No products in the cart.

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

Ask Question
  • 1 answers

Naitik Nama 1 year, 9 months ago

(2,1), (p,−1) & (−1,3) are collinear  ⇒p−2−1−1​=−1−p3−(−1)​ ⇒p−2−2​=−1−p4​ ⇒2(1+p)=4(p−2)⇒2+2p=4p−8                                                 ⇒10−2p        ∴   p=5
  • 3 answers

Priyanshu Singh 1 year, 10 months ago

assume that √5 is rational number √5=a/b (here 'a' and 'b' are co prime) b√5= a Square on both sides (b√5)^2 =(a)^2 5b^2 = a^2.............(i) If a^2 is divisible by 5 than it means that 5 is also divisible by a. Now, let there be another integer 'c' therefore, a=5c square on both side a^2 =25c^2 5b^2 =25c^2 {from equation (i)} then, b^2 =5 c^2 b^2/5 = c^2 If b^2 is divisible by 5 than it means b is also divisible by 5 . Therefore 'a' and 'b' has common factor 5 . But , we ensure that 'a' and 'b' are co prime. Therefore, our assumption is wrong that √5 is rational number. Contradiction occurs. Therefore √5 is irrational numbers.

Gunpreet Kaur 1 year, 10 months ago

lets assume that √5 is rational number √5=a/b (here 'a' and 'b' are co prime) b√5= a Square on both sides (b√5)^2 =(a)^2 5b^2 = a^2.............(i) If a^2 is divisible by 5 than it means that 5 is also divisible by a. Now, let there be another integer 'c' therefore, a=5c square on both side a^2 =25c^2 5b^2 =25c^2 {from equation (i)} then, b^2 =5 c^2 b^2/5 = c^2 If b^2 is divisible by 5 than it means b is also divisible by 5 . Therefore 'a' and 'b' has common factor 5 . But , we ensure that 'a' and 'b' are co prime. Therefore, our assumption is wrong that √5 is rational number. Contradiction occurs. Therefore √5 is irrational numbers.

Gunpreet Kaur 1 year, 10 months ago

lets assume that √5 is rational number √5=a/b (here 'a' and 'b' are co prime) b√5= a Square on both sides (b√5)^2 =(a)^2 5b^2 = a^2.............(i) If a^2 is divisible by 5 than it means that 5 is also divisible by a. Now, let there be another integer 'c' therefore, a=5c square on both side a^2 =25c^2 5b^2 =25c^2 {from equation (i)} then, b^2 =5 c^2 b^2/5 = c^2 If b^2 is divisible by 5 than it means b is also divisible by 5 . Therefore 'a' and 'b' has common factor 5 . But , we ensure that 'a' and 'b' are co prime. Therefore, our assumption is wrong that √5 is rational number. Contradiction occurs. Therefore √5 is irrational numbers.
  • 2 answers

Dhwani Panchal 1 year, 9 months ago

SecA+tanA = 1- sinA

Sriya Sharma 1 year, 9 months ago

Cos A/ 1+ sinA Now put cos A as (1- sinA) And put 1+ sinA as ( cos A ) Then it will be (1/ cosA - sinA /cos A) Since 1/cos A is (secA ) And SinA / cosA is( tan A ) Therefore (secA - Tan A ) hence proved ...
  • 2 answers

Gunpreet Kaur 1 year, 10 months ago

AP- 2,7,12,17,22.......227 an=227 a=2 d= 5 , n=? an= a+ (n-1) d 227= 2+ (n-1)(5) 227-2= (n-1)(5) 225/5 = (n-1) 45 +1 =n 46 = n

Ayushi Angra 1 year, 10 months ago

Reverse the series...so that a=227..and then solve using formula an=a+(n-1)d
  • 1 answers

Raghav Jindal 1 year, 10 months ago

Volume=4πr³/3 4×22/7×35/10×35/10×35/30 179.67mm³
  • 4 answers

Abirami N 1 year, 9 months ago

60

Meet Jadhwani 1 year, 10 months ago

HCF OF 15=3×5 HCF of 20=2×2×5 HCF OF 15,20=5 LCM OF 15,20=60

Pritam Panghal 1 year, 10 months ago

15= 3 × 5 20 = 2×2×5 5 is common in both then, HCF = 5 LCM is product of HCF and other numbers LCM = 5×2×2×3 LCM = 60

Shubham Pathak 1 year, 10 months ago

60
  • 1 answers

Priyanshu Singh 1 year, 10 months ago

Ncert ya rs Aggarwal
  • 2 answers

Raghav Jindal 1 year, 10 months ago

Circumference of circle=2πr 4π=2πr 2=r Similarly, 8π=2πr 4=r Area = πr² π×4×4 16π unit²

Gurleen Dhillon 1 year, 10 months ago

2πr=4π πr=4π2/2 πr=2π Put the values =r=2 Area of circle πr2 Put values Ans -88/7
  • 2 answers

Yatharth Yadav 1 year, 10 months ago

Bhosale🤣🤣🤣

Raghav Jindal 1 year, 10 months ago

2.23606798
  • 3 answers

Shubham Pathak 1 year, 10 months ago

√3 sinΦ = cosΦ => tanΦ = √3 Φ = 60° ,then 3cos²+ 2 cos/3cos +2 => 3 * 1/4 + 2* 1/2 // 3* 1/2 +2 => 3/4 + 1 // 3/2 +2 => 7/4 * 2/7 = 1/2

