In order to identify two lines as perpendicular, there must be at least     (blank) right angels?
88745
one, the one forming by the projection of both lines.
779641
Ahh, a triangle. Since you know the two long sides of the triangle and the angle between them, use the Lore of Co-signs to get the third side (Law of Cosines).
478296
Fermat's Last Theorem has been solved ashas the Four Color Theorem. The Poincare conjecture has recently been solved. Trisecting a general angle with unmarked straightedge and compass has been shown to be unsolvable (although it is possible with marked straightedge and compass).\n\nHigh on the list of open problems:\n\nRiemann Hypothesis.\nP=NP?\nTwin prime conjecture.\nAre there any odd perfect numbers?\nProof of existence of solutions to Navier-Stokes equations.
572729
When two lines are not perpendicular, the types of angles that can be formed are:\n\n(1) Acute - one of the angles formed by the two lines is less than 90 degrees.\n(2) Obtuse - one of the angles formed by the two lines is more than 90 degrees.\n(3) Straight - the two lines are somewhat inclined in the same direction\n(4) Reflex - one of the angles formed by the two lines is either greater than 180 degrees, whilst the other is less than 180 degrees (which the exception that the two lines form a 270 degree, and a 90 degree angle). This applied to angles of a revolution.
874225
Sounds like they are looking for a *right* trapezoid.  \n\nYou can't just say trapezoid, because that doesn't require the shape to have any right angles.  A picture of a right trapezoid is given below:\n\n(Note: with their clues, it could actually be any polygon with at least 4 sides.  For example it could be a five sided polygon, with a couple right angles and one set of parallel sides.  (You know, the little house with a pointy roof we all drew when we were little)... and there are other shapes too.\n\nBut for this question, I'd go with *right* trapezoid.
715400
Angle of line from horizontal position.
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in the plane,\nthere are two kinds of curvature,\nthe regular one, which is the norm of the second derivative, if the curve has unit tangent.\nand the signed curvature, which in addition measures if the first derivative and the second derivative preserve orientation.\ns
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to draw lines u hav to select 2 points at a time (as no 3 r collinear)\n\nthus no of selections of such points is nC2= n!/ [ 2! (n-2)!]
1162204
The space between two lines.
144689
there is no answer, as the lines are parellel.
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Here's a quasi-logical answer:\nBy definition, two lines are parallel if they have the same slope (s1=s2) and they are perpendicular if the product of their slopes is -1 (s1*s2=-1)\n\nSo we have s1=s2 and s1*s2=-1\nTherefore s1^2=-1, so s1=s2=i=sqrt(-1)\nIn conclusion, any two lines y=ix+a and y=ix+b are both parallel and perpendicular at the same time.
965470
The answer is neither. If they are parallel they must have the same slope and if they are perpendicular the slopes cannot have the same sign. The slopes of the lines are both negative and unequal.\n\nfind the slopes of both lines.\n\n5x+2y =3\ny = (3-5x)/2 = -5x/2 +3/2\nslope 1 = -5/2\n\n2x+5y =8\ny= (8-2x)/5 = -2x/5 + 8/5\nslope 2 = -2/5
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For a line to be perpendicular all you need is the negative inverse slope.\ny=2x-3 has a slope of 2 so the negative inverse would be (-1/2)\n\nSo y=(-1/2)x-3 would cross perpendicular at the y axis at -3
628857
Not a, since the congruent sides could be adjacent sides.\n\nNot b, because that would be a trapezoid.\n\nDefinitely c - there's a way to prove this using properties of parallel lines, interior angles, and triangles.\n\nMaybe d, since congruent diagonals means that your angles could be right angles, but it also could mean that you have an isosceles trapezoid.\n\nBisecting diagonals always means that your quadrilateral is a parallelogram.  Congruent diagonals can also yield a parallelogram, albeit a specialized one - a rectangle.  Now if you have diagonals that both bisect each other and are congruent, then you have a square.
775751
You need something more, unless you are talking about a right triangle (since you'd then know an angle).
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To answer this question, you need to know the shape that these points make.  Are they on a line? or are they some two-dimensional shape?
43253
Depending on what you need to solve, you need two angles and a side, or two sides and an angle. Hit me with an E-Mail if you are stuck with something.\n\nkeptcompany@yahoo.com