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Yogita Ingle 7 years, 3 months ago
The smallest composite number is 4
The smallest prime number = 2
To find HCF between 2 and 4, we write
2 = 2 × 1
4 = 2 × 2
HCF = 2
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Sia ? 6 years, 4 months ago
Let a = 4q + r, when r = 0, 1, 2 and 3
{tex}\therefore{/tex}Numbers are 4q, 4q + 1, 4q + 2 and 4q + 3
{tex}( a ) ^ { 2 } = ( 4 q ) ^ { 2 } = 16 q ^ { 2 } = 4 ( 4 q ) ^ { 2 } = 4 m{/tex}
{tex}( a ) ^ { 2 } = ( 4 q + 1 ) ^ { 2 } = 16 q ^ { 2 } + 8 q + 1 = 4 \left( 4 q ^ { 2 } + 2 q \right) + 1 = 4 m + 1{/tex}
{tex}( a ) ^ { 2 } = ( 4 q + 2 ) ^ { 2 } = 16 q ^ { 2 } + 16 q + 4 = 4 \left( 4 q ^ { 2 } + 4 q + 1 \right) = 4 m{/tex}
{tex}( a ) ^ { 2 } = ( 4 q + 3 ) ^ { 2 } = 16 q ^ { 2 } + 24 q + 9 = 4 \left( 4 q ^ { 2 } + 6 q + 2 \right) + 1 = 4 m + 1{/tex}
{tex}\therefore {/tex} the square of any +ve integer is of the form 4q or 4q + 1
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Sonali Aggarwal 7 years, 3 months ago
Vietnam's capital city has been Hanoi since the reunification of North and South Vietnam in 1976, with Ho Chi Minh City as the most populous city.
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Sia ? 6 years, 4 months ago
GIVEN : A circle C (0, r) and a tangent AB at a point P.
TO PROVE : {tex}O P \perp A B{/tex}
CONSTRUCTION Take any point Q, other than P, on the tangent AB. Join OQ. Suppose OQ meets the circle at R.
PROOF : We know that among all line segments joining the point O to a point on AB, the shortest one is perpendicular to AB.
To prove that {tex}O P \perp A B{/tex}, it is sufficient to prove that OP is shorter than any other segment joining O to any point of AB.
Clearly, {tex}OP = OR{/tex}
Now, {tex}OQ = OR + RQ{/tex}
{tex}\Rightarrow \quad O Q > O R{/tex}
{tex}\Rightarrow \quad O Q > O P \quad [ \because O P = O R ]{/tex}
{tex}\Rightarrow \quad O P < O Q{/tex}
{tex}\therefore{/tex} {tex}O P \perp A B{/tex}
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Mehak . 7 years, 3 months ago
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