Mathematical Challenge

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

 

oooo purty pentagram

 

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:

 

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?

 

Convert the shit to real units and I might try to solve it

*mumbles about stupid people using stupid units*

 

units don't matter, :mno: you'd get the same answer if you convert all the distances to km, and all speeds to kmph.

 

it looks like the top of a cut diamond

 

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

 

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....

 

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?

 

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.

 

Would a straight line sitting atop a circle help one to get started?