Melody

Tues 17/3/15

1)    The Monty Hall problem.                   Yes again LOL    I found a cool clip for it.

2)    Finding volume of an odd shape.       Melody

3)    What is the last digit of pi?               Thanks TayJay, anon and Heureka    

4)     Manipulating algebra                        Thanks CPhill and anon

5)     Physics - Force and Power.               Thanks Alan

6)     Probability  (secret santa)                 Melody

7)     Fining an angle using sine rule           Thanks melody and CPhill.

         

               ♫♪  ♪ ♫                                ♬ ♬ MELODY ♬ ♬                                 ♫♪  ♪ ♫

Thanks CPhill, I just want to see if I can do it too.        (๑‵●‿●‵๑)

 

Mmm that looks tricky.

 

 

let

$$y= -6*sec(sin(5x^2+3x+2))$$

let

$$\\g = sin(5x^2+3x+2)\\\\
\frac{dg}{dx}=(10x+3)[cos(5x^2+3x+2)]\\\\\\
y=-6sec(g)\\\\
y=-6(cos(g))^{-1}\\\\
\frac{dy}{dg}=6(cos(g))^{-2}(-sin(g))\\\\
\frac{dy}{dg}=\frac{-6sin(g)}{cos^2(g)}\\\\\\
\frac{dy}{dx}=\frac{dy}{dg}\times \frac{dg}{dx}\\\\
\frac{dy}{dx}=\frac{-6sin(g)}{cos^2(g)}\times (10x+3)[cos(5x^2+3x+2)]\\\\$$

 

$$\\\frac{dy}{dx}=\frac{-6sin(g)(10x+3)[cos(5x^2+3x+2)]}{cos^2(g)}\\\\ \frac{dy}{dx}=\frac{-6sin(sin(5x^2+3x+2))(10x+3)[cos(5x^2+3x+2)]}{cos^2(sin(5x^2+3x+2))}\\\\ \frac{dy}{dx}=-6tan(sin(5x^2+3x+2)sec(sin(5x^2+3x+2))(10x+3)[cos(5x^2+3x+2)]}\\\\
\frac{dy}{dx}=-(60x+18)tan(sin(5x^2+3x+2)sec(sin(5x^2+3x+2))[cos(5x^2+3x+2)]}\\\\$$

 

WOW this is the same as CPhill's answer     (๑‵●‿●‵๑)