Elements of GeometryHilliard and Metcalf, 1825 - 224 pages |
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Page v
... problems by means of algebraic signs 3 Explanation of algebraic formulas Of Equations . To solve questions by the assistance of algebra Explanation of the words , equation , members , and terms 8 11 ib . ib . Resolution of Equations of ...
... problems by means of algebraic signs 3 Explanation of algebraic formulas Of Equations . To solve questions by the assistance of algebra Explanation of the words , equation , members , and terms 8 11 ib . ib . Resolution of Equations of ...
Page vii
... problem Examples for illustration Recapitulation of the above remarks · 62 63 - 64 65 66 67 Of Negative Quantities Demonstration of the rules for the calculus of insulated negative quantities The combination , with respect to their ...
... problem Examples for illustration Recapitulation of the above remarks · 62 63 - 64 65 66 67 Of Negative Quantities Demonstration of the rules for the calculus of insulated negative quantities The combination , with respect to their ...
Page viii
... problems of the second degree become absurd Expressions called imaginary An equation of the second degree has always two roots 117 · ib . 118 121 - 122 - 123 Resolution of certain problems · 124 To divide any number into two parts , the ...
... problems of the second degree become absurd Expressions called imaginary An equation of the second degree has always two roots 117 · ib . 118 121 - 122 - 123 Resolution of certain problems · 124 To divide any number into two parts , the ...
Page xi
... problem 190 192 193 - 194 Inconvenience of the successive elimination of the unknown quantities when there are more than two equations and in- dication of the degree of the final equation 198 Of Commensurable Roots , and the equal Roots ...
... problem 190 192 193 - 194 Inconvenience of the successive elimination of the unknown quantities when there are more than two equations and in- dication of the degree of the final equation 198 Of Commensurable Roots , and the equal Roots ...
Page 2
... problem , but which becomes very complicated in others , consists in gen- er of a certain number of expressions , such as added to , dimin ished by , is equal to , & c . often repeated . These expressions relate to the operations by ...
... problem , but which becomes very complicated in others , consists in gen- er of a certain number of expressions , such as added to , dimin ished by , is equal to , & c . often repeated . These expressions relate to the operations by ...
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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.