex  is (-4, 2)

The  focus is given by :    ( -4 , 2+ p)  ⇒  (-4 , 2 + 1/4) ⇒ ( -4, 9/2)

The directrix is given by :

y   =  ( 2 - 1/4)  =  7/4

Does the last graph have the point (5,2) on it?

Slight typo in CP's last answer   

SUM  :    -4 * 3/2 = -6

a.   -4+  (-4/3) +  (-4/9) +  (-4/27)

b. converges......l r l < 1      ⇒   { r  = 1/3 }

c.  The sum is

   -4                    -4

_____      =      ____    =     -4 * (3/2)   =   -6

1 - 1/3                2/3

EDIT TO CORRECT AN ERROR  !!!

10.   The sum is

26000[ 1 - (1.03)^5 ]  /  [ 1 - 1.03 ]  =  $138,037. 53

11.  Converges....the common ratio is  (1/2)

The sum is

     First term                        (1/5)                  (1/5)

_______________    =       ______     =       _____   =   2/5

 1 - common ratio                  1 - 1/2               1/2

The third answer is correct

7.  Sum of a geometric series =

First term  * [ 1  - common ratio^ (number of terms) ] / [ 1 - commo ratio ]  =

1 [ 1 - 4^6] / [ 1 - 4 ]  =  1365

8.   Sum of the first seven terms is

6 [ 1   - (-4)^7 ] / [ 1 - (-4)  ]  =

6 [ 1 - (-4)^7 ] / 5   =  19662

9. Sum of 1st 10 terms  =

12 [ 1 - 2^10] / [ 1 - 2 ]   =  12276

The radius of the circle  can be found as

4pi = pi*r^2   ⇒  r    = 2

(1/2) of the side, m,  of the triangle can be found as

2 / sin(30)  =  m / sin( 60)

2sin(60) / sin( 30)  = m

4 * sin 60    =

4 *√3/2  =  m  = 2√3

So.....the side of the triangle is  twice this = 4√3  = √48

And the area  is 

(1/2)( √48)^2 sin 60  =

24 * √3 / 2

12√3   cm^2

7 bags  each have 29 coins to start    .

Let x be the final coin coint in the EIGHT bags     8x > 200   so x >25

Now    53 - x    has to be divisible by the remaining 7 bags

trial and error 

26     53 - 26 = 27   nope

27     53 - 27 = 26   nope

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32    53-32 = 21   Yes !       x = 32 = final count in the 8 bags

(8 * 32   -53  ) / 7 = inital coin count in the 7 bags = 29 coins/bag

5. The first  answer is correct, NSS....to see why this is so....note that we have

 2(1) + 2(2) + 2(3) + 2(4) + ........=

2  + 4  +   6  + 8  + .....

6.  Note that the first term is

5(3)  = 15

And the last term is  

5(8)  = 40

And the number of terms is  [ 8 - 3] + 1   = 6

So....the sum is

 [ First term + Last term ] *  number of terms / 2   =

[ 15 + 40 ] * 6 / 2   =

55 * 3   =

165

The equation of BA  is

y = -2 (x - 4) + 4

y = -2x + 8 + 4

y   = -2x + 12

So..."B"  =  -2(0) + 12  =  12

And "A"  can be found as

0  = -2x + 12

2x  = 12

x = 6

So..."A"  =  (6,0)

So.....the area of the shaded quadrilateral   =  

Area of triangle OBA  +  area of triangle EAC  =

{Note that EAC has a base of 2  and a height of 4 }

(1/2)(12*6)  + (1/2)(4*2)    =

36 + 4   =

40 units^2 

Angle at Q is 45 degrees also so RT = RQ

sin 60 =  sqrt 3 / 2 = opp/hyp =  2*sqrt 3 / hyp        solve for hyp

sqrt3/2 = 2*sqrt3/hyp

hyp = (2*sqrt3)/ (sqrt3/2) = 4   = RT       

The two legs are the same length so   x = 4

cos 30

  30 degrees is a reference angle on the unit circle      cos(30)  = sqrt3/2