Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Kamlesh Vijay 6 years, 4 months ago
- 0 answers
Posted by Kamlesh Vijay 6 years, 4 months ago
- 0 answers
Posted by Vidyanshu Dubey 6 years, 4 months ago
- 0 answers
Posted by Yare Yare Dazze 6 years, 4 months ago
- 1 answers
Chitra Verma 6 years, 4 months ago
Posted by Yogesh Pandey 6 years, 4 months ago
- 4 answers
Posted by Ajit Naik 6 years, 4 months ago
- 0 answers
Posted by Yash Panchal 6 years, 4 months ago
- 2 answers
Ayush Vishwakarma?? 6 years, 4 months ago
Posted by Akansha Singh 6 years, 4 months ago
- 0 answers
Posted by Diksha Garg 6 years, 4 months ago
- 0 answers
Posted by Pratham Gupta 6 years, 4 months ago
- 1 answers
Karan Kakati 6 years, 4 months ago
Posted by Ashish Kumar Mandal 6 years, 4 months ago
- 1 answers
Gaurav Seth 6 years, 4 months ago
Let A1, A2, A3, A4, and A5 be five numbers between 8 and 26 such that
8, A1, A2, A3, A4, A5, 26 is an A.P.
Here, a = 8, b = 26, n = 7
Therefore, 26 = 8 + (7 – 1) d
⇒ 6d = 26 – 8 = 18
⇒ d = 3
A1 = a + d = 8 + 3 = 11
A2 = a + 2d = 8 + 2 × 3 = 8 + 6 = 14
A3 = a + 3d = 8 + 3 × 3 = 8 + 9 = 17
A4 = a + 4d = 8 + 4 × 3 = 8 + 12 = 20
A5 = a + 5d = 8 + 5 × 3 = 8 + 15 = 23
Thus, the required five numbers between 8 and 26 are 11, 14, 17, 20, and 23.
Posted by Vivek Patidar 6 years, 4 months ago
- 0 answers
Posted by Karan Kakati 6 years, 4 months ago
- 4 answers
Jheel Tippan 6 years, 4 months ago
#Aditi~ Angel???? 6 years, 4 months ago
Posted by Renu Yadav 6 years, 4 months ago
- 1 answers
Posted by Ankit Yadav 6 years, 4 months ago
- 6 answers
Priyanshu Kumar 6 years, 4 months ago
Posted by Yuvika Gupta 4 years, 4 months ago
- 0 answers
Posted by Mudit Mahar 6 years, 4 months ago
- 0 answers
Posted by Snighdha Das 6 years, 4 months ago
- 0 answers
Posted by Ashish Jaiswal 6 years, 4 months ago
- 4 answers
Yuvika Gupta 6 years, 4 months ago
Posted by Srini Kketan 6 years, 4 months ago
- 1 answers
Posted by Prabhudayal Katre 6 years, 4 months ago
- 0 answers
Posted by Jheel Tippan 6 years, 4 months ago
- 0 answers
Posted by Dikshit Jbl 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
There are 8 letters in the word DAUGHTER including 3 vowels and 5 consonants. We have to select 2 vowels out of 3 vowels and 3 consonants of 5 consonants.
{tex}\therefore {/tex} Number of ways of selection {tex}{ = ^3}{C_2}{ \times ^5}{C_3} = 3 \times 10 = 30{/tex}
Now each word contains 5 letters which can be arranged among themselves in 5! Ways.
So total number of words {tex} = 5! \times 30 = 120 \times 30 = 3600{/tex}
Posted by Ashish Gautam 6 years, 4 months ago
- 0 answers
Posted by Kamal Sahni 6 years, 4 months ago
- 0 answers
Posted by Mohit Khandelwal 6 years, 4 months ago
- 2 answers
Kamal Sahni 6 years, 4 months ago
Yogita Ingle 6 years, 4 months ago
Definition 1: If two sets A and B have same cardinality, if there exists a objective function from Set A to B.
Definition 2: Two sets A and B are said to be equivalent if they have the same cardinality i.e n(A) = n(B)
A = {1, 2, 3} Here n(A) = 3
B = {p, q, r} Here n(B) = 3
Therefore, A ↔ B
Posted by Ajay Sharma 6 years, 4 months ago
- 0 answers
Posted by Mohit Yadav 6 years, 4 months ago
- 1 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
