Riddle: What is the best way to pass a Geometry test?
818443
I think it is to know all the angles!!!! funny joke
417055
OK:\na good way to study geometry is as follows.\nin class, take meticulous notes. try using Cornell notes technique. works great,. google Cornell notes, for that.\nsecond. always study geometry right away, either after class, or before other subjects. its a hard subject therefore it should be done first.\nHome work problems should be done, while reviewing notes. if you get stuck, review notes or textbook. always try to memorize the important postulates and theorems. but don't just read them, try to understand them.\nexample, the area of a Rhombus is 1/2times the product of the diagonals.. if you don't understand what the names are, you won't understand the problem. diagonals means, segment from corner to corner. product is what you get when you multiply.. etc.. vocabulary is the most important part of geometry, specially the first part of geometry. usually the second semester is more algebraic in nature. \nthe first semester involves proofs, so language is extremely important.\nand after you done homework, always review, review. daily... give yourself enough time everyday for math.... do this on a regular basis and you can an A.. very possible.\nGood luck.
534530
If its all true or false you can just guess because you have a 50/50 chance of getting each question correct.  If its multiple choice you can just randomly choose and your bound to get a few right.\nIf its short answer you can just answer to the best of your ability.\nIf its a math test you can just make something up but the chances of getting anything right on that are pretty slim.\nOr you could cheat off someone else, but if they dont know what they're doing either then you are out of luck on that one too.\nYou could pretend you're sick on test day so you could study.\nHow about next time you just study ahead of time and be prepared.
626899
solve as many questions as you can
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The best way to study math and is to make sure you understand all the fundamentals. Relate them to things you already understand. \nDo lots of different types of problems to figure out what to don't understand.
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a: width\nb: length\nperimeter = 2 x (a+b)= 420 ---> a+b = 210\nwe need to find max (a x b)\n\na= 210-b ---> area = a x b = b x (210-b) = 210 - b2 = - b2 + 210b = - b2 + 2 x 105 x b - 105squared + 105squared\n= - (b2 - 2 x 105 x b + 105squared) + 105 squared\n= - (b - 105)squared + 11025\n\nSince (b - 105) squared is always >= 0,\n - (b - 105)squared is always <= 0\n  - (b - 105)squared + 11025 is always  <= 11025\n\nthe largest area will be 11025 squared feet. This happens when - (b - 105)squared = 0, which means b = 105.\n\na = 210 - b = 105\n\n\nthe shortest way is to say that a rectangle has the largest area when it's a square, so each side would be 420/4 = 105.
563121
a right angled isosceles triangle\n\nplease give me 10 points now
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Hint, try to work out the radius!  Do your homework yourself.
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take a straw (cylinder) and ram it through a potato.  If you do it right, the straw will go right through the potato!  The cylinder is the strongest and most stable of the shapes and you can see why soups, veggies, fruits, etc. all come in cans!\n\nRight triangels, or right angles are all around you.  The floor and the wall of you room for example.  The bottom of a pool is a good example of an obtuse triangle, suspension bridges - trapezoid, if you look at the tip of a pencil - it's a cone but the shape of a cone when you look at it head on is a scalene triangle
586493
Quadrilateral..........D is the correct answer.....
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Learn the prefixes and suffixes -- this will help you weed out incorrect answers on the test as well as choose correct answers.\n\nAlso, I always found it useful to post short lists of words I needed to memorize (both for the GRE and when studying a foreign language) over the sinks in the kitchen and bathroom -- that way I can review when I brush my teeth and do the dishes.  \n\nFlash cards also work wonders.  I like to color-code mine.  You could do yours by parts-of-speech or into groups of words with similar meanings.   That way if you come across a word and don't remember the definition, you might remember the color of its flash card and be able to guess.\n\nGood luck!
