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Ask QuestionPosted by Krati Sharma 7 years, 9 months ago
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Posted by Arav Arav 7 years, 9 months ago
- 1 answers
Posted by Sanjeev Malhotra 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
Let us assume that <m:omath><m:r>√7</m:r></m:omath> is rational.
That is, we can find integers a and b (<m:omath><m:r>≠0)</m:r></m:omath> such that a and b are co-prime
{tex}\style{font-family:Arial}{\begin{array}{l}\sqrt7=\frac ab\\b\sqrt7=a\\on\;squaring\;both\;sides\;we\;get\\7b^2=a^2\end{array}}{/tex}
Therefore, a2 is divisible by 7,
it follows that a is also divisible by 7.
So, we can write a = 7c for some integer c.
Substituting for a, we get 7b2 = 49c2, that is, b2 = 7c2
This means that b2 is divisible by 7, and so b is also divisible by 7
Therefore, a and b have at least 7 as a common factor.
But this contradicts the fact that a and b are co-prime.
This contradiction has arisen because of our incorrect assumption that <m:omath><m:r>√7</m:r></m:omath> is rational.
So, we conclude that <m:omath><m:r>√7</m:r></m:omath> is irrational.
Posted by Jisna Fathima 7 years, 9 months ago
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Posted by Arav Arav 7 years, 9 months ago
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Posted by Aman Rajput 7 years, 9 months ago
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Posted by Gourav Kumar 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Let radius of circle be r cm
Its area = {tex}\pi{/tex}r2 sq units
Side of an equilateral triangle = diameter of the circle = 2r
Area of the equilateral triangle
{tex}= \frac { \sqrt { 3 } } { 4 } \times ( 2 r ) ^ { 2 }{/tex}
{tex}= \frac { \sqrt { 3 } } { 4 } \times 4 r ^ { 2 } = \sqrt { 3 } r ^ { 2 }{/tex}
{tex}\frac { \text { Area of the circle } } { \text { Area of the triangle } } = \frac { \pi r ^ { 2 } } { \sqrt { 3 } r ^ { 2 } } = \frac { \pi } { \sqrt { 3 } }{/tex}
Posted by Vikash Jaiswal 7 years, 9 months ago
- 2 answers
Posted by Bhawna Kumari 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
First term = a = 114
Common difference = a2 - a1
d = 109 - 114
d = -5
Let n th term of A.P = 0
a + ( n - 1 )d = 0
114 + ( n - 1 ) ( -5 ) = 0
114 - 5n + 5 = 0
119 - 5n = 0
-5n = - 119
n = 119/5
n = 23.4
Therefore ,
24 th term of given A. P is first negative term
Posted by Yogesh Kumar 7 years, 9 months ago
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Deepanshi Lovely 7 years, 9 months ago
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Posted by Nikhil Kumar 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let A(a, a2), B(b, b2) and C(c, c2) be the given points.
{tex}\therefore{/tex}Area of {tex}\Delta A B C{/tex}
{tex}= \frac { 1 } { 2 } \left\{ a \left( b ^ { 2 } - c ^ { 2 } \right) + b \left( c ^ { 2 } - a ^ { 2 } \right) + c \left( a ^ { 2 } - b ^ { 2 } \right) \right\}{/tex}
{tex}= \frac { 1 } { 2 } \left\{ a b ^ { 2 } - a c ^ { 2 } + b c ^ { 2 } - b a ^ { 2 } + c a ^ { 2 } - c b ^ { 2 } \right\}{/tex}
{tex}= \frac { 1 } { 2 } \times 0{/tex} [if a = b = c]
=0
i.e., the points are collinear if a = b = c
Hence, the points can never be collinear if {tex}a \neq b \neq c.{/tex}
Posted by Vivek Pal 7 years, 9 months ago
- 1 answers
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Vikram Viku 7 years, 9 months ago
1Thank You