Mathematics: Compiled from the Best Authors, and Intended to be the Text-book of the Course of Private Lectures on These Sciences in the University at Cambridge, Volume 1W. Hilliard, 1808 - Mathematics |
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Page 358
... tangent B C. † n * For a tangent is perpendicular to a radius , drawn to the point of contact . + For if a right line be drawn from n to O , the angle An O , being an angle in a semicircle , is a right angle . Therefore A is a tangent ...
... tangent B C. † n * For a tangent is perpendicular to a radius , drawn to the point of contact . + For if a right line be drawn from n to O , the angle An O , being an angle in a semicircle , is a right angle . Therefore A is a tangent ...
Page 369
... tangent of the inscribed circle at the point of contact , must , if continued far enough , pass through the cen- tre . The same may be said of O n . Therefore their intersec- tion at O is the centre . PROBLEM XXXI . In any given ...
... tangent of the inscribed circle at the point of contact , must , if continued far enough , pass through the cen- tre . The same may be said of O n . Therefore their intersec- tion at O is the centre . PROBLEM XXXI . In any given ...
Page 371
... tangents , each parallel to the other diameter , and they will form the outer square m n o p . m P D NOTE . If any quadrant , as A C , be bisected in q , it will give one eighth of the circumference , or the side of the octa- gon ...
... tangents , each parallel to the other diameter , and they will form the outer square m n o p . m P D NOTE . If any quadrant , as A C , be bisected in q , it will give one eighth of the circumference , or the side of the octa- gon ...
Page 372
... tangents to the circle be drawn through all the angular points of any inscribed figure , they will form the sides of a like circumscribing figure . PROBLEM XXXVI . In a given circle to inscribe a pentagon , or a decagon .. Draw the two ...
... tangents to the circle be drawn through all the angular points of any inscribed figure , they will form the sides of a like circumscribing figure . PROBLEM XXXVI . In a given circle to inscribe a pentagon , or a decagon .. Draw the two ...
Page 373
... Tangents , being drawn through the angular points , will form the circumscribing pentagon or decagon . PROBLEM XXXVII . To divide the circumference of a given circle into twelve equal parts , each being 30 degrees . Or to inscribe a ...
... Tangents , being drawn through the angular points , will form the circumscribing pentagon or decagon . PROBLEM XXXVII . To divide the circumference of a given circle into twelve equal parts , each being 30 degrees . Or to inscribe a ...
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Common terms and phrases
affirmative amount of 11 angle annuity annum arithmetical Bisect carats cent centre chord circle circumference coefficient common denominator completing the square compound interest compound quantity consequently cube root debt decimal denoted diameter difference Divide dividend division divisor draw equal equation EXAMPLES exponent figure fourth gallons geometrical progression geometrical series give given number greater greatest common measure half improper fraction infinite series less number logarithm manner Multiply negative NOTE nth root number of combinations number of terms number of things payment perpendicular polygon present worth PROBLEM proportion quadratic equation quotient radius ratio Reduce remainder repetend required to find right line RULE sides simple interest sine square root subtract Suppose surd taken tangent third unknown quantity vulgar fraction Whence whole number yards
Popular passages
Page 175 - RULE.* — Multiply each payment by the time at which it is due; then divide the sum of the products by the sum of the payments, and the quotient will be the true time required.
Page 140 - Multiply the divisor, thus augmented, by the last figure of the root, and subtract the product from the dividend, and to the remainder bring down the next period for a new dividend.
Page 255 - Divide the first term of the dividend by the first term of the divisor, and write the result as the first term of the quotient. Multiply the whole divisor by the first term of the quotient, and subtract the product from the dividend.
Page 198 - A man was hired 50 days on these conditions. — that, for every day he worked, he should receive $ '75, and, for every day he was idle, he should forfeit $ '25 ; at the expiration of the time, he received $ 27'50 ; how many days did he work...
Page 149 - To the remainder bring down the first figure in the next period, and call it the dividend. 4. Involve the root to the next inferior power to that which is given, and multiply it by the number denoting the given power, for a divisor.
Page 315 - If A and B together can perform a piece of work in 8 days, A and C together in 9 days, and B and C in 10 days : how many days would it take each person to perform the same work alone ? Ans.
Page 124 - As the sum of the several products, Is to the whole gain or loss ; So is each man's particular product, To his particular share of the gain or loss.
Page 139 - ... 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 120 - When it is required to find how many of the first sort of coin, weight or measure, mentioned in the question, are equal to a given quantity of the last.
Page 132 - When one of the ingredients is limited to a certain quantity. RULE. Take the difference between each price and the mean rate, as before ; then,