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Ask QuestionPosted by Jyoti Shinde 2 years, 11 months ago
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Na. Shakthi 2 years, 11 months ago
Posted by Ritik Sharma 2 years, 11 months ago
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Preeti Dabral 2 years, 11 months ago
In non-linear editing, edits are specified and modified by specialized software. A pointer-based playlist, effectively an edit decision list (EDL), for video or a directed acyclic graph for still images is used to keep track of edits.
Posted by Tushar Goswami 2 years, 11 months ago
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Abhishek Meena Saxena 2 years, 11 months ago
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Na. Shakthi 2 years, 11 months ago
Posted by Jyoti Shinde 2 years, 11 months ago
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Na. Shakthi 2 years, 11 months ago
Tushar Goswami 2 years, 11 months ago
Posted by Shivam Malhotra 2 years, 11 months ago
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Posted by Amit Rawat 2 years, 11 months ago
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Preeti Dabral 2 years, 11 months ago
माल्थस के अनुसार, 'जनसंख्या दोगुनी रफ्तार (1, 2, 4, 8, 16, 32) से बढ़ती है, जबकि संसाधनों में सामान्य गति (1, 2, 3, 4, 5) से ही वृद्धि होती है। परिणामतः प्रत्येक 25 वर्ष बाद जनसंख्या दोगुनी हो जाती है।
Posted by Manshi Bansal 2 years, 11 months ago
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Posted by Manshi Bansal 2 years, 11 months ago
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Ashish Kumar 2 years, 11 months ago
Posted by Malik Ujaiv 2 years, 11 months ago
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Punya Yadav 2 years, 11 months ago
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Posted by Malik Ujaiv 2 years, 11 months ago
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Diana Chaudhary 2 years, 11 months ago
Posted by Monu Dahiya 2 years, 11 months ago
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Mayank Prajapati 2 years, 11 months ago
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Posted by Aarzoo Saini 2 years, 11 months ago
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Posted by Aarzoo Saini 2 years, 11 months ago
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Posted by Tanushri Bhayani 2 years, 11 months ago
- 2 answers
Preeti Dabral 2 years, 11 months ago
Distance covered by a boat in 5 hours = 36 km
Rate upstream of boat {tex}=\frac{36}{5}{/tex} 7.2 kmph
Speed of the stream = 2.4 kmph
{tex}\therefore{/tex} Speed of the boat in still water
= (7.2 + 2.4) kmph
= 9.6 kmph
{tex}\therefore{/tex} Rate downstream of the boat
= (9.6 + 2.4) kmph
= 12 kmph
{tex}\therefore{/tex} Time taken in covering 78 km distance
{tex}=\frac{78}{12}{/tex}
= 6.5 hours or 6 hour 30 minutes.
Posted by Tanushri Bhayani 2 years, 11 months ago
- 2 answers
Preeti Dabral 2 years, 11 months ago
Let P be the principal at any time t. According to the given problem, {tex}\frac{d p}{d t}=\left(\frac{5}{100}\right) \times \mathrm{P}{/tex}
or {tex}\frac{d p}{d t}=\frac{\mathrm{P}}{20}{/tex} ...(i)
separating the variables in equation (i), we get
{tex}\frac{d p}{\mathrm{P}}=\frac{d t}{20}{/tex} ...(ii)
Integrating both sides of equation (ii), we get
log P = {tex}\frac{t}{20}{/tex} + C1
or P = {tex}e^{\frac{t}{20}} \cdot e^{\mathrm{C}_{1}}{/tex}
or P = {tex}\mathrm{C} e^{\frac{t}{20}}{/tex} (where eC1 = C) ...(iii)
Now, P = 1000, when t = 0
Substituting the values of P and t in (iii), we get C = 1000. Therefore, equation (iii), gives
P = 1000 et/20
Let t years be the time required to double the principal. Then 2000 = 1000 et/20 {tex}\Rightarrow{/tex} t = 20 loge2.
Posted by Barun Saha 8 months, 1 week ago
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Posted by Sahil Jakhar 3 years ago
- 3 answers
Ansil Mathew 2 years, 11 months ago
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Ansil Mathew 2 years, 11 months ago
Posted by Sapanarai Sapanarai 3 years ago
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Posted by Jyoti Shinde 3 years ago
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Bava Samyuka 2 years, 11 months ago
Suryansh Srivastav 3 years ago
Posted by Ravi Gurung 3 years ago
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Aarzoo Saini 2 years, 11 months ago
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Posted by Charan Teja 3 years ago
- 5 answers
Mannat Thukral 2 years, 9 months ago
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Ayush Singh 2 years, 11 months ago
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Satyam Pratap 2 years, 11 months ago
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