Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Aysha Shayan 5 years, 8 months ago
- 1 answers
Posted by Ashwani Yadav 6 years, 10 months ago
- 1 answers
Posted by Surendrapal Singh Singh Gaur 6 years, 10 months ago
- 3 answers
Surendrapal Singh Singh Gaur 6 years, 10 months ago
Posted by Abhishek Bandane 6 years, 10 months ago
- 1 answers
Achal Bharadwaj 6 years, 10 months ago
Posted by Quazi Aqmar 6 years, 10 months ago
- 1 answers
Honey ??? 6 years, 10 months ago
Posted by Aasiya Mulani 6 years, 10 months ago
- 0 answers
Posted by Abhishek Pandey 6 years, 10 months ago
- 0 answers
Posted by Ashwani Yadav 6 years, 10 months ago
- 0 answers
Posted by Reena No 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
A(1, -2), B(2, 3), C(a, 2) and D(-4, -3)are the vertices of a parallelogram.
{tex}\Rightarrow{/tex}AB = CD and AD = BC
Consider AD = BC
{tex}\Rightarrow{/tex}AD2 = BC2
{tex}\Rightarrow{/tex}(-4 -1)2 + (-3 + 2)2 = (a - 2)2 + (2 - 3)2
{tex}\Rightarrow{/tex}(-5)2 = (a - 2)2
{tex}\Rightarrow{/tex} a - 2 = -5
{tex}\Rightarrow{/tex} a = -3
Area of {tex}\Delta A B C = \frac { 1 } { 2 } | 1 ( 3 - 2 ) + 2 ( 2 + 2 ) - 3 ( - 2 - 3 ) |{/tex}
{tex}= \frac { 1 } { 2 } | 1 + 8 + 15 |{/tex}
{tex}= \frac { 1 } { 2 } \times 24{/tex}
= 12 sq. units.
Diagonal of a parallelogram divides it into two equal triangles.
Area of parallelogram ABCD = 2 {tex}\times{/tex}12=24 sq. units
{tex}\Rightarrow {/tex} AB {tex}\times{/tex}Height = 24
{tex}\Rightarrow {/tex} AB{tex}\times{/tex}Height = 24
{tex}\Rightarrow \left[ \sqrt { ( 2 - 1 ) ^ { 2 } + ( 3 + 2 ) ^ { 2 } } \right] \times \text { Height } = 24{/tex}
{tex}\Rightarrow [ \sqrt { 1 + 25 } ] \times \text { Height } = 24{/tex}
{tex}\Rightarrow \text { Height } = \frac { 24 } { \sqrt { 26 } } = \frac { 12 \times \sqrt { 2 } \times \sqrt { 2 } } { \sqrt { 13 } \times \sqrt { 2 } }{/tex}
{tex}= \frac { 12 \sqrt { 2 } } { \sqrt { 13 } }{/tex}
Posted by Nitesh Sanwal 6 years, 10 months ago
- 4 answers
Posted by Devika Sharma 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
Let n-1 and n be consecutive positive integers,Let P be their product
Then {tex}\style{font-family:Arial}{\style{font-size:12px}{\mathrm n(\mathrm n-1)=\mathrm n^2-1\;.................(1)}}{/tex}
We know that any positive integers is of the form 2q or 2q + 1, where q is a positive integer
Case I: When n = 2q, then
P=n2 - n = (2q)2 - 2q = 4q2 - 2q = 2q(2q - 1)-----(2)
Case II: When n = 2q + 1, then
P=n2 - n = (2q + 1)2 - (2q + 1)
= 4q2 + 4q + 1 - 2q - 1
= 4q2 + 2q
P = 2q(2q + 1)..........(3)
From (2) and (3) we conclude that
The product of n-1 and n is divisible by 2
Posted by Sabita Das 6 years, 10 months ago
- 3 answers
S.P Singh 6 years, 10 months ago
Let A(2,-3)is Mid-point of PQ.
