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Zegroes i think Melody told us not to post out here!!!Remember!

Bye DragonSlayer!!!!

Heh, I got here a bit late it seems. Now it is past midnight, but a day later!

New riddle!

Whether it be on prison windows or on your phone

You can grab a drink, and I am often raised.

What am I?

1 m³ = 1000 L so 1 m³ = 1 kL

Here's the graph:

Can you answer the question now?

GEE, that's brilliant.

Thanks Melody and CPhill!!

The answers are the same but they look different.  It will take me a while to work it all out!

There are 100cm in a meter...so, in a cubic meter, there are (100 cm)³

So we have.......

7.73 * (100cm)³  / .27cm  ≈ 28,629,629.6 cm² ≈ 2862.96 m²

-6q + 1.2 = -21.8 - 1.4q

Add 6q to each side

1.2 = 4.6q - 21.8

Add 21.8 to each side

23 = 4.6q    

Divide both sides by 4.6

5 = q

This is just a fancy way of asking..."What is the power (exponent) on "10" when we write this number using scientific notation??"

So we have......

1.178 x 10⁵   

So  "5"  is our answer !!!!

Call the number of tuna pies "x."

Then the number of chicken pies = x + 185

If he sold .6 of the chicken pies, then he had .4 of those left, and if he sold 1/2 = (.5) of the tuna pies he must have had .5 of those left.

So......

.4(x + 185) + .5(x) = 146

.9x + 74 = 146

.9x = 72

x = 72/.9  = 80   and that's the number of tuna pies he started with

So, he sold 1/2 of these = 40

And the number of chicken pies he sold is given by:

.6(80 + 185)  = 159

So the total pies sold =   40 + 159   = 199

This is an old problem from my calculus class days....let me see if I remember how to solve it.

Let x be the distance that the tip of his shadow is from the pole. Let xp be the distance the person is from the pole, and let xs be the length of his shadow.

So......x = xp + xs

We want to determine x' when the person is 6m from the pole.

Note that x'p  = 1m/s

Let's eliminate xs from the above equation.

Using similar triangles, we have

1.8/6 = xs/x = xs/(xp + xs)  so

1.8/6 = xs/(xp + xs)     (  note that 1.8/6   = .3   )

(.3) (xp + xs)  = xs

.3xp = .7xs

xs = (3/7)xp

Now.....substituting back into the original equation, we have

x = xp + (3/7)xp  = (10/7)xp    ....   So.....differentiating both sides, we have....

x' = (10/7)x'p  = (10/7)(1m/s)  = (10/7)m/s

And that's how fast the tip of the shadow is moving away from the pole.

The second part is easy.....we want to find x's when xp = 6

Again, we can use

x = xp + xs      and differentiating, we have

x' = x'p  + x's

But, we know x'p = 1m/s  and x'= (10/7)m/s.....so we have

(10/7)m/s  = 1m/s  + x's

So  x's  =  (3/7)m/s  ....   And that's how fast the length of the shadow is changing!!

Here's a diagram of the problem......

EDIT:  This answer is correct just rediculously long.  I have redone it much more efficiently on a later post.

The two triangles are similar because they each have a right angle and they have a common angle.

Let the third angle be  $$\theta$$

You know that      $$\frac{dx}{dt}=1m/s$$

a) you are being asked to find  $$\frac{dy}{dt}$$  when x=6m

b) you are being asked to find   $$\frac{d(y-x)}{dt}=\frac{dy}{dt}-\frac{dx}{dt}$$

a)  I think it will be easier for me to get y and x in terms of $$\theta$$

$$\begin{array}{rllrll}
tan\theta&=&\frac{y}{6}\qquad \qquad \qquad \qquad & tan\theta&=&\frac{y-x}{1.8}\\\\
y&=&6tan\theta}\qquad& y-x&=&1.8tan\theta\\\\
&&&x&=&y-1.8tan\theta\\\\
&&&x&=&6tan\theta -1.8tan\theta\\\\
&&&x&=&4.2tan\theta \\\\
\frac{dy}{d\theta}&=&6sec^2\theta}\qquad& \frac{dx}{d\theta}&=&4.2sec^2\theta\\\\
\end{array}$$

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

              $$\begin{array}{rll}
\frac{dy}{dt}&=&\frac{dy}{d\theta}\times \frac{d\theta}{dx}\times \frac{dx}{dt}\\\\
\frac{dy}{dt}&=&6sec^2\theta}\times \dfrac{1}{4.2sec^2\theta}\times 1\\\\
\frac{dy}{dt}&=&\frac{6}{4.2}\\\\
\frac{dy}{dt}&=&\frac{10}{7}\\\\
\frac{dy}{dt}&=&1\frac{3}{7}\;\;m/s\\\\
\end{array}$$

b)  $$\frac{dy}{dt}-\frac{dx}{dt}= 1\frac{3}{7}-1=\frac{3}{7}\;\;m/s$$

So the end of the shadow is moving away at a constant rate of      $$1\frac{3}{7}m/s$$

and

The length of the shadow is changing at a constant rate of     $$\frac{3}{7}\;\;m/s$$

Since both of these are constants, it makes no difference how foar away from the pole the man is.

The answers will always be the same.     

See if you had a cool avatar like godzilla ya wouldnt be switching

See yah DS

1+1=2

1+1+1+1+1+1+1+1=8

one plus one equals two

one plus one plus one plus one plus one plus one plus one plus one equals eight.

Hope it helps!

Okay!So this seems to be a lil long process in memorizing but still here we go

as u have said we will write the numbers as

1+1+1+1+1+1+1+1 = 8 ANSWER

I think that the answer is 42. Check the calculator.

Yep.....that's what I got, too!!!

Good work !!!

You're correct bioschip.......if you interpret the first thing as "32"

I think it's supposed to be   3²

I'm gonna' give you points, anyway !!!

is it 19

There are about 39.37 inches in a meter

And 3 feet = 36 inches = 1 yd

So......

(39.36 / 36)yds  * 3feet/1yd   = about 3.28 feet in a meter

i dont know if it's right but is it 42

Chris, as I was reading, the numbers are the solutions for the variable. Just think of it...

For who is this question, Chris?