CPhill

lim x→ 1     [  (5 - x)^(1/2) - 2 ] / [(10 - x)^(1/2) -3 ]

Here, we have a  0/0 situation when we try to evaluate the limit. In this case, we can use something called "L'Hopital's Rule" to evaluate the limit.

Take the derivative of the numerator and the denominator.....this gives

 lim x→ 1   (-1/2)(5-x)^(-1/2) / (-1/2)(10 - x)^(-1/2) =

 lim x→ 1     (10-x)^(1/2)/(5-x)^(1/2)   now, take the limit and we get

(10-1)^(1/2) / (5 -1)^(1/2) = 9^(1/2) / 4^(1/2) =  3 / 2

If this is

(3x-1)/(3x+2)= 4         multiply both sides by 3x + 2

3x -1 = 4(3x + 2)        simplify

3x -1 = 12x + 8          subtract 12x from both sides and add 1 to both sides

-9x =  9                     divde both sides by - 9

x = -1

Assuming we know the radius (or diameter) and the area of the sector....we have

Area of sector  = (1/2)(r^2)Θ   multiply both sides by 2 and divide by r^2

2*Area of sector/(r^2) =  Θ       

Then using the value you obtain for Θ , we have

Arc length of sector = r*Θ 

The tangent(A) is = 1 at  45 degrees, 225 degrees, 405 degrees and 585 degrees

So, let A = 2Θ + 15

And setting 2Θ + 15  equal to the above values, we have

2Θ + 15 = 45          and solving for Θ we have Θ = 15 degrees

2Θ + 15 = 225         Θ = 105 degrees

2Θ + 15 = 405         Θ = 195  degrees

2Θ + 15 = 585         Θ = 285 degrees

And those are all the values that make the equation true for    0 ≤ Θ ≤ 360

If the ratio is 4:5, then 4/9 are boys and 5/9 are girls.

Then....5/9 x 1620 = 900 girls

And 4/9 x 1620 = 720 boys

18x^4 - 9x^2 +12x / 3x^2      we can just simplify this term by term

18x^4 /3x^2 - 9x^2/3x^2 + 12x/3x^2  =

6x^2 - 3 + 4/x

Let x be the tank's capacity.......And the difference between being 5/8ths full and 1/6th full is 180 (gallons??)

So

(5/8)x - (1/6)x = 180

(30 - 8)x /48 = 180

(22/48)x = 180     multiply both sides by 48/22

x = (48.22)180 = about 392.72 (gallons??)

Very nice, Melody.