A Treatise of Algebra: Wherein the Principles are Demonstrated ... To which is Added, the Geometrical Construction of a Great Number of Linear and Plane Problems ...M. Carey & sons, 1821 - 408 sider |
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Resultat 1-5 av 16
Side 100
... square root ; but , first of all , so much is to be added to both sides thereof as to make that where the unknown quantity is a perfect square ; this is usually called completing the square , and is always done by taking half the ...
... square root ; but , first of all , so much is to be added to both sides thereof as to make that where the unknown quantity is a perfect square ; this is usually called completing the square , and is always done by taking half the ...
Side 101
... square of the second : from whence it is manifest , that , if the first and second terms of the square be given or ... completing the square : thus , the equation + 2ax - b , by completing the square , will become + 2ax2 + a2 = b + a2 ...
... square of the second : from whence it is manifest , that , if the first and second terms of the square be given or ... completing the square : thus , the equation + 2ax - b , by completing the square , will become + 2ax2 + a2 = b + a2 ...
Side 102
... squares , added to- gether , make 1666 ( c ) ? Let the lesser of the two required numbers be x ; then , bx a : b :: x : = the greater ; therefore , by the question , x2 + a b2x2 a2 c ... completing the square , 102 The Application of Algebra.
... squares , added to- gether , make 1666 ( c ) ? Let the lesser of the two required numbers be x ; then , bx a : b :: x : = the greater ; therefore , by the question , x2 + a b2x2 a2 c ... completing the square , 102 The Application of Algebra.
Side 103
... squares 1424 . Let the lesser be x , and the greater will be x + 12 ; therefore , by the problem , x + 12 ] 2 + x2 = 1424 , or 2x2 + 24x + 144 = 1424 ; this , ordered , gives x2 + 12x = 640 ; which , by completing the square , becomes ...
... squares 1424 . Let the lesser be x , and the greater will be x + 12 ; therefore , by the problem , x + 12 ] 2 + x2 = 1424 , or 2x2 + 24x + 144 = 1424 ; this , ordered , gives x2 + 12x = 640 ; which , by completing the square , becomes ...
Side 104
... completing the square , sax + x2 9a2 ― 9a2 4 4 = ( — a2 + 942 ) 543 ; of which the root being extracted , there ... squares , of two numbers be- ing given ; to find the numbers . - Let half the sum of the two numbers be denoted by a ...
... completing the square , sax + x2 9a2 ― 9a2 4 4 = ( — a2 + 942 ) 543 ; of which the root being extracted , there ... squares , of two numbers be- ing given ; to find the numbers . - Let half the sum of the two numbers be denoted by a ...
Vanlige uttrykk og setninger
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Populære avsnitt
Side 241 - The circumference of every circle is supposed to be divided into 360 equal parts, called degrees ; and each degree into 60 equal parts, called minutes ; and each minute into 60 equal parts, called seconds ; and these into thirds, &c.
Side 53 - Multiply the numerators together for a new numerator, and the denominators together for a new denominator.
Side 64 - ... then, by adding, or subtracting, the two equations thus obtained, as the case may require, there will arise a new equation, with only one unknown quantity in it, which may be resolved as before.
Side 251 - ... the sum of the segments of the base is to the sum of the sides as the difference of the sides to the difference of the segments of the base.
Side 87 - A composition of copper and tin containing 100 cubic inches weighed 505 ounces. How many ounces of each metal did it contain, supposing a cubic inch of copper to weigh of ounces, and a cubic inch of tin to weigh 4т ounces ? Ans. 420 of copper, and
Side 88 - ... half of what he had left, and half a sheep over ; and, soon after this, a third party met him, and used him in the same manner, and then he had only five sheep left. It is required to find what number of sheep he had at first, Ans, 47 sheep.
Side 254 - The following particular directions, however, may be of some use. 1st, In preparing the figure, by drawing lines, let them be either parallel or perpendicular to other lines in the figure, or so as to form similar triangles. And if an angle be given, it will be proper to let the perpendicular be opposite to that angle, and to fall from one end of a given line, if possible.