Trapezium :Area(A) = 1/2 h(a + b)where a and b are the length of the parallel sidesh= distance between two parallel lines .Perimeter(P) = a + b + c + dwhere a,b,c and d are the length of the sides.
It is possible.
To prove a ring is commutative, one must show that for any two elements of the ring their product does not depend on the order in which you multiply them. For example, if p and q are any two elements of your ring then p*q must equal q*p in order for the ring to be commutative. Note that not every ring is commutative, in some rings p*q does not equal q*p for arbitrary q and p (for example, the ring of 2x2 matrices).
Possibly the letters with no lines of symmetry.
Only in certain circumstances:The probability of success, p, in each trial must be close to 0.Then, for the random variable, X = number of successes in n trials, the mean is npand the variance is np(1-p). But since p is close to 0, (1-p) is close to 1 and so np(1-p) is close to np.That is, the mean of the distribution is close to its variance. This is a characteristic of the Poisson distribution.Furthermore, the other characteristics of the distribution: constant probability, independence are met so the Binomial can be approximated by the Poisson.It is possible to prove this analytically but the limitations of this browser - especially in terms of mathematical notation - preclude that.Only in certain circumstances:The probability of success, p, in each trial must be close to 0.Then, for the random variable, X = number of successes in n trials, the mean is npand the variance is np(1-p). But since p is close to 0, (1-p) is close to 1 and so np(1-p) is close to np.That is, the mean of the distribution is close to its variance. This is a characteristic of the Poisson distribution.Furthermore, the other characteristics of the distribution: constant probability, independence are met so the Binomial can be approximated by the Poisson.It is possible to prove this analytically but the limitations of this browser - especially in terms of mathematical notation - preclude that.Only in certain circumstances:The probability of success, p, in each trial must be close to 0.Then, for the random variable, X = number of successes in n trials, the mean is npand the variance is np(1-p). But since p is close to 0, (1-p) is close to 1 and so np(1-p) is close to np.That is, the mean of the distribution is close to its variance. This is a characteristic of the Poisson distribution.Furthermore, the other characteristics of the distribution: constant probability, independence are met so the Binomial can be approximated by the Poisson.It is possible to prove this analytically but the limitations of this browser - especially in terms of mathematical notation - preclude that.Only in certain circumstances:The probability of success, p, in each trial must be close to 0.Then, for the random variable, X = number of successes in n trials, the mean is npand the variance is np(1-p). But since p is close to 0, (1-p) is close to 1 and so np(1-p) is close to np.That is, the mean of the distribution is close to its variance. This is a characteristic of the Poisson distribution.Furthermore, the other characteristics of the distribution: constant probability, independence are met so the Binomial can be approximated by the Poisson.It is possible to prove this analytically but the limitations of this browser - especially in terms of mathematical notation - preclude that.
P
p = a. In a regular polygon with an even number of sides, every side is parallel to the one opposite it. So all a sides are parallel.
Lines are parallel if they are perpendicular to the same line. Since the lines m and l are parallel (given), and the line l is perpendicular to the line p (given), then the lines m and p are perpendicular (the conclusion).
P = Parallel lines S = Sides 2/S x 4S = 2P P/2 = Parallel lines |*|*|*|*|*|*|*|*|*| Works Everytime cos' I am the inventor of maths
after a TON of research we came p with alternate exterior angles.
Assume there are no lines through a given point that is parallel to a given line or assume that there are many lines through a given point that are parallel to a given line. There exist a line l and a point P not on l such that either there is no line m parallel to l through P or there are two distinct lines m and n parallel to l through P.
Two parallel lines.
Both!! gezz that's easy :P
There are letters in the alphabet with both parallel and perpendicular lines. In alphabetical order, they are E, F, and H. If the joining point can be considered perpendicular and parallel, then B, D, P, and R also match the criterion.
A quadrilateral with one pair of parallel lines would be a trapezoid. : P
lines m and n are parallel if x= 12 and y= 54
Parallel means lines that are avoiding each other (in other words, they don't touch.) So the problem would have to be the way the lines are going. If the lines were intersecting (or crossing) then that would be the problem.For example: You have a problem on a paper that asks you this same question:What problem would there be in your picture if the lines were not parallel?Answer: Then your lines would be an intersecting pair.Slight definition: If your lines are not parallel, then they would have to be intersecting. Intersecting is when two lines (line segments, angles, etc.) cross on each other. For example... they might make an X by forming together in this pattern. Parallel is when your lines are NOT touching. If they are, they would be considered our P/Intersecting. Standing for "Parallel and Intersecting lines."SRW!!