Pure mathematics, Volum 11874 |
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Resultat 1-5 av 63
Side 20
... contain the multiplier as a factor , we use the first method , thus : -- 35 ( 1. ) z x 5 = 7 X 5 = 13 13 13 ( 2. ) × 10 9 X 10 . 55 = 55 Here we see that the numerator and denominator have a common factor 5 , and therefore , by Art . 7 ...
... contain the multiplier as a factor , we use the first method , thus : -- 35 ( 1. ) z x 5 = 7 X 5 = 13 13 13 ( 2. ) × 10 9 X 10 . 55 = 55 Here we see that the numerator and denominator have a common factor 5 , and therefore , by Art . 7 ...
Side 29
... contain each of the given denominators as a factor , and our problem is therefore to obtain the least of such numbers . DEF . - The least common multiple ( L.C.M. ) of two or more numbers is the least number which contains each of the ...
... contain each of the given denominators as a factor , and our problem is therefore to obtain the least of such numbers . DEF . - The least common multiple ( L.C.M. ) of two or more numbers is the least number which contains each of the ...
Side 30
... contained in the L.C.M. of the deno- minators , when the fractions are all in their lowest terms , it is unne- cessary to reduce the fraction to its lowest terms . We strongly recommend the beginner , however , to always com- mence by ...
... contained in the L.C.M. of the deno- minators , when the fractions are all in their lowest terms , it is unne- cessary to reduce the fraction to its lowest terms . We strongly recommend the beginner , however , to always com- mence by ...
Side 56
... contained in 216 sq . dekam . 56 sq . met . 70 sq . decim . ? Sq . Dekam . Sq . Dekam . 12-0315 ) 216-5670 ( 18 Sq . Decim . Sq . Decim . 120315 ) 2165670 ( 18 120315 120315 962520 or thus , by integers- 962520 962520 962520 Ans . 18 ...
... contained in 216 sq . dekam . 56 sq . met . 70 sq . decim . ? Sq . Dekam . Sq . Dekam . 12-0315 ) 216-5670 ( 18 Sq . Decim . Sq . Decim . 120315 ) 2165670 ( 18 120315 120315 962520 or thus , by integers- 962520 962520 962520 Ans . 18 ...
Side 58
... contained in 36759.50 metres ? 20 ( 2. ) 12 sq . decim . 75 sq . centim . contained in 10 sq . met . sq . decim . ? ( 3. ) 13 sq . met . 25 sq . decim . contained in 318 sq . dekam . ? ( 4. ) 31 cub . met . 725 cub 58 ARITHMETIC .
... contained in 36759.50 metres ? 20 ( 2. ) 12 sq . decim . 75 sq . centim . contained in 10 sq . met . sq . decim . ? ( 3. ) 13 sq . met . 25 sq . decim . contained in 318 sq . dekam . ? ( 4. ) 31 cub . met . 725 cub 58 ARITHMETIC .
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Pure Mathematics: Including the Higher Parts of Algebra and Plane ..., Volum 1 Edward Atkins Uten tilgangsbegrensning - 1877 |
Vanlige uttrykk og setninger
a²b a²b² a²x² a³b ab² ab³ ABCD adjacent angles algebraical algebraical quantity angle ABC angle BAC angle BCD angle EDF angle equal base BC BC is equal bisect brackets cent centim centre circle ABC coefficient common Const cube root decimal figures denominator distance divided divisor equation expression exterior angle factor Find the value fraction given rectilineal given straight line greater Hence join kilom Let ABC logarithm metres millig Multiply opposite angles parallel parallelogram perpendicular PROOF.-Because Q. E. D. Proposition quotient ratio rectangle contained remainder right angles segment side BC square on AC square root subtraction term triangle ABC x²y x²y² x³y xy² xy³
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