CPhill

5.

We  have this

y  = Asin (Bx) + C

A is the amplitude  =  5

The midline here is shifted down 8 units from normal.....thus  C  = -8

To find B  we have

B  = 2pi / period    =  2pi / 16  =  pi / 8

So  the function is

y  = 5 sin ( ( pi/ 8)  x )  - 8

Here's the graph :    https://www.desmos.com/calculator/vhovcnbzy3

4.  

y  = Asin(Bx + C )  

A  =  the amplitude  =  [ difference in high/low/point ] / 2  =  6 / 2  =   3

The period is   20sec...so B  =

2pi / period  = 2pi / 20  = pi / 10

Since the sine graph starts at its highest position...the is a shift of pi/2 units to the left....thus  C  = pi/2

The function is :

y = 3sin ((pi/10) x   +  pi/2 )

The only question I have about this one is  at the beginning of the observation, x = 0 and the weight will be at its  highest point......thus.....the graph's first point will not be at the midline......that occurs 5 seconds later.... 

3.

In the form   y  = Asin (Bx)  + C

Period   =  4     we need to  find "B"  thusly

B  = 2pi / period   = 2pi/ 4  =  pi / 2

A  is the amplitude  =  4

C  is the shift of the midline from y  = 0.....thus.....the shift is 1 unit upward.....so C  = 1

The function is

y  = 4sin (pi * x / 2)  + 1

Here's the graph : 

2.

In the form Asin(Bx)  + C

Amplitude  = 2  = A

Midline  y  = 3  =  C

To find B   =   2pi/ period  =  2pi / 4pi   =  1/2

y intercept  (0, 3)

The function  is :

y  =  2sin( (1/2) x ) + 3

Note  that the parent function is :

y  = -2sin ( (1/2) x )  -  3

Here's the graph of both : https://www.desmos.com/calculator/xnb8aumkek

1.  Freq  =  1 / [ 6pi] 

The frequency  is  1 / period....so the period is 6pi....

In the form   y  = Asin (Bx)  + C

Amplitude  = 2  = A

Midline     .....y  = 3    =  C

To find B

B  = 2pi/ period  =   2pi / 6pi   =  1/3

y -intercept  =  (0, 3)

The function is : 

y  = 2sin ( (1/3) x)  + 3

Here's the graph :   https://www.desmos.com/calculator/8hza0yorxy

f(x)  = 3 (x - 5)^2   + 9

1.  f(x) is shifted 6 units to right to obtain function g(x). What is the equation of g(x) in terms of f(x)?

g(x)  = f(x - 6)

2.  What is the equation of g(x) in terms of x?

g(x)  = 3 (x - 11)^2  +  9

3.  Then, g(x) is shifted 7 units down to obtain function j( x) . What is the equation of j(x) in terms of

f(x)?

j(x)  =  f(x - 6)  - 7

4.  What is the equation of j(x) in terms of x? 

j(x)  = 3 (x - 11)^2  + 2

EDIT to correct a previious omission....

For...the second.....the  bisected apex   angle   GEH sets up the following proportion :

GD / GE  =  DH / HE

x + 4  / 44.8   =  35 / 56    multiply both sides by 44.8

x + 4  =   44.8 * 35  / 56       simplify

x + 4  =  28      subtract 4 from both sides

x  =  24

Check for youself that  GD / GE  = DH / HE   =  5/8

Hope that helped......!!!!

The first and third one use the fact that  a line segment drawn parallel to a base splits the sides proportionally

So on the first

MA  =  46.2 - 14  =  32.2

And  we have the following proportion :

MA / MN  =  MB / MP

32.2 /  46.2  =  MB / 72.6       multiply both sides by 72.6

72.6 * 32.2 / 46.2  =  MB   =  x   =  50.6

For the third  one

5x + 10 / 10  =  121 / 11

5x + 10 / 10  =  11    multiply both sides by 10

5x + 10  = 110     subtract 10 from both sides

5x  = 100      divide both sides by 5

x  = 20     CORRECT!!!

There may be other ways of doing this....but....we can represent this as a piecewise function :

                  5.25x                       if    0 ≤ x ≤ 200

y   = 

                 5.25x  +  480           if      200 < x 800

The domain is  [0, 800]

The range  is [ 0, 4680 ]

Here's the graph :https://www.desmos.com/calculator/wfww9jpzwu

Let M  =  number of months in which deposits are made

Let T  be the total amount  in the account after M months  [ assuming no withdrawals ]

So...we have

T  = 723M  

And this is linear