CPhill

Let the number of days  = 365

Then each day  must be  2pi/365  rads

The midline is  [ as before ]    12      and the amplitude here is  15. 8  - 12  =   3.3

I'm going to use the sine function here

I'm letting January 1st   =  "Day 0"...that is......where x  = 0

Using the basic function   y  =  3.3sin [  (2pi* x / 365 ) ]     we need to find the phase shift

Using this function.....one minimum  occurs when x ≈  -91.25 

And we need this minimum to occur on Dec 21st, i.e., 11 units (days) before x = 0 ( Jan 1st) ....so....the phase shift, P,  can be found thusly :

-91.25  + P  =  -11   ⇒    P  =  80.25....so.....this means that we need to shift the graph 80.25 units (days) to the right ........

So......the phase shift  can be expressed as  80.25 *2 *pi  / 365

So....... the [approximate] function  is    y  =  3.3sin [  (2 pi * x / 365)  -  (80.25 * 2pi / 365) ] + 12 

March 27th is the 86th  day of the year..so x  = 85 days after Jan 1st...so the hours of daylight when  x  = 85  are ≈ 12.27  hrs

And October 2nd is the 275th day of the year.....x = 274 days after Jan 1st....so the hours of daylight when   = 274 ≈ 11.365 hrs 

Here is the graph  : https://www.desmos.com/calculator/kyp7tucydf

cos (60)  =  JL / KL

cos (60)  = JL / 20  ⇒   JL   =  20*cos (60)

And the area can be found as

(1/2) JL * KL * sin (60)   =

(1/2) 20* cos (60) * 20 * sin (60)  =

(1/2) 20 * (1/2) * 20 * sin (60)  =

(1/4)*400 * √3 / 2  =

(1/8)*400*√3  =

50√3    units^2

'A"  is the correct answer for No. 4 

Both of the inequalities  have negative slopes, but the top two graphs have  a line with a positive slope, so the answer cannot be either of these

Looking at the bottom left  graph,  if this one is correct, the point  (0,4)   should make both inequalities true....but notice that this would mean that  4  < (-1/2) (0)  + 2 ⇒  4 < 2     which isn't true

So......the bottom right graph must be correct.....

Draw  perpendicular bisectors to AB, FE and CD....since these are equal chords in the larger circle, by Euclid, equal chords in a circle are equidistant from the center of the larger circle.......

Let O  be this center and let the points where the perpendicular bisectors intersect AB, FE and CD  be labeled as X, Y  and Z, respectively......and....as before, OX  = OY  = OZ  and a smaller circle can  be constructed with  these as radii............but.....this must mean that O  is also the center of this circle, as well....and...  the largest circle that will fit inside the triangle will touch all three sides, so, by definition,  O  is also the incenter of PQR

A function f has a horizontal asymptote of y = -4, a vertical asymptote of x = 3, and an x-intercept at (1,0). PART A: Let f be of the form

                       f(x)=(ax+b)/(x+c)

                       Find an expression for f(x).

Since we have a vertical asymptote at x = 3, c  = - 3.....and since we have a horizontal asymptote at y  = -4, a  = -4......and since we have an  x intercept at (1,0), we can solve this for b ⇒  -4(1)  +  b =  0 ⇒  b  = 4......here's the graph : https://www.desmos.com/calculator/1pmoaiuapl

PART B: Let f be of the form

                       f(x)=(rx+s)/(2x+t)

Find an expression for f(x).

Since the vertical asymptote is at 3 , we can solve this for t.....2(3) + t  = 0 ⇒  t  = -6

And since the horizontal asymptote is at -4, r  = -8

And since we have an x intercept at (1,0), we can solve this for s......-8(1) + s  = 0 ⇒  s  = 8

Here's a graph : https://www.desmos.com/calculator/gyqo2l6jdl

log x   - log y  = a

log y  - log z  =  15

log z   - log x   =   -7

Add  the first two equations and we get that

log x  - log z   =  a  + 15

Multiplying the third equation by  -1 on both sides we have that

log x  -  log z  =  7

This implies that

a + 15  =  7         subtract 15 from both sides

a  = - 8

x^(1/2)  +  x^(1/4)   = 30        subtract 30 from both sides

x^(1/2)   +  x^(1/4)  - 30    =  0     we can factor this directly as

[  x^(1/4 )  +  6  ]    [ x^(1/4)  - 5  ]   =  0

Settting both of these factors to 0  and solving for x, the first produces no solution

For the second, we have that

x^(1/4)  - 5  = 0     add 5 to both sides

x^(1/4)  = 5           take both sides to the 4th power

x  = 5^4   =    625

The denominators that you have indicate that this is probably  with replacement

The number of possible sums  without replacement are  12 * 11   =  132

Sums of 10  =

1   9

9   1

2   8

8   2

7   3

3   7

4   6

6   4

So.......the probability that the numbers sum to 10  without replacement is just    8  / 132   =  2 / 33

However.....with replacement we have    12  * 12   =  144 possible sums

1   9

9   1

2   8

8   2

7   3

3   7

4   6

6   4

5   5

So....the probability  that we have a sum of 10 with replacement is just   9  / 144   =  1  / 16

Pefectly fine, hectictar.....!!!!