need help with one geometry question?
956626
Figure out how many square inches on the cube.  One side is 10" x 10" = 100 inches square.\n\n6 sides makes 600 inches square.\n\nA square foot is 12 x 12 inches = 144 inches square\n\nSquare footage is 600 / 144 = 4.17
808242
Yes. These are intriguing geometry problems. What is YOUR question? Be specific. Post ONE question and be specific about where you are stuck. Right now, this just looks like you want someone to DO your homework for you. But you wouldn't cheat on your homework like that, would you?\n\nHere's a hint -- supplemental angles add up to 180\n\nFor example, if one angle is 60 degrees, its supplement is 120
137113
Draw the two points (a,b) and (c,d) on a graph so that the segment connecting them is NOT horizontal or vertical. Then make a right triangle with one vertical leg and one horizontal leg where the connecting segment is the hypotenuse.  \n\nThe length of the horizontal leg is c-a\nThe length of the vertical leg is d-b\n\nSo the length of the hypotenuse, which is the distance from (a,b) to (c,d), is\n\nsquare root of (c-a)^2+(d-b)^2
902666
This may possibly be the easiest question anyone has ever asked on this forum. If you can't figure this out yourself, you really need to do more homework!! BTW, that is one freakin' tall sheep!!
903397
You might try asking this question under "Mathematics."
357173
This should help.\n\nhttp://www.geomatters.com/products/category.asp?CID=1
1106432
Diameter == 10.\nRadius == 5.\nCenter == (1,7)\n\nMethod: use Pythagorean Theorem.
821495
a) Assume the following:\n\nAngles a= 10 deg, b= 80 Deg, c=90 deg \nHypotenuse length = 2.5 m\nheight = 2.5 * cos 80 Deg= .434 m\nlength = 2.5 * sin 80 Deg= 2.462 m\nacceleration = "g" * Sin 10 Degree
957591
line = 1 side\nparallel lines = 2 sides\ntriangle = 3 sides\nsquare/ rectangle = 4 sides\npentagon = 5 sides\nhexagon = 6 sides\nheptagon = 7 sides\noctagon = 8 sides\nnonagon = 9 sides\ndecagon = 10 sides\n\ni hope i have helped you.... good luck
414316
Hi,\nThe answer is C.\n\nYou could make a ray-diagram and see for yourself...
247689
Use analytical trig (hint: sin function).  That should give you the answer you are looking for.
689719
The mode of presentation of data in the question is not sufficient to draw a single quadrilateral, there are two such quadrilaterals and hence two different answers.\n\nFIRST ANSWER: \nEqual sides i.e. sides of length 5*sqrt(2) are adjacent.\nLet the quadrilateral be ABCD. Then AB = BC = 5*sqrt(2), CD = 6, <DAB and <ABC are right angles. Drop a perpendicular to side AD from C intersecting AD at P.\n\nTotal area = area of rectangle ABCP + area of triangle CPD\n= [5*sqrt(2) x 5*sqrt(2)] + [0.5*{5*sqrt(2)-6}*5*sqrt(2)]\n= 50 + 3.7868 sq units\n= 53.7868 sq units\n\nSECOND ANSWER: \nEqual sides i.e. sides of length 5*sqrt(2) are opposite to each other.\nLet the quadrilateral be ABCD. Then AB = CD = 5*sqrt(2), BC = 6, <BCD and <ABC are right angles. \nClearly, quadrilateral ABCD is a rectangle. \nSo area = product of lengths of adjacent sides = AB x BC = 5*sqrt(2) x 6\n= 42.4264 sq units
87753
Just add up the lengths of the three sides. Works for any triangle - wether it's equilateral, isosceles or obtuse.
1107584
|\n  | \n  |  A\n  |\n  |_______\n......B\n\nThis is what "A" perpendicular to "B" looks like.  I hope this helps!
323670
measure each diagonal, come up with the area of one triangle X it by 4, then work out the area of the base simple!
65888
a parabola facing up!\n\nwith center (0,1)
1169010
Find d?\n\nWhere are a, b, and c?\n\nwhy don't you write out the entire problem if you want help?