CPhill

Let Matt's savings  = m

So

10  + m  = 70

Since ROB  and SOT  are central angles....their measures = the measures of their intercepted arcs

So....We have  that

m RTB  =  (1/2) ( m ROB  - m SOT)

28°  = (1/2) ( 3m SOT - m SOT)   multiply both sides by 2

56°  =  2mSOT       divide both sides by 2     

28°  = mSOT

So....mROB  =  3 * 28   =  84°

So  RS  =  180 -  28  -   84   =   68°

Sorry for the earlier error!!!

This becomes a little easier if we factor up front

sin^4x -sin^2xcos^2x -2cos^4x      factor as

(sin^2x  - 2cos^2x) (sin^2x + cos^2x)

(sin^2x - 2cos^2x)  (1)

sin^2x  - 2cos^2x

sin^2x  - 2 (1 - sin^2x)

3sin^2x -  2   or

3(1 - cos^2x) - 2

3 - 3cos^2x - 2

1 - 3cos^2x

     x - 3

_____________

 x^2 + 7x  - 30

Notice that the denominator factors as

(  x  - 3 )                                          1

______________    =                 _____

 (x - 3) ( x + 10)                          x + 10

We will have a vertical asymptote at  x + 10  = 0  ⇒  x  = -10

And we will have a "hole"  at  x -3  = 0   ⇒  x  = 3

See the graph, here :  https://www.desmos.com/calculator/skml1smvwr

[The hole doesn't show....but. trust me...there is one at x  = 3   !!!  ]

    4x                                  A                   B

__________     =          ______   +       _____

x^2 - 8x + 15                  x  - 3               x - 5

This is known as a partial-fraction decomposition

Note that the denominator on the left factors as  ( x - 3)  (x  - 5)

Multiply both sides by this and we have

4x   =  A(x - 5)   +  B ( x - 3)     simplify

4x  = Ax - 5A + Bx - 3B

4x + 0 =  (A + B)x  + (- 5A -3B)

Equating coefficients, we have this system

A + B  = 4         ⇒  B  = 4 - A      (1)

-5A  - 3B  =  0   ⇒  5A  + 3B  =  0     (2)

Sub (1)  into (2)   and we have that

5A + 3(4 - A)  = 0

5A + 12 - 3A  = 0

2A  =  -12

A  = -6

And  B  =  4 - A   =   4 - (-6)   =  10

Verify  that

-6/ ( x - 3)  +  10/ (x  - 5)     =   the left hand side of the original equation

\(0\leq x \leq10 and 0\leq y \leq10\)

\(2x+5y \geq 20 \)

The first  two conditions set  up  a square that is  10 units  on a side ....so the total area  is  10^2  =

100 units^2  (1)

The area below  the  inequality   2x  + 5y  ≥ 20  [ but still within the square ]  forms a triangle  with a base  of 10 and a height of 4

Its  area  =  10  * 4 / 2  = 20 units^2

Then  the area  above the inequality  [ but still within the square is 100 - 20 ]  =  80 units^2   (2)

So....the probability that  the coordinates of a point satisfy all three conditions  is

(2) / (1)    =     80 / 100    =   4/5

Here's a graph : https://www.desmos.com/calculator/ome3jb01pa

The total number of pairs  =  C(8,2)  =  28

Adjacent pairs

(1,8) (1,2) (2, 3) (3,4) (4, 5) (5, 6) (6, 7) (7, 8)

So....the probability  is

8 / 28    =

2 / 7

sin^4x  - sin^2xcos^2x  -cos^4x

sin^4x  -  (1 - cos^2x)cos^2x - cos^4x

sin^4x - cos^2x + cos^4x - cos^4x

sin^4x  - cos^2x

Note :  other answers may be acceptable....!!!

The area  below l  and still in the first quadrant  is a triangle with a base of 6 and a height of 6

So...its area  is    base * height /2  =  6 * 6 /2  =   18 units^2   (1)

The area between the two lines also forms a triangle with a height of 6  and a base of 4.5

So...its area   is   6 * 4.5  / 2  =  13.5  units^2    (2)

So...the probability of a point being selected in the specified area  is   (2)  / (1)  =

13.5  / 18   = 0.75

Connecting the centers  of the smaller circles  forms a hexagon  with  sides  of  6  units

But segments drawn from the center of the larger circle to each of these  6 centers  will form 6 equilateral triangles......so....a side of one of these triangles =  the side of the hexagon  =  6 units  (1)

And the distance from  a smaller circle center to the edge of the larger circle  = the radius of a smaller circle   = 3 units   (2)

So....the radius of the larger circle  =  (1)  + (2)  =    6 + 3  =   9 units

So...its diameter  =  2 * 9    =    18 units