CPhill

I'll use x instead of "theta"  [ easier to type.....LOL!!! ]

csc x + 1

_______    =

csc x -  1

1/sin x  + 1

_________  =

 1/sinx - 1

(1 + sin x ) / sinx

______________  =

(1 - sin x) / sin x

(1 + sin x)           sin x

________   *  ________  =

   sin x              1 - sin x

1 + sin x

________  =   the left hand side

1 - sin x

See the following, SydSu.....

A perpendicular bisector drawn to the chord from the center of the circle will be the shortest distance from the center to the chord.....

Let the equation of the circle be x^2 + y^2  = 16

The equation of the line containing the chord  is  y  = (1/sqrt(3)) ( x + 2)

Call the point of intersection of the bisector and the chord, M

Then....AMP  will form a right triangle.....with angle AMP  = 90°   and AP the hypotenuse  = 2

Then...AM  will be the distance we are looking for.....and since this side of the right triangle is opposite the 30°  it will be (1/2) of AP  =  1

The period  = 40 seconds

The ampllitude  is  (50 - 10 ) / 2   =  40/2  = 20

The midline  =  (50 + 10) / 2  = (60/2)  =  30

We have the following form

y  = A sin (B t) + C

A = the amplitude = 20

B =  2pi / period  =  [ 2pi / 40 ]  =  pi/20

C =  the midline value = 30

So....the  function is

y  = 20sin [ pi * t / 20 ] + 30

Here's the graph  : https://www.desmos.com/calculator/w7j1brsd5f

f(x)  =  l x + 3 l  - 1

d   absolute value

ALSO

if this function were reflected along the x axis, translated 4 units left and 2 units up, what would the new equation become??

1st - Reflected about  the x axis  = - f(x)  = - l x + 3 l + 1

2nd - translated 4 units left  =   - l x + 7 l + 1

3rd - 2 units up  =  - l x + 7 l + 3

10.9^2  =  5.8^2 + 10.5^2  - 2(5.8)(10.5) cos θ

 ( 10.9^2 - 5.8^2 - 10.5^2)  / [ -2(5.8)(10.5) ]  = cos θ

Cupcake......I get  that   cos  θ   ≈  .21

https://www.wolframalpha.com/input/?i=(+10.9%5E2+-+5.8%5E2+-+10.5%5E2)%C2%A0+%2F+%5B+-2(5.8)(10.5)+%5D

2. Given  N(-5, 3) and M (5, -2). Find the point P on MN such that MP = 3·PS

N    1   P   3      M

I'm assuming that you  mean    MP  = 3PN

Like the last one, MN can be divided into 4 equal parts.....and  MP  is 3 of them......so...P is (1/4) of the way between M and N......so we have

P = 

[ -5 + (1/4)(5 - -  5) , 3 + (1/4)(-2 - 3) ]   =

[ -5 + (1/4)(10) , 3 + (1/4)(-5) ]  =

[ - 5 + 10/4 , 3 - 5/4 ]   =

[ -20/4 + 10/4, 12/4 - 5/4 ]  =

[ - 10/4, 7/4 ]  =

(-5/2, 7/4 )

1. Given J(-2, 5) and K(2, -3). Find the point P on JK such that 4·JP = PK.

J    1      P     4           K

We can divide JK  into 5 equal segments......and  PK is 4 of these....so....P is (1/5) of the distance from J to K

So....we can find P as follows

P  =  [  ( -2 + (1/5)(2 - -2)  ,    5 + (1/5) (-3 - 5)   ]  =   

[ -2  + (1/5) (4) ,  5 + (1/5)(-8) ]   =

[ -2 + 4/5 , 5 - (8/5) ]  =

( - 10/ 5 + 4/5 , 25/5 - 8/5  )  =

( - 6/5 , 17/ 5)

If    x - y  = 6   then   (x - y)^2  = ( 6 )^2  = 36

If  x - y  = - 6 then  (x - y)^2  = (-6)^2  = 36

4x + 7y = 6

9x - 2y = 49 

Let's choose to eliminate y.....we can do this by multiplying the first equation by 2   and the second by 7.....do you see why this might work, SMM???.....so we have

8x + 14y = 12

63x - 14y = 343          add these   and notice that "y" disappears

71x  =  355       divide both sides by 71

x  = 5       

Can you take it from here and find y ???

3x - y = 4

y = 2x -2      easier than the first one......we have an explicit expression for y......so....sub the second equation into the first for y   and we have

3x - [ 2x - 2 ]  = 4       simplify

3x - 2x + 2  = 4

x + 2  = 4

x = 2

Sub this into the second equation to find y

There is no "set" way to do these, SMM......the idea is to multiply one equation by something so that when we add the result to the other equation, a variable will "drop out".....