A Course of Mathematics: Composed for the Use of the Royal Military Academy |
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Page 99
... hence , 2 + 0 2 + 0 - 4 + 0 - - 3+ 5 5 6 +10 - 100 + 15 — 25 4 + 0-16 + 10 + 15 — 25 - Hence 4a5 16a3 b2 + 10 a2 b3 + 15ab1 — 2565 = product . - The coefficient of a being zero in the product , causes that term to dis- appear . - · 3x2 ...
... hence , 2 + 0 2 + 0 - 4 + 0 - - 3+ 5 5 6 +10 - 100 + 15 — 25 4 + 0-16 + 10 + 15 — 25 - Hence 4a5 16a3 b2 + 10 a2 b3 + 15ab1 — 2565 = product . - The coefficient of a being zero in the product , causes that term to dis- appear . - · 3x2 ...
Page 107
... hence , by writing 4 for n in formula ( 1 ) , we have = a3 + b . a1 - b1 a - b a3 - b3 a- -b = a3 + b ( a2 + ab + b2 ) by ( 3 ) = a3 + a2b + ab2 + b3 ... ( 4 ) . Hence , generally , a " -b " will always be exactly divisible by a − b ...
... hence , by writing 4 for n in formula ( 1 ) , we have = a3 + b . a1 - b1 a - b a3 - b3 a- -b = a3 + b ( a2 + ab + b2 ) by ( 3 ) = a3 + a2b + ab2 + b3 ... ( 4 ) . Hence , generally , a " -b " will always be exactly divisible by a − b ...
Page 112
... hence a ± b = hd ± k'd = d ( h ‡ k ) and therefore d measures both a + b and a − b , the quotient being h + k in ... hence ( Lemma 1 ) d measures nc , and it likewise measures itself ; therefore ( Lemma 2 ) d mea- sures nc + d , which ...
... hence a ± b = hd ± k'd = d ( h ‡ k ) and therefore d measures both a + b and a − b , the quotient being h + k in ... hence ( Lemma 1 ) d measures nc , and it likewise measures itself ; therefore ( Lemma 2 ) d mea- sures nc + d , which ...
Page 113
... hence d measures n d and b ; therefore it measures b - nd , or kď by ( 2 ) ; but again , d does not measure k . and hence it must measure d ' ; but d is greater than d ' , and cannot therefore measure it ; hence d ' is the greatest ...
... hence d measures n d and b ; therefore it measures b - nd , or kď by ( 2 ) ; but again , d does not measure k . and hence it must measure d ' ; but d is greater than d ' , and cannot therefore measure it ; hence d ' is the greatest ...
Page 114
... Hence x + y is the greatest common measure sought , and = 22 - y ( x3 + y3 ) ÷ ( x + y ) x2 - xy + y2 ( x2 — y2 ) + ( x + y ) = x - y = reduced fraction . ( 3. ) Required the greatest common measure of the two polynomials 6a3- 6a2y + ...
... Hence x + y is the greatest common measure sought , and = 22 - y ( x3 + y3 ) ÷ ( x + y ) x2 - xy + y2 ( x2 — y2 ) + ( x + y ) = x - y = reduced fraction . ( 3. ) Required the greatest common measure of the two polynomials 6a3- 6a2y + ...
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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...