Mathematical Challenge

  • Thread starter Thread starter Shai Gar
  • Start date Start date
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Shai Gar

Okay, the external lines form up to become a perfect Pentagon.
You are to find the dimensions asked for.

A is given as 200kilometers.

zombie_city.png
 
42 for all of them.
 
ugh is this uni stuff? I figure it has to go along with breaking it down with pythagorean theorem but probably not now that I look at it more.

I think J equals 6000km2 the 2 is squared

e=108
 
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oooo purty pentagram
 
Ah the golden ratio. Simple problem really, but I don't have a scientific calculator on me and no way I'm spending time on this cuz I hate math.
 
Oh I know how to do this, actually very easy (I like geometry!). Lots of pthagorean therom and sin cos tan stuff! ...but my quackulator died so I will do it once I can get new batteries. Not to mention it is like 6am, and I am running off 2 hours of sleep.

Me: Ok time to do some trigonometry!
Brain: Trig? Oh no you don't! You did not give me enough sleep! Now where is that "dumb blonde" button... ah! *push* :bump2:
Me: ok so its sin(108/3......) ...errrrrrrpplllhhhhh... OH LOOK! SHINY OBJECT!!! YAY!!!!1! *runs across room* :bounce:
 
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WARNING!! SPOILER BELOW!!
A=200
B=323.6
C=123.0
D=76.0
E=29.0
F=161.7

(these are in degrees)
G=36
H=108
I=72

J=68819.1

Like I said, I'm a math geek.
 
Here's one: An army jeep left its base at 12 noon for a destination 300 miles distant across a flat terrain. The jeeps average rate the first hour was 40 mph, during the second hour 45 mph, during the third 50 mph, ect. At 1:40 pm, a helicopter, traveling at a uniform rate of 75 mph, set out to over take the jeep. At what time did the helicopter reach the jeep?
 
VindurNótt said:
Here's one: An army jeep left its base at 12 noon for a destination 300 miles distant across a flat terrain. The jeeps average rate the first hour was 40 mph, during the second hour 45 mph, during the third 50 mph, ect. At 1:40 pm, a helicopter, traveling at a uniform rate of 75 mph, set out to over take the jeep. At what time did the helicopter reach the jeep?
Click to expand...
Convert the shit to real units and I might try to solve it

*mumbles about stupid people using stupid units*
:m080:
 
Seeker said:
Convert the shit to real units and I might try to solve it

*mumbles about stupid people using stupid units*
:m080:
Click to expand...
units don't matter, :mno: you'd get the same answer if you convert all the distances to km, and all speeds to kmph.
 
VindurNótt said:
Here's one: An army jeep left its base at 12 noon for a destination 300 miles distant across a flat terrain. The jeeps average rate the first hour was 40 mph, during the second hour 45 mph, during the third 50 mph, ect. At 1:40 pm, a helicopter, traveling at a uniform rate of 75 mph, set out to over take the jeep. At what time did the helicopter reach the jeep?
Click to expand...

250 miles out at 5 PM I would guess, given constant directions.
 
VindurNótt said:
Here's one: An army jeep left its base at 12 noon for a destination 300 miles distant across a flat terrain. The jeeps average rate the first hour was 40 mph, during the second hour 45 mph, during the third 50 mph, ect. At 1:40 pm, a helicopter, traveling at a uniform rate of 75 mph, set out to over take the jeep. At what time did the helicopter reach the jeep?
Click to expand...

250 miles out at 5 PM I would guess, given constant directions.

edit to add given no time change zones were crossed.
 
Of course, if the jeep stopped to take a bathroom break.........:hungry:
or to eat......or both....
or....
 
VindurNótt said:
Here's one: An army jeep left its base at 12 noon for a destination 300 miles distant across a flat terrain. The jeeps average rate the first hour was 40 mph, during the second hour 45 mph, during the third 50 mph, ect. At 1:40 pm, a helicopter, traveling at a uniform rate of 75 mph, set out to over take the jeep. At what time did the helicopter reach the jeep?
Click to expand...

The helicopter will reach the jeep at 5:00pm.

The helicopter will reach the base at 5:40pm, the jeep will reach the base at ~5:46:09.23pm.
 
Eye level will be at 60 inches above sea level at each end.
Location: on the ocean where nothing is between both sets of eyes
Conditions: calm; no wind
What is the greatest distance, in linear feet, one set of eyes can look toward the horizon
and see the other set of eyes?
 
just me said:
Eye level will be at 60 inches above sea level at each end.
Location: on the ocean where nothing is between both sets of eyes
Conditions: calm; no wind
What is the greatest distance, in linear feet, one set of eyes can look toward the horizon
and see the other set of eyes?
Click to expand...


I hate math so I am not going to solve it with numbers, but this is how (I assume) you solve it.

Figure out the curvature of the earth, and take a tangential point to the surface, and see how long out the line goes before it is 5 feet from the surface. Using sin, cos, and tan and all that good stuff. The radius of the earth could be related to the unit circle and scaled acordingly.
 
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