Key to System of practical mathematics. 2 pt. No.xvii |
Inni boken
Resultat 1-5 av 50
Side 28
... PROBLEMS PRODUCING EQUATIONS CON- TAINING TWO OR MORE UNKNOWN QUANTITIES . 1. Let x the first , and y = the second ; then 1 2 ( 1 ) x2 ( 3 ) - ( 2 ) ( 4 ) ÷ 3 ( 2 ) x4 ( 6 ) — ( 1 ) ( 7 ) ÷ 7 2. Let x 678 x + 2y = 21 , 2a + 1 = 14 ...
... PROBLEMS PRODUCING EQUATIONS CON- TAINING TWO OR MORE UNKNOWN QUANTITIES . 1. Let x the first , and y = the second ; then 1 2 ( 1 ) x2 ( 3 ) - ( 2 ) ( 4 ) ÷ 3 ( 2 ) x4 ( 6 ) — ( 1 ) ( 7 ) ÷ 7 2. Let x 678 x + 2y = 21 , 2a + 1 = 14 ...
Side 36
... problem . 8. Let x = the first ; then since the sum of the first and second is 10 , the second will be ( 10 - x ) ; again , since they are in geometrical progression , x : ( 10 — x ) :: ( 10 — x ) : ( 10 - x ) 2 the third ; therefore by ...
... problem . 8. Let x = the first ; then since the sum of the first and second is 10 , the second will be ( 10 - x ) ; again , since they are in geometrical progression , x : ( 10 — x ) :: ( 10 — x ) : ( 10 - x ) 2 the third ; therefore by ...
Side 107
... problem solved by logarithms gives the answer 16 years 138 days .. 132d . L.4500 at 6 per cent . in 9 years will amount ( Table 7 ) to 1.689479 × 4500 = L.7602 · 6555 ; this being divided by the amount of L.1 at 4 per cent . for 15 ...
... problem solved by logarithms gives the answer 16 years 138 days .. 132d . L.4500 at 6 per cent . in 9 years will amount ( Table 7 ) to 1.689479 × 4500 = L.7602 · 6555 ; this being divided by the amount of L.1 at 4 per cent . for 15 ...
Side
... Problem IV . Do. , Problems V. and VI . , 31 38 45 49 Problems VII . and VIII . , Problems IX . and X. , Problem XI . , 51 54 56 Problem XII . , Problem XIII . , Problem XIV . , 59 61 62 Problem XV . , 64 Mensuration of Solids , 65 Problem ...
... Problem IV . Do. , Problems V. and VI . , 31 38 45 49 Problems VII . and VIII . , Problems IX . and X. , Problem XI . , 51 54 56 Problem XII . , Problem XIII . , Problem XIV . , 59 61 62 Problem XV . , 64 Mensuration of Solids , 65 Problem ...
Side
Scottish school-book assoc. Page . Problem III . 88 Problem IV . , Problem V. , 89 91 Problem VIII . , 92 Problems IX . and XI . 94 Problem XII . 95 Miscellaneous Exercises , 96 Specific Gravities , 99 Problems II . and III . , 100 Problem ...
Scottish school-book assoc. Page . Problem III . 88 Problem IV . , Problem V. , 89 91 Problem VIII . , 92 Problems IX . and XI . 94 Problem XII . 95 Miscellaneous Exercises , 96 Specific Gravities , 99 Problems II . and III . , 100 Problem ...
Vanlige uttrykk og setninger
a+b+c AABC ABCD acres base binomial theorem bisected centre changing the signs chord circle circumference coefficients collecting the terms completing the square cosec denominator diameter difference distance dividing divisor equal extracting the root feet find the area find the differential fraction given equation gives greater segment half the sum height hence the area hypotenuse inches inverted latitude least common multiple Let ABC Log.cosec logarithm miles Mult Multiply number sought perp perpendicular poles Problem XI Prop question radius rectangle semiperimeter sine slant slant height solidity square root substituting Subt Subtract surf Tabular area tangent Theorem third side transp transposing transposition triangle Trig value of x wherefore whole arc whole surface yards دو
Populære avsnitt
Side 74 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz.
Side 75 - If the vertical angle of a triangle be 'bisected 'by a straight line which also cuts the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square on the straight line which bisects the angle.
Side 9 - Let x measure у by the units in n, then it will measure cy by the units in nc. 2d. If a quantity measure two others, it will measure their sum or difference. Let a be contained...
Side 15 - ... sin(a + b + c). Again (a) represents the coarse ROM, and bands b and c are two controls of the fine-tuned ROMs so that a < 90°, b < 90 • 2~a and c < 90 • 2~(a + 6). This is shown in Fig. 7-7. Sunderland showed that the trigonometric identity can be written as sin(a + b + c) = sin(a + 6) cos c + cos a cos b sin...
Side 10 - The truth of this rule depends upon these two principles ; 1". If one quantity measure another, it will also measure any multiple of that quantity. Let x measure y by the units in n, then it will measure cy by the units in nc.
Side 139 - Arc, on the Sine and Cosine of an Arc in terms of the Arc itself, and a new Theorem for the Elliptic Quadrant.
Side 137 - The differential of the logarithm of a function is equal to the differential of the function, divided by the function itself.
Side 149 - The pyramid may be conceived to be made up of an infinite number of planes parallel to ABC.
Side 81 - ... sum of any number of quantities is equal to the sum of the corresponding functions of each of these quantities, will be called distributive
Side 86 - We thus derive the following method for multiplying two binomials which have a common first term : The first term of the product is the square of the common first terms of the binomials.