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There are several answers possible  ......

Using the form  y  =  A sin ( Bx + C)  + D

The period of the graph  is 15 units in length.....then each quarter point  = 15/4 units

The amplitude of the graph = A  =  4

The midline of the graph is y = 1   = D

We can solve for B  thusly

pi/2 =  (15/4)B

pi/ 2  * (4/15)  = B

(2/15)pi  = B

And  we can find C thusly......note that we are shifting the normal cosine curve to the right

1   =  sin [ (2/15)pi (15/4) - C] + 1       subtract 1 from both sides

0  =  sin  [ ( pi/2)  -  C ]     →  cC  = pi /2 

So....our function is

y  = 4  sin [ (2/15)pi* x - pi/2 ]  + 1

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

The previous didn't find the final distance from (-2,-4) to (9,5), so I will:

\(d=\sqrt{(9-(-2))^2+(-5-(-4))^2}\)

\(=\sqrt{(9+2)^2+(-5+4)^2}\)

\(=\sqrt{11^2+(-1)^2}\)

\(=\sqrt{121+1}\)

\(=\sqrt{122}\approx11.0454\)

The square root of 122 has no perfect square factors, so it is already in simplest radical form.

First, know the distance formula:

\(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

Now, simply substitute the values into this formula and simplify:

\(d=\sqrt{(-6-(-2))^2+(8-(-4))^2}\)

\(=\sqrt{(-6+2)^2+(8+4)^2}\)

\(=\sqrt{(-4)^2+12^2}\)

\(=\sqrt{16+144}\)

\(=\sqrt{160}\)

Of course, put all radicals in simplest radical form. I also included an approximate decimal to the ten thousandths place:

\(\sqrt{160}=\sqrt{16*10}=\sqrt{16}\sqrt{10}=4\sqrt{10}\approx12.6491\)

Let's continue this process for the rest of the points. I'll do the distance from (-2,-4) to (-6,8):

\(d=\sqrt{(-5-(-2))^2+(2-(-4))^2}\)

\(=\sqrt{(5+2)^2+(2+4)^2}\)

\(=\sqrt{7^2+6^2}\)

\(=\sqrt{49+36}\)

\(=\sqrt{85}\approx9.220\)

The square root of 85 is already in simplest radical form, so no need to do anymore.

OK......go ahead and post them....

Since  the vertex is halfway between the focus and the directrix, the vertex must be at (-2,3)

So....the distance between the vertex and the focus  =  1  = p

So.....using   4py  =  (x - h)^2 + k       where p = 1, h = -2  and k = 3

And we have

4(1)y  = (x - -2)^2 + 3

4y  = ( x + 2) ^2   + 3          multiply both sides by (1/4)

y =  (1/4) ( x + 2)^2  + 3

Here's the graph : https://www.desmos.com/calculator/dfundaxw40

But i have other questions

OK......good job....!!!!!

Here's the graph you need : https://www.desmos.com/calculator/e3rve2bfvr

Using  y  =  Asin (Bx + C)  + D

We can divide the normal  period of the sine  (2 pi)  into  6  equal parts....so that  each part (one minute)  is  [2pi / 6] rads   =  pi/ 3  rads  =  B.......

After 1.5 minutes  the normal sine will reach its max height  = 1

Note..... sin [  pi (1.5)/3]  =

sin (pi/2)  =  1 

And it will return to the midline point of y= 0  after 3 minutes since sin (pi (3)/ 3)  = sin (pi)  = 0

And it will reach a minimum of y = -1 after 4.5 minutes  because sin (pi (4.5)/3)  = sin (3pi/2)  = -1

And it returns to the midline  (the starting height) of y = 0  at  6 minutes since sin (pi * 6/ 3)  =

sin (2pi)  =  0 

There is no shift left or right, so  "C"  = 0 

Next we need to make some adjustments

We know  that the amplitude  is 8  because the graph rises 8 units up and down from the midline of y = 0 ....so "A'  = 8

But......the midline  of the graph starts at 8  so  we need to append 8 to the function.....so, "D'"  = 8  will give us that shift

I hope that helps   ....!!!!

I have it now Cphill thanks though!! 

whats wrong?

I guess we have

√(x + 2)  + 4   =  6          subtract 4 from both sides

√(x + 2)  =  2             square  both sides

x + 2    =  4              subtract 2 from both sides

x  =  2

Not extraneous

Can you post a picture, NSS???  .....it's a little difficult to tell what you have.... 

The smaller area between the chord and the edge of the circle is given  by :

(1/2)*6^2 (pi/3)- (1/2)*6^2 sin (60)  =  36 ( pi/3 - √3/2)  (1)

The area  of the whole circle  =  pi (6^2)  =  36pi     (2)

So  the area  of the  larger part   =  (2)  - (1)  =

36pi  - 36 ( pi/3 - √3/2)  =

36   [  pi  - pi/3 + √3/2 ]  =

36 [ 2pi / 3  + √3/2]  =

[ 24pi + 18√3 ]  units^2 

 5 mil ​​​​

need more information

Don’t worry about it! Five seconds after your last post, we forgot about you.

Since AB is parallel to CD, then

A + D   = 180   and

B + C  = 180

Then

2D + D  =  180

3D  =  180

D  =  60°

And 

B + 3B  = 180

4B  =  180

B = 45°

18st=2s

18st/2=2s/2

9st=1s.

nine stamps on one sheet