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Rakesh Kumar Tiwari 7 years, 3 months ago
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Vaibhav Goel 7 years, 3 months ago
Posted by Radhika :) Mishra:} 7 years, 3 months ago
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Brahmani Sharma 7 years, 3 months ago
Shubham Carbon 7 years, 3 months ago
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Uttam Kumar 7 years, 3 months ago
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Sanjivani Gawade 7 years, 3 months ago
Rakesh Kumar Tiwari 7 years, 3 months ago
Posted by Mita H Shukla Shukla 7 years, 3 months ago
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Radhika :) Mishra:} 7 years, 3 months ago
Posted by Siddu Sundara 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
The given equation is:
{tex}\frac { x - 2 } { x - 3 } + \frac { x - 4 } { x - 5 } = \frac { 10 } { 3 }{/tex}
By cross multiplication method we have,
{tex} \frac { ( x - 2 ) ( x - 5 ) + ( x - 4 ) ( x - 3 ) } { ( x - 3 ) ( x - 5 ) } = \frac { 10 } { 3 }{/tex}
{tex}\Rightarrow \frac { \left( x ^ { 2 } - 7 x + 10 \right) + \left( x ^ { 2 } - 7 x + 12 \right) } { \left( x ^ { 2 } - 8 x + 15 \right) } = \frac { 10 } { 3 }{/tex}
{tex}\Rightarrow \frac { \left( 2 x ^ { 2 } - 14 x + 22 \right) } { \left( x ^ { 2 } - 8 x + 15 \right) } = \frac { 10 } { 3 }{/tex}
{tex}\Rightarrow{/tex} 3(2x2 - 14x + 22) = 10(x2 - 8x + 15) [ by cross multiplication method ]
{tex}\Rightarrow{/tex} 6x2 - 42x + 66 = 10x2 - 80x +150
{tex}\Rightarrow{/tex} 4x2 - 38x + 84 = 0 {tex}\Rightarrow{/tex} 2x2 - 19x + 42 = 0
{tex}\Rightarrow{/tex} 2x2 - 12x - 7x + 42 = 0 {tex}\Rightarrow{/tex} 2x(x -6 ) - 7(x - 6) = 0
{tex}\Rightarrow{/tex} (x - 6)(2x - 7) = 0 {tex}\Rightarrow{/tex} x - 6 = 0 or 2x - 7 = 0
Either x = 6 or x = {tex}\frac{7}{2}{/tex}.
Hence, the roots of given equation are 6 and {tex}\frac{7}{2}{/tex}.
Posted by Shreyas Jayaram 7 years, 3 months ago
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Radhika :) Mishra:} 7 years, 3 months ago
Posted by Vaijnath Mitkari 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
By using year by year calculation
S.I. on Rs. 18000 at 10% per annum for 1 year
{tex} = \frac{{18000 \times 10 \times 1}}{{100}} = Rs.1800{/tex}
∴ Amount at the end of 1st year
= Rs. 18000 + Rs. 1800
= Rs. 19800
= Principle for 2nd year.
S.I. on Rs. 19800 at 10% per annum for 1 year
{tex} = \frac{{19800 \times 10 \times 1}}{{100}}{/tex}
= Rs. 1980
∴ Amount at the end of 2nd year
= Rs. 19800 + Rs. 1980
= Rs. 21780
= Principle for 3rd year
S.I. on Rs. 21780 at 10% per annum for {tex}\frac{1}{2}{/tex} year
{tex} = \frac{{21780 \times 10 \times 1}}{{2 \times 100}}{/tex}
= Rs. 1089
∴ Amount at the end of {tex}2\frac{1}{2}{/tex} years
= Rs. 21780 + Rs. 1089
= Rs. 22869
this is the required amount.
Now,
C.I. = Rs. 22869 – Rs. 18000
= Rs. 4869.
Posted by Anjali Pal 7 years, 3 months ago
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Posted by Arun Banna 7 years, 3 months ago
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Yogita Ingle 7 years, 3 months ago
Cos 45 = 1/√2
Sin 30 = 1/2
Cosec 30 = 2
Cos45 Sin30 + Cosec 30 = (1/√2) (1/2) + 2 = 1/2√2 + 2 = (1 + √2)/2√2 =
Posted by Megha Kumari 7 years, 3 months ago
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Posted by Isha Pareek 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Check NCERT Solutions here : https://mycbseguide.com/ncert-solutions.html
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Yogita Ingle 7 years, 3 months ago
Sodium metal reacts rapidly with water to form a colourless solution of sodium hydroxide (NaOH) and hydrogen gas (H2).
2Na + 2H2O → 2NaOH + H2
Anjali Singh 7 years, 3 months ago
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Aditya Kumar Tiwary 7 years, 3 months ago

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Brahmani Sharma 7 years, 3 months ago
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