Note that Density = $\frac{\text{population}}{\text{area}}$

As stated in the question, there are 15 lizards, and the area of the circle that surrounds the boulder is simply $14^2\times\pi$or $196\pi$Thus the answer is $\frac{15}{196\pi} \approx \frac{1}{41.0293333}$

Density  = Population  / Area   =

63555 /  3.1  km ^2    =

about  20501.6  people per km^2

a^2  - 36    factors  as  ( a + 6)  ( a - 6)

a^2 + 5a  - 6      factors as  ( a + 6 )  (a - 1)

So.....the lcd    =   ( a - 6 ) ( a + 6) ( a - 1)

Could you post a specific example???....I'm a little confused as to what you mean....

Hi Coltsrock,

You forgot to give the address of the tutorial that you spoke of :(

This isn’t a sustainable population. It’s probably 1/4 mile radius.

15/(3.14*.25²) = 76.43

76 lizards per square mile.

We have

2P  = Pe^(r * t )     where r  =   .065   and t is the time to double

2P = Pe^( .065 * t)     divide both sides by P

2  =  e^(.065 * t )        take the Ln  of both sides

Ln 2  = Ln  2^(.065 * t)     and by a log property we can write

Ln 2  =  ( .065 * t)  Ln e         and Ln e   = 1   so we can disregard this

Ln 2   = .065 * t      divide both sides by .065

Ln 2 / . 065  = t  ≈  10.7  years

AGAIN,SORRY

HERE!THIS IS FOR YOU!

Looks pretty accurate to me, hectictar....!!!!.....I have to admit....the concept of "nets"  is foreign to me, too......but....it does seem like a good way to visualize exactly what is going on....

sine

$= \frac{15\ lizards }{ 3.14 \times {14}^2 square\ miles} = ( \frac{15}{615.44}=0.0244)\frac{ lizards }{square\ mile}$

$\approx 1 \frac {1 lizard }{square \ mile}$

  asinus

The  side of the triangle with a length of 3959 + 1.4   is the hypotenuse of a right triangle with the Earth's radius forming another leg.....the distance from the skydiver to the horizon, "x" is given by :

sqrt[ (3969 + 1.4)^2  - 3969^2 ]  ≈ 105.296 miles

A biologist is researching the population density of lizards near a boulder in a desert. The biologist counts 15 lizards within a radius of 14 mi of the boulder.

What is the population density of lizards?

 Use 3.14 for pi and round only your final answer to the nearest whole number.

$The \ population \ density \ of \ lizards \ is$

$= \frac{15\ lizards }{2\times 3.14 \times {14}^2 square\ miles} = ( \frac{15}{1230.88}=0.012)\frac{ lizards }{square\ mile}$

$\approx 1 \frac {1 lizard }{square \ mile}$

  !

I solved it 

$tan(\theta)=\frac{y}{x}=\frac{b}{a}=\frac{4}{4\sqrt{3}}$

$\theta=\frac{\pi}{6}$

$r=8$

In polar form

$8[cos(\frac{\pi}{6})+isin(\frac{\pi}{6})]$

Now to find the roots.

$\sqrt[n]{8}[cos(\frac{\frac{\pi}{6}+2\pi k}{n})+isin(\frac{\frac{\pi}{6}+2\pi k}{n})]$

This question asks for cube roots so n=3

$\sqrt[3]{8}[cos(\frac{\frac{\pi}{6}+2\pi k}{3})+isin(\frac{\frac{\pi}{6}+2\pi k}{3})]$

Now that this is all filled in, lets change the k to each root we need.

$k=0: 2[cos(\frac{\frac{\pi}{6}}{3})+isin(\frac{\frac{\pi}{6}}{3})]$

$k=1: 2[cos(\frac{\frac{\pi}{6}+2\pi}{3})+isin(\frac{\frac{\pi}{6}+2\pi}{3})]$

$k=2: 2[cos(\frac{\frac{\pi}{6}+4\pi }{3})+isin(\frac{\frac{\pi}{6}+4\pi }{3})]$

It's 0

A woodchuck can chuck approximately 700 pounds of wood!

The More You Know!

-MagikalRedNiteTiger

You could set up the relation as a table of ordered pairs. Then, test to see if each element in the domain is matched with exactly one element in the range. If so, you have a function! Watch this tutorial to see how you can determine if a relation is a function.

Get real guest!

This is not a question. This is just a page copied from a text book!

But that has been answered Adam.   

If you think the answer might be wrong or you want more explanation (or whatever) could you please explain this on the original thread.  ://  

Bananas are priceless!

If you pay a nickel for a penny, how much do you pay for a banana?

21 is the answer