Key to Ray's Algebra: Parts First and Second : Containing Statements and Solutions of Questions ... |
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Page 40
... 2 , 3 , 4 , and 5 , so that the results will be true in an arithmetical sense . . What number must be added to 20 , that the sum may be 25 ! Ans . 5 . 3. What number must be subtracted from 11 , that 40 KEY TO PART FIRST.
... 2 , 3 , 4 , and 5 , so that the results will be true in an arithmetical sense . . What number must be added to 20 , that the sum may be 25 ! Ans . 5 . 3. What number must be subtracted from 11 , that 40 KEY TO PART FIRST.
Page 103
... true in an arithmetical sense . 2. What number must be subtracted from the number 30 , that the remainder shall be 19 ? " Ans . 11 . 3. The difference of two numbers is 9 , and their sum 25 ; re- quired the numbers . Ans . 17 and 8 . 4 ...
... true in an arithmetical sense . 2. What number must be subtracted from the number 30 , that the remainder shall be 19 ? " Ans . 11 . 3. The difference of two numbers is 9 , and their sum 25 ; re- quired the numbers . Ans . 17 and 8 . 4 ...
Page 104
... true at the same time . EXAMPLES INVOLVING THE SECOND POWER OF THE UNKNOWN QUANTITY . Article 171 . ( 9 ) First divide both members by xm . ( 11 ) Let x the number , then X х х X x 4 X = XX or 3 3 27 . Multiplying both members by 27 and ...
... true at the same time . EXAMPLES INVOLVING THE SECOND POWER OF THE UNKNOWN QUANTITY . Article 171 . ( 9 ) First divide both members by xm . ( 11 ) Let x the number , then X х х X x 4 X = XX or 3 3 27 . Multiplying both members by 27 and ...
Page 155
... true in an arithmetical sense . * ( 9 ) Let x and y represent the numbers , then x2 + y2 + x + y = 42 , xy = 15 . ( 1 ) ( 2 ) If we add twice the second equation to the first , the resulting equation is ( x + y ) 2 + ( x + y ) = 72 ...
... true in an arithmetical sense . * ( 9 ) Let x and y represent the numbers , then x2 + y2 + x + y = 42 , xy = 15 . ( 1 ) ( 2 ) If we add twice the second equation to the first , the resulting equation is ( x + y ) 2 + ( x + y ) = 72 ...
Page 174
... true proportion α Second , nd = d but since a : b :: c : d , we have nb b d = bc ( Art . 267 ) , hence d bc с α α a α which is the ratio of a to c , therefore ( 2 ) is a true propor- tion . ( Art . 263. ) b_d ( 19 ) Since a : b :: c : d ...
... true proportion α Second , nd = d but since a : b :: c : d , we have nb b d = bc ( Art . 267 ) , hence d bc с α α a α which is the ratio of a to c , therefore ( 2 ) is a true propor- tion . ( Art . 263. ) b_d ( 19 ) Since a : b :: c : d ...
Common terms and phrases
a²+b² a²x² arithmetical progression arithmetical sense Article assumed numbers cents clearing of fractions coefficients completing the square cost denominator derived polynomial Dividing the given equa equal equation Art equation becomes example expression extracting the square factors Find a number find the number find the values find x=5 formula given equation gives greater number greatest common divisor hence the numbers integral least common multiple Let x Limit of positive logarithms method Multiplying both members Multiplying both sides negative roots number of balls number of days number of miles number of solutions positive roots pupil Quotients Add radix ratio readily find remainder render represent the numbers result second equation solved square number square root substituting the value subtracting three numbers tion transposing unknown quantities Whence x=2 whole number
Popular passages
Page 323 - It is required to find three numbers in arithmetical progression, such, that the sum of every two of them may be a square.
Page 335 - Calcutta by the payment of £5025 16s. 8d. in London ? 2. Show why it follows from our system of notation, that a number when divided by 9 leaves the same remainder as the sum of its digits will leave when divided by 9. Write down all the numbers that can be composed of the four digits 3, 4, 5, 6, which will each be exactly divisible by 11. 3. At what rate per cent. simple interest will £7433 6s. 8d. amount to £9942 Is. 8d. in 7£ years ? 4. Show that the rule for dividing...
Page 6 - 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.
Page 240 - The coefficient of the second term of any equation is equal to the sum of all the roots with their signs changed. 2. The coefficient of the third term is equal to the sum of the products of all the roots taken two and two. 3. The coefficient of the fourth term is equal to the sum of the products of all the roots taken three and three, with their signs changed.
Page 341 - I rejoice that a Western House has been able to meet the increasing wants of the West in this great field.
Page 153 - ... and each boy got as many nuts as there were boys in his company. Moreover all the boys in the larger company got 225 nuts more than all the boys in the smaller company ; and the whole number collected was 1025. How many boys were there ? Ans.
Page 314 - Divide a line into two parts, such that the sum of their squares shall be double the square on another line.
Page 282 - E, represent the errors which result from these substitutions. "We assume that the errors of the results are proportional to the errors of the assumed numbers. This supposition is not entirely correct ; but if we employ numbers near to the true values, the error of this supposition is generally not very great, and the error becomes less and less the further we carry the approximation. We have then E:E'::xr:xr'. ~Whence, Art. 305, EE...
Page 215 - It is evident that the terms of a proportion may undergo any change which will not destroy the equality of the ratios ; or which will leave the product of the means equal to the product of the extremes.
Page 325 - Problems, the first of which finds "three square numbers, such that the sum of every two of them may be a square number;" the second determines " values for the sides of a triangle in whole numbers, such that the lengths of the three lines from the angles to the middle of the opposite sides may be expressed by rational whole numbers ;