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Yogita Ingle 6 years, 2 months ago
- Find the prime factorization of 25
25 = 5 × 5 - Find the prime factorization of 91
91 = 7 × 13 - To find the HCF, multiply all the prime factors common to both numbers:
Therefore, HCF = 1
Posted by चौधरी सुभाष 6 years, 2 months ago
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Posted by Shivam Prajapati 6 years, 2 months ago
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Yogita Ingle 6 years, 2 months ago
When three points are collinear then the condition is
x1 (y 2 - y3) + x2(y3 - y1) + x3(y1 - y2) = 0
Where, x1 = 2, x2 = 4, y1 = 3, y2 = k, x3 = 6, y3 = -3
Substituting the values,
2(k + 3) + 4(-3 - 3) + 6(3 - k) = 0
2k + 6 + 4(-6) + 18 - 6k = 0
2k + 6 - 24 + 18 - 6k = 0
-4k + 0 = 0
k=0
Therefore, The value of k is 0.
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Yogita Ingle 6 years, 2 months ago
Let the two consecutive positive integers be x and x + 1
Product of two consecutive positive integers = x (x + 1)= x2 + x
Case (i): x is even number
Let x = 2k
⇒x2 + x = (2k)2 + 2k
= 4k2 + 2k
= 2k ( 2k + 1)
Hence the product is divisible by 2
Case (ii): x is odd number
Let x = 2k + 1
⇒ x2 + x = (2k + 1)2 + (2k + 1)
= 4k2 + 4k + 1 + 2k + 1
= 4k2 + 6k + 2
= 2 (2k2 + 3k + 1)
Clearly the product is divisible by 2
From the both the cases we can conclude that the product of two consecutive positive integers is divisible by 2.
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Yogita Ingle 6 years, 2 months ago
When we are asked to calculate the distance between a point and any of the axes, we calculate the perpendicular distance between them.
Therefore, the point (-3,4) is 3 units away from the y-axis and 4 units from the x-axis. This is because in case of either axes the co-ordinate of the other axis is zero.
Therefore the distance between the point P(-3,4) and x-axis is 4 units.
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Ashish Singh 6 years, 2 months ago
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