The worlds first riddle!

appointed?
The other one I got the same answer as Stormwatch

Lzzie:

AZETTPPAECLK = ZEPPELIN ATTACK

Get well soon sweetie!


TTH:

tapepod = appointed

The other one I got the same answer as Lzzie

You Get well soon too sweetie!




You have a 6 foot high fence and at 15 foot board. The sun is shining directly overhead. You want to lean your board against the fence so that as large a shadow as possible covers your neighbor's yard. (Don't ask)

What is the maximum length of the shadow you can form

Note: The ladder must rest against the fence and touch the ground on your side of the fence.

[size=7]Create a right triangle with height of h, hypotenuse of 15 and a base of x+y, similar to this right triangle is a triangle with a height of 6, a hypotenuse of (36+x^2)^(1/2), and a base of x.

Since these triangles are similar
x/(34+x^2)^(1/2)=(x+y)/15
Separating the variables
y=15x/(36+x^2)^(1/2)-x
Since the board is a constant cross section maximizing the length of the shadow (y) maximizes the shadow. At yhr maximum dy/dx=0
dy/dx=0=((15(36+x^2)^(1/2)-15x*1/2*2x(36+x^2)^(-1/2))/(36+x^2)-1
solving for x
x=((15*36)^(2/3)-36)^(1/2)= 5.505685 ft
substitute x in terms of y for the length of dhadow over the fence
y=15*5.505685 /(36+5.505685 ^2)^(1/2)- 5.505685 =4.635873 ft
Interesting intuition told me that the angle of the board with the ground would be 45 deg, although close, that is not the case, the angle is slightly more, just greater than 47 degrees.[/size]

Rap

Try wrote: "Get well soon sweetie!"

Hi Try,,, it's me,,, you know who


Are you cheating on me handsome? ~ hi.

Try,,, check out the color in this beautiful picture ~


http://www.seds.org/messier/Pics/Png/m51close.png

Rap:

Shadow

Create a right triangle with height of h, hypotenuse of 15 and a base of x+y, similar to this right triangle is a triangle with a height of 6, a hypotenuse of (36+x^2)^(1/2), and a base of x.

Since these triangles are similar
x/(34+x^2)^(1/2)=(x+y)/15
Separating the variables
y=15x/(36+x^2)^(1/2)-x
Since the board is a constant cross section maximizing the length of the shadow (y) maximizes the shadow. At yhr maximum dy/dx=0
dy/dx=0=((15(36+x^2)^(1/2)-15x*1/2*2x(36+x^2)^(-1/2))/(36+x^2)-1
solving for x
x=((15*36)^(2/3)-36)^(1/2)= 5.505685 ft
substitute x in terms of y for the length of shadow over the fence
y=15*5.505685 /(36+5.505685 ^2)^(1/2)- 5.505685 =4.635873 ft

Interesting intuition told me that the angle of the board with the ground would be 45 deg, although close, that is not the case, the angle is slightly more, just greater than 47 degrees.




Great answer Rap!

Let me help out here:
The angle formed is approximately 47.46 degrees.


Jovay wrote. "Are you cheating on me…?"

No way Jovay; there is only one cutie I dream about! BTW: There is no color in space; NASA fakes the pictures!

"The quintessential Hubble photograph is a 1995 image of the popular Eagle Nebula, also known as M16 or the Pillars of Creation. The soaring structures had one of their red emissions converted to green -- by the astronomers who took the picture -- in order to highlight scientific detail.

In "reality," no green was detected coming from the Pillars.
Interestingly, all Hubble images are created with black-and-white cameras. Ones and zeros are sent to Earth. Color is dropped in later with the popular Photoshop program."

Anyhoo; I hope you are keeping well!

You are in a race in which the starting line is at a certain point on a straight beach. The finish line is in the water.

One way to arrive at the finish line is to run 4 kilometers down the beach, make a 90 degree turn and swim 1 kilometer.

However, you may cut into the water at any point. Your speed on land is 6 k.p.h and your speed in water is 2 k.p.h.

At what point, measured from the starting line, should you cut into the water

My Dog can do Calculus

I've seen this problem before, or something very similar. I'm at a beach and playing catch with my dog. I throw his stick out in the water and my dog runs down the beach and dives in the water at a point that minimizes his retrieval time. The strange thing is that my dog uses something pretty close to the most optimal path based upon his land running speed and his swimming speed.. And my dog inherently knows where this minimum path is without having to do a bunch of differential equations.

All in all it's pretty much the same as my contention that Willie Mays was the best physicist I've ever seen---he could just glance at the arc of a fly ball and just predict where he had to put his glove to make the out. Amazing.

[size=7]Any who here's the problem as I see it

Let T be the time for the path to the finish line. The optimal path is to run down the beech to some point (x) and then swim to the finish line. The Time required is given by
T=(4-x)/6+(x^2+1)^(1/2)/2=((4-x)+3(x^2+1)^(1/2))/6
Find dt/dx=0 (minimum)
0=(-1+3*(1/2)*2x*(x^2+1)^(-1/2))/6
so
(x^2+1)^(1/2)=3x
x^2+1=9x^2
x^2=1/8
x=1/(2*2^(1/2))=2^(1/2)/4=sqrt(2)/4
So to minimize time run down the beach (16-sqrt(2))/4 km and then swim 3*sqrt(2)/4 km to the finish line[/size]

Just as my dog would do.

