CPhill

25sinxcosx-5sinx+20cosx= 4

This is known as a "see-saw" problem.....that is, the product of one side of the equation must equal the product of the other side

So, two people dig a hole in 5 hrs. = 300 minutes

So

2 x 300  = N * 5    where N is the number of people we're looking for

600 = N * 5      divide by 5

120 = N   

So it takes 120 people to dig the same hole in 5 minutes

To see the logic of this, we need to dig the hole 60 times as fast......therefore....we need 60 times as many people....  and 60 x 2 = 120.....!!!

[9x^ + 12x + 4] /[12x^2 + 5x -2] * [16x^2 -1] / [4x^2 + 5x + 1]   factor

[(3x + 2)(3x + 2)] / [(4x- 1)(3x + 2)] * [(4x + 1)(4x -1)] / [(4x + 1)(x + 1)]  "cancel" where possible

(3x + 2)/(4x -1)  * (4x -1)/(x + 1) =

(3x + 2) / (x + 1)

{ I think you only made a slight mistake in factoring [9x^2 + 12x + 4] }

((-3/) /(x² - 9)) + ((5) / (4x + 12))    factor, where possible

(-3)/[(x + 3)(x - 3)] + (5)/[4(x + 3)]  get common denominators

[(-3)(4) + (5)(x - 3)]  / [(4)(x -3)(x + 3)]   simplify

[-12 + 5x - 15] /[4(x -3)(x + 3)] =

[5x - 27] / [4(x -3)(x + 3)]

Not infinity.......just REALLY big  !!!

5+2(x-1)=12   subtract 5 from both sides

2(x - 1)  = 7   divide both sides by 2

x - 1  = 7/2     add 1 to both sides

x = 7/2 + 1   = 7/2 + 2/2  = 9/2

[x^2 - 4x - 5] / [x^2 - 2x - 3]  * [2x^2 - 7x + 3] / [x^2 -25]

Factor everything

[(x -5)(x+ 1)] / [ (x -3)(x+1)] * [(2x -1)(x -3)] / [(x + 5)(x -5)] =

(x-5)/(x -3) * [(2x -1)(x-3)] /[ (x+5)(x-5)] =

(2x -1) / (x+5)

Notice that (x^2 +1)  can never equal 0

This is a right triangle.

And the sine of (abc) = 5/√41

if y=x^2 + 2x - 3 and x= -3

So we have

y = (-3)^2 + 2(-3) - 3

y = 9 - 6 - 3  = 0