Easy Introduction to Mathematics, Volume 1

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Barlett & Newman, 1814
 

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Page xxii - Just so it is in the mind; would you have a man reason well, you must use him to it betimes, exercise his mind in observing the connection of ideas and following them in train. Nothing does this better than mathematics, which therefore I think should be taught all those who have the time and opportunity, not so much to make them mathematicians as to make them reasonable creatures...
Page 48 - LIQUID MEASURE 4 gills (gi.) = 1 pint (pt.) 2 pints = 1 quart (qt...
Page 98 - Then multiply the second and third terms together, and divide the product by the first term: the quotient will be the fourth term, or answer.
Page 42 - AVOIRDUPOIS WEIGHT. 16 drams, dr. make 1 ounce, - - - - oz. 16 ounces - - - 1 pound, - - - - Ib. 28 pounds - - - 1 quarter, - - - qr. 4 quarters - - - 1 hundred weight, - cwt. 20 hundred weight, 1 ton, T.
Page 448 - What number is that, which, being divided by the product of its digits, the quotient is 3 ; and if 18 be added to it, the digits will be inverted ? Ans.
Page 54 - M. 60 minutes, 1 hour, h. 24 hours, 1 day, d. 7 days, . 1 week, w. 4 weeks, 1 month, mo. 13 months, 1 day and 6 hours, 1 Julian year, yr. Thirty days hath September, April, June and November ; February twenty-eight alone, all the rest have thirtyone.
Page 106 - State and reduce the terms as in the Rule of Three Direct. 2. Multiply the first and second terms together, and divide the product by the third ; the quotient will be the answer in the same denomination as the middle term was reduced into.
Page 234 - ... and to the remainder bring down the next period for a dividend. 3. Place the double of the root already found, on the left hand of the dividend for a divisor. 4. Seek how often the divisor is contained...
Page 432 - A hare is 50 leaps before a greyhound, and takes 4 leaps to- the greyhound's 3, but 2 of the greyhound's leaps are as much as 3 of the hare's ; how many leaps must the greyhound take to catch the hare ? Ans. 300.
Page 138 - To reduce a whole number to an equivalent fraction, having a given denominator. RULE. Multiply the whole number by the given denominator, and place the product over the said denominator, and it will form the fraction required.

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