A Course of Mathematics: In Two Volumes. Composed for the Use of the Royal Military Academy, Volume 1Longman, Orme & Company, 1841 - Mathematics |
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Page 136
... hence M = ( x3 x2 + x − 1 ) ( x3 + x2 + x + 1 ) --- x2 + 1 = x4 - 1 . A x2 Ex . 3. Required the least common multiple of a2 + ab , a1 + a2b2 and a2 — b2 . Here V = a , and hence XX , ( a2 + ab ) ( a1 + a2b2 ) M = = V a = a2 ( a + b ) ...
... hence M = ( x3 x2 + x − 1 ) ( x3 + x2 + x + 1 ) --- x2 + 1 = x4 - 1 . A x2 Ex . 3. Required the least common multiple of a2 + ab , a1 + a2b2 and a2 — b2 . Here V = a , and hence XX , ( a2 + ab ) ( a1 + a2b2 ) M = = V a = a2 ( a + b ) ...
Page 158
... hence , u + v = a , and ± 2√uv = ± √b . In resolving these equations , we shall have successively , u2 + 2uv + v2 = a2 4uv = u2 2uv + v2 = -- u + v = a and u - v = ± √ a2 u = + √a2 2 b 22 } or subtracting - - b b hence , Ja + √a2 ...
... hence , u + v = a , and ± 2√uv = ± √b . In resolving these equations , we shall have successively , u2 + 2uv + v2 = a2 4uv = u2 2uv + v2 = -- u + v = a and u - v = ± √ a2 u = + √a2 2 b 22 } or subtracting - - b b hence , Ja + √a2 ...
Page 160
... hence , by formula ( 5 ) , we have d 1+ 41 2 X 21 441 . 2. A decreasing arithmetical series has its first term 199 , its difference - and its number of terms 67 : what is its sum ? 3 , Here z 199 , d : 66 × 31 , and = 3 , and n = 67 ; hence ...
... hence , by formula ( 5 ) , we have d 1+ 41 2 X 21 441 . 2. A decreasing arithmetical series has its first term 199 , its difference - and its number of terms 67 : what is its sum ? 3 , Here z 199 , d : 66 × 31 , and = 3 , and n = 67 ; hence ...
Page 177
... hence 5 and 2 are the multipliers . Hence we get 20x + 30y = 10 , and 20x6y = 118 . = - - Subtracting , we have 36y · 108 , or y —— -3 ; and hence x = 5 . Or again , multiply ( 2 ) by 2 , and ( 1 ) by 1 , then 4x + 6y = 2 , and — 6y + ...
... hence 5 and 2 are the multipliers . Hence we get 20x + 30y = 10 , and 20x6y = 118 . = - - Subtracting , we have 36y · 108 , or y —— -3 ; and hence x = 5 . Or again , multiply ( 2 ) by 2 , and ( 1 ) by 1 , then 4x + 6y = 2 , and — 6y + ...
Page 178
... hence the equations are prepared for subtraction . Let ( 1 ) be taken from ( 2 ) , and ( 2 ) from ( 3 ) , then y + 2z = 7 , and y + z = 5 . The coefficients of y are here equal , hence subtracting , z = 2 ; and hence ≈ and y may be ...
... hence the equations are prepared for subtraction . Let ( 1 ) be taken from ( 2 ) , and ( 2 ) from ( 3 ) , then y + 2z = 7 , and y + z = 5 . The coefficients of y are here equal , hence subtracting , z = 2 ; and hence ≈ and y may be ...
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ABCD algebraic altitude arithmetical arithmetical progression base bisect breadth centre chord circle circumference coefficients common cone cosec cube root decimal denominator denoted diagonal diameter difference dihedral angle distance divided divisor draw drawn equal equation equiangular EXAMPLES expression figure fraction frustum geometrical given line greater hence inscribed integer intersection join length less lineation logarithms mantissa measure meeting method multiplied parallel parallel ruler parallelogram perpendicular plane polygon prism PROBLEM proportional quantity quotient radii radius ratio rectangle Reduce right angles rule Scholium segment sides sine solid angle solution square root straight line subtraction tangent THEOREM third trapezium triangle ABC u₁ vulgar fraction Whence