Elements of GeometryHilliard and Metcalf, 1825 - 224 pages |
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Page vii
... changing the signs of quantities to comprehend several ques- tions in one 77 78 Solation of the preceding questions by employing only one un- known quantity 79 To resolve equations of the first degree , when there are two un- known ...
... changing the signs of quantities to comprehend several ques- tions in one 77 78 Solation of the preceding questions by employing only one un- known quantity 79 To resolve equations of the first degree , when there are two un- known ...
Page 4
... change in the language by which they are expressed ; whereas , by considering the numbers as deter- minate , we perform ... signs which it contains , gives the rule found before , according to which , in order to obtain the less of two ...
... change in the language by which they are expressed ; whereas , by considering the numbers as deter- minate , we perform ... signs which it contains , gives the rule found before , according to which , in order to obtain the less of two ...
Page 33
... change the signs ( 20 ) . We have then a c bc adbd for the result required . - - 31. Agreeably to the above examples ... sign , as the corresponding part of the multiplicand , when the multiplier has the sign + , and the contrary sign ...
... change the signs ( 20 ) . We have then a c bc adbd for the result required . - - 31. Agreeably to the above examples ... sign , as the corresponding part of the multiplicand , when the multiplier has the sign + , and the contrary sign ...
Page 46
Adrien Marie Legendre. different signs . Multiplying it by all the terms of the divisor and changing the signs , we obtain the quantity 20 a b 8 as b2 + 16 a b3 , - which taken with the dividend and reduced , gives for a remainder +10 a1 ...
Adrien Marie Legendre. different signs . Multiplying it by all the terms of the divisor and changing the signs , we obtain the quantity 20 a b 8 as b2 + 16 a b3 , - which taken with the dividend and reduced , gives for a remainder +10 a1 ...
Page 47
... changing the signs , we have — a b2 + b2 ; the first term ab3 destroys the first term of the dividend , and the second b3 destroys the other — b3 . The mechanical part of the operation will be better under- stood , if we look for a ...
... changing the signs , we have — a b2 + b2 ; the first term ab3 destroys the first term of the dividend , and the second b3 destroys the other — b3 . The mechanical part of the operation will be better under- stood , if we look for a ...
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Common terms and phrases
a² b³ algebraic Algebraic Quantities Arith arithmetic becomes binomial changing the signs coefficient common divisor consequently contains courier cube root decimal deduce denominator denoted divided dividend division employed entire number enunciation equa evident example exponent expression extract the root figures follows formula fraction given in art given number gives greater greatest common divisor last term letters logarithm manner method multiplicand multiplied negative number of arrangements observed obtain operation perfect square polynomials preceding article proposed equation proposed number quan question quotient radical quantities radical sign reduced remainder represented resolve result rule given second degree second member second term simple quantities square root subtract suppose taken tens third tion tities units unity unknown quantity vulgar fractions whence whole numbers
Popular passages
Page 9 - If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles are congruent.
Page 44 - 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 63 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. A D A' Hyp. In triangles ABC and A'B'C', To prove AABC A A'B'C' A'B' x A'C ' Proof. Draw the altitudes BD and B'D'.
Page 101 - Which proves that the square of a number composed of tens and units, contains the square of the tens plus twice the product of the tens by the units, plus the square of the units.
Page 8 - Any side of a triangle is less than the sum of the other two sides...
Page 122 - ... is negative in the second member, and greater than the square of half the coefficient of the first power of the unknown quantity, this equation can have only imaginary roots.
Page 180 - CD, &c., taken together, make up the perimeter of the prism's base : hence the sum of these rectangles, or the convex surface of the prism, is equal to the perimeter of its base multiplied by its altitude.
Page 54 - The sum of the squares on the sides of a parallelogram is equal to the sum of the squares on the diagonals.
Page 185 - The convex surface of a cone is equal to the circumference of the base multiplied by half the slant height.
Page 164 - If two triangles have two sides and the inchtded angle of the one respectively equal to two sides and the included angle of the other, the two triangles are equal in all respects.