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A log is a power (exponent) and powers can be negative

$$5^{-2}=\frac{1}{25}\\\\
\mbox{therefore}\\\\
log_5 \:\left(\frac{1}{25}\right)=-2$$

Hey Chris, 

Why do you think it is base 10???

My interpretation

$$log_a 75+log_a 2
=log_a 150$$

And that is the end of the story 

I think that this is your question - you should have put an ^ (hat) between the r and the 2. ^ is the symbol used for power of when you don't know how to put it in the correct position. 

$$\frac{1}{3}\times3.14\times r^2\times20=18149.2$$

I'm sure CPhill's answer is perfect although I will admit i haven't thoroughly checked it.

I really like the first answer too.  I can't give it a tick because someone might take it seriously but it is very funny.

I am really pleased that more people are starting to have fun in the forum    

Look at this - it will help you remember the quadratic formula!    

2.3 x 10-2 = 23-2 = 21  Just like our anonymous guest told you 

CPhill and I both  think that you probably mean 2.3 x 10^-2

this is in scientific notation - to change it to usual notation you wil need to move the point 2 places but the minus sign means that it will be a little answer - less than 1 so move the point in the correct direction to make the number little.

2.3 = 0002.300000 I can add as many zeros as I want so long as the point says in the same place between the 2 and the 3.

so now when I jump the point it will jump over the 2 and over one zero and the answer will be 0.023 Just like CPhill told you.  

You probably should work through this unit then you will understand properly.

If you draw a graph of this function you will see there are two real roots at x = 2.561553 and x = 3.117176 (approximately).

Very pithy word-play, "Pi" thag !!!

cotx/secx + sinx = ?

Let's change everything to sines and cosines

(cosx/sinx)/(1/cosx) + sinx

(cosx/sinx)*(cosx/1) + sin x

(cos²x)/sinx + sinx

(1-sin²x)/sinx + sinx

1/sinx - sin²x/sinx + sinx

1/sinx - sinx + sinx

1/sinx

cscx

If you are going to write in Portuguese, at least use proper syntax. Morde-me

Como se presenta intervalos aqui

Are you psychic or something? How’d you know my name? It’s hyphenated, and I spell it with an “i” and a “y” Pithy-Agoras? My friend’s name is Archie Medes. He’s kind of screwy. My other friend is Rene Horas Descartes. Everyone makes fun of his name. Not because he a guy named Rene, but because he has Horas before Descartes. I don’t get it though.

Re: loga75+loga2

Mmmmmm....after I looked at this again , I think you MIGHT have meant

log 75 + log 2

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

So, log 75 + log 2 = log(75)(2) = log(150) = log(10)(15) = log 10 + log 15

But, log 10 = 1

So, we have........ 1 + log 15

And that's it  (I hope........ )

$$\[\huge a^2 + b^2 = c^2\]$$

Everyone knows that.

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Apparently, "everyone" doesn't include  Mr. Pythagoras - (AKA, Anonymous No. 1)

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.

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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??"

I like them too. Especially when prepared in a slow cooker with salt and pepper.

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

-b±√(b²-4ac)

____________

         2a

$$\[\huge a^2 + b^2 = c^2\]$$

Everyone knows that.

You need to reflash your bios, chip

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

21

is the answer

Just learn from the teacher dood!

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 !!!!

hellokitty