CPhill

I don't have any "magic" formula for doing these, DS......it's mainly just factoring and "cancelling"......plus a few algebraic "tricks".......

x² = 2011x - 1  →  x² + 1 = 2011x

So

x^2-2010x+【2011/(x^2+1)] =

[2011x - 1 ] - 2010x + 2011/(2011x) =

x - 1 + 1/x =

[x² - 1x + 1] / x =

[x² + 1 - x ] /x =

[2011x - x] / x =

2010x / x =

2010      (with the restriction that x ≠ 0)

3a³(16- a²) / [12a⁶ (a² - 9a + 20)]   factoring everything, we have

3a³ (4 - a)(4+a) /[ 12a⁶ (a - 4)(a-5)] =

-(1/4a³ )(a-4)(a+4)/[(a-4)(a-5)] =   (we can factor a "-" out of (4 - a) = - (a - 4))

-(1/4a³)(a+4)/(a-5) = (5-a)(4+a)/(4a³)

If 790 is in feet, we have

790/9 = 87.7 ft per min = 1.462 ft per second

Note that, every minute, the wheel rotates through 40° of arc

Huh ???????

We have two vectors here, v₁ and v₂

v₁ = <165cos180, 165sin180>  and v₂ = <95cos(225), 95sin(225)>

Summing the "x" components we have

165cos(180) + 95cos(225) = about -232.18

Summing the "y" components we have

165sin(180) + 95sin(225) = about -67.18

And the magnitude is given by   √[(-232.18)² + (-67.18)²] = about 241.7 km

And the direction is given by

tan^-1 (-67.18/-232.18) = about 16.14° in the 3rd quadrant = about 196.14°

V_s = (4/3)* pi * r³ =  (4/3) * pi * (6)³     and you can take it from here....

[ 6a² -30a + 36 ] / [4a - 12 ]   factor out the common factor of 6 at the top and 4 at the bottom

(6/4) [a² - 5a + 6 ] / [a - 3]   factoring the trinomial, we have

(6/4) [(a -3)(a-2)] / (a-3)      "canceling the "(a-3)'s "  and reducing (6/4) we have

(3/2)(a - 2) or   (3a - 6) / 2    ...whichever you prefer !!!

Assuming the tank is a cylinder, the "formula" for its volume is given by

V_c = pi*r²*h   and, if the diameter = 60, then the radius = 30

So we have

V_c = pi* 30² * 72 = about 203,575.2  in³

We can't give a definitive distance, because it depends upon the rebound height  of the ball. We can model the situation as follows:

The ball travels 10 ft and then strikes the ground. It then rebounds to a height of 10r, where r is  some value between 0 and 1. And it falls this distance again before striking the ground.

So, the ball has traveled 10 + 2*10r feet.

And on the next bounce it rises to a height of (10r)r = 10r² feet ..and it also falls this far before striking the ground again.

So, after two bounces, the ball has traveled  10 + 2*10r + 2*10r²  feet = 10 + 20 (r + r²) feet

So, the total distance traveled after 30 bounces is just  10 + 20(r + r² + r³ +......+ r²⁹ + r³⁰) feet

And we can rewrite this as:     -10 + 20 (1 + r + r² + r³ +......+ r²⁹ + r³⁰ )  ...  and this can be simplified to.......

Total Distance  =  -10 + [ 20 / (1 - r) ]  where  0 < r < 1