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Diksha ? 6 years, 3 months ago
Diksha ? 6 years, 3 months ago
Posted by Monu Kumar 6 years, 3 months ago
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Posted by Prarthana Prasad 6 years, 3 months ago
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Yogita Ingle 6 years, 3 months ago
Smallest prime number = 2
Smallest composite number = 4
2 = 2 × 1
4 = 2 × 4
HCF(2, 4) = 2
Miss Zainab 6 years, 3 months ago
Posted by Varsha Narwariya 6 years, 3 months ago
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Varsha Narwariya 6 years, 3 months ago
Yogita Ingle 6 years, 3 months ago
2x + y + 3 = 0
4x + ky + 6 = 0
For a pair of lines to be parallel, the ratio of the respective coefficients and the constant terms must be the same.
{tex}\therefore \dfrac{2}{4}=\dfrac{1}{k}=\dfrac{3}{6}\\\Rightarrow k=\dfrac{4}{2}\\\Rightarrow k=2{/tex}
hence, the value of k must be 2 in order to have the given pair of lines parallel to each other.
Posted by Alok Dubey 6 years, 3 months ago
- 2 answers
Yogita Ingle 6 years, 3 months ago
On dividing 398 by the required number, there is a remainder of 7. This means that 398 7 = 391 is exactly divisible by the required number. Similarly, 436 -11 = 425 and 542 15 = 527 are exactly divisible by the required number.
The HCF of two positive integers is the largest positive integer that divides both the integers.
So, the required number will be the HCF of 391, 425 and 527. And that can be found by using Euclids division algorithm.
425 = 391 x 1 + 34
391 = 34 x 11 + 17
34 = 17 x 2 + 0
Thus, HCF = 17
Hence, the required number is 17.
Posted by Laxman Phadake 6 years, 3 months ago
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Vimal Singh 6 years, 3 months ago
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Gaurav Seth 6 years, 3 months ago
LCM- “Lowest Common Multiple” Smallest integer that are the multiple of both numbers.
HCF- “Highest Common Factor” Largest whole number that divides both numbers.
LCM of two numbers is 1200.
Factorizing 1200, we get,
∴ Prime Factors
Since we cannot find 500 or multiples of 500 in the factors of 1200, hence 500 cannot be the HCF of the two numbers.
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