# A key to the National Arithmetic, exhibiting the operation of the more difficult questions in that work: for the use of teachers only

Robert S. Davis & Co., 1860 - Arithmetic - 252 pages

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### Contents

 Contractions in Division 12 Practical Questions by Analysis 18 Addition of Compound Numbers 38 Difference of Latitude 45 Factoring 51 Common Fractions 57 Division 66
 4 68 Proposed Numerator or Denomi 75 Subtraction of 83 Decimal Fractions 90 1 92 11 124

### Popular passages

Page 246 - DA, we have the semi-diameter of the larger circle, 5.7735027 -(- 5 = 10.7735027. By multiplying this last number by 2, we have the diameter of the larger circle, 10.7735027 X 2 = 21.5470054. As the area of a circle may be found by multiplying the square of the diameter by .785398, therefore, by dividing the area by .785398, the quotient will be the square of the diameter. The area of the lady's field is 500 acres = 80000 square rods ; 80000 -=-.785398 = 101859.28 square of the diameter ; V 101859.28...
Page 248 - Find the altitude of an equilateral triangle each side of which is 12 inches.
Page 206 - ... 112 small squares. But these exceed the area of the 6 pieces of which the box is made, by the strips and small squares in the edges of the top and bottom at the sides, and of the top, bottom, and sides, at the ends. Referring to Fig. 3, and noticing that the strips marked off by dotted lines are part of the pieces in which they are found, and not to be deducted, we see that we must deduct, for each side, 8 strips and 16 small squares ; and for an end, 6 strips and 8 small squares ; or 28 strips...
Page 2 - Court of the District of Massachusetts. Entered according to Act of Congress, in the year 1868, by CATHERINE WREATON, in the Clerk's Office of the Distzicb Court of the District of Massachusetts.
Page 179 - If now 18 be divided into two parts, one of which shall be twice as great as the other, we have 12 and 6 ; hence, if one note of...
Page 245 - Since regular solid bodies are to each other as cubes of their homologous sides, I say, As the quantity of water it requires to immerse the sphere in the given cone is to the cube of the diameter of the sphere, so is any other quantity of water in the conical glass to the cube of the diameter of a sphere that may be immersed in it. Now, the quantity of water given to immerse the required sphere is -J of the content of the conical glass, being £ a^jm =7.85398 &c.
Page 244 - The distance between the rings being 5 feet, it follows that the diameter of the inner ring will be 10 feet, and the diameter of the outer ring 20 feet.