No products in the cart.

Ask questions which are clear, concise and easy to understand.

Ask Question
This content has been hidden. One or more users have flagged this content as inappropriate. Once content is flagged, it is hidden from users and is reviewed by myCBSEguide team against our Community Guidelines. 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.
  • 5 answers

Sayak Mondal 1 month ago

This content has been hidden. One or more users have flagged this content as inappropriate. Once content is flagged, it is hidden from users and is reviewed by myCBSEguide team against our Community Guidelines. 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

This content has been hidden. One or more users have flagged this content as inappropriate. Once content is flagged, it is hidden from users and is reviewed by myCBSEguide team against our Community Guidelines. 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

This content has been hidden. One or more users have flagged this content as inappropriate. Once content is flagged, it is hidden from users and is reviewed by myCBSEguide team against our Community Guidelines. 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

This content has been hidden. One or more users have flagged this content as inappropriate. Once content is flagged, it is hidden from users and is reviewed by myCBSEguide team against our Community Guidelines. 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

This content has been hidden. One or more users have flagged this content as inappropriate. Once content is flagged, it is hidden from users and is reviewed by myCBSEguide team against our Community Guidelines. 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.
  • 0 answers
  • 0 answers
  • 0 answers
  • 0 answers
  • 0 answers
  • 1 answers

Mahitesh Rawat 7 months, 2 weeks ago

Continuity ?
  • 1 answers

Tanushri Bhayani 11 months, 3 weeks ago

Kindly provide complete question !!
  • 1 answers

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

  • 2 answers
Ohh' alright

Nishant Dogra 1 year ago

Using the properties of matrix algebra, we can expand the expression (I + A)3 as follows: (I + A)3 = (I + A)(I + A)(I + A) = (I + A)(I2 + 2IA + A2) = (I + A)(I + 2A + A) = I2 + 3IA + 3A2 + A3 = I + 3A + 3A2 + A3 (since A2 = A) Substituting this expression into the original equation, we get: (I + A)3 - 7A = (I + 3A + 3A2 + A3) - 7A = I + 2A + 3A2 - 7A = I - 5A + 3A2 Therefore, (I + A)3 - 7A is equal to I - 5A + 3A2.
  • 2 answers
Hmm' got it 👍 thanks

Nishant Dogra 1 year ago

Solution We can calculate the product AB as follows: AB = [[-i, 0], [0, i]] [[1, 0], [0, -1]] = [[-i, 0], [0, -i]] Next, we can calculate the determinant of AB as follows: det(AB) = (-i) * (-i) - 0 * 0 (determinant of a 2x2 matrix) = 1 Now, we can use the formula for the product of determinants to find the determinant of A multiplied by the determinant of (i^2-1)B: det(A) * det((i^2-1)B) = det(AB) det(A) * (i^2-1)^2 * det(B) = 1 Substituting the known values, we get: det(A) * (i^2-1)^2 * (-1/1) = 1 det(A) * 4 = 1 det(A) = 1/4 Since A is a 2x2 matrix, its determinant can be calculated as follows: det(A) = (-i) * i - 0 * 0 (determinant of a 2x2 matrix) = -1 Therefore, we have a contradiction: det(A) = -1 and det(A) = 1/4. Hence, there is no value of a that satisfies the given conditions.
  • 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.

Thank you bro
  • 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.

Thanks a lot dude 😇
  • 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} + 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.

Thames bro 😘

myCBSEguide App

myCBSEguide

Trusted by 1 Crore+ Students

Test Generator

Test Generator

Create papers online. It's FREE.

CUET Mock Tests

CUET Mock Tests

75,000+ questions to practice only on myCBSEguide app

Download myCBSEguide App