CPhill

4/12  = 1/3

So

(1/3) (1/3) (1/3)   =  1/27  ≈ 0.037037037037037

Bakar's Distance divided by his rate  = his total time   =  6km / [ 4km/h ]  =  3/2 hr

And Ali's Distance  divided by his unknown rate , R,  rate  = his total time  = 6km / [ R km / h ]

And Bakar's time  plus 1/2 hr   =   Ali's time .......so

(3 / 2)   + (1/ 2)  = 6 / R   simplify

[ 6 + 2] / 4=   6 / R

[ 8 / 4]   = 6 / R

2 = 6 / R     and we can write

1/2 =  R / 6    multiply both sides by 6

3  = R      and Ali's rate  = 3 km / hr

We can solve this with a proportion

( 3/8) / (3/4)  =   x  / (1/3)         where x is what we need ....cross-multiply

(1/3) (3/8)  =  (3/4) x

(3 /24)  =  (3/4)x

(1/8)   = (3/4)x     multiply both sides by 4/3

(4/3) (1/8)   = x  =   4 / 24   =   1 / 6  cups of blueberries

Let   f(x)  = 2 u * v    where

u  = sin (4x)

v= sin^-1 (x)   =  arcsin (x)

Using the product and chain rules :

f ' (x)  =   2 [  u ' * v +  u * v ' ]

So

f  ' (x)  =  2  [   4cos (4x)* arcsin (x)  +  sin (4x) *   1 / √ (1 - x^2 ) ] =

8cos(4x)arcsin (x)  +  2 sin (4x) / √ (1 - x^2 )

Here's one other method based on hectictar's pic......

The intersection of two grid lines represents a "point" in the following theorem

So.......the area of this figure  =

(Number of interior points in the figure) + (number of boundary points ) / 2  - 1    =

( 23)   + (4 ) / 2   - 1   =

23 +  2  - 1   =

24  sq units

[ This is known as Pick's  Theorem ]

The derivative of the inverse function is given by :

g'(x)  =   1 / [ f ' (g (x) ) ]

Note that  ( 3, 4)  is on  " f " so   (4, 3)  is on   " g "    so.....g (4)  = 3

So

g ' ( 4)  =  1 / [ f ' ( g (4) ) ]  =   1 / [ f ' (3) ]  =   1 / [ 5 ]   =   1 / 5 

K(H)  =  - ( x^2 + 1)^2 + 4  ....so

-(x^2 + 1)^2  + 4  = 0      

-(x^2 + 1)^2   = -4   multiply through by -1

(x^2 + 1)^2   = 4     take the pos/neg square roots

x^2 + 1  =  ± 2

So...either

x^2 + 1  = 2          subtract 1 from both sides

x^2  = 1         take both square roots

x = ± 1

Or

x^2 + 1  = -2    subtract 1 from both sides

x^2  =   -3    take both square roots

x = ± (√3) i

And those are the 4 solutions

We can use the inverse sine to solve this

arcsin (32 / 58)  = x ≈ 33.49°

32  = .40x      divide both sides by .40

32 / .40  = x  = 80

y  = -x ( x + 2) (x - 3)

The zeroes are  at x = -2, x = 0  and x  = 3

From (-inf, -2) the graph will be > 0

From (-2, 0) the graph will be < 0

From (0, 3) the graph will be > 0

And from (3, inf) the graph will be < 0

A minimum turning point will occur on ( -2, 0)  and a maximum turning point will occur between (0, 3)

We could use Calculus to find these turning points but look at the graph here : https://www.desmos.com/calculator/1qkjsbo1r3

It shows that the max turning point occurs at x ≈ 1.786  ≈ 1.8