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Yeah....I am a confused wolfy

It depends what form your equation is in.

If it is of the form

\((x-h)^2=4a(y-k)\)

Then the vertex is  (h,k) 

the focal length is |a|

If a is positive then it is concave up, if a is negative then it is concave down.

The focus is (h,k+a)  the axis of symmetry is x=h   and the directrix is   y=k-a

The definition of a parabola is the locus of all points equidistant from a line called the directrix and a point called the focus.

Give the directrix and the focus you can determine the equation of the parabola using locus (which is a fun things to do when you get used to the method)

What would you use the directrix for ? ... not sure what this question is after but if you google it the answer may become clear :/

You may well be correct Chris but Panda told me that ALL numbers and expressions  were in degrees. 

See my questions #2 and  her answer  #3

But now that i re-read Pandas comments I can see she does not know what degrees actually are!

This never occured to me before now!    Silly me!

Well it seems you must be correct. nice work Chris.  

Next time I better have my mathematics forensic hat better afixed to my head!

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RainbowPanda,   a degree is an amount of turning.  

You cannot say things like "one of the arcs and 5x-7 is another degree" That makes no sense!

You are either talking about a length (eg arc length) OR you are talking about an angle. (measured in degrees) 

The 2 things are not the same at all!

Mmm interesting

\(\frac{13t^2 - 34t + 12}{3t - 2 } + 5t = 6t - 1\\ \frac{13t^2 - 34t + 12}{3t - 2 } = t - 1\\ 13t^2 - 34t + 12=(t-1)(3t-2)\)

1) Now expand the right hand side

2) Take all the terms to the left hand side so the right hand side is 0

3)  Solve using the quadratic formula (or any other method)

4) The answer you want is the biggest one.

Now I will leave the algebra for you.

The answer is quite simple, you can check it by plugging it into the original formula and seeing that the right hand side=left hand side.

If you get in a mess you can ask for help but be specific about what your problem is.    

Have fun :)

Please no one answer over the top of me. The detail in this answer is perfect as it is.

The equation of the circle that passes through (-1,6) and which has a center at (2,3)
can be written as x^2 + y^2 + Ax + By + C = 0.
Find A*B*C

\(\text{Let $x_c=2,\ y_c =3,\ x_p=-1,\ y_p=6$} \)

\(\begin{array}{|rcll|} \hline (x-x_c)^2+(y-y_c)^2 &=& r^2 \quad | \quad r^2 = (x_p-x_c)^2+(y_p-y_c)^2 \\ (x-x_c)^2+(y-y_c)^2 &=& (x_p-x_c)^2+(y_p-y_c)^2 \\ x^2-2xx_c+ \not{x_c^2}+y^2-2yy_c+\not{y_c^2} &=& x_p^2-2x_px_c+\not{x_c^2}+y_p^2-2y_py_c+\not{y_c^2} \\ x^2-2xx_c +y^2-2yy_c &=& x_p^2-2x_px_c +y_p^2-2y_py_c \\ x^2+y^2+(-2x_c)x+(-2y_c)y &=& x_p^2-2x_px_c +y_p^2-2y_py_c \\ x^2+y^2+ (-2x_c) x+ (-2y_c)y -(x_p^2-2x_px_c +y_p^2-2y_py_c ) &=& 0 \\ x^2+y^2+\underbrace{(-2x_c)}_{=A}x+\underbrace{(-2y_c)}_{=B}y +\underbrace{(2x_px_c+2y_py_c -x_p^2-y_p^2)}_{=C} &=& 0 \\\\ A&=& -2x_c\\ &=& -2\cdot 2 \\ \mathbf{A} &=& \mathbf{-4} \\\\ B&=& -2y_c\\ &=& -2\cdot 3 \\ \mathbf{B} &=& \mathbf{-6} \\\\ C &=& 2x_px_c+2y_py_c -x_p^2-y_p^2 \\ &=& 2(-1)2+2\cdot 6 \cdot 3-(-1)^2-(6)^2 \\ &=& -4+36-1-36 \\ \mathbf{C} &=& \mathbf{-5} \\\\ A\cdot B\ \cdot C &=& (-4)(-6)(-5) \\ &=& -4\cdot 5 \cdot 6 \\ \mathbf{A\cdot B \cdot C } &=& \mathbf{-120} \\ \hline \end{array} \)

\(r^2 = (-1 - 2)^2 + (6-3)^2 = 18\\ (x-2)^2 + (y-3)^2 = 18\\ x^2 - 4x + 4 + y^2 - 6y + 9 = 18\\ x^2 + y^2 -4x -6y-5 = 0\\ (-4)(-6)(-5) = -120\)

Because the denominator can be equal to 0 in any circumstance. And anything divided by 0 is undefined, therefore that value of x is extraneous. 

