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  • 1 answers

Yogita Ingle 5 years, 2 months ago

(i) 140   = 2 x 2 x 5 x 7 =2² x 5 x 7
(ii)156 =2 x 2 x 3 x 13 =2² x 3 x 13
(iii)3825 =3 x 3 x 5 x 5 x 17 =3² x 5² x 17
(iv)5005  =5 x 7 x 11 x 13  =5 x 7 x 11 x 13
(v)7429  =17 x 19  x 23  =17 x 19 x 23

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Riya Philip 5 years, 2 months ago

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Diksha Dangi 5 years, 2 months ago

Given : 12, 15 , 21 Prime factor of 12 : 2*2*3 Prime factor of 15 : 3*5 Prime factor of 21 : 3*7 HCF of 12, 15 ,21 : 3 LCM of 12 , 15 , 21 : 420
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Diksha Dangi 5 years, 2 months ago

140 : 2*2*5*7
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Riya Philip 5 years, 2 months ago

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Adesh Shitole 5 years, 2 months ago

Hwgshs
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Riya Philip 5 years, 2 months ago

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Sachin Ray 5 years, 2 months ago

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Gagandeep Singh 5 years, 2 months ago

1

Sunidhi Singh 5 years, 2 months ago

1

Riya Philip 5 years, 2 months ago

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  • 4 answers

Yangzee Sherpa 5 years, 2 months ago

“a^2+2ab+b^2” HoPe It HeLpS yOu??

Ajin Ajin 5 years, 2 months ago

a2 + 2ab + b2

Ajin Ajin 5 years, 2 months ago

a²+b²+2ab I HOPE IT HELPS YOU

Nidhi Jain 5 years, 2 months ago

a²+b²+2ab
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Riya Philip 5 years, 2 months ago

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Yogita Ingle 5 years, 2 months ago

Δ ABC is a Right Triangle at ∠C = 90°

To Find:

sin A × cos B + cos A × sin B = ?

Solution:

Δ ABC is a Right Triangle at ∠C = 90° .......Given

Also

Therefore,

∠ A = 30°

Now in ΔABC

 .........angle sum Triangle property

substituting the values we get

So the value of

sin A × cos B + cos A × sin B  = 

Therefore

sin A × cos B + cos A × sin B  = 1

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Riya Philip 5 years, 2 months ago

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Dhyey Sonani 5 years, 2 months ago

In valid question
  • 5 answers

Pinky Singh 5 years, 2 months ago

It's 0

Aprita Pal 5 years, 2 months ago

Undefined

Praveen Singh 5 years, 2 months ago

0

Riya Philip 5 years, 2 months ago

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Riya Philip 5 years, 2 months ago

Undefined
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Riya Philip 5 years, 2 months ago

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Riya Philip 5 years, 2 months ago

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Dhyey Sonani 5 years, 2 months ago

From LHS (CosecA-cotA)² =(1/sinA - cosA/sinA)² ={(1-cosA) /sinA}² =(1-cosA) ² /sin²A =(1-cosA) ²/ (1-cos²A) =(1-cosA) (1-cosA) / (1-cosA) (1+cosA) =(1-cosA) /(1+cosA) =RHS

Aditya Kumar 5 years, 2 months ago

From LHS (CosecA-cotA)² =(1/sinA - cosA/sinA)² ={(1-cosA) /sinA}² =(1-cosA) ² /sin²A =(1-cosA) ²/ (1-cos²A) =(1-cosA) (1-cosA) / (1-cosA) (1+cosA) =(1-cosA) /(1+cosA) =RHS

Riya Philip 5 years, 2 months ago

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Riya Philip 5 years, 2 months ago

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Tejasvini S 5 years, 2 months ago

Stactis
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Riya Philip 5 years, 2 months ago

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Yogita Ingle 5 years, 2 months ago

The number series 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, .  .  .  .  , 240.

The first term a = 8

The common difference d = 8

Total number of terms n = 30

 

step 2 apply the input parameter values in the AP formula

Sum = n/2 x (a + Tn)

= 30/2 x (8 + 240)

= (30 x 248)/ 2

= 7440/2

8 + 16 + 24 + 32 + 40 + 48 + 56 + 64 + 72 + 80 + 88 + 96 + .  .  .  .   + 240 = 3720

Therefore, 3720 is the sum of first 30 positive integers which are divisible by 8

  • 3 answers

Manish Sharma 5 years, 2 months ago

Thanks

Riya Philip 5 years, 2 months ago

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Riya Philip 5 years, 2 months ago

Of which subject?
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Raveena Beniwal 5 years, 2 months ago

What is in this site

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  • 5 answers

Dhyey Sonani 5 years, 2 months ago

1032=408×2 + 216 408 = 216×1 + 192 216 = 192×1 + 24 192=24×8 + 0 so the hcf will be 24 now 24 = 1032 - 408×5 1032m - 2040 = 24 m=2

Dasaratha Barik 5 years, 2 months ago

Statistick

Riya Philip 5 years, 2 months ago

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Yogita Ingle 5 years, 2 months ago

 

1032=408×2 + 216

408 = 216×1 + 192

216 = 192×1 + 24

192=24×8 + 0

so the hcf will be 24
now

24 = 1032 - 408×5

1032m - 2040 = 24

m=2

Anshuman Omm 5 years, 2 months ago

m=2/5
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Riya Philip 5 years, 2 months ago

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Shubham Pujara 5 years, 2 months ago

Ratios of corresponding sides.
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Riya Philip 5 years, 2 months ago

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Rimple Kumari Kumari 5 years, 2 months ago

Liner equation and two variyebal

Shubham Pujara 5 years, 2 months ago

By putting the value of x as -1 and 2 one by one the given polynomial 1) when x=-1 (-1)^3-5(-1)2+2(-1)+d=0 # -1-5-2+d=0 # d=8 Similarly on putting the value of x as 2 we will get d=8. Therefore the required value of d=8. Hope it will be helpful

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