The most famous impossible problem from Greek Antiquity is doubling the cube. The problem is to construct a cube whose volume is double that of a given one. It is often denoted to as the Delian problem due to a myth that the Delians had look up Plato on the subject. In another form, the story proclaims that the Athenians in 430 B.C. consulted the oracle at Delos in the hope to break the plague devastating their country. They were advised by Apollo to double his altar that had the form of a cube. As an effect of several failed attempts to satisfy the god, the plague only got worse and at the end they turned to Plato for advice. (According to Rouse Ball and Coxeter, p 340, an Arab variant asserts that the plague had wrecked between the children of Israel but the name of Apollo had been discreetly gone astray.) According to a message from the mathematician Eratosthenes to King Ptolemy of Egypt, Euripides mentioned the Delian problem in one of his (now lost) tragedies. The other three antiquity are: angle trisection, squaring a circle, and constructing a regular heptagon.
Answer:
15 to 40
Step-by-step explanation:
First, you would add the total number of cars and trucks together. Secondly, you would take the total, which is all the vehicles, and you would take 15, which is the number of cars, and put them together.
Answer:
30√3 ≈ 51.96 miles
Step-by-step explanation:
The distance between the two ships can be found using the Law of Cosines, or using your knowledge of the side relationships in special triangles.
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Each ship is traveling at 10 mph, so after 3 hours will have traveled 30 miles.
The triangle OS1S2 formed by the harbor and the two ship locations is an isosceles triangle with base angles of 30°. Each half of OS1S2 is a 30-60-90 triangle whose longer leg is √3 times half the hypotenuse. The sum of those two "longer legs" is the distance between the ships.
The distance between ships is 2×15√3 = 30√3 ≈ 51.96 miles.
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<em>Additional comment</em>
If you prefer to use the Law of Cosines, you are looking for the length of the side opposite the 120° angle in a triangle with sides of 30 miles.
c² = 30² +30² -2·30·30·cos(120°) = 30²(2-2·(-0.5)) = 3·30²
c = 30√3 . . . . . take the square root (miles)
Solution is where the lines intersect. It's is (-2 , 1)
Answer is B.(-2,1)
