if d is HCF of 40and 65,find the
value of integers X and Y which
satisfy d=40x 65y
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since hcf of 40,65 is 5...value of d=5.
5=40 x 65y (since d=5)
5=2600y
5/2600 = y
y = 1/520 (0.001923 )
5=40 x 65y (since d=5)
5=2600y
5/2600 = y
y = 1/520 (0.001923 )
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