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What do you mean, it is?

It is

We can us the tangent to find  the length of a side. So we have

tan(60) = (2/√3) / (s/2)   where s   = side length

√3 = 4/ (√3s)

3s = 4

s = 4/3

I can't believe you zegroes.

Yes

Its a way of writing a number that is way to small or too big to write regularly 

Really zegroes?

That isnt really a math question?but ok 4,132 miles gosh i learned something from that!

Cuales son los grados?

141,907 is the answer

Thank you.

Just saw it sorry about that Melody was scrolling by the messages too quickly. Gave you a +5, appreciate the explanation! See the edit

Quazars, please tell zegroes to answer my messages.

What about my detailed answer?  

Thanks for the points Quazars - We all like those. 

Thank you Phil for the detailed answer! Appreciate everyone trying to help as well!

EDIT: Sorry missed your post Melody as I was scrolling too fast!

Sorry that I got the wrong answer Quazars.

Let's take it from what you have, Quazars......

We have

1000 = 100000(.8)ⁿ

Divide both sides by 100000

1/100 = (.8)ⁿ

Take the log of both sides

log(1/100) = log (.8)ⁿ

And by a property of logs, we can bring the "n" out front

log(1/100) =n log(.8)

Divide both sides by log (.8)

log(1/100) / log(.8)  = n = about 21 wars are required

Every war, 20% of the population dies/leaves the army. How many wars would it take for the population to drop below 1000? Original population is 100 000

0.8 since 1-20%=0.8

There are 5 wars because 100,000*20%=20,000. Then you multiply 20,000 by 5 and your answer will be 100,000. So, there would be 5 wars.

Zegroes, the problem with your logic is that you are taking 20% off the original amount each year.

NOT 20% of the the NEW amount.

sorry - I was called a way on a phone call. 

$$\begin{array}{rlll}
100000\times 0.8^n &<&1000&\\\\
100000\times 0.8^n &<&1000&\mbox{ Divide both sides by 100000}\\\\
0.8^n &<&0.01&\mbox{ When you are looking for a power you must take the log (a log IS a power)}\\\\
log0.8^n &<&log0.01&\mbox{ }\\\\
nlog0.8&<&-2&\mbox{Remember: log 0.8 is negative }\\\\
n&>&\frac{-2}{log0.8}&\mbox{ }\\\\
n&>&20.63&\mbox{ }\\\\


\end{array}$$

So it will take 21 wars.

Can you please explain?i still think its three wars.

There are going to be 5 wars.

n = amount of wars

What is n?

If it's any help I'm stuck at 100 000 x 0.8^n = 1000 where I'm trying to find n

0.8^n= 0.1 how do I calculate for n?

I think the answer is 21 but how would I calculate for it without using trial and error?