A Course of Mathematics: Composed for the Use of the Royal Military Academy |
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Page 66
... PROB . L. Given the extremes , and the number of terms ; to find the sum of all the terms . RULE . ADD the extremes together , multiply the sum by the number of terms and divide by 2 . EXAMPLES . 1. The extremes being 3 and 19 , and the ...
... PROB . L. Given the extremes , and the number of terms ; to find the sum of all the terms . RULE . ADD the extremes together , multiply the sum by the number of terms and divide by 2 . EXAMPLES . 1. The extremes being 3 and 19 , and the ...
Page 67
... PROB . II . Given the extremes , and the number of terms ; to find the common difference . RULE - Subtract the less extreme from the greater , and divide the remainder by 1 less than the number of terms , for the common difference ...
... PROB . II . Given the extremes , and the number of terms ; to find the common difference . RULE - Subtract the less extreme from the greater , and divide the remainder by 1 less than the number of terms , for the common difference ...
Page 68
... PROB . IV . To find an arithmetical mean proportional between two given terms . RULE . Add the two given extremes or terms together , and take half their sum for the arithmetical mean required . Or , subtract the less extreme from the ...
... PROB . IV . To find an arithmetical mean proportional between two given terms . RULE . Add the two given extremes or terms together , and take half their sum for the arithmetical mean required . Or , subtract the less extreme from the ...
Page 69
... on , each succeeding payment being double the last ; and what will the last payment be ? Ans . the debt 40954 , and the last payment 20484 PROB . I. To find one geometrical mean proportional between Geometrical Proportion.
... on , each succeeding payment being double the last ; and what will the last payment be ? Ans . the debt 40954 , and the last payment 20484 PROB . I. To find one geometrical mean proportional between Geometrical Proportion.
Page 70
... PROB . IL To find two geometrical mean proportionals between any two numbers . RULE . Divide the greater number by the less , and extract the cube root of the quotient , which will give the common ratio of the terms . Then multiply the ...
... PROB . IL To find two geometrical mean proportionals between any two numbers . RULE . Divide the greater number by the less , and extract the cube root of the quotient , which will give the common ratio of the terms . Then multiply the ...
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Other editions - View all
A Course of Mathematics: Composed for the Use of the Royal Military Academy Charles Hutton No preview available - 2015 |
A Course of Mathematics, Vol. 3: Composed for the Use of the Royal Military ... Charles Hutton No preview available - 2016 |
A Course of Mathematics, Vol. 3: Composed for the Use of the Royal Military ... Charles Hutton No preview available - 2018 |
Common terms and phrases
algebraic algebraic quantities axis binomial Binomial Theorem bisected centre circle circumference coefficient Completing the square contained Corol cosec cube root decimal degree denote diameter difference distance Divide dividend division divisor draw equal Example exponent expression extract factors feet figure fraction gives greater greatest common measure Hence inches least common multiple letters logarithm manner monomial multiplied negative nth root number of terms parallel parallelogram perfect square perpendicular plane polynomial positive Prob problem Prop proportional proposed equation quadratic quotient radical radius ratio rectangle reduce remainder result right angles rule second term sides sine square root straight line Substituting subtract tangent THEOREM triangle ABC unknown quantity VULGAR FRACTIONS Whence whole number yards
Popular passages
Page 332 - EC, have also the same altitude ; and because triangles of the same altitude are to each other as their bases, therefore the triangle ADE : BDE : : AD : DB, and triangle ADE : CDE : : AE : EC.
Page 352 - angle in a segment' is the angle contained by two straight lines drawn from any point in the circumference of the segment, to the extremities of the straight line which is the base of the segment.
Page 326 - Proportion, when the ratio is the same between every two adjacent terms, viz. when the first is to the second, as the second to the third, as the third to the fourth, as the fourth to the fifth, and so on, all in the same common ratio.
Page 17 - OF TIME. 60 Seconds = 1 Minute. 60 Minutes = 1 Hour. 24 Hours = 1 Day. 7 Days = 1 Week. 28 Days = 1 Lunar Month.
Page 338 - CD. conseconsequently the whole polygon, or all the triangles added together which compose it, is equal to the- rectangle of the common altitude CD, and the halves of all the sides, or the half perimeter of the polygon. Now, conceive the number of sides of the polygon to be indefinitely increased ; then will its perimeter coincide with the circumference of the circle, and consequently the altitude CD will become equal to the radius, and the whole polygon equal to the circle. Consequently the space...
Page 297 - The Height or Altitude of a figure is a perpendicular let fall from an angle, or its vertex, to the opposite side, called the base.
Page 26 - Multiply the number in the lowest denomination by the multiplier, and find how many units of the next higher denomination are contained in the product, setting down ,what remains.
Page 62 - From these theorems may readily be found any one of these five parts ; the two extremes, the number of terms, the common difference, and the sum of all the terms, when any three of them are given, as in the following Problems : PROBLEM I.
Page 326 - Proportional, when the ratio of the first to the second, is equal to the ratio of the second to the third.
Page 62 - SUBTRACT the less extreme from the greater, and divide the difference by 1 more than the number of means required to be found...