CPhill

See  here :

https://web2.0calc.com/questions/two-equilateral-triangles#r1

5)Given that AB = 2, CD = 4 and that AB and CD are diameters of the respective semicircles shown, find the area of the shaded region.

This one is a little difficult

Call the mid-point of CD,, M..... connect MA  and MB

Triangle  MAB will be equilateral with sides = 2  and will have and area of  

(1/2)(2^2) sin (60)   =   √3  units^2  (1)

And the area of sector  AMB  is    pi (2)^2 (60/360)  =  (2/3) pi  units ^2   (2)

So... the area between chord AB and the circle's edge is just   (2) - (1)   =

[ (2/3)pi  - √3 ] units^2      (3)

And the area of the small semi-circle is  (1/2) pi ( 1)^2  =  pi / 2    (4)

So the shaded area is just  (4)  - (3)   =

pi/ 2  -  [ 2/3 pi - √3 ]  =

pi/2 + √3  - (2/3) pi   =  

[ √3  - pi/6 ]  units^2  ≈   1.208  units^2

f(x)   =  3x + p

g (x)  =  px + 4

f (g(x))  =  6x + q

Note that

f(g(x)  =   3 [px + 4] + p   =   3px + 12 + p

And this equals  6x + q

So

3px  + 12 + p   =  6x +  q

Equating terms we have that

3px   =   6x 

12 + p  = q

In the first equation, it's obvious that p  must   =  2

So.....

12 + 2  = q  ⇒    14  =   q

So   { p, q }  =  { 2, 14 }

-14. 5   =  - (14 + 5/10)  =     - (14 + 1/2)  =  - 29/2

4. 4)Minor arc AB in the diagram is 30 degrees and the radius of the circle is 6. Find the area of circular segment AB.

The area is

pi (6)^2  (30/ 360)  - (1/2)6^2sin (30)  =

36 pi (1/12)  - (1/2) * 36 * (1/2) =

3pi  -  9  =

0.4248  units^2

1)Let A, B, and C, be three points on a line such that AB = 2 and BC = 4. Semicircles are drawn with diameters AB, AC, and BC. Find the area of the shaded region.

BC  has a diameter of 4  so the radius is 2

AB  has a diameter of 2  so the radius  = 1

And AC has a diameter of 6  so the radius  = 3

The shaded region  is jusi

Area of semi-circle AC   -   area of semi-circle BC  - area of semi-circle AB

So we have

(1/2) pi  ( 3^2 - 2^2  - 1^2)  =

(1/2) pi  ( 9 - 4 - 1 )  =

(1/2) pi (4)  =

2pi  units^2

8 * (a + b + c)  =  49a + 7b + c

8a + 8b + 8c =  49a  + 7b + c

41a - b - 7c  = 0

41a  - b  =  7c

a  =  1

b = 6

c = 5

165₇

Proof

8 ( 1 + 6 + 5 )  =  1(7)^2  + 6(7)  + 5

8 * 12   =   49    +  42   + 5

96  =  49 + 47

96  =  96

⁴√[25c⁶ ]  =     

c ⁴√ [5 c² ]

f(-1)  = 15

f(0)  = 0

f(1)  =  -5

f(2)  = 12

A cubic  has the form  ax^3  + bx^2 + cx + d

Since (0,0)  is on the graph, then d  must equal 0

So.....we have this system

-a + b - c  =  15

a + b + c  =  -5

8a + 4b + 2c  = 12

-a + b - c  =  15

a + b + c    = -5      add these and we get that   2b = 10  ⇒ b = 5

And

8a + 5(4) + 2c  = 12

8a + 2c  =  -8   using this  and   -a + b - c  = 15   we have that

8a + 2c  =  -8

-a + 5 - c  = 15

8a + 2c  =  -8    (1)

-a - c   = 10       ⇒  -2a - 2c = 20  (2)     add (1)  and (2)

6a  =  12      ⇒  a  =  2

And using   - a + b - c  = 15

2 + 5 + c  = -5 ⇒   c = - 12

So.... the function is  

2x^3  + 5x^2  - 12x   .......to find the x intercepts.....set this to 0

2x^3  + 5x^2 - 12x  =  0      factor

x (2x^2 + 5x - 12)  = 0

x (2x - 3) (x + 4)  = 0

Setting the factors to 0   and solving for x we get that the x intercepts are  

 x = -4 , x  = 0   and x  =  3/2   = 1.5