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  • 1 answers

Sachin Mahto 5 years, 11 months ago

SecA(1-sinA) (secA+tanA) =1/cosA(1-sinA) (1/cosA +sinA/cosA) =1/cosA(1-sinA) (1+sinA/cosA) =1/cosA(1-sinA) 1/cosA(1+sinA) =1/cos squareA (1-sin square A) =1/cos square A × cos square A =1
  • 0 answers
  • 1 answers

Ranjan Jha Jonki 5 years, 11 months ago

Yes
  • 5 answers
Khud kro easy h

Aviral Goyal 5 years, 11 months ago

The correct answer underroot3 +1/4
Hi

Aadya Singh ? 5 years, 11 months ago

Or √3/2..

Aadya Singh ? 5 years, 11 months ago

(2√3)/4
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  • 1 answers

Harsh Raj 5 years, 11 months ago

L.H.S ; S12= 12/2{2a + (12-1)d} = 6{2a + 11*d} = 12a + 66d R.H.S ; 3(S8 - S4) =3[n/2{2a+(n-1)d} - n/2{ 2a+(n-1)d}] = 3{8/2(2a+7d) - 4/2{2a+3d)} =3{4(2a+7d) - 2(2a+3d)} =3(8a+28d - 4a-6d) =3(4a+22d) =12a+66d Therefore ; L.H.S = R.H.S Proved......?????????????????????????????????????????????????
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  • 3 answers

Aviral Goyal 5 years, 11 months ago

Last term should be 95 not 100 bcos in question it is less than 100 so sum will be 950

Aviral Goyal 5 years, 11 months ago

Parishi ur answer is wrong

Parishi Thada 5 years, 11 months ago

AP formed will be- 5,10,15.......100 So, a=5 d=5 an=100 n=? an= a+(n-1)d 100=5+(n-1)5 95= 5n-5 5n = 100 n=20 Putting these values in the sum formula we get, Sn=n/2 [ 2a+(n-1)d] S20= 20/2 [ 2(5) + (19)5] S20=10[10+95] S20=10[105] S20=1050 ANS. Hope it helps
  • 3 answers

Mohammad Saifullah 5 years, 11 months ago

a^3+b^3+3a^2b+3ab^2

Raj Clasher 5 years, 11 months ago

a3+b3 3ab(a+b)

Shreya ... 5 years, 11 months ago

a^3+b^3+3ab(a+b)
  • 1 answers

Gaurav Seth 5 years, 11 months ago

The largest number by which x , y divisible and gives the remainder a , and b is the HCF of ( x - a ) and ( y - b)
__________________________

According to the given problem ,

The largest number which divides 70 and 125 leaving remainders 5 and 8 respectively are

HCF of ( 70 - 5 ) = 65 and

( 125 - 8 ) = 117

65 = 5 × 13

117 = 3 × 3 × 13

HCF ( 65 , 117 ) = 13

Required number is 13.

  • 1 answers

Harsh Raj 5 years, 11 months ago

726 = 275*2 + 176 275 = 176*1 + 99 176 = 99*1 + 77 99 = 77*2 + 22 77 = 22*3 + 11 22 = 11*2 +0 Therefore ; H.C.F = 11.. ??????
  • 1 answers

Harsh Raj 5 years, 11 months ago

7x-15y=2.........1 7(X+2y)=3*7..........2 From eq.1 &eq.2 7x - 15y = 2 7x +14y = 21 ...↑-....↑-........↑-........... 0 -29y = -19 Therefore; y = -19/-29 Y = 19/29???????? Putting the value of "Y" in eq.2 X + 2y = 3 X + 2(19/29)=3 X +2*19/29 = 3 X = 3 - 38/29 X = 49/29??????? Hence; X = 49/29 and Y = 19/29....
  • 3 answers

Adarsh Adarsh 5 years, 11 months ago

Theorem 6.1, 6.2, 6.6, 6.8, 6.9, 10.1, 10.2

Vikas Saini 5 years, 11 months ago

Important theorem is 6. First five are in the chapter triangles and one is in the chapter construction

Ashish Athghara 5 years, 11 months ago

Triangle me acha theorem hai
  • 1 answers

Yogita Ingle 5 years, 11 months ago

Let p(x) = 3x²-2kx+2m

Given, 2 & 3 are zeroes of p(x).
Then , p (2) = 0 & p(3)= 0

Now, p(2)= 3×2²-2k(2)+2m=0

3× 4-4kx+2m=0

12-4k+2m=0………..(1)

p(3) =  3×3²-2k(3)+2m=0

3× 9-6kx+2m=0

27-6k+2m=0………..(2)

On Subtracting Equation 1 from equation 2.

27-6k+2m=0
12-4k+2m=0
(-)  (+) (-)
------------------
15-2k =0

15=2k

k= 15/2

On putting k= 15/2 in eq 1

12-4k+2m=0

12-4(15/2)+2m=0

12-30+2m=0

-18+2m=0

2m=18
m=18/2
m=9

Hence, k=15/2, m= 9

  • 5 answers

Piyush Yadav 5 years, 11 months ago

Put the number given

Piyush Yadav 5 years, 11 months ago

Don't add anyone

Aadya Singh ? 5 years, 11 months ago

(11)+(15)+(17-13)= 30....Ab sahi h...?

Vikas Saini 5 years, 11 months ago

Where is 15adya singh all numbers are compulsory ??

Aadya Singh ? 5 years, 11 months ago

(11) + (13)+(17-11) = 30...
  • 3 answers

Sambhav Maheshwari 5 years, 11 months ago

X+y=5 (1) 2x-3y=4. (2) Eq(1) multiply by 2 2x+2y=10 (3) From eq(3) and eq(1) 2x+2y=10. (3) 2x-3y=4. (1) (-) (+) (-) ................................ 5y=6 y=6/5 Put the value of y in eq(1) X+6/5=5 X=5-6/5 X=25-6/5 X=19/5 (x=19/5, y=6/5 ) ans

Yogita Ingle 5 years, 11 months ago

X + y = 5 ( equation 1 )
2x - 3y = 4 ( equation 2 )
Now,
multiplying (eq.1) by 2 and subtracting (eq.2) from ( eq.1 )
(2x + 2y = 10) - (2x - 3y = 4 )
=> 2x + 2y - 2x + 3y = 6
=> 5y = 6
=> y = 6/5
Putting the value of y in ( eq.1 )
=> x+ y = 5
=> x + 6/5 = 5
=> x = 5 - 6/5
=> x = 19/5

Piyush Yadav 5 years, 11 months ago

X:13/5 y: 6/5

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