The elements of algebraOliver & Boyd, 1857 - 95 sider |
Fra bogen
Resultater 1-3 af 3
Side 41
... number of reapers to be a , the wages of the men m shillings , and of the women n shillings per day , and the total amount p shillings per day . Ans . P -an men , and am- mn mn α P women . ( 49. ) Divide the fraction into two parts ...
... number of reapers to be a , the wages of the men m shillings , and of the women n shillings per day , and the total amount p shillings per day . Ans . P -an men , and am- mn mn α P women . ( 49. ) Divide the fraction into two parts ...
Side 77
... total number of combinations of n things taken 1 , 2 , 3 ... n together 2 " — 1 . = ( 11. ) The total number of combinations of 2 n things 33 times the total number of combinations of n things ; find n . Ans . n5 . ( 12. ) Out of a ...
... total number of combinations of n things taken 1 , 2 , 3 ... n together 2 " — 1 . = ( 11. ) The total number of combinations of 2 n things 33 times the total number of combinations of n things ; find n . Ans . n5 . ( 12. ) Out of a ...
Side 84
... total number of balls = 1o + 2o + 3o + ... or ( § 87. Exam . 2. ) = } n ( n + 1 ) ( 2n + 1 ) . n2 , 90. If the base be an equilateral triangle , and if n denote the number of balls in one of its sides , the number in the bottom layer ...
... total number of balls = 1o + 2o + 3o + ... or ( § 87. Exam . 2. ) = } n ( n + 1 ) ( 2n + 1 ) . n2 , 90. If the base be an equilateral triangle , and if n denote the number of balls in one of its sides , the number in the bottom layer ...
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Almindelige termer og sætninger
a+b+c a²+2ab+b² annex arithmetical series ax² binomial binomial theorem coefficient common denominator common difference common ratio compound quantity consecutive numbers cube root deno denote the number Divide divisor double equal equation contains EXERCISES expressed Extract the square factor Find a number find the Greatest Find the number Find the sum following fractions following quantities geometric means given quantity Greatest Common Measure Hence Least Common Multiple letters miles per hour mixed quantities multiplied negative number of balls number of combinations number of permutations number of terms numbers whose sum numerator and denominator Prove QUADRATIC EQUATIONS quotient remainder shillings side sign changed simple quantity square root subtracting surd t₁ things taken third total number transposing travelled unknown quantity ах
Populære passager
Side 61 - To divide a given straight line into two parts, so that the rectangle contained by the -whole, and one of the parts, may be equal to the square of the other part.
Side 88 - The fore-wheel of a carriage makes six revolutions more than the hind-wheel in going 120 yards ; but if the circumference of each wheel be increased one yard, it will make only four revolutions more than the hind-wheel in going the same distance.
Side 60 - Divide the number 24 into two such parts, that their product shall be to the sum of their squares, as 3 to 10.
Side 16 - If the numerator and denominator of a fraction be both multiplied or both divided by the same number, the value of the fraction is not altered.
Side 44 - Divide the first term of the remainder by three times the square of the first term of the root, and write the result as the next term of the root.
Side 9 - A power of a quantity is divided by any other power of the same quantity by subtracting the index of the divisor from that of the dividend, the quotient being that power of the quantity whose index is the remainder so obtained.
Side 66 - From the preceding, it appears, that the sum of the extremes is equal to the sum of any other two terms equally distant from the extremes.
Side 41 - ... be divided by the number of terms to that place, it will give the coefficient of the term next following.
Side 44 - Take the root of the first term, for the first term of the required root...
Side 40 - A man was hired 50 days on these conditions. — that, for every day he worked, he should receive $ '75, and, for every day he was idle, he should forfeit $ '25 ; at the expiration of the time, he received $ 27'50 ; how many days did he work...