\(\text{let }s_n = \sum \limits_{k=1}^n~a_k = n(n+1)(n+2)\\ s_{10}=s_9+a_{10}\\ a_{10} = s_{10}-s_9 = (10)(11)(12) - (9)(10)(11) = 330\)

\(a+b+c=5,~b = 5-a-c\\ \dfrac{1}{a+b} + \dfrac{1}{b+c} = \dfrac{1}{5-c}+\dfrac{1}{5-a}\\ \text{If we want the minimum of this expression it seems clearly that }a,~c\\ \text{should be made as small as possible}\\ \text{Further, as the problem is symmetric in }a \text{ and }c\\ \text{we expect these two to be equal}\\ (a,b,c)=\left(\dfrac 3 2,~2,~\dfrac 3 2\right)\\ \dfrac{1}{a+b}+\dfrac{1}{a+c} =\dfrac 4 7\)

Purplemath  explains this well :

https://www.purplemath.com/modules/ellipse2.htm

1.   This seems that, at first glance, it would lead to huge numbers...it doesn't....

Note the pattern   .......

Term

   1     13

   2     1^3 + 3^3  = 28

   3     2^3 + 8^3 =  520

   4     5^3 + 2^3 + 0^3  = 133

   5     1^3 + 3^3 + 3^3  = 55

   6     5^3 + 5^3 = 250

   7     2^3 + 5^3 + 0^3 = 133

   8     1^3 + 3^3 + 3^3 = 55

   9      5^3 + 5^3   = 250

Note

4mod 3 = 1 =  1st number in the repeating cycle   = 133

5 mod 3 = 2 =  2nd number in the repeating cycle  = 55

6 mod 3 = 0  =  3rd number in the  repeating cycle  = 250

So..for the 100th term.....

100 mod 3  = 1  =  the 1st number in the repeating pattern =  133

If I'm not mistaken, the point of maximum growth should be the point at which the bluegill population stops growing(presumably), so I guess the best bet would be either to skip that problem, or graph it out somehow and then approximate it :P 

The mulitplication property of equality, which holds that when you have a given equation, say :

a/5 + 6 = b/5 + 3,  multiplying on both sides by the same constant will provide the same result as the original equation, thus making it the multiplication property of equality. 

In the case of this problem, you would multiply by 5 on both sides, to get W = -50, and there's your answer!

presumably because the LCD would be factors of everything on both sides, meaning all the denominators are "canceled"?

Let's take an example

5 / (x - 1)    +  3 / (x + 2 )   =  5

The LCM  is  (x - 1) ( x + 2)

So

5 (x - 1) ( x + 2) / (x - 1)  + 3(x - 1) ( x + 2) / (x + 2)   =  5(x - 1) ( x + 2)

5(x + 2)  + 3(x - 1)  = 5(x - 1) ( x + 2)

So...multiplying by the LCM    "cancels out" the denominators

I know a few

1.  y intercept....let t = 0    and we have that

y =   1000 / [ 1 + 9e^(-.4(0)) ]    =  1000/ [ 1 + 9(1) ]  =  1000/ 10   = 100

2.     asymptotes

No  vertical asymptotes since  the exponential in the denominator is never  negative....so....the denominator is never = 0

Horizontal asymptote........as t gets larger and larger, the exponential at the bottom approaches 0

So.....the horizontal asymptote is

y =  1000 /  [ 1 + 0 ]   =  1000

3. Don't know this one

4. After 15 years...there will be   1000 / [ 1 + 9e^(-.4 * 15) ]  ≈  978  bluegills

4y=-2x-17

y=-1/2 x-17/4

The slope of a line parallel is the same slope, so it is -1/2.

You are very welcome!

:P

a^2 + b^2 + c^2  = 13

ab + bc + ca = 6

Just by inspection....an integer solution is  a = 0, b = 2   and c = 3

So

0^2 + 2^2 + 3^2 =   13

And

0*2 + 2*3  + 3*0   =   6

And their sum = 5

Another possible integer  solution is  (a, b, c) =   (0, 3, 2 )

a)    y = -(1/8) ( x + 3)^2 - 2

The vertex is (-3, -2)

The parabola turns downward

In this form

(1/8) =   1/[4p]    so     p   =  2

The focus is    (-3, -2 - p)  =  (-3 , - 2 - 2)  =  (-3, -4)

The directrix  is  y  =  -2 + p  =  -2 + 2  = 0   

b )    4y = x^2 - 20

4y + 20  = x^2

4 ( y + 5) = (x - 0)^2

The vertex is  (0, -5)   and the parabola turns upward

In this form

4p = 4  ⇒  p = 1

The focus is   (0, -5 + p) =  ( 0, -5 + 1)   = (0, -4)

The directrix is

y =  -5 - p =  -5 - 1  = -6