Vedant Bhosale 1 year, 10 months ago

Wrong spelling

Vedant Bhosale 1 year, 10 months ago

Byr
  • 2 answers

Pooja Naval 1 year, 9 months ago

Same search 🔍 on Google you definitely find it

Vipin Khola 1 year, 10 months ago

Ufnk
  • 1 answers

Naitik Nama 1 year, 9 months ago

Let's solve your equation step-by-step. �+1�−1+�−2�+2=4−(2�+3�−2) 2�2+�−1�=−2�2+6�−3� Step 1: Multiply both sides by x. 2�2+�−1=−2�2+6�−3 Step 2: Multiply both sides by x. 2�3+�2−�=−2�3+6�2−3� 2�3+�2−�−(−2�3+6�2−3�)=−2�3+6�2−3�−(−2�3+6�2−3�)(Subtract -2x^3+6x^2-3x from both sides) 4�3−5�2+2�=0 �(4�2−5�+2)=0(Factor left side of equation) �=0or4�2−5�+2=0(Set factors equal to 0) �=0 Check answers. (Plug them in to make sure they work.) �=0(Doesn't work in original equation)
  • 3 answers

Rohan Baisoya 1 year, 10 months ago

20th term is200
Here, S14 = 1050, n = 14, a = 10. We know that Sn = (n2)[2a+(n−1)d] Substituting the values we have, 1050=(142)[20+13d] ⇒1050=140+91d ⇒910=91d ⇒d=10 Therefore, a20=10+(20–1)×10=200 i.e. 20th term is 200.

D.Thrishanth Reddy 1 year, 10 months ago

200 is the 20th term
  • 4 answers

Piyush Sahani 1 year, 10 months ago

24•24-20•20=√176=14

Aditya Bansal 1 year, 10 months ago

26 cm ( by using Pythagoras theorem )
Here, AB = 20cm and OC = 24cm Now, AC=CB=2AB​=10cm (perpendicular drawn from centre of the circle to the chord,bisects the chord) Then using Pythagorean Theorem: OA=(AC2+OC2)​=(102+242)​=676​=26

D.Thrishanth Reddy 1 year, 10 months ago

26cm is radius
  • 4 answers

Piyush Sahani 1 year, 10 months ago

3+3√4

Navneet Kaur Sekhon 1 year, 10 months ago

3+4√2

Diva Shah 1 year, 10 months ago

Therefore 3(1+√3)

Diva Shah 1 year, 10 months ago

It's 3+3√3
  • 0 answers
  • 1 answers

Diva Shah 1 year, 10 months ago

1/2+√3
  • 5 answers

Navpreet Bal 1 year, 10 months ago

Sn=n/2[2a+(n–1)d] 192=8/2[2×3+(8–1)d] 192=4[6+7d] 192-42=6+7d 48=6+7d 48-6=7d 42=7d d=42/7 d=6

Pratiksha Yadav 1 year, 10 months ago

D = 6

Aman Khan 1 year, 10 months ago

Kaise

Jatin Bansal 1 year, 10 months ago

6

Nandini . 1 year, 10 months ago

D=6
  • 5 answers

Ashish Yadav 1 year, 10 months ago

Tagri

Aman Khan 1 year, 10 months ago

Achhe se chal rahi h

Saumya Singh 1 year, 10 months ago

Baith gyi hai

Shubham Pathak 1 year, 10 months ago

Bas chal hi rhi hai , kash daurti !!

Priya Yadav 1 year, 10 months ago

Thik chal rahi hai
  • 5 answers

Jatin Bansal 1 year, 10 months ago

4

Shubham Pathak 1 year, 10 months ago

69

Om Kumar 1 year, 10 months ago

4

Nikhil Ranjan 1 year, 10 months ago

4

Salim Siddiqui 1 year, 10 months ago

4
  • 5 answers

Abhishek Dubey 1 year, 10 months ago

a²-b²
a²-b²

Pratha Shrivastava 1 year, 10 months ago

a²-b²

Radhika Bhardwaj 1 year, 10 months ago

a²-b²

Dattatrey Mangavati 1 year, 10 months ago

a²-b²
  • 1 answers

Preeti Dabral 1 year, 10 months ago

Value of {tex}\frac{x}{\sqrt{a^2+x^2}}{/tex}

 

{tex}\tan \theta=\frac{\mathrm{a}}{\mathrm{x}} Since, \begin{aligned} & \frac{x}{\sqrt{a^2+x^2}} \\ & \Rightarrow \frac{\frac{a}{\tan \theta}}{\sqrt{a^2+\left(\frac{a}{\tan \theta}\right)^2}} \\ & \Rightarrow \frac{\frac{a}{\tan \theta}}{\sqrt{a^2+\frac{a^2}{\tan ^2 \theta}}} \end{aligned} {/tex}

{tex}\begin{aligned} & \Rightarrow \frac{\frac{a}{\tan \theta}}{a \sqrt{\frac{1+\tan ^2 \theta}{\tan ^2 \theta}}} \\ & \Rightarrow \frac{\frac{1}{\tan \theta}}{\sqrt{\frac{1+\tan ^2 \theta}{\tan ^2 \theta}}} \\ & \Rightarrow \frac{1}{\sqrt{\sec ^2 \theta}} \\ & \Rightarrow \frac{1}{\sec \theta} \end{aligned}{/tex}

{tex}\Rightarrow \cos \theta {/tex}

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