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Rearrange  as

y^2 + 5y  - x^2  =   25

HYPERBOLA......  the signs on y^2  and x^2 are opposite    and the  right side  of the equation is positive

Here's a graph : https://www.desmos.com/calculator/ke4orxoybx

Rearrange as

(x - 3)^2  + y^2  = -10

This is a  "circle"  with a negative radius......not possible....EMPTY  !!!

Each  smaller triangle is  similar  to the larger isosceles  right triangle

Each leg of these two triangles  =   (6 - 5)   =1  unit

So.....their combined areas =  2 (1/2) (1) ( 1)  =  1

Area of  isosceles triangle =  (1/2) (6)(6)   =18

Area of overlap =   18 -  1  =   17  units^2  

 \(x = {-1 \pm \sqrt{41} \over4}\)

LOL! Thank you so much CPhill. 

CORRECT   !!!!    5/pi/6 is a 2Q angle....sine  is +  here

CORRECT  !!!       tan (n *pi)   = 0

-11p/3  =    -11/pi/3 +  4pi  =    pi/3   =   1Q angle....all trig  functions are  positive here...so....

CORRECT  !!!!! 

Hey, Spring....you don't really need me....LOL!!!!

-4x^2  +  y^2  - 2y  =  3       re-write  as

y^2 - 2y   - 4x^2   = 3     complete the square on  y

y^2  - 2y + 1   - 4x^2  = 3  + 1

( y - 1)^2   - 4x^2    =  4      divide  both sides  by 4

(y - 1)^2            x^2

_______   -     _____   =    1

      4                  1

a^2 =  4        b^2  =1

a = 2              b  =1

The center is  (0, 1)  = (h, k)

Equation  of asymptote with a positive  slope   = 

y = ( a/b) ( x - h)  +  k

y= (2/1) ( x -  0)  +  1

y = 2x  + 1

See the graph here :  https://www.desmos.com/calculator/bbtkhsfpum

Thanks, CPhill! :)

CORRECT !!!

Limit  = 2/6   =   1/ 3      good job  !!!

x^2 + 2y^2   - 6x   - 8y  + 21  = 0    complete the square on x, y

(x^2 -6x  + 9 ) +  2(y^2 - 4y  + 4)  =  -21 + 9 + 8

( x - 3)^2  + 2 ( y - 2)^2  = -4     

This is empty......both  squared quantities  added together will have a minimum of  0  when (x,y)= ( 3,2)

And this sum will never =   -4

I would say the graph is an ellipse.

Here is the graph:

Points A(0,0), B(9,6) and C(6,12) are vertices of triangle ABC. Point D is on segment AB such that 3(AD) = DB, point E is on segment BC such that 3(BE) = EC, and point F is on segment CA such that 3(CF) = FA. What is the ratio of the area of triangle DEF to the area of triangle ABC? Express your answer as a common fraction.

Area of ΔABC = 36 u²             area of ΔDEF = 15.75 u²

So, the ratio...         [DEF] / [ABC] = 7/16 

Here is a computer code that uses "successive approximations" to arrive at the answer:

c=0; a=listfor(n, 1, 15, listforeach(b,reverse (d=12), c=2#(b - c);printa

Answer =(3.464101615, 2.921625983, 3.013034022, 2.997826876, 3.000362165, 2.999939638, 3.00001006, 2.999998323, 3.000000279, 2.999999953, 3.000000008, 2.999999999, 3, 3, 3) == 3

Thanks!

The graph is a circle.

Here is the graph:

Since 3(AD) = DB, then D  must  be (1/4)  the distance from  A to B

We  can  figure its  coordinates as

[ 0 +  (1/4)(9-0) , 0 + (1/4)(6-0) ] =  ( 9/4, 6/4)  = (9/4, 3/2)

Likewise, we can find E  as

[ 9 + (1/4)(6-9) , 6 + (1/4)(12 - 6) ] =   [ 9 - 3/4 ,  6 + 6/4 ]  =  [ 33/4 , 6 + 3/2 ] = [33/4 , 15/2 ] 

Finally, we note  that F is  3/4  of the distance  from A to C

To find its coordinates we have

[ 0 + (3/4)(6-0) , 0  + (3/4) (12-0) ]  =  [ 18/4, 36/4 ]  =  [ 9/2, 9]

To find the area of ABC we  can use this formula  where

Let (0,0)  =  (x1, y1)    (9,6)  = (x2, y2)    (6,12)  = (x3, y3)

Area = (1/2)  [ x1 (y2 - y3)  + x2 (y3 - y1) + x3(y1 - y2) ]

(1/2) [ 0 ( 6 - 12)  + 9(12 -0) + 6(0 - 6) ]  =  (1/2 [ 9 *12 + 6(-6) ] = 

(1/2) [ 108 - 36 ]  =

(1/2) 72 =

36

Likewise

Let  (9/4, 3/2) =  (x1,y1)     (33/4, 15/2 )  =    (x2, y2)        (9/2, 9)  = (x3, y3)

Area = (1/2)  [ (9/4) ( (15/2) - 9)  + (33/4) (9 - 3/2)  + (9/2)(3/2 - 15/2) ]  =   63/4

Area of DEF  to  Area of ABC  =

(63/4)  / 36  =   

63  / 144  = 

7 / 16

Thanks!

The graph is a parabola.

Here is the graph: 

1! + 2! + 3! + 4! + 5!.........etc.=....... 0420940313

n(n - 1) = 12, what is this going to give you? It gives you:

n = - 3  and   n = 4

a = 2   b = 1   c = -5   in the Quadratic Formula

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)   

                                which results in x =  - 1/4  + -  sqrt(41)/4

Okay. Thank you for the clarification.

So to solve the equation \(\sqrt{12-\!\sqrt{12-\!\sqrt{12-\cdots}}}\) we will use the following equation:

n(n-1)=12; with the answer to the infinite nested radical function (x) being equal to n-1.

So,

n(n-1)=12

n = 4

n-1=?

4-1=3

Thus, the solution is 3.

Hope that answered your question!

Also, just a tip, don't place your questions in the header. It makes it a bit difficult to know the question that you are asking. You are supposed to place your question where you said "help please." Just switch the places, and you'll be perfectly fine!

Good job, Guest!

infnate amouts of 12's square roots.