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CPhill's answer is correct of course.  I am not going to do all of it all over again.  I just want to look at the beginning.

The trajectory is an upside down parabola.

I am going to let time be represented by t because x is used for distance and I think it is confusing (sorry)

Height can be h that is fine.

$$h(t)= -at^2+bt+c\qquad where\:\: a>0$$      

Already it looks different Cphill had a but his a was negative so my answer will work out the same.

I have used -a because I know the parabola is upside down (negative concavity)

NOW when time=0 height =0 therefore aber you substiture thesw values in you get c=0 so

$$h(t)= -at^2+bt\qquad where\:\: a>0$$

NOW it is the same as CPhill's so you can keep going from there.

A great answer anonymous 

I think if it is a circle you drop everything down by one because there is no specific endpoint.

So the original answer was  2*23!/24! = 2/24=1/12

and this becomes 2*22!/23! = 2/23

so yes I think anonymous is right also and 

Of course Rosala is correct. she is a genious too!!!!   

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But he hasn't given you a 'way' and you have used your question up. 

I would ask the brothers if the ground is dirt. If a brother says no, then he is lying and I shouldn't go his way.

CPhill has covered it well but I will stress something here.

I specifically want to look at    $$f(x)=5^{-x}$$

If x is negative this becomes  

$$5^{--number}=5^{+number}=positive\:\:number$$

If x=0 then 5⁰=1 = positive number

If x is postitive then

 $$5^{-number} = \dfrac{1}{5^{+\:number}}=\dfrac{1}{+\:\: number}=positive\:\: number$$ 

When the + number on the bottom gets very big this will approach 0 but it won't ever actually get there.

So f(x)=0 is an ASYMTOTE    f(x)>0

so the range is         $$(0,\infty)\quad \mbox{ or put differently }\quad 0

I have written some posts on negative indices it may be useful for you to revise.

You question could mean many different things.  Would you like to give an example?

In order to add any fractions, you must first make the bottom parts the same.  

They must have a common denominator.

Both good answers - thank you. 

This is modeled by.....

h(t) =  ax^2 + bx  

and

h(1) =  a(1)^2 + b(1) =  110  →     110 = a + b    (1)

and

h(2) =  a(2)^2 + b(2) =  200  →     200 = 4a + 2b   (2)

Multiplying (1) by -2 and adding it to (2) gives us

-20 = 2a       so   a = -10

And substituting this back into (1) we can find b

110= -10 + b       so b = 120

So our function is

h(t) = -10x^2 + 120x

So....after 4 seconds

h(4) =   -10(4)^2 + 120(4)  =  320 ft.

0.30 times 250   =  75

3 * 10% so

3*25=75

Here's the homepage if for some reason you can't find it. :)

I think you accidently posted this twice.

OK....

f(x) = 5^-x   can be wrtten as  1/5^x

Note that we can put any real number in for "x" because  an exponential function never = 0 , so the denominator never = 0   So "domain" relates to x....and the domain is just (-∞ , ∞).....just as you suspected........!!!

The range - which relates to "y" - is trickier....as x is more and more negative, the function grows larger and as x gets more positive, the function gets very close to 0 (but not actually 0).  So the range is (0 , ∞). Here's where a graph may help!!

Note how the graph gets "close" to 0 on the right, but is unbounded on the left!!

Hope this helps !!!

yes. f(x) = 5^−x 

I'm thinking that it is -infinity, positive infity for the domain but i have no idea if this is right. and i do not know how to find the range.

Actually...I'm starting to think that anonymous's answer hit it right on the dot.

I believe he assumed this was a circle, which would give you a little more of a chance of being next to your friend.

That's for your imput anonymous. You should create an account here, you seem to have some nice answers in your brain. :)

Graphing is sometimes easier, but we can often analyze the function, too.......

Sometimes....I use both....just to verify my answer

Do you have a particular function in mind??

Thanks for your imput guys.

I forgot to meantion, this was a circle of pedestals, I think some of you might have assumed that this was a line. (sorry about that)

I didn't really need a completely accurate answer either, just was a bit curious.

Points are awarded where they're due. :)

OK.....I guess all of us can put ourselves on pedestals   .....!!!!

After all....maybe we can't ALL be smart eggs.....but at least we're "colorful"......!!!!!!

justin is using 120 inches of wire to biuld two structures one rectangle that is five times as long as it it is wide and one square whose side lenght is the same as the width of the rectangle. what will be the exact area of his square?

So ......the perimeter of the rectangle is given by 2(L + W)= 2(5W + W) = 2(6W) = 12W

And the perimeter of the square is given by 4W

So

12W + 4W = 120

16W = 120

W = 7.5 inches

So...the area of the square is just   (7.5)^2 =  56.25 in²

The perimeter of a rectangle is two times its length (L) plus two times its width (W).

The perimeter of a square is four times its side (S).

The total perimeter (P) is: $$P = 2L + 2W + 4S$$

We also know that:

$$P = 120$$ (120 inches of wire)
$$L = 5W$$ (five times as long as it is wide)

$$S = W$$ (whose side lenght is the same as the width of the rectangle)

We get:

$$120 = 10W + 2W + 4W$$

$$120 = 16W$$

$$W = \frac{120}{16}$$

$$W = 7.5 inches$$

The area of the square is given by its side length squared (S²). Since $$S=W$$, the area of the square is W².

$$Area = W^2$$

$$Area = (7.5 inches)^2$$

$$Area = 56.25 inches^2$$

Assuming that your initial location doesn't matter and that you just want to know the chance of your friend being placed on a pedestal right next to you, we can do the following:

Place you in a random pedestal.

Place your friend in another of the 23 remaining pedestals.

Your friend can be put in 23 different pedestals, but only two of those pedestals will be next to you.

We have 23 events of which 2 are positive events. The chance of your friend being next to you is $${\frac{{\mathtt{2}}}{{\mathtt{23}}}}$$, which is aproximately equal to 8,7%. Basically that will happen once in every 12 rounds.

If the orc is 3 times as tall as Marvin, Marvin's height (M) times 3 equals the orc's height (O).

$$3*M=O$$

Knowing that the orc's height = 9 feet, we have:

$$3*M=9 feet$$

$$M = \frac{9 feet}{3}$$

$$M = 3 feet$$

Thus, Marvin's height equals 3 feet.

If c(x) is a quadratic function, then $${c}{\left({\mathtt{x}}\right)} = {\mathtt{A}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{B}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{C}}$$, where A, B, and C are constants. Our job here is to find the constants.

$$c(2) = 23 = A*2^2 + B*2 + C$$

$$c(4) = 103 = A*4^2 + B*4 + C$$

$$c(10) = 631 = A*10^2 + B*10 + C$$

So we have:

$$23 = 4A + 2B + C$$

$$103 = 16A + 4B + C$$

$$631 = 100A + 10B + C$$

If we solve that, we get A = 6, B = 4, C = -9.
Thus, since $${c}{\left({\mathtt{x}}\right)} = {\mathtt{A}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{B}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{C}}$$:

$$c(x) = 6x^2+4x-9$$

Now we just use the function for x=6.
$${c}{\left({\mathtt{6}}\right)} = {\mathtt{6}}{\mathtt{\,\times\,}}{{\mathtt{6}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{9}}$$
$$c(6) = 231$$ dollars.

Melody if you are saying " WE all are GENIUSES" then this means all of us so i am also one of those "all of us" so are u calling me also a genius or just you,CPhll and Alan are GENIUSES!if u r calling me a genius (too) then thank you very much!