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Take the average between the side of an inscribed square\nand that of an inscribed hexagon. With a hexagon you\nget an isosceles triangle with base as a side and a 60\ndegree angle at the apex, leaving 120 degrees to be split\nbetween the equal pair which are then also each 60 degrees.\nSo you have an equilateral triangle whose base is equal\nto the radius of the circle and of course the base is a side\nwhich has a length of 1. \n  Now inscribe a square. There are now 4 isosceles\ntriangles with a 90 degree angle at the apex and a pair\nof 45's at the base. Draw an altitude from the apex, \nperpendicular to the base, thus bisecting the 90 at\nthe apex making another 90, 45, 45 triangle whose\nside is 1/2 the side of that of the square. The hypotenuse\nis a radius of the circle so the side of the last mentioned\ntriangle is 1/2^2 (.707). This makes the side of the\nsquare 1.414. So averaging 1 and 1.414 you get 1.207.\n Using trig we find the real answer to be 1.1755.\nSo the percentage of error is 2.67 percent from 1.207.\nI wouldn't win an engineering prize with that kind of\naccuracy, but I think I could build a pretty good\ncatapult.
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When you are trying to find out various measurements of a right triangle (angle measurements or lengths of sides). Try this mnemonic (memory device):\n\nSaddle Our Horses\nCanter Away Happily\nTo Other Adventures
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just practice by doing more math problems\n\nThe more you practice, the more you'll get the hang of it.
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We can do this 2 different ways.\n\nFirst: Use the Pythagorean Theorem. Since all angles are right angles, the diagonal (measure from corner to corner) is the hypotenuse of the right triangle formed.\n\nBy the Pythagorean Theorem, 10^2 + 10^2 =x^2, where x=diagonal\n\nx^2=200\nx=sqrt(200)\nx=10 sqrt 2\n\n2nd way: use 45-45-90 Triangles\n\nWhen you draw any diagonal in a square, you get a 45-45-90 degree right triangle. This is because the 2 sides are the same, so the 2 base angles must be also. The top is 90, so the others must be 45 and 45 to equal each other.\n\nThe sides to a 45-45-90 right triangle are always in the ratio 1:1: sqrt(2) where sqrt(2) is the hypotenuse.\n\nYou know that the sides are 10 feet. Therefore the hypotenuse is 10 sqrt2 feet
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Navigation with trig uses the basic principals of the patagorean theorem (a squared + b squared = c squared), sin (pronounced sine), cosin (pronounced co-sine), and tan (tangent).  If you remember the helpful hint SOH CAH TOA (pronounced so-cah-toe-uh) it will help you to remember sin = opposite over hypotenuse, cosin = adjacent over hypotenuse, and tan = opposite over adjacent.  These refer to the sides of a traingle.  Draw a right trinagle (that is a traingle with a right angle in it).  To find any one of the angles, you use what I described above (if you know the length of the sides).  You can mix the equations up and solve for any part of it that is unknown if you know the other two.  Now, back to your question...  A navigator will usually have a chart (map) with a scale on it (measurement on the paper).  With this he / she can tell the distance between any two points, thus giving the length of the side of a traingle.  If they do this from a couple of different points, it is called "triangulation".  Now they have the "triangle", they use the stuff above to figure out the angles, distances or whatever.  Finally, they use a formula of distance = rate x time (D=RT) and variations of it (T=D/R, and R=D/T).  Let's say the navigator want to know how long it will take to get from point A to point B.  They use what I've described above to measure the distance (D), and they know how fast they are traveling (with a speedometer or airspeed indicator, etc).  Now they can use T=D/R to figure out how long it will take to get there.  In "old" times, the accuracy of charts and maps was not quite up to par, so they used the "stars" as thier maps.  these were "known" quantities, with specific angles to the horiozon (which remained fixed).  The device they used was called a "sextant".  This gave them a reading of the angle.  They would also used an hourglass for time, and a specially-made board with a rope with knots in it.  As they let out the board and it hit the water, they would "time" how many knots passed through thier fingers.  This gave them "rate" (or speed).  Using what I described above, they could figure out the rest.  Hope this helped.
1039521
Dear Lord,\n\nPlease help me circle the wagons!  There is no parallel for my situation.  I'm surrounded by squares who will ace this test and I forgot to study a segment.  Be my ray of light, beaming through a kite of confusion.  Send down your angles to assist me.  I need points on this test.  Please, send me a sine.