The Elements of AlgebraHarper & Bros., 1856 - 268 σελίδες |
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Αποτελέσματα 6 - 10 από τα 85.
Σελίδα 15
... + 20 + 5 · 12 + 28—3 = 42—17—9 . Ex . 2 . Ex . 3 . 4 42 Ex . 4 . 3 + 25-7 = -5 + 19 . 6 QUEST . — What is the sign of equality ? ( 10. ) When a problem is proposed for solution PRELIMINARY DEFINITIONS . 15 Exercises upon the Symbols.
... + 20 + 5 · 12 + 28—3 = 42—17—9 . Ex . 2 . Ex . 3 . 4 42 Ex . 4 . 3 + 25-7 = -5 + 19 . 6 QUEST . — What is the sign of equality ? ( 10. ) When a problem is proposed for solution PRELIMINARY DEFINITIONS . 15 Exercises upon the Symbols.
Σελίδα 16
... Problem 1. A boy bought an apple and an orange for 6 cents ; for the orange he gave twice as much as for the apple . How much did he give for each ? Let x represent the number of cents he gave for the apple , then 2x will represent the ...
... Problem 1. A boy bought an apple and an orange for 6 cents ; for the orange he gave twice as much as for the apple . How much did he give for each ? Let x represent the number of cents he gave for the apple , then 2x will represent the ...
Σελίδα 17
... problem ; that is , x + 3x = 60 . But three times x , added to once x , makes four times x ; that is , 4x = 60 ; and if four times x is equal to 60 , once x must be equal to 15 ; that is , 60 x = - = 15 . 4 Therefore the cow was worth ...
... problem ; that is , x + 3x = 60 . But three times x , added to once x , makes four times x ; that is , 4x = 60 ; and if four times x is equal to 60 , once x must be equal to 15 ; that is , 60 x = - = 15 . 4 Therefore the cow was worth ...
Σελίδα 19
... the horse was worth 81 dollars , the sum of which is 90 dollars Prob . 9. A cask which held 143 gallons was filled with a mixture of brandy and water , and there was How much was ten times as much brandy as water PROBLEMS . 19.
... the horse was worth 81 dollars , the sum of which is 90 dollars Prob . 9. A cask which held 143 gallons was filled with a mixture of brandy and water , and there was How much was ten times as much brandy as water PROBLEMS . 19.
Σελίδα 20
... problem is proposed for solution , the first thing to be done is to find an expression which shall contain the unknown quantity , and which shall be equal to a given quanti- ty . Then , from this expression , by arithmetical opera ...
... problem is proposed for solution , the first thing to be done is to find an expression which shall contain the unknown quantity , and which shall be equal to a given quanti- ty . Then , from this expression , by arithmetical opera ...
Συχνά εμφανιζόμενοι όροι και φράσεις
A's share apples arithmetical means arithmetical progression bers binomial binomial theorem brandy cents coefficient common denominator common difference complete equation Completing the square cube denotes digits Divide the number dividend divisor dollars equal Examples exponent extract the square factor Find the square Find the sum find the value Find two numbers following RULE four quantities fourth power gallons geometrical progression Give the rule Given greater number Hence horse last term least number Let x represent letters lowest terms metical miles per hour mixed quantity monomial multiplied number of terms obtain paid pears perfect square polynomial preceding problem Prob QUEST.-Give the rule QUEST.-How QUEST.-What quotient radical quantities receive Reduce remainder represent the number Required the square result second degree second power second term shillings solved square root subtract tion Transposing twice units unknown quantity values of x Whence worth yards
Δημοφιλή αποσπάσματα
Σελίδα 75 - The square of the difference of two quantities is equal to the square of the first minus twice the product of the first by the second, plus the square of the second.
Σελίδα 96 - To reduce fractions to a common denominator. RULE. Multiply each numerator into all the denominators except its own for a new numerator, and all the denominators together for a common denominator.
Σελίδα 73 - In the multiplication of whole numbers, place the multiplier under the multiplicand, and multiply each term of the multiplicand by each term of the multiplier, writing the right-hand figure of each product obtained under the term of the multiplier which produces it.
Σελίδα 94 - To reduce a mixed number to an improper fraction, — RULE : Multiply the whole number by the denominator of the fraction, to the product add the numerator, and write the result over the denominator.
Σελίδα 85 - ... the first term of the quotient ; multiply the divisor by this term, and subtract the product from the dividend. II. Then divide the first term of the remainder by the first term of the divisor...
Σελίδα 134 - To divide the number 90 into four such parts, that if the first be increased by 2, the second diminished by 2, the third multiplied...
Σελίδα 76 - ... the product of the two, plus the square of the second. In the third case, we have (a + b) (a — 6) = a2 — b2. (3) That is, the product of the sum and difference of two quantities is equal to the difference of their squares.
Σελίδα 225 - Three quantities are in proportion when the first has the same ratio to the second, that the second has to the third ; and then the middle term is said to be a mean proportional between the other two.
Σελίδα 192 - Multiply the divisor thus increased by the last term of the root, and subtract the product from the last remainder.
Σελίδα 174 - RULE. 1. Separate the given number into periods of two figures, each, beginning at the units place.