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Maybe a picture will help:

-2,-4  is a  POINT......the DIRECTRIX is a LINE.....  y= -6     the direct  distance between the two only involves the y distances.....

I see that, but why was -2 not included.

Did you miss this solution by Max Wong?  https://web2.0calc.com/questions/help_21404#r4

The focus was GIVEN as  -2,-4   in your question.....

How did you know the focus was -4?

So, you want me to find f(2x)-2f(x), huh?

In the above equation, you can just plug in 2x, which yields:

2(2x)^2-5(2x) = 8x^2 - 10x

Now we find 2f(x), which is just 2(2x^2-5x) = 4x^2 - 10x

subtracting 4x^2 - 10x from 8x^2 - 10x, we get 4x^2.

x^3/27 - 3x^2 + 81x -729 = 25+30+9

multiply everything by 27 to clear the fraction

x^3 - 81x^2+2187x - 19683 = 1728

subtract by 1728

x^3 - 81x^2 + 2187x - 21411=0

factoring yields:

x^3 - 8x^2 + 2187x -21411 = (x-39)(x^2-42x+549)

since two of the solutions are complex/imaginary numbers, the only solution is 39.

Since the y coordinate of the focus is -4  and the directrix is y = -6  .....the focus lies above the directrix .....so....this parabola opens upward

The vertex lies 1/2 of the way between the focus and directrix......to find this    we have ( -2,  [ -4 +  - 6] /2 ) =

(-2, -10/2 )   =   (-2, -5)  = (h, k)

"p' is the distance between the vertex and focus  =  l -5 - (-4) l = l -5 + 4 l =  l -1 l  = 1

So....we have the form

4p[ y - k ] = (x - h)^2

4(1) [ y - - 5 ] =  (x - - 2)^2

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

y + 5 = (1/4)(x + 2)^2           subtract 5 from both sides

y =  (1/4)(x + 2)^2  - 5

boi 

Nice, heureka!!!

Nice, itsyaboi...!!!!

Thanks, Melody.....this is a clever little problem....

Floor   [ (10.3)^3 - (10.2)^3 ]   =  31 integers

P(T)=3/4 =0.75     find P of 2tails  in 7 flips

7C2 * (0.75)^2 * (0.25)^5

just algebra q2

\(2^{2x}+7*2^x=8\)

\(\begin{array}{|lrcll|} \hline & 2^{2x}+7*2^x &=& 8 \\ & 2^{2x}+7*2^x -8 &=& 0 \\ & (2^x+8)(2^x-1) &=& 0 \\\\ 1. & 2^x+8 &=& 0 \\ & 2^x &=& -8 \quad x \text{ is an imaginary number},\ x = \dfrac{2 i \pi n + i \pi + 3 \ln(2)}{\ln(2)}, n \in \mathbb{ Z} \\\\ 2. & 2^x-1 &=& 0 \\ & 2^x &=& 1 \\ & 2^x &=& 2^0 \\ & \mathbf{x} & \mathbf{=} & \mathbf{0} \\ \hline \end{array} \)

LOL