CPhill

 (c+1)^2= 21/2     take pos/neg roots 

c + 1   =  ±√(21/2)      subtract 1 from both sides

c  =   ±√(21/2)  -  1

+√(21/2)  -  1  ≈  2.24

-√(21/2)  -  1  ≈  -4.24

Assuming that Bus B travels at the same speed as Bus A, let's find out how long it will take Bus A to make the trip

Distance / Rate  = Time

50 miles / 30miles per hour  =  5/3 hrs  =  1 hr, 40 min

So......if Bus B leaves at 5:30PM, it will reach Station B  at  7:10 PM

{ B' ∩ C' }   = {  {2,3,4,5} ∩  {2, 4, 5 }  }  =  {2, 4, 5 }

So

{ B' ∩ C' }'   =   { 1, 3}

And

{ A  ∩  { B' ∩ C' }' }   =  { { 1, 2, 4} ∩ {1 , 3 }  }   =   { 1 }

(a)    

| √9 − 9 |    =   l 3 - 9 l  =  l - 6 l   =  6

(b)

| 10 − π |   =  10  - π ≈   6.8584

Unit rate   =   price  / quantity  =  2.34 / 6  =  39 cents per ounce

OK.....I believe that the equation of the two axis  are   x  = pi/ 2 and  y  = 80

They represent the horizontal and vertical shifts of the graph

We have  the form   y  = Asin [Bx + C] + D

A  = the amplitude  = 2

D is the midline  = -4 

We can find B  thusly

period  =   regular period / B  .... so.....  2  =  2pi / B →    B =  2pi/ 2  =  pi  

So we have

y =  2sin [ pi*x ] - 4

We have no shift so "C"  = 0

I haven't heard of the equation  of the axis before.....but I assume that we have  x = 0   and y = -4

Here's the graph : https://www.desmos.com/calculator/4ijjishjps

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

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