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Navya Singh 7 years, 1 month ago
let the denominator be x then numerator becomes x-4, further your question is incomplete what is added to both ? whatever the no. is add it with the numerator x-4 and denominator x which will be equal to 1/2 than solve the equation
Posted by Ayush Lamba 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Steps of Construction.
- Draw MO = 4.5 cm
- With M as centre and radius ME = 6 cm, draw an arc.
- With O as centre and radius OE = 7.5 cm, draw an arc to intersect the arc of step 2 at E.
- With O as centre and radius OR = 6 cm, draw an arc on the side of OE opposite to that of M.
- With E as centre and radius ER = 4.5 cm, draw another arc to intersect the arc of step 4 at R.
- Join OR, RE, EM and EO
Then, MORE is the required parallelogram.
Posted by Rahil Khan 7 years, 1 month ago
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Posted by Sujal Rai 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Let a be the first term and d be the common difference of the given AP. Then,
Sm = sum of first m terms of the given AP;
Sn = sum of first n terms of the given AP.
{tex}\frac { S _ { m } } { S _ { n } } = \frac { m ^ { 2 } } { n ^ { 2 } } \Rightarrow \frac { \frac { m } { 2 } [ 2 a + ( m - 1 ) d ] } { \frac { n } { 2 } [ 2 a + ( n - 1 ) d ] } = \frac { m ^ { 2 } } { n ^ { 2 } }{/tex}
{tex} \Rightarrow \frac { 2 a + ( m - 1 ) d } { 2 a + ( n - 1 ) d } = \frac { m } { n }{/tex}
{tex}\Rightarrow{/tex} 2an+mnd-nd= 2am+mnd-md
{tex}\Rightarrow{/tex} 2an-2am=nd-md
{tex}\Rightarrow{/tex} 2a(n - m) = d(n - m) {tex}\Rightarrow{/tex}2a=d...(i)
{tex}\therefore \quad \frac { T _ { m } } { T _ { n } } = \frac { a + ( m - 1 ) d } { a + ( n - 1 ) d } = \frac { a + ( m - 1 ) \cdot 2 a } { a + ( n - 1 ) \cdot 2 a }{/tex} [from (i)]
{tex}= \frac { a + 2 a m - 2 a } { a + 2 a n - 2 a } = \frac { 2 a m - a } { 2 a n - a } = \frac { a ( 2 m - 1 ) } { a ( 2 n - 1 ) } = \frac { 2 m - 1 } { 2 n - 1 }{/tex}.
Posted by Sujal Rai 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Let units digit and tens digit of the two digit number be 'x' and 'y' respectively.
{tex}\therefore{/tex} Number is 10y + x
According to question, The two digit number = 4 {tex}\times{/tex} The sum of the digits
{tex}\therefore{/tex}10y + x = 4(y + x)
10y + x = 4y + 4x
10y - 4y = 4x - x
6y = 3x
2y = x....(i)
Also, The two digit number = 3 {tex}\times{/tex} The produce of the digits
10y + x = 3xy......(ii)
From equation...(i) 2y=x
{tex}\therefore{/tex}10y + 2y = 3(2y)y
12y = 6y2
6y2 - 12y = 0
6y(y - 2) = 0
{tex}\therefore y=0 \quad or \quad y=2{/tex}
The number cannot be zero, {tex}\therefore{/tex} y=0 is not possible
{tex}\therefore y=2 \quad and \quad x=4{/tex}
{tex}\therefore{/tex} The required number is 24.
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Rounak Kumar 7 years, 1 month ago
1Thank You