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Ask QuestionPosted by Palai Taposwini 7 years, 2 months ago
- 0 answers
Posted by Xyz Z 7 years, 2 months ago
- 1 answers
Yogita Ingle 7 years, 2 months ago
x2 - 3x + 2 = 0
x2 - 2x - 1x + 2= 0
x(x - 2) - 1( x - 2) = 0
(x -1) ( x -2) = 0
x - 1 = 0 or x - 2 = 0
x = 1 or x = 2
Posted by Diwakar Kumar 7 years, 2 months ago
- 1 answers
Posted by Diwakar Kumar 7 years, 2 months ago
- 0 answers
Posted by Rohan Kumar 7 years, 2 months ago
- 2 answers
Posted by Sonadhan Chakma 7 years, 2 months ago
- 0 answers
Posted by Kiran Yadav 7 years, 2 months ago
- 2 answers
Diksha Y 7 years, 2 months ago
Posted by Shubham Shankar 7 years, 2 months ago
- 4 answers
Kritika Mishra 7 years, 2 months ago
Posted by Vikash Kumar 7 years, 2 months ago
- 2 answers
Anshika Mittal 7 years, 2 months ago
Posted by Sp Mishra 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
Given, {tex}\theta{/tex} = 30° and r = 4 cm
Area of sector OAPB ={tex}\frac { \theta } { 360 } \times \pi r ^ { 2 }{/tex}
Let 'A' be the area of corresponding major sector.
Then, A = Area of sector OAQB
{tex}\Rightarrow{/tex}A = Area of the circle - Area of the corresponding minor sector
{tex}\Rightarrow A = \pi r ^ { 2 } - \frac { \theta } { 360 } \times \pi r ^ { 2 }{/tex}
{tex}\Rightarrow A = \pi r ^ { 2 } \left( 1 - \frac { \theta } { 360 } \right){/tex}
{tex}\Rightarrow A = 3.14 \times 4 \times 4 \left( 1 - \frac { 30 } { 360 } \right) \mathrm { cm } ^ { 2 }{/tex}
{tex}\Rightarrow A = 3.14 \times 4 \times 4 \times \frac { 11 } { 12 } \mathrm { cm } ^ { 2 }{/tex}{tex}= \frac { 3.14 \times 44 } { 3 } \mathrm { cm } ^ { 2 }{/tex}= 46.05 cm2
Posted by Piyush Giri Goswami 7 years, 2 months ago
- 0 answers
Posted by Shivani Deshmukh 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
Given: In ABC, {tex}\frac { \mathrm { AP } } { \mathrm { BP } } = \frac { \mathrm { AQ } } { \mathrm { QC } }{/tex}
To prove: PQ {tex}\|{/tex} BC
Construction: Let PE {tex}\|{/tex} BC(construction).
{tex}\therefore \quad \frac { \mathrm { AP } } { \mathrm { PB } } = \frac { \mathrm { AE } } { \mathrm { EC } }{/tex} ..(i)
Also, {tex}\frac { \mathrm { AP } } { \mathrm { BP } } = \frac { \mathrm { AQ } } { \mathrm { QC } }{/tex} [given (ii)]
From (i) and (ii) {tex}\frac { \mathrm { AE } } { \mathrm { EC } } = \frac { \mathrm { AQ } } { \mathrm { QC } } \Rightarrow{/tex} E and Q coincides
{tex}\therefore{/tex} PQ {tex}\|{/tex} BC.
Posted by Aditya Tiwari 5 years, 8 months ago
- 1 answers
Yogita Ingle 7 years, 2 months ago
3(median) = 2(mean) + mode
Let the mean be 'x'
So, median will be x + 3
Then, according to the question
3(x+3) = 2(x) + mode
3x + 9 = 2x + mode
3x - 2x + 9 - mode
x+9 = mode
Now, mode - mean
⇒ x+9 - x
= 9
So, mode exceeds the mean by 9.
Posted by Aditya Tiwari 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
Value of tan A increases when A increases from {tex}0^o{/tex} to {tex}90^o{/tex}{tex}(0,\frac{1}{\sqrt{3}},1,\sqrt{3},{/tex} not defined at {tex}90^{\circ}{/tex})
Posted by Raushan Kumar 7 years, 2 months ago
- 2 answers
Anshika Mittal 7 years, 2 months ago
Anshika Mittal 7 years, 2 months ago
Posted by Pankaj Chauhan 7 years, 2 months ago
- 4 answers
Posted by Jasmeet Kaur 7 years, 2 months ago
- 3 answers
Posted by Aniket Dalmia 7 years, 2 months ago
- 4 answers
Posted by Abhishek Dwivedi 7 years, 2 months ago
- 5 answers
Pallavi Balotiya 7 years, 2 months ago
Devanshi Grover 7 years, 2 months ago
Posted by Arun Dip 7 years, 2 months ago
- 3 answers
Saurabh Singh Harariya 7 years, 2 months ago
Posted by Abhi Rajput 7 years, 2 months ago
- 3 answers
Anjali Rastogi 7 years, 2 months ago
Posted by Ashish Ashish 7 years, 2 months ago
- 2 answers
Khushal Rathore 7 years, 2 months ago
Posted by Legend Raj 7 years, 2 months ago
- 1 answers
Hardik Panwar 7 years, 2 months ago
Posted by Gokulram Jeyaraman 7 years, 2 months ago
- 1 answers
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