CPhill

I'll opt for a graphical solution, here :

https://www.desmos.com/calculator/ehnmjndhom

It appears that these two functions are equal at  x =  [ 60 * n ] °     where n is an integer

Here's one possibility :

7 - 7/7  - 1* 2 * 3 + 6 + 2  = 8

7 - 1 - 6 + 6 + 2   =

6 - 6 + 6 + 2  =

0 + 6 + 2  =

8

Thanks........we try hard on here  [ even though we don't always get the correct answer  !!!]

triangle was A,B,C angle B 56 and side length little c is 14 little b is 17 solve triangle using Law of sines (SSA case)

Solving for angle C we have

sin C / 14  = sin 56 / 17

sin C  = [14 * sin 56] / 17  

arcsin [ (14 * sin 56) / 17 ]  ≈ 43°

And angle A   =   180 - 56 - 43  ≈   81°

And  side a  =

a / sin 81  =  17/ sin 56   →  a = 17sin 81/sin 56 ≈ 20.25

To see if we have a second triangle, C might also  be  180 - 43  = 137°

However......when we add this to the known angle of 56°, we get more than 180°....so....only one triangle is possible

a = 20.25        b = 17       c  =  14

A = 81            B = 56       C = 43

Volume =  [ Floor area ] x [height]

270  = 72 * Height    divide bboth sides by 72

270/72  = Height = 3.75 ft

We have that  D/T  = R

Let the speed in still water be S   and let the current speed = C

So....the rate with the current is S + C   and the rate going against the current is S - C

And we know that

30/ 2.5  = S + C    

12  = S + C    (1)     and

30/4   =  (S - C) 

7.5 + C  = S   (2)

Put (2)  into (1)

12 =  7.5 + C + C

12  = 7.5 + 2C

4.5 = 2C     divide both sides by 2

2.25  mph  = C  = rate of the current

And, using (2), the speed of the boat in still water = 7.5 + 2.25 =  9.75mph

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

4sinxcosx  =  2*2sinxcosx = 2sin2x

2cos^2(2x) - 1 =   cos(4x)

But.....as the graph below shows,

sin (4x)   is not equal to   2sin(2x)cos(4x)

So....this is not an identity.....

x^2 - 6x + 5 ≥ 0 

Set this to 0

x^2 - 6x + 5  = 0       factor

(x - 5) ( x - 1)  = 0   set each factor to  0   and solve  and x = 5 and x =1

The solution to the original inequality will come from either of these two intervals (-inf,1] and [5, inf) or from the interval  [1, 5]

Note that testing any point between x = 1 and x = 5 will make the inequality false

So.....the solution is   x = (-inf, 1] U [5, inf )