Rap

Here's an interesting article about canine calculus.

http://www.sciencenews.org/articles/20030607/mathtrek.asp

Thanks markr, that the problem I saw. It made the rounds of secondary school math sites.

Rap

dog ed f it


Q?

4 men and 4 women randomly walk arm in arm 8 abreast into a bar.

What is the probability that a woman is touching a man?

Q?

c
-
tempstempstemps

MEN/WOMEN
[size=7]The probability that at least one woman is touching at least one man is 1. Otherwise they could not all be connected. The probability that a man is touching abreast is high.

Do you want the probability that a given woman is touching any man or a given man?[/size]

Hi Try,,, <waves>,,,<waved>. In real life i'm boring ,,, if only it were the 'new order',,, for all eh's,,, then the world and everything would be better off,,, it would be like this -


(~U~)


and you know what that means! <what>,,, thanks for tutoring my lazy brain.


In my head,,, there was a terrific ~beautiful~ parade in Haverhill today.

I'm poor and was raised that way (financially speaking), but one wouldn't know it if they looked at my taste in,,, most thing's.


And yourself?

Rap:

My Dog can do Calculus

Any who here's the problem as I see it:

Let T be the time for the path to the finish line. The optimal path is to run down the beech to some point (x) and then swim to the finish line. The Time required is given by
T=(4-x)/6+(x^2+1)^(1/2)/2=((4-x)+3(x^2+1)^(1/2))/6
Find dt/dx=0 (minimum)
0=(-1+3*(1/2)*2x*(x^2+1)^(-1/2))/6
so
(x^2+1)^(1/2)=3x
x^2+1=9x^2
x^2=1/8
x=1/(2*2^(1/2))=2^(1/2)/4=sqrt(2)/4
So to minimize time run down the beach (16-sqrt(2))/4 km and then swim 3*sqrt(2)/4 km to the finish line

Just as my dog would do.


Damn clever dawg!




Mark, great link, a very interesting story!



8 abreast

25%



Hi Jovay;

"I'm poor and was raised that way (financially speaking), but one wouldn't know it if they looked at my taste in,,, most thing's."

And yourself?


I'm poor and was raised that way (financially speaking), but now I'm stinking rich; but one wouldn't know it if they looked at my taste in,,, most thing's. :wink:




CPRE


TEABRSEARS

PRINCE

BRAINTEASERS

Try,,, i have (thanks btw) sensitive hearing. About 5 years ago i was at an AA meeting,

(after my hearing, taste and eye-sight had improved,,, getting sober heightens Thee Senses ~yhis~ me > )

and i could hyper-hear a man breathing through his nose from across a room filled with people,,, i was ultra sensitive and straight,,, it was as if that was all i could hear,,, not the worst thing that has happened to me.

(silly story^)

Lzzie:

CPRE = PRINCE

TEABRSEARS = BRAINTEASERS


Looks like you are feeling better!



Jovay wrote, "…About 5 years ago i was at an AA meeting…"

Hey! When I joined the Auto Association all they wanted was the cash!





Stormy's school has 200 students. These are special students: some of them ALWAYS tell the truth, and the rest of them, alas, NEVER tell the truth.

Among the subject areas of math, science, and social studies, each student has one favorite.

A survey was conducted; each student was asked three yes or no questions:

"Do you like math the most?" "Do you like science the most?" "Do you like social studies the most?"

The results were as follows:

104 students said "yes," they like math the most.
86 students said "yes," they like science the most.
60 students said "yes," they like social studies the most.

How many students tell the truth, and how many do not

[size=7]Students that tell the truth will answer that there favorite subject is math, science, or social studies only---students that lie will say that there favorite subject is any thing but their favorite subject and answer the other two. So no student will answer all three since that will be both a lie and the truth, and the students either lie or tell the truth not both. So if I were to draw three circles (sets) that represent the answers then (I'll call then A, B, & C) Then

A+B+C-A∩B-A∩C- B∩C+ A∩B∩C=200
A+B+C=250
So
A∩B+A∩C+ B∩C-A∩B∩C=50
But
A∩B∩C=0
Since these students lie and tell the truth
So
A∩B+A∩C+B∩C=50

Results 150 of the students tell the truth and 50 lie.[/size]

Rap

"all they wanted was the cash!"

What you said! ^(that makes me laugh) :wink:

I was told,,, that the basis for all bad behavior is fear,,, and, if one has a problem, then they should face their fears!

Face
everything
and
recover



I'm afraid to Gamble rotflmao!

STUDENTS
[size=7]liars will answer yes twice; truthers will answer yes once

2L + T = 250 (number of responses)
L + T = 200 (number of students)
L = 50 (liars)
T = 150 (truthers)[/size]

Try,,, my daughter's dad and i were watching a movie,,, neither one of us were paying much attention to what was being said, a mentally challenged character uttered ~

"he's in the lap of Jesus"

I looked at Steve and said "what did he say?"

Steve quickly replied

"he's eating bread and cheezit's"

i was eating bread and Steve was eating cheezits

I said "no he didn't", and then the truth sunk in "he's in the lap of Jesus"

That popped into my head Amen.