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I assume you want it in this form

\(x=ay^2 +by+c\)     where a b and c are constants.

Although I am not really so sure that it is correct to call this general form.

Simply expand your equation and move it into this form.

If you have specific problems doing this then say what they are and we will help you further. 

Please no one answer over me. 

If 2x+7 then 2(x+7/2) are factors 

and thes means that

f(-7/2)=0

Do the substitution and find the answer for yourself.

Please no one do more of the working.

I anyone, including Lightning has a question, they will ask.

Thanks.

THX, hectictar  !!!!

Here's the "b"  part

Since the hperbola has the same foci its center is an equal distance from both foci = (0,0)....and one branch cuts the x axis at x  = 1 , then  a  =  1  and c = 2

Therefore....b  =   sqrt(c^2 - a^2)  =   sqrt (2^2 - 1^2)  = sqrt (4 - 1)  = sqrt (3)

So a^2  = 1  and b^2  = 3

So the equation of the hyperbola is

x^2          y^2

__    -     _____  =     1

 1              3

Here is the graph of the ellipse and hyperbola :  https://www.desmos.com/calculator/ur21asuwwk

very clever dr. Jones!

2x^2001 + 3x^2000 + 2x^1999 + 3x^1998 + .....+ 2x + 3    note that we can write

x^2000 ( 2x + 3) + x^1998(2x + 3) + x^1996(2x + 3) +  ....+ x^2(2x + 3) + x^0 (2x + 3) =

x^2000(2x+ 3) +x^!998(2x + 3) + x^1996(2x + 3) + ......+ x^2(2x + 3)  + 1(2x + 3)  =

(2x + 3) ( x^2000 + x^1998 + x^1996 + .....+ x^2 + 1)

Since the second polynomial is > 0  for all x  the only real root  is

2x + 3  =0

2x = - 3

x = -3/2

a)

x  =  2√2 cos θ

y  =  2 sin θ

x  =  2√2 cos θ

                            Square both sides of this equation

x²  =  8 cos² θ

                            Divide both sides by  8.

x² / 8  =  cos² θ

y  =  2 sin θ

                                   Square both sides of this equation.

y²  =  4 sin² θ

                                   By the Pythagorean identity, we can substitute  (1 - cos²θ)  in for  sin²θ

y²  =  4(1 - cos²θ)

                                   Substitute  x² / 8  in for  cos²θ

y²  =  4(1 -  x² / 8)

                                   Divide both sides of the equation by  4 .

y² / 4  =  1 - x² / 8

                                   Add  x² / 8  to both sides of the equation.

x² / 8  +  y² / 4  =  1

Here is what the ellipse looks like:

See the answer here:  https://web2.0calc.com/questions/help_8662#r3

yes, I'm sure

This question doesn't make a ton of sense.

a is clearly less than 1 and thus a divided by any integer m will be 0m + a,

i.e. a itself will be the remainder.  Are you sure a isn't supposed to just be 3²ⁿ + 4?

Maybe you should show the working of these stages:

  1  Find the midpoint of AB, and thus, the coordinates of point D.

  2. Find the midpoint of AC, and thus, the coordinates of point E.

  3. Calculate the slope of DE.

  4. Calculate the slope of BC.

Show how you find each of those things.

Then, at the end, add the same concluding sentence which you have written.

y^2 + 4x - 8y - 4  = 0

y^2 - 8y  =   -4x + 4            complete the square on y

y^2 - 8y + 16  = -4x + 4 + 16

(y - 4)^2  = - 4x + 20

-4x + 20  = (y - 4)^2

-4( x - 5)  = (y - 4)^2     multiply both sides by (-1/4)

x -5  = (-1/4)(y - 4)^2     add 5 to both sides

x = (-1/4)(y - 4)^2 + 5

We have a set cost of $2.25   to replace the card

So....it must be that   the fines themselves must  be just 6.25 - 4.25  = $4.00

So the fine per day must be just   [ 4.00] / [21 - 5 ] =  4.00 / 16   = .25

a)  Recursive  formula

a_n = a_n-1  + .25   where  a₁  = .25

b) Explicit formula

a_n  =  a₁ + .25(n - 1)    =   .25 + .25n - .25    =  .25n 

(c) Explicit.....it would be tedious to calculate the fine on the 60th day using the recursive formula....we would need to calculate 59 terms !!!!

(d)  Fine on Day 60  =  ,25(60)  =  $15

ok thank you very much

Midline                 Amplitude             Range

y = 2                           1                      [1, 3 ]

y  = -1                         2                      [-3, 1 ]

y =  1                          3                      [ -2, 4 ]

y =  2                          4                      [-2, 6 ]

We need to  write   x^2 + 2  as    x^2  + 0x  + 2

So we have

            x   +  2

x - 2  [ x^2   + 0x   + 2   ] 

           x^2   - 2x

          _______________

                     2x    + 2

                     2x    - 4

                    ________

                               6

Quotient  =  x + 2

Remainder  =   6 / (x - 2)

This is a trig identity

sin (2 θ)  =  2sin θ cos  θ

x^4 = x^2 + y^2       [x^2 + y^2  = r^2,   x = r cos θ, y = r sin θ  ]

(r cos θ)^4   =  r^2

r^4 (cos  θ)^4  = r^2

how does sin20 become 2sin0 cos0?

sorry im terrible at this topic

r^2 [sin2 θ]  = 6

r^2 (2sin θ cos θ)  =  6     divide through by 2

r^2 (sin θ cos  θ)  =  3        [ sin θ = y/r ,  cos θ= x / r ]

So....

r^2 ( y/r)(x/r)  = 3

r^(2) ( xy) / r^2   = 3

xy  =  3

y =  3 / x       

For the ellipse shown below, find the distance between the foci.

The formula to find the distance from the center to the foci is c^2= a^2-b^2. So once when find this distance we can double it to find the total distance between the foci. The variable a is 6 and b is 2. Therefore c^2= 36-4 which is c= the square root of 32. But to get the final answer you must double the square root of 32. Which is the square root of 64 also known a

\(8\)

(c)   Parametric form.....this is new to me.....but.....the  parametric form is given by :

( h + asec t , k + b tan t )   =   (3 + 2sec t, -2 + 3 tan t )

(d)  I don't know what this one is asking.....sorry!!!!

Thanks I understand now

It took me awhile

Thank you so much. :) 

I can do some of this

(a)  9x^2 -54x -4y^2 -16y + 29 = 0

9x^2 - 54x - 4y^2 - 16y  =  -29        complete the square on x and y

9(x^2 - 6x + 9) - 4(y^2 + 4y + 4)  =  -29 + 81 - 16

9(x - 3)^2  - 4(y + 2)^2  =  36        divide both sides by 36

(x - 3)^2  -   (y + 2)^2      

______      ________   =       1

    4                  9

The center is  ( 3, -2)   = (h, k)

a^2 = 4     so  a  = 2

b^2  = 9    so   a  = 3

c  = sqrt ( 4 + 9)  =  sqrt (13)

So....the eccentricity , e   =   c  / a   =   sqrt (13) / 2

(b)  The foci   are   ( h + c, k)  and (h - c, k)  =      ( 3 + sqrt (13)  , -2)  and  (3 - sqrt(13), -2)

The equation of the asymptotes   is    y  =  ±b/a ( x - h) + k    =    ±(3/2)(x - 3) - 2

Here's the graph : https://www.desmos.com/calculator/ojovjqb27k