CPhill

y'+y* tanx=1/cosx

y' + y [tan x]  = sec x          multiply both sides by sec x

y'sec x  + y [  sec x tan x ]   =  sec^2 x

Let u  =  y sec x   →  du   =  y'sec x  + y*  sec x tan x

So

du  = sec^2 x     integrate both sides

∫ du   = ∫ sec^2 dx

u =  tan x + C₁

y sec x   =  tan x + C₁   

y [ 1/cos x ]   = sin x / cos x  + C₁      multiply through by cos x

y =   sin x + C₁ cos x

Write an equation of the ellipse with focus at (0,+-9) and vertices at (0,+-13)

  [ The major axis is along  the  y  axis]

We need to find " b"  = √[a^2  − c^2] =  √ [ 13^2 − 9^2]  = √ [169 − 81]  = √88

So.... the equation is

y^2 / a^2     +     x^2 / b^2      =   1

y^2 / 169   +     x^2 / 88      =   1

Here's  the graph :   https://www.desmos.com/calculator/wsiick7axh

If you use 3.14 for pi.....your answer is good......

C = pi * [ 9 in ]  ≈  28.27  in

That's nothing, Hawkeye.....I can type the Roman "0"...............

See???

Call the original amount    $30 + x....and x is the amount bet on Hot Feet

After the first race  he has  $2x  =  twice what he bet on Hot Feet

In the second race......he bets $30 on the Favorite   and [ 2x − 30]  on Lucky Horseshoe

And after the second race he has $[2x  − 30] *3   = $[6x  −  90]

In the 3rd race he bets $30  on  Wish Me Luck and $ [ 6x −  90 ] − 30   = $[ 6x − 120]  on  Bred for Speed

And after the last race he has

4 *  [ 6x − 120]   = 240

24x  −  480   = 240

24x  = 720

x  = $30

So.......the original amount  =  $30 + x  =  $30  + $30  =  $60

Dear GA...... 

Originally....Melody's standards were much too high for her to be "bought off" with  cheap alcohol........however......things being what they are "Down Under,"  she's had to recently "revise" her moral codes........now???......money and/or cheap alcohol are perfectly acceptable forms of bribery.......

Sincerely,

Sir CPhill

I'll give this a shot

Afer taking 12 red b***s from the first bag and putting them into the second :

The first bag conains 18 b***s  [all red ]

And the second bag now contains :

 72 b***s..... 60 black,  and 12 red  =   5/6 black and 1/6 red

Assuming the second bag is well−mixed and taking 12 b***s from it and putting hem back into the  first bag.:

We should select 10 black b***s  (5/6)*12 and 2 red ones (1/6)*12 to be put back into the first bag

So the second bag [ theoreically] now contains   50 black and 10 red

And the first bag should now contain 10 black and 20 red

So.......the first bag should contain as many black b***s as the second bag contains red ones !!!

You have to have a solid knowledge in Mathematics, and more specifically, in a number of branches of Mathematics such as Algebra, Calculus, Trigonometry, Geometry, Number Theory,Statistics, Probability theory.....etc, and the ability to explain your solutions as thoroughly as possible. Also, the ability to reason using good logic, such as when solving "word problems". As well, a good and solid understanding of the Sciences, such as Physics, Chemisty....etc., would be very helpful. At least, that is my understanding! Somebody could correct me on this!!.

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All of that helps......but not as much as promising  to slip  Melody a few bucks under the table......

Here's another approach to Melody's answer:

Number of ways to choose a  committee composed of all girls or boys  = 2C(10,4)

Number of possible committees   = C(20,4)

So......the probability that at least one girl or boy is on the committee  =

1   −   2C(10,4)  / C (20,4)    =  1  − 420 / 4845  =  1  − 28.323  = 295 / 393