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Posted by Prt Nr 7 years, 8 months ago
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Posted by Kusum Antil 7 years, 8 months ago
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Swadha Thakkar 7 years, 8 months ago
Posted by Manvi Tiwari 7 years, 8 months ago
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Posted by Sia Rai 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Let R and I are rational number and irrational number respectively.
Assume that sum of R and I is a rational number and equal to P<u1:p></u1:p>
So R + I =P<u1:p></u1:p>
or I =P - R......., (1)<u1:p></u1:p>
As P and R both are rational number so P - R is also a rational number.<u1:p></u1:p>
Hence from (1) I is a rational number<u1:p></u1:p>
But this contradict that I is an irrational number.<u1:p></u1:p>
This contradiction has come because we assumed that R+ I is a rational number.<u1:p></u1:p>
Therefore the sum of irrational number and rational number is always an irrational number.<u1:p></u1:p>
Posted by Rajesh Rajesh 7 years, 8 months ago
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Posted by Rajesh Rajesh 7 years, 8 months ago
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Posted by Shubham Pandey 7 years, 8 months ago
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Anand Jain 7 years, 8 months ago
Posted by Pushp Bhardwaj 7 years, 8 months ago
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Manan Khandelwal 7 years, 8 months ago
Posted by Naman Kaushsl 7 years, 8 months ago
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Posted by Shifana Km 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Let a and b are numbers and HCF = x
Then LCM = 14x
Now sum of HCF and LCM
x+14x =600
15x = 600
x = 40
Hence HCF=40 and LCM=14{tex}\times{/tex}40
Given a=280 and b=?
We know that
a {tex}\times{/tex} b = HCF {tex}\times{/tex} LCM
So {tex}\mathrm b=\frac{40\times14\times40}{280}=2\times40=80{/tex}
Hence the other number = 80
Posted by Anamika Kashyap 7 years, 8 months ago
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Posted by Riya Sharma 7 years, 8 months ago
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Posted by Shifana Km 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Let a and b are numbers and HCF = x
Then LCM = 14x
Now sum of HCF and LCM
x+14x =600
15x = 600
x = 40
Hence HCF=40 and LCM=14{tex}\times{/tex}40
Given a=280 and b=?
We know that
a {tex}\times{/tex} b = HCF {tex}\times{/tex} LCM
So {tex}\mathrm b=\frac{40\times14\times40}{280}=2\times40=80{/tex}
Hence the other number = 80
Posted by Thrupthi N 7 years, 8 months ago
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Posted by Abhishek Kumar 7 years, 8 months ago
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Posted by Aishwarya Aishu 7 years, 8 months ago
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Posted by Safiya Memon 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Here we have function p(x) = 2x3 - 5x2 - 4x + 3
By using remainder theorem, we get
{tex}\therefore{/tex} p(3) = 2(3)3 - 5 {tex}\times{/tex}(3)2 - 4 {tex}\times{/tex}3 + 3
= 54 - 45 - 12 + 3 = 0
{tex}\because{/tex} p(3) =0
Therefore, x = 3 is a zero of p(x).
Posted by Nikita Gupta 7 years, 8 months ago
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Posted by Anuj Singh 5 years, 8 months ago
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Ayush Jain 7 years, 8 months ago
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Raagini Pandey 7 years, 8 months ago
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