# Table of Quarter-squares of All Integer Numbers Up to 100,000, by which the Product of Two Factors May be Found by the Aid of Addition and Subtraction Alone

C. and E. Layton, 1856 - Mathematics - 214 pages
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### Contents

 Section 1 1 Section 2 4 Section 3 11 Section 4 64 Section 5 77 Section 6 110 Section 7 118 Section 8 122
 Section 10 129 Section 11 140 Section 12 168 Section 13 185 Section 14 197 Section 15 205 Section 16 211 Section 17 214

 Section 9 128

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Page 207 - Ans. 56.64+ feet. 5. A line 81 feet long, will exactly reach from the top of a fort, on the opposite bank of a river, known to be 69 feet broad ; the height of the wall is required. Ans. 42.42 6 feet. 6. Two ships sail from the same port, one goes due east 150 miles, the other due north 252 miles; how far are they asunder ? Ans. 293.26 miles. 269. To find a mean proportional between two numbers. RULE. — Multiply...
Page 11 - Given in numbers any two sides of a right angled triangle, the third side can be found, for it is the square root of the sum or difference of the squares of the given lines, according as the given sides contain the right angle or not. Cor. 2. To find a square equal to the sum of two F<g.
Page 208 - ... area. PROBLEM III. To find the area of a triangle. RULE. Multiply one of its sides as a base by a perpendicular let fall from the opposite angle, and take half the product for the area. Or, from half the sum of the three sides subtract each side separately, and multiply the three remainders so obtained and the half sum together, and the square root of the product will be the area. EXAMPLE 1. Required the area of a triangle ABC, whose busa AB =; 16-5, and perpendicular DC = 10-25.
Page 214 - Ans. 9.05952 feet. 5. The diameter of a well is 3 feet 9 inches, and its depth 45 feet ; what did it cost sinking at 7s.
Page 210 - X 17.75 =. 2183.25; and = 1091.625 feet = 1091 fi. 7 in. 6 pa. the area required. 2. What is the area of a -court-yard in the form of a regular pentagon, whose side measures 92 feet 6 inches, and perpendicular 63 feet 8 inches ? Am.