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9 inches abscissa acres added angle area corresponding area segment answering axis base body circle circular circumference cone conjugate convex surface cube cubic feet cubic inches cylinder denotes diameter difference dimensions distance divided double ellipse equal feet 6 inches find the area find the solidity follows foot formed frustum gallons Given gives greater half the sum height Hence hypothenuse length lesser mean breadth mean proportional MENSURATION miles multiplied number of balls Once ordinate ounces parallel perches perpendicular pile placed portion Problem IX Problem XV Problem XVIII quantity radius rood rule second remainder Section shutters side signifies similar triangles slant smaller square square root square yards surface Table tabular versed sine third remainder three sides timber transverse triangle VIII yards zone Х Х
Seite 81 - Parallelograms upon the same base, and between the same parallels, are equal to one another.
Seite 27 - RULE. — Proceed, as in the last rule, for this, or any other solid, formed by the revolution of a conic section about an axis, namely : Add together the squares of the greatest and least diameters, and the square of double the diameter in the middle between the two; multiply the sum by the length, and the product by '1309, and it will give the solidity.
Seite 24 - Ana. 2094-4. PROBLEM XIX. To find the solidity of the segment of a spheroid, the base of the segment being parallel to the revolving axis of the spheroid. CASE I. RULE. From three times the fixed axis, deduct twice the height of the segment, multiply the remainder by the square of the height, and that product by -5236. Then...
Seite 82 - ... principle that similar figures are to each other as the squares of their homologous sides.
Seite 11 - RULE.* As the conjugate diameter is to the transverse, So is the square root of the difference of the squares of the ordinate and semi-conjugate, To the distance between the ordinate and centre.