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The content is presented as a series of questions pertaining to the subject of the particular chapter of the books. Amid the questions, pictures and photographs, there are details from established comic strips and complete comic strips, occasionally with its dialogue adjusted to the chapter's theme.
Malfatti's assumption that the two problems are equivalent is incorrect. Lob and Richmond (), who went back to the original Italian text, observed that for some triangles a larger area can be achieved by a greedy algorithm that inscribes a single circle of maximal radius within the triangle, inscribes a second circle within one of the three remaining corners of the triangle, the one with the ...
Langley's Adventitious Angles Solution to Langley's 80-80-20 triangle problem. Langley's Adventitious Angles is a puzzle in which one must infer an angle in a geometric diagram from other given angles. It was posed by Edward Mann Langley in The Mathematical Gazette in 1922. [1] [2]
If D > 1, no such triangle exists because the side b does not reach line BC. For the same reason a solution does not exist if the angle β ≥ 90° and b ≤ c. If D = 1, a unique solution exists: γ = 90°, i.e., the triangle is right-angled. If D < 1 two alternatives are possible. If b ≥ c, then β ≥ γ (the larger side corresponds to a ...
Francis answers the call of a female in heat. When he inquires about her breed, she states that her breed has no name, but is both old and new. That night, Francis experiences another dream, this time involving a white cat who calls himself Felidae. Joker goes missing and Francis finds him dead amongside some porcelain statues.
A parabolic segment is the region bounded by a parabola and line. To find the area of a parabolic segment, Archimedes considers a certain inscribed triangle. The base of this triangle is the given chord of the parabola, and the third vertex is the point on the parabola such that the tangent to the parabola at that point is parallel to the chord.
As a special case, a complete CAT(0) space is also known as a Hadamard space; this is by analogy with the situation for Hadamard manifolds. A Hadamard space is contractible (it has the homotopy type of a single point) and, between any two points of a Hadamard space, there is a unique geodesic segment connecting them (in fact, both properties ...
The apparent triangles formed from the figures are 13 units wide and 5 units tall, so it appears that the area should be S = 13×5 / 2 = 32.5 units. However, the blue triangle has a ratio of 5:2 (=2.5), while the red triangle has the ratio 8:3 (≈2.667), so the apparent combined hypotenuse in each figure is actually bent. With the bent ...