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Ask QuestionPosted by Sayak Mondal 1 month ago

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Sayak Mondal 1 month ago

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Sayak Mondal 1 month ago

**If content is found in violation, the user posting this content will be banned for 30 days from using Homework help section.**Suspended users will receive error while adding question or answer. Question comments have also been disabled. Read community guidelines at https://mycbseguide.com/community-guidelines.html

**Few rules to keep homework help section safe, clean and informative.**

- Don't post personal information, mobile numbers and other details.
- Don't use this platform for chatting, social networking and making friends. This platform is meant only for asking subject specific and study related questions.
- Be nice and polite and avoid rude and abusive language. Avoid inappropriate language and attention, vulgar terms and anything sexually suggestive. Avoid harassment and bullying.
- Ask specific question which are clear and concise.

Remember the goal of this website is to share knowledge and learn from each other. Ask questions and help others by answering questions.

Sayak Mondal 1 month ago

**Few rules to keep homework help section safe, clean and informative.**

- Don't post personal information, mobile numbers and other details.
- Don't use this platform for chatting, social networking and making friends. This platform is meant only for asking subject specific and study related questions.
- Be nice and polite and avoid rude and abusive language. Avoid inappropriate language and attention, vulgar terms and anything sexually suggestive. Avoid harassment and bullying.
- Ask specific question which are clear and concise.

Remember the goal of this website is to share knowledge and learn from each other. Ask questions and help others by answering questions.

Sayak Mondal 1 month ago

**Few rules to keep homework help section safe, clean and informative.**

- Don't post personal information, mobile numbers and other details.
- Ask specific question which are clear and concise.

Remember the goal of this website is to share knowledge and learn from each other. Ask questions and help others by answering questions.

Posted by Kajal Verma 2 months, 1 week ago

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Posted by Tanushri Bhayani 11 months, 2 weeks ago

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Preeti Dabral 11 months, 2 weeks ago

The effective rate is the actual rate compounded annually. Therefore, the required sum S is given by

S = 12000 (1 + {tex}\frac{3}{100}{/tex})^{10} (1 + {tex}\frac{4}{100}{/tex})^{4} (1 + {tex}\frac{5}{100}{/tex})^{2}

{tex}\Rightarrow{/tex} S = 12000 (1.03)^{10} (1.04)^{4} (1.05)^{2}

{tex}\Rightarrow{/tex} S = 12000 (1.34391638) (1.16985856) (1.1025) = 20800.10

Hence, the amount is ₹20,800.10

Posted by Tanushri Bhayani 1 year ago

- 0 answers

Posted by Tanushri Bhayani 1 year ago

- 2 answers

Nishant Dogra 1 year ago

Posted by Bhumika Jain 1 year ago

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Posted by Tanushri Bhayani 1 year ago

- 2 answers

Nishant Dogra 1 year ago

Posted by Tanushri Bhayani 1 year ago

- 2 answers

Tanushri Bhayani 1 year ago

Posted by Tanushri Bhayani 1 year ago

- 2 answers

Preeti Dabral 1 year ago

Let the number of guests be x

If x > 100, then the amount received by the company will be-

{4800 - 200/10(x - 100)}x

Now, we need to maximize this amount so let us differentiate with respect to x as follows-

P = 4800x - 20(x - 100)x

P = 4800x - 20x² + 2000x

P = 6800x - 20x²

dP/dx = 6800 - 40x

differentiating again-

d²P/dx² = -40

As d²P/dx² > 0, the maximum value will occur at dP/dx = 0

⇒ 6800 - 40x = 0

40x = 6800

x = 6800/40

x = 170

Now, the amount received will be = 6800(170) - 20(170)²

= 1156000 - 578000

= 578000

Thus, the maximum amount that the company will receive is 5,78,000 when the number of guests are 170.

Posted by Tanushri Bhayani 1 year ago

- 2 answers

Preeti Dabral 1 year 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 1 year ago

- 2 answers

Preeti Dabral 1 year 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} + C_{1}

or P = {tex}e^{\frac{t}{20}} \cdot e^{\mathrm{C}_{1}}{/tex}

or P = {tex}\mathrm{C} e^{\frac{t}{20}}{/tex} (where e^{C1 }= 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 e^{t/20}

Let t years be the time required to double the principal. Then 2000 = 1000 e^{t/20} {tex}\Rightarrow{/tex} t = 20 log_{e}2.

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Sayak Mondal 1 month ago

If content is found in violation, the user posting this content will be banned for 30 days from using Homework help section.Suspended users will receive error while adding question or answer. Question comments have also been disabled. Read community guidelines at https://mycbseguide.com/community-guidelines.htmlFew rules to keep homework help section safe, clean and informative.Remember the goal of this website is to share knowledge and learn from each other. Ask questions and help others by answering questions.

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