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  • 1 answers

Preeti Dabral 1 year, 10 months ago

Given, a4 = 0
a + (n - 1)d = 0 [n = 4]
a + 3d = 0
a = -3d
From LHS, a25 = a + (n - 1)d
= a + (25 - 1)d
= a + 24d [∵∵ a = -3d]
= -3d + 24d = 21d
From RHS, a11 = a + (n - 1)d
= a + (11 - 1)d = a + 10d [∵∵ a = -3d]​
= -3d + 10d
= 7d
∴∴ Hence, a25 = 3 ×× a11

  • 1 answers

Suhani Verma 1 year, 10 months ago

Qp= √( X-x )² + (Y-y)² = √(5-0)² + (-3-1)² = √ (5)² + (-4)² = √ 25+16 = √ 41 QR= √ (X-x)² + (Y-y)² = √ (x-0)² + (6-1)² = √ (x)² + (5)² = √ x² + 25 Qp =QR √41 = √x² +25 Both side squaring (√41)² = (√x²+25) 41 = x² + 25 41- 25=x² 16 = x² 4 = x Distance between, QR=√( 4-0)² + (6-1)² = √(4)² + (5)² =√16+25 =√41 PR=√(5-4)² + (-3-6)² =√(1)² + (-9)² =√ 1+81 =√82
  • 3 answers

Tilak Raj 1 year, 10 months ago

3median = mode +2mean

Guri Mundi 1 year, 10 months ago

2median= mode -2mean

Naman Saxena 1 year, 10 months ago

Mode = 3median - 2mean
  • 1 answers

Saksham Dogra 1 year, 10 months ago

1−cotθ tanθ ​ + 1−tanθ cotθ ​ = 1− tanθ 1 ​ tanθ ​ + 1−tanθ tanθ 1 ​ ​ = tanθ tanθ−1 ​ tanθ ​ + tanθ(1−tanθ) 1 ​ = tanθ−1 tan 2 θ ​ + tanθ(1−tanθ) 1 ​ = tanθ−1 tan 2 θ ​ − tanθ(tanθ−1) 1 ​ = tanθ(tanθ−1) tan 3 θ−1 ​ = tanθ(tanθ−1) (tanθ−1)(tan 2 θ+tanθ+1) ​ = tanθ tan 2 θ+tanθ+1 ​ = tanθ tan 2 θ ​ + tanθ tanθ ​ + tanθ 1 ​ =tanθ+1+cotθ =1+ cosθ sinθ ​ + sinθ cosθ ​ =1+ sinθcosθ sin 2 θ+cos 2 θ ​ =1+ sinθcosθ 1 ​ =1+secθcosecθ = RHS
  • 1 answers

Gayatri Pathade 1 year, 10 months ago

Median=l+[n/2-cf] h ----------- F
  • 4 answers

Piyush Maheshwari 1 year, 10 months ago

There are 7 months that have 31 days January, March, May, July, August, October, December

Vivek Prajapati 1 year, 10 months ago

Nahi 32

Aditya Raj 1 year, 10 months ago

H

Anshika Sharma 1 year, 10 months ago

July, August January, March, May, July, August, October, December months have 31 days each. So, here July and august are the consecutive months having 31 days each. 
  • 3 answers

Mangifera Indica 1 year, 10 months ago

7msq

Mangifera Indica 1 year, 10 months ago

9msq

Anshika Sharma 1 year, 10 months ago

Total surface area of the cabinet = 2(l × b + b × h + h × l) Area of the cabinet that was painted = (Total surface area of the cabinet) - (Area of bottom of the cabinet) = 2(l × b + b × h + h × l) − l × b = 2(b × h + h × l) + l × b = 2 × (1 × 1.5 + 1.5 × 2) + 2 × 1 m² = 2 × 4.5 + 2 m² = 11 m²
  • 2 answers

Hamza Ansari 1 year, 9 months ago

b/a

Zaid Agha Khan 1 year, 10 months ago

√b²-a²
  • 2 answers

Gayatri Pathade 1 year, 10 months ago

462msq

Vanshika Zade 1 year, 10 months ago

If the circumference of the road is 88 cm then its radius would be 44 cm
  • 1 answers

Ankit Class 1 year, 10 months ago

a+d=10 a+4d=31 Subtracting we get 3d=21 d=7, a=3 hence a=3,b=17,c=24
  • 0 answers
  • 1 answers

Utkarsh Yadav 1 year, 10 months ago

Exercise 12. 1
  • 5 answers

Shourya Kumar 1 year, 10 months ago

an=99 a=12 d=3 an=a+(n-1)d 99=12+(n-1)×3 (n-1)×3=99-12 (n-1)=87÷3 n-1=29 n=29+1 [n=30].

