CPhill

What is the law of cosines?

That ‘s the law where you co-sign as a promise to pay if the main signer decides to be a deadbeat.

---------------------------------------------------------------------------------------------------

LOL!!!........... I believe that's known as the Law of "Co-Signs."

I like that answer, Anonymous. A little "board" humor is never a bad thing !!

Actually........you can read all about it here:

http://en.wikipedia.org/wiki/Law_of_cosines

Find the number of different signals consisting of 9 flags that can be made using 3 white, 3 red, and 3 blue.

9! / (3! * 3! * 3!) =1680

This may not be clear at first, but let me give a simple example of why it's true.

Suppose we had three white flags and two red flags.

The number of "arrangements" that we could make could make of these flags is just 5! = 120.

But note the arrangement of

R R W W W would look just like the arrangement of R R W W W where the position of the  white flags at the end were interchanged in some manner. Thus, I could choose any of the three white flags to be in the 3rd position, any of the two remaining ones to be in the 4th position and the lone remaining one to be in the 5th position. And the number of arrangements of white flags at the end would just be 3 x 2 x 1 = 3!

Thus, the same thing would be true, no matter which positions the white flags occuppied. Thus, I've got to "divide away" the "repeats" of white flags (3!) in the total number of arrangements (120). The same holds true of the red flags. I have to divide away their repeats as well = 2! 

Therefore, in any arrangement of n things with repeats of one (or more) indistinguishable items, I must divide the total number of arrangements, (n!), by the factorial number of each repeated thing!!!

I hope this helps........

For extra credit, how many distinguishable "words" can you make from the word "MISSISSIPPI??"

whats the basic number of 2.3x 10-2

We have 2.3 x 1 /100

Move the decimal two places to the left on both "top" and "bottom," and we have

.023/ 1  = .023

How can a log be negative?

A "log" can be negative....we just can't take the "log" of a negative number (or 0)

For instance

Log(.001) = -3

But, Log(-3)....... is undefined......This says, what power can I raise 10 to get "-3"  ???

Answer....there is none !!!!

how do i find a length if i have the value

I'm assuming that you have the "value" of the area and the width

Since  ...   Length x Width = Area

Then divide both sides by the Width to find the Area. i.e.,   L = A /W

1/3 *3.14*r2*20=18149.2

It appears that we have the volume of a cone with a height of 20 and we'd like to find its radius.

OK....let's start by multiplying both sides by 3 to "clear" the fraction on the left...this gives

3.14*r2*20 = 54447.6   Now, divide both sides by the "20" and the "3.14" and we have

r^2 = (54447.6)/(20*3.14)

r^2 = 867

Now, just take the square root of both sides and you're done.......you'll have r = (something)

I assume we have this

log(a)(75) + log (a)(2)

By a property of "logs" ......    log(a)(b) = log(a) + Log (b)

So this gives us

log(a)(75) = log(a) + log(75)

And log(a)(2) = log(a) + log(2)

So, putting everything together, we have

log(a)(75) + log (a)(2) = log(a) + log(75) + log(a) + log(2)

= 2log(a) + log(75) + log(2)

But note, by "reversing" the log property above, I can write

2log(a) + [log(75) + log(2)]  as

2log(a) + [log(75)(2)] =

2log(a) + log(150)

Note, now that log(150) = log(10)(15) = log(10) + log(15)

And log(10) = 1

So we really have is..... 2log(a) + log(15) + 1

And another "log" property says that alog(b) = log(b)^a

So 2log(a) becomes log(a)²

And the final result is just

log(a)² + log(15) + 1

he sum of two numbers is 19. three times the first number minues four times the second number is eight. find the two numbers

OK...call the first number "x,' and the second number "y"

Then, x + y = 19

And three times the first (3x) plus four times the second (4y) = 8

Mathematically, we have

x + y = 19

3x - 4y  = 8

We can solve this by one of two methods (besides just guessing, of course !!) - either sustitution or elimination. I'll use the substitution method.

If I subtract the "x" from both sides in the first equation, I have y = 19 - x

So, in the second equation I can "substitute" "19 - x" for "y."

So we have

3x - 4(19-x) = 8    Note that this gives us an equation in one variable that should be easy to solve.  We have

3x - 76 + 4x = 8    And simplifying gives us

7x - 76 = 8

Add the "76" from both sides and we get

7x = 84   dividing both sides by 7   we have

x = 12  

Now...... knowing that y = 19 - x, I'll let you find  "y".

if Mike runs 2miles. How many millimeters has he ran?

Well....let's see

There are about 1609 meters in a mile

There are 1000 millimeters in each meter

So, the number of meters in one mile (1609) x the number of millimeters in each meter (1000) = 1,609,000 mm in one mile

To find the number for 2 miles, I'd just double this.

(We should adopt the metric system here in the U.S., huh ?? It's way easier to work with!!)

Unfortunately not......I copied the formula straight from Wikipedia (I actually knew it....but I'm too lazy to type !! LOL!!)

I found a LaTex "primer" on the Internet that I'm trying to work through, but it contains a lot of technical mumbo-jumbo that is indecipherable (particularly about saving/displaying output).  I learned this much, at least....."LaTex" isn't pronounced as it seems !!!

I think I'm going to watch the video(s) that you suggested.....I'm sure they're clearer

P.S. - I know we're not supposed to be passing messages back and forth in the forum, but I don't fully trust the "message" service, yet!!!