Elements of GeometryHilliard and Metcalf, 1825 - 224 σελίδες |
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Αποτελέσματα 1 - 5 από τα 77.
Σελίδα v
... solution 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 ...
... solution 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 ...
Σελίδα viii
... solutions In what cases 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 ...
... solutions In what cases 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 ...
Σελίδα 1
... solution of which is composed of two parts ; the one having for its object to find to which of the four fundamental rules the determination of the unknown number by means of the numbers given belongs , and the other the application of ...
... solution of which is composed of two parts ; the one having for its object to find to which of the four fundamental rules the determination of the unknown number by means of the numbers given belongs , and the other the application of ...
Σελίδα 3
... solution , we shall see that the one is only a translation of the other . The number 2 , the result of the preceding operations , will answer only for the particular example which is selected , while the course of reasoning considered ...
... solution , we shall see that the one is only a translation of the other . The number 2 , the result of the preceding operations , will answer only for the particular example which is selected , while the course of reasoning considered ...
Σελίδα 5
... solution of the question proposed ; but Algebra does more ; it furnishes a rule for calculating the greater part without the aid of the less as follows ; b — 2 being the value of this , augmenting it by the excess h , we have for the ...
... solution of the question proposed ; but Algebra does more ; it furnishes a rule for calculating the greater part without the aid of the less as follows ; b — 2 being the value of this , augmenting it by the excess h , we have for the ...
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Άλλες εκδόσεις - Προβολή όλων
Συχνά εμφανιζόμενοι όροι και φράσεις
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
Δημοφιλή αποσπάσματα
Σελίδα 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.
Σελίδα 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.
Σελίδα 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'.
Σελίδα 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.
Σελίδα 8 - Any side of a triangle is less than the sum of the other two sides...
Σελίδα 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.
Σελίδα 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.
Σελίδα 54 - The sum of the squares on the sides of a parallelogram is equal to the sum of the squares on the diagonals.
Σελίδα 185 - The convex surface of a cone is equal to the circumference of the base multiplied by half the slant height.
Σελίδα 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.