CPhill

You can divide by the commonly used value for pi  = 3.14

Cool, Melody!!.....I like that compact "box" diagram.....very nice....

OK, Rosala...let's examine these statements one-by-one and find a solution.

First. let the total number of deer = x

And half of them are grazing, so that just must be !/2 of x = (1/2)x

And "the remainder" must just be the other half = (1/2)x

And (3/4) of the remainder - (1/2)x -  are playing..... "of" usually means "multiply"...so we have..... (3/4) x (1/2)x = (3/8)x ..

....and 9 others are drinking

So we have

(1/2)x + (3/4)(1/2)x + 9 = x

(1/2)x + (3/8)x + 9 = x     ........ Note that..... (1/2)x = (4/8)x 

(4/8)x + (3/8)x + 9 = x         Add the fractions

(7/8)x + 9 = x        Subtract (7/8)x from both sides

9 =  x - (7/8)x         Note that..... x = 1x = (8/8)x

9 = (8/8)x - (7/8)x

9 = (1/8)x                  Multiply by the reciprocal of (1/8)  =  (8/1)  on both sides

(8/1)*9 = (8/1)* (1/8)x   

72 = x

And there's our answer.....72 deer

By a log property we can write

Log 4x + Log 2x = Log (8x^2)       So this gives us

Log(8x^2) - 4 = 0      Subtract 4 from both sides

Log (8x^2) = 4          Which says that, exponentially

10^4 =  8x^2       

10,000 = 8x^2           Divide by 8 on both sides

1250 = x^2                Take the square root of both sides

±√(1250) = x =       ± 25√(2).....we must reject the negative root because it would mean that we would have the log of a negative number in the original equation, and that's undefined.

Let h be the length of the hypoteneuse. Then by the Pythagorean Theorem, we have

(h-4)^2 +  (h-8)^2 = h^2       Expanding the terms, we have

h^2 - 8h + 16 + h^2 - 16h + 64 = h^2        And simplifying

2h^2  - 24h + 80 = h^2         Subtract h^2 from both sides

h^2 - 24h + 80 = 0               And factoring, we have

(h -20) (h-4) = 0             Reject h =4, since it would make one side 0 and the other -4

So the hypoteneuse is 20 and, one side = (20-4) = 16, and the other side = (20 - 8) = 12

It's just the scale factor squared.

To see this, note that one square foot = 1 ft X 1ft

Now, let us suppose that we increased or decreased every side of the square foot by some scale factor, "p."  So now, the "new" area is just (1p) X (1p) = 1p^2 sq ft.

And every square foot of the object would behave in the very same way.......

I'll take a run at the pearl problem......although I kind of like the idea that Anonymous had...snatching the bag and running!!!

I also believe that drawing all the pearls out of the bag, except one, is the correct strategy. Since, after any number of "even" draws, we never know how many white or black ones we have, we can never be sure that their totals are equal...if we could, "fabulous" Steve would soon be "broke" Steve.

Notice that after just 2 draws, there are 3 possibilities.....0W 2B, 1W 1B, 2W 0B  ........Thus only one thing out of 3 wins for us.

Notice that after just 4 draws, there are 5 possibilities.....0W 4B, 1W 3B , 2W 2B, 3W 1B, 4W 0B........Thus only one thing out of 5 wins for us

And after 100 draws, again, only one thing out of 101 wins for us - 50W 50B, and 100 other combinations lose for us.

But, consider the scenario in which two pearls remain in the bag......If we draw either a white or black one on the next draw and a white one remains, we win. Thus, (draw white)(white remains) or (draw black)(white remains) are two winning scenarios.

But, if we draw either a black or white one and a black remains, we lose no matter what. Thus, out of the four remaining scenarios, 2 win for us and 2 lose, so we have a 50% chance of winning. And drawing all but one pearl out of the bag results in a greater chance of winning than stopping at any other point of even draws. Also note, that when we draw 2 pearls out ot the bag at first, we have a 33% chance of winning, but when two pearls remain, we are even money to win !!!

One final note....you could turn the tables on "fabulous" Steve and tell him that you will take one pearl out of the bag, and whatever remains in the bag is your "draw." Thus, if you draw a white one out, you win......otherwise, no.

That's my story, and I'm sticking to it!! (But, it might be incorrect, too......)

I think you might be talking about finding some angle, given a ratio.....use the tangent inverse for this.......hit the "2nd" button on the calculator, and then "atan" and key in your value.......then hit enter or the "=" key

Let's try it this way, rosala....

Call the total amount of money "x"

Then the son gets (1/3)x, the daughter gets (1/4)x, and the wife gets $20000

And the total of those three things must be the total amount.....so......

(1/3)x + (1/4)x + 20000 = x     Simplifying

(7/12)x + 20000 = x      Subtract (7/12)x from both sides

20000 = (5/12)x             Multiply by (12/5) on both sides

20000 (12/5) = x            The 20000 and the 5 "cancel" to 4000

4000(12) = $48000

And that's the answer.........