Using Mid-point formula:
A(2,-3)= 6-2/2, b-2+4/2
Comparing y Co-ordinate
-3=b-2+4/2
-6=b+2
b=-6-2
b=-8
<font color="blue "> value of b is -8
Posted by Durga Appa 6 years, 10 months ago
- 4 answers
Shivam Tripathi 6 years, 10 months ago
Posted by Anju Thakur 6 years, 10 months ago
- 0 answers
Posted by Danish Warsi 6 years, 10 months ago
- 2 answers
Posted by Shivkami Pillai 6 years, 10 months ago
- 1 answers
Madhur Dubey ??? 6 years, 10 months ago
Posted by Sova Suravi 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
{tex}\frac{1}{2a + b + 2x}{/tex} = {tex}\frac{1}{2a}{/tex} + {tex}\frac{1}{b}{/tex} + {tex}\frac{1}{2x}{/tex}
{tex}\Rightarrow{/tex} {tex}\frac{1}{2a + b + 2x}{/tex} - {tex}\frac{1}{2x}{/tex} = {tex}\frac{1}{2a}{/tex} + {tex}\frac{1}{b}{/tex}
{tex}\Rightarrow{/tex}{tex}\frac { 2 x - 2 a - b - 2 x } { ( 2 a + b + 2 x ) ( 2 x ) }{/tex} = {tex}\frac{b + 2a}{2a \times b}{/tex}
{tex}\Rightarrow{/tex} {tex}\frac { - ( 2 a + b ) } { ( 2 a + b + 2 x ) 2 x }{/tex} = {tex}\frac{b + 2a}{2ab}{/tex}
{tex}\Rightarrow{/tex}{tex}\frac { - 1 } { 4 a x + 2 b x + 4 x ^ { 2 } }{/tex} = {tex}\frac{1}{2ab}{/tex}
{tex}\Rightarrow{/tex} {tex}4x^2 + 2bx + 4ax = -2ab{/tex}
{tex}\Rightarrow{/tex} {tex}4x^2 + 2bx + 4ax + 2ab = 0{/tex}
{tex}\Rightarrow{/tex} {tex}2x(2x + b) + 2a(2x + b) = 0{/tex}
{tex}\Rightarrow{/tex} (2x + b)(2x + 2a) = 0
{tex}\Rightarrow{/tex} x = -{tex}\frac{b}{2}{/tex} or x = -a
Posted by Ashwani Yadav 6 years, 10 months ago
- 2 answers
Ashwani Yadav 6 years, 10 months ago
Posted by Vishal Kumar 6 years, 10 months ago
- 1 answers
Ashwani Yadav 6 years, 10 months ago
Posted by Shaarif Siddiqui 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
According to the question,
L.H.S. = {tex}\frac { 1 } { ( \sec \theta - \tan \theta ) } - \frac { 1 } { \cos \theta }{/tex}
{tex}= \frac { 1 } { ( \sec \theta - \tan \theta ) } \times \frac { ( \sec \theta + \tan \theta ) } { ( \sec \theta + \tan \theta ) } - \sec \theta{/tex}
{tex}= \frac { ( \sec \theta + \tan \theta ) } { \left( \sec ^ { 2 } \theta - \tan ^ { 2 } \theta \right) } - \sec \theta{/tex}
= (sec{tex}\theta{/tex} + tan{tex}\theta{/tex}) - sec{tex}\theta{/tex} [{tex}\therefore{/tex} {tex}sec^2\theta - tan^2\theta = 1{/tex}]
= tan{tex}\theta{/tex}.
R.H.S. = {tex}\frac { 1 } { \cos \theta } - \frac { 1 } { ( \sec \theta + \tan \theta ) }{/tex}
{tex}= \sec \theta - \frac { 1 } { ( \sec \theta + \tan \theta ) } \times \frac { ( \sec \theta - \tan \theta ) } { ( \sec \theta - \tan \theta ) }{/tex}
= sec{tex}\theta{/tex} - (sec{tex}\theta{/tex} - tan{tex}\theta{/tex}) [{tex}\therefore{/tex}{tex} sec^2\theta - tan^2\theta = 1{/tex}]
= tan{tex}\theta{/tex}.
{tex}L.H.S. = R.H.S.{/tex}
Posted by Virender Bhardwaj 6 years, 10 months ago
- 0 answers
Posted by Ateek Khan 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
Let n =3k
then n + 2 = 3k + 2
and n + 4 = 3k + 4
Case 1: When n=3k ,n is divisible by 3 ............(1)
n + 2 = 3k + 2
or, n + 2 is not divisible by 3
n + 4 = 3k + 4
= 3(k + 1) + 1
or, n + 4 is not divisible by 3
Case 2:When n=3k+1, n is not divisible by 3
n + 2 = (3k + 1) + 2
=3k + 3 = 3(k + 1)
{tex} \Rightarrow{/tex} n+ 2 is clearly divisible by 3..........................(2)
n + 4 = (3k + 1) + 4
= 3k + 5
= 3(k + 1) + 2
{tex}\Rightarrow{/tex} n + 4 is not divisible by 3
Case 3:When n=3k+2,n is not divisible by 3
n + 2 = (3k + 2) + 2
= 3k + 4
(n + 2) is not divisible by 3
x + 4 = 3k + 6 = 3(k + 2)
{tex}\Rightarrow{/tex} n + 4 is divisible by 3........................(3)
Hence, from (1),(2) and (3) it is clear that exactly one of the numbers n, n + 2, n + 4, is divisible by 3.
Posted by Anmol Khanna 6 years, 10 months ago
- 2 answers
Posted by Chirayu . 6 years, 10 months ago
- 3 answers
Posted by Shriyanshi Gupta 6 years, 10 months ago
- 2 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
Pruthviraj Gurav 6 years, 10 months ago
4Thank You