I have altitude relative to sea level so I only need 3 more. this is for my homework.
199913
Carbondioxide gas, Oxygen gas and Nitrogen gas
369
Here's a list.  There's more than one possible answer since there is more than one distance.
672819
3 helpers and Charles will start the trip so collectively they are having 4*5=20 dayz supply.\n\nAfter day one: one helper will start back with one supply for himself, leaving 15 for other three (20 - 4 -1)\nAfter day two: second helper will start back with two for himself,\nleaving 10 for other two(15- 3 -2)\nAfter day three: third helper will start back with three for himself leaving 5 for Charles (10 - 2 - 3)\nWith these 5 supplies Charles can cross the desert.\n\nThis is the minimum possible result, you can have any number more than that.
468074
let the height of cliff be h\nthen \ndistance of first ship =x\ndistance of first ship =x+500\nnow \n1.h/x=tan 24\n2.h/(x+500)=tan 15\nfrom 1\nx=hcot24\nfrom 2\nx=hcot15-500\ntherefore\nh(cot15-cot24)=500\nh=336.47
542547
It's a geometry problem.  Suppose you're high up in a building on the beach overlooking the ocean.  You're at height h above sea level.\n\nNow draw a circle which represents the earth, with radius r.  Your distance from the center of the earth is r+h.  The horizon is some distance away, x.  Your line of sight to the horizon is <tangent> to the radius of the earth, forming a right triangle, where the hypoteneuse is r+h.\n\nUsing the Pythagorean Theorem,\n\nr^2 + x^2 = (r+h)^2 = r^2 + 2rh + h^2\nx^2 = 2rh + h^2\nx = sqrt(2rh + h^2)\n\nThat's the exact answer,  But for practical purposes, 2rh is <much, much> bigger than h^2.  To see this, suppose you're 500 feet in the air.  That's about a tenth of a mile, and using miles, h^2 = 0.1^2 = 0.01.  On the other hand, the earth's radius is about 4000 miles, so 2rh = 2 x 4000 x 0.1 = 800.  This is 80,000 times bigger than h^2.\n\nSo we can safely ignore the h^2, and say x = sqrt(2rh).  That can be broken down further.  Again using miles (you can also do this in kilometers),\n\nx = sqrt(2rh) = sqrt(2r) sqrt h = sqrt(8000) sqrt h = 90 sqrt h\n\n(rounding off, since sqrt 8000 = 89.44; also, h must be in miles).\n\nUsing the formula x = 90 sqrt h, suppose you're in an airplane 4 miles up in the air (21,000 feet).  Then the horizon is 90 sqrt 4 = 180 miles away.\n\nOkay, now let's work it out for h in feet instead of miles.  We'll go back to the formula x = sqrt(2rh), and use r = 3960 miles (a more accurate figure).  Then\n\nx = sqrt(2 x 3960 x h / 5280) = sqrt(3h/2) = 1.225 sqrt h\n\nwhere h is in feet and x is in miles.\n\nTwo examples:  (1) For h = 6 feet, use x = sqrt(3h/2) = 3 miles.  (2) For x = 100 feet, use x = 1.225 sqrt h = 12.25 miles.\n\nTo recap:  For h in miles, use x = 90 sqrt h.  For h in feet, either use x = sqrt(3h/2) or x = 1.225 sqrt h.\n\nIn all cases, x will be in miles.  And of course, all of this can be done using meters and kilometers.
96162
try surfrider.com   they may have a link on that site to check out the tide levels/times.
187382
I know it has already been answered but I"ll take a chance with something different. Ok 3 to the third power plus two would be 3 to the fifth power which is 243.\n :)
196608
c) shows the elevation of the land.
657457
since P1.V1 / T1 = P2.V2 / T2 ... u don't actually care about the altitude. so V2 = 8.88 cm^3 (btw those degrees are °C, °F ?? cuz u gotta take them to Kevin before using them)\n\n1*10^5 Pa * 2cm^3 / 270 °K = 2*10^4 Pa * V2 / 240 °K\nV2 = 8.88 cm^3
268350
Still not going to answer your homework...dork...thanks again, for the 3rd time! I could really use more points...keep asking.
454331
The diver would be experiencing 3 atmoshpheres at 99ft.\n\n33 x 3= 99
78186
I'm not sure if the sea level rise is that dramatic. There are a lot of different predictions being bandied about. The estimate from the telegraph is that there will be a 20ft rise by 2600. The rise this century is expected to be about 1m.
355201
Try www.3gforfree.com
176253
go to http://earth.google.com
230134
3853 feet (1175 meters)?
353201
I own a very fine acre of the light side of the moon. It's kinda left a bit from the middle. I could send you a map.
174755
Another homework question.