A Course of Mathematics: Composed for the Use of the Royal Military Academy |
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Page 88
... Hence 5a + 3a = a + a + a + a + a + a + a + a = 8a . Similarly , -5a = ( -a ) + ( - a ) + ( —a ) + ( —a ) + ( —a ) -3a = ( -a ) + ( -a ) + ( -a ) . = Hence 5a + ( −3α ) ( -a ) + ( -a ) + ( −a ) + ( -a ) + ( − ...
... Hence 5a + 3a = a + a + a + a + a + a + a + a = 8a . Similarly , -5a = ( -a ) + ( - a ) + ( —a ) + ( —a ) + ( —a ) -3a = ( -a ) + ( -a ) + ( -a ) . = Hence 5a + ( −3α ) ( -a ) + ( -a ) + ( −a ) + ( -a ) + ( − ...
Page 93
... Hence the reason for the change of the signs in the quantity to be subtracted . Or thus : Since cd is to be subtracted from a + b ; then , if c be subtracted , we shall have subtracted too much by d ; hence the remainder a + b - c is ...
... Hence the reason for the change of the signs in the quantity to be subtracted . Or thus : Since cd is to be subtracted from a + b ; then , if c be subtracted , we shall have subtracted too much by d ; hence the remainder a + b - c is ...
Page 99
... hence , 2 + 0 3+ 5 2 + 0 5 4 + 06 + 10 - 100+ 15 — 25 4 + 0-16 + 10 + 15 — 25 Hence 4a16a3 b2 + 10 a2 b3 . + 15ab1 — 25b5 = product . The coefficient of a being zero in the product , causes that term to dis- appear . ( 4. ) Multiply 3 ...
... hence , 2 + 0 3+ 5 2 + 0 5 4 + 06 + 10 - 100+ 15 — 25 4 + 0-16 + 10 + 15 — 25 Hence 4a16a3 b2 + 10 a2 b3 . + 15ab1 — 25b5 = product . The coefficient of a being zero in the product , causes that term to dis- appear . ( 4. ) Multiply 3 ...
Page 100
... hence = a3 -- 5 = a — 2 ; but = 2 ... = a a2 Similarly , And 1 a + x = 1 ( a + x ) ; ( x2 + y2 ) 3 ( x2 — y2 ) + 1 1 -2 = ( x + y ) a2 = ( x + y ) 3 = ( x2 + y2 ) ̄3 ( x2 — y2 ) — * ; and so on . For more information on negative ...
... hence = a3 -- 5 = a — 2 ; but = 2 ... = a a2 Similarly , And 1 a + x = 1 ( a + x ) ; ( x2 + y2 ) 3 ( x2 — y2 ) + 1 1 -2 = ( x + y ) a2 = ( x + y ) 3 = ( x2 + y2 ) ̄3 ( x2 — y2 ) — * ; and so on . For more information on negative ...
Page 107
... hence , by putting 3 for n in formula ( 1 ) , we get a2 - b2 a3 - b3 = a2 + b . a -b a - b = a2 + b . ( a + b ) by ( 2 ) = a2 + ab + b2 . ( 3 ) . Again , a1 - b1 must be exactly divisible by a - b , since a3 - b3 is divisible by a - b ; ...
... hence , by putting 3 for n in formula ( 1 ) , we get a2 - b2 a3 - b3 = a2 + b . a -b a - b = a2 + b . ( a + b ) by ( 2 ) = a2 + ab + b2 . ( 3 ) . Again , a1 - b1 must be exactly divisible by a - b , since a3 - b3 is divisible by a - b ; ...
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algebraic axis bisected called centre circle circumference coefficient contained Corol cosec cosine cube root curve decimal denominator denote diameter difference differential co-efficient distance Divide dividend division divisor draw dy dx equal EXAMPLES exponent expression extract factors feet figure fraction given number greater greatest common measure Hence hyperbola inches latus rectum least common multiple logarithm manner monomial multiply negative nth root number of terms parallel parallelogram perpendicular plane polynomial positive Prob problem Prop proportional proposed equation quotient radius ratio rectangle Reduce remainder right angles rule sides sine square root straight line Substituting subtract tangent Taylor's theorem THEOREM unknown quantity VULGAR FRACTIONS whole number yards