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Ask QuestionPosted by Shabeeha Khadeeja 6 years, 10 months ago
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Posted by Shabeeha Khadeeja 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel be d km.
Since, Speed = {tex}\frac{{{\text{Distance travelled}}}}{{{\text{Time taken to travel that distance}}}}{/tex} {tex} \Rightarrow x = \frac{d}{t} \Rightarrow{/tex} d = xt ....(1)
According to the question,
x + 10 = {tex}\frac{d}{{t - 2}} \Rightarrow{/tex} (x + 10)(t - 2) = d
{tex}\Rightarrow{/tex} xt + 10t - 2x - 20 = d
{tex}\Rightarrow{/tex} -2x + 10t = 20 .....(2) [Using eq. (1)]
Again, x - 10 = {tex}\frac{d}{{t + 3}} \Rightarrow{/tex} (x - 10)(t + 3) = d
{tex}\Rightarrow{/tex} xt - 10t + 3x - 30 = d
{tex}\Rightarrow{/tex} 3x - 10t = 30 .....(3) [Using eq. (1)]
Adding equations (2) and (3), we obtain:
x = 50
Substituting the value of x in equation (2), we obtain:
(-2) {tex}\times{/tex} (50) + 10t = 20 {tex}\Rightarrow{/tex}-100 + 10t = 20
{tex}\Rightarrow{/tex}10t = 120
t = 12
From equation (1), we obtain:
d = xt = 50 {tex}\times{/tex} 12 = 600
Thus, the distance covered by the train is 600 km.
Posted by Shabeeha Khadeeja 6 years, 10 months ago
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【-】⭕ 【\】 € ¥ (@_@;) 6 years, 10 months ago
Posted by Shabeeha Khadeeja 6 years, 10 months ago
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Posted by Shabeeha Khadeeja 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Since {tex}\left(1,\frac p3\right){/tex}, is the mid point of the line segment joining the points (2,0) and {tex}\left( 0 , \frac { 2 } { 9 } \right){/tex}.
{tex}\therefore \quad \quad \frac { p } { 3 } = \frac { 0 + \frac { 2 } { 9 } } { 2 }{/tex}
{tex} \frac { 2 p } { 3 } = \frac { 2 } { 9 }{/tex}
{tex}p = \frac { 1 } { 3 }{/tex}
Now the given points are (-1, 3p). Let x = -1 , y = 3p
Hence, y = 3 {tex}\times{/tex} {tex}\frac13{/tex} = 1
The equation to prove is: 5x + 3y + 2 = 0
L.H.S.
= 5x + 3y + 2
Put the values of x & y,
= 5(-1) + 3(1) + 2 = -5 + 3 + 2 = -2 + 2 = 0 = R.H.S.
Hence, the line 5x + 3y + 2 = 0 passes through the point (–1, 1) as 5(–1) + 3(1) + 2 = 0.
Posted by Satyam Singh 6 years, 10 months ago
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【-】⭕ 【\】 € ¥ (@_@;) 6 years, 10 months ago
Posted by Lakshay Choudhary 6 years, 10 months ago
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Posted by Atul Chaudhary 6 years, 10 months ago
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Posted by Sw Kkk 6 years, 10 months ago
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Posted by Vikas Sharma 6 years, 10 months ago
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Posted by Sparsh Bansal 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago
Since cos A + cos2 A=1
Now,cosA= 1-cos2A=sin2A
Then taking L.H.S. sin2A+sin4A
=Sin2A(1+sin2A)
=CosA(1+cosA)
CosA+cos2A=1(given)
L.H.S =1 Hence proved
Posted by Ashi Mishra 6 years, 10 months ago
- 1 answers
Gaurav Seth 6 years, 10 months ago
Question: Show that n (m2-1 ) = 2m , if sin theta + cos theta = m and sec theta + cosec theta = n
Answer:

Posted by Sarthak Varshnay 6 years, 10 months ago
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Somu Sinha 6 years, 10 months ago
Posted by Neel Shah 6 years, 10 months ago
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Posted by Priya Darshni 6 years, 10 months ago
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Jogendra Nonia 6 years, 10 months ago
Posted by Kiran Yadav 6 years, 10 months ago
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Posted by Rakesh Pal 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
As, volume of hemisphere = 2425{tex}\frac { 1 } { 2 } c m ^ { 3 }{/tex}
{tex}\Rightarrow \frac { 2 } { 3 } \pi r ^ { 3 } = 2425 \frac { 1 } { 2 }{/tex}
{tex}\Rightarrow \frac { 2 } { 3 } \times \frac { 22 } { 7 } r ^ { 3 } = \frac { 4851 } { 2 }{/tex}
{tex}\Rightarrow r ^ { 3 } = \frac { 4851 \times 3 \times 7 } { 2 \times 2 \times 22 }{/tex}
{tex}\Rightarrow r ^ { 3 } = \frac { 441 \times 3 \times 7 } { 2 \times 2 \times 2 }{/tex}
{tex}\Rightarrow r ^ { 3 } = \frac { 21 ^ { 3 } } { 2 ^ { 3 } }{/tex}
{tex}\Rightarrow r = \frac { 21 } { 2 } c m{/tex}
So, the curved surface area of the hemisphere ={tex}2 \pi r ^ { 2 }{/tex}
{tex}= 2 \times \frac { 22 } { 7 } \times \frac { 21 } { 2 } \times \frac { 21 } { 2 } = 693 \mathrm { cm } ^ { 2 }{/tex}
Posted by Preeti Rahi 6 years, 10 months ago
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Posted by M Nehruji 6 years, 10 months ago
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Garima Choudhary 6 years, 10 months ago
Posted by Sw Kkk 6 years, 10 months ago
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Posted by @Saraswathi Gowda??? 6 years, 10 months ago
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Lucky? Dangi 6 years, 10 months ago
Posted by Lucky? Dangi 6 years, 10 months ago
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Pallavi Verma 6 years, 10 months ago
Posted by Ashish Sharma 6 years, 10 months ago
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Posted by Rishabh Banthia 6 years, 10 months ago
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Abhishek Rai 6 years, 10 months ago
Sw Kkk 6 years, 10 months ago
Posted by Ujjwal Kumar 6 years, 10 months ago
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Abhay Modi 6 years, 10 months ago

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Kamboj Ji 6 years, 10 months ago
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