Poornima Garg 1 year, 10 months ago

a=12 d=15-12=3 an=99 an=a+(n-1)d 99=12+(n-1)3 99=12+3n-3 99=9+3n 99-9=3n 90=3n n=30

🥱🥱 ... 1 year, 10 months ago

itna to kr dia bro... solve to khud hi kr loge...

Nikhil Gautam 1 year, 10 months ago

Sir solve kar do

🥱🥱 ... 1 year, 10 months ago

use 99 as an... put terms in formula an = a + ( n - 1 ) ( d ) 99 = 12 + ( n - 1 ) ( 3 )
  • 1 answers

Shubham Pathak 1 year, 10 months ago

(a + 24d) - (a + 11d) = -52. => 13d = -52. => d = - 4
  • 5 answers

Sarika Maurya 1 year, 10 months ago

156

Harsh Jadon 1 year, 10 months ago

156

Bhavna Vimal 1 year, 10 months ago

16464733393392

Omya Aryan 1 year, 10 months ago

156

Vijay Makwana 1 year, 10 months ago

LCM of 12 and 13
  • 4 answers

Saumya Singh 1 year, 10 months ago

3median =mode + 2 mean

Lakshmi Kanth T 1 year, 10 months ago

2mean = 3median - mode

Shubham Pathak 1 year, 10 months ago

3median = mode + 2mean Mean = (3median - mode)/2

Vishal Kumar 1 year, 10 months ago

Gnbs'wb
  • 3 answers

Lakshmi Kanth T 1 year, 10 months ago

P+q = 7/2, pq = 3/2, P+q - PQ= 7/2 - 3/2 =(7-3)/2 =4/2=2

Manish Sharma 1 year, 10 months ago

p+q=7/2. pq=3/2. p+q-pq=7/2-3/2=2

Samreen Khanam 1 year, 10 months ago

Given: p and q are the zeros of the polynomial f(x)=2xsq.-7x+3=2xsq.-(6x+1x) =2xsq.-6x-x+3=2x(x-3) -1(x-3) =(2x-1) (x-3) X=1/2 X=3 Now, P+Q=7/2 Pq=3/2 Answer
  • 4 answers

Gayatri Pathade 1 year, 10 months ago

It's -4

Mohd Arsalan 1 year, 10 months ago

-7

Prem Kumar 1 year, 10 months ago

Let An term be 1 A1=7-4*1 =3 Let An term be 2 A2=7-4*2 =7-8 =-1 Now Common difference=A2-A1 =-1-3 =-4 Hence the required common difference is -4.

Nikhil Yadav 1 year, 10 months ago

-7
  • 3 answers

Mohd Arsalan 1 year, 10 months ago

12.4

Nikhil Yadav 1 year, 10 months ago

12.3 and 12.4

Sumit Sharma 1 year, 10 months ago

12.4
  • 2 answers

Dhwani Panchal 1 year, 10 months ago

1

Ekesh Sharma 1 year, 10 months ago

5(sec^2- tan^2) =5(1) =5
  • 2 answers

A B 1 year, 10 months ago

Option:-D is correct

Omya Aryan 1 year, 10 months ago

Since, side of the square is 6cm, therefore diameter of the circle is also equal to 6cm. Then radius is 3cm. Area of circle= πr^2, then area =π×3×3= 9πcm^2. Hence, (D) is the correct option.
  • 1 answers

Preeti Dabral 1 year, 10 months ago

Construct EC || AB AB = 15 cm = EC AD = 12 cm CB = 4 cm DE = 12 - 4 = 8 Tangent is perpendicular to radius ABCE is rectangle and DEC is a right angled triangle So DC ^ 2 = DE ^ 2 + EC ^ 2

DC = 17

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