Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Giaa Manek 6 years, 10 months ago
- 4 answers
Safeeq Ahamed 6 years, 10 months ago
S N Mahato 6 years, 10 months ago
Posted by Mahek Agrawal 6 years, 10 months ago
- 4 answers
Mahek Agrawal 6 years, 10 months ago
Posted by Rohan Sarkar ⚽⚽ 6 years, 10 months ago
- 2 answers
S N Mahato 6 years, 10 months ago
Posted by Antia Singh 6 years, 10 months ago
- 2 answers
Purvanshi Yadav 6 years, 10 months ago
Lionel Messi⚽️ 6 years, 10 months ago
Posted by Gulpreet Kaur 6 years, 10 months ago
- 3 answers
Amisha Sharma ? 6 years, 10 months ago
Posted by Gulpreet Kaur 6 years, 10 months ago
- 0 answers
Posted by Shub Tyagi 6 years, 10 months ago
- 2 answers
Ram Kushwah 6 years, 10 months ago
The pair of eqn does not have any solution if
{tex}\begin{array}{l}\frac{a1}{a2}=\frac{b1}{b2}\neq\frac{c1}{c2}\\\mathrm{Putting}\;\mathrm{all}\;\mathrm{values}\;\mathrm{we}\;\mathrm{get}\\\frac23=\frac{-\mathrm k}2\neq\frac{-3}{-1}\\\;\mathrm{or}\;-3\mathrm k=4\\\mathrm k=-\frac43\\\\\end{array}{/tex}
Ram Krishna Chouksey 6 years, 10 months ago
Posted by Ayush .Jhagta 6 years, 10 months ago
- 4 answers
Ram Kushwah 6 years, 10 months ago
Here A3= a+2d=16.....(1)
A17-A5=(a+16d)-(a+4d)=12
or 12d=12
Hence d=1
from (1)
a+2=16, a=14
Thus the series is 14,15,16,17,18.......
Drashti Vaish 6 years, 10 months ago
Drashti Vaish 6 years, 10 months ago
Posted by Aman Maurya 6 years, 10 months ago
- 3 answers
Posted by Piyush Kumar 6 years, 10 months ago
- 2 answers
Puja Sahoo? 6 years, 10 months ago
Gaurav Seth 6 years, 10 months ago
CosA +sinA = √2cosA
squaring
⇒(cosA + sin A)² = (√2cosA)²
⇒cos²A + sin²A + 2sinAcosA = 2cos²A
⇒1 - sin²A + 1 - cos²A + 2sinAcosA = 2cos²A
⇒2 - 2cos²A = cos²A + sin²A - 2sinAcosA
⇒2(1 - cos²A)= (cosA - sinA)²
⇒ cosA - sinA = √[2sin²A]
⇒cosA-sinA = √2sinA
Posted by Jayesh Patil 6 years, 10 months ago
- 1 answers
Gaurav Seth 6 years, 10 months ago
Given, equation is:
(c 2 – ab) x 2 – 2 (a 2 – bc) x + (b 2 – ac) = 0
To prove: a = 0 or a 3 + b 3 + c 3 = 3abc
Proof: From the given equation, we have
a = (c2 – ab)
b = –2 (a 2 – bc)
c = (b 2 – ac)
It is being given that equation has real and equal roots
∴ D = 0
⇒ b 2 – 4ac = 0
On substituting respective values of a, b and c in above equation, we get
[–2 (a 2 – bc)]2 – 4 (c 2 – ab) (b 2 – ac) = 0
4 (a 2 – bc)2 – 4 (c 2 b 2 – ac 3 – ab 3 + a 2 bc) = 0
4 (a 4 + b 2 c 2 – 2a 2 bc) – 4 (c 2 b 2 – ac 3 – ab 3 + a 2 bc) = 0
⇒ a 4 + b 2 c 2 – 2a 2 bc – b 2 c 2 + ac 3 + ab 3 – a 2 bc = 0
⇒ a 4 + ab 3 + ac 3 –3a 2 bc = 0
⇒ a [a 3 + b 3 + c 3 – 3abc] = 0
⇒a = 0 or a 3 + b 3 + c 3 = 3abc
Posted by Abhijot Matharoo 6 years, 10 months ago
- 1 answers
Jayesh Patil 6 years, 10 months ago
Posted by Raj Oraon 6 years, 10 months ago
- 2 answers
Posted by Abhishek Maran 5 years, 8 months ago
- 2 answers
Puja Sahoo? 6 years, 10 months ago
Posted by Aman Pandey 6 years, 10 months ago
- 1 answers
Gaurav Seth 6 years, 10 months ago
The rational number 7 / 75,can not be written in form 2m ,5n .So it is non-terminating repeating decimal expansion
Posted by Royal Ssoy Alok 6 years, 10 months ago
- 3 answers
Posted by Abhishek Jain 6 years, 10 months ago
- 2 answers
Lol No 6 years, 10 months ago
Posted by Yash Prajapati 6 years, 10 months ago
- 2 answers
Amrish Chaubey 6 years, 10 months ago
@ Aashu 6 years, 10 months ago
Posted by Rajendra Yogi 6 years, 10 months ago
- 0 answers
Posted by Kajal Dogra 6 years, 10 months ago
- 3 answers
Sidharth Jha 6 years, 10 months ago
Shiva Kumar 6 years, 10 months ago
Posted by Kundan Sharma 6 years, 10 months ago
- 4 answers
Anuj Jain 6 years, 10 months ago
Posted by Syed Mohd Hussain 6 years, 10 months ago
- 0 answers
Posted by Yashwant Kumar 6 years, 10 months ago
- 1 answers
Posted by Harshita Swami 6 years, 10 months ago
- 3 answers
Posted by Rajput Singh 6 years, 10 months ago
- 3 answers
Geetanand Yadav 6 years, 10 months ago
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
Ram Kushwah 6 years, 10 months ago
{tex}\begin{array}{l}48=16\;\times3=2^4\times3\\72=8\times9=2^3\times3^2\\108=4\times27=2^2\times3^3\\LCM=2^4\times3^3=16\times27=432\end{array}{/tex}
Lights will change simultaneous after 432 sec
0Thank You