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Ask QuestionPosted by Kshitij Prakash 6 years, 10 months ago
- 1 answers
Yash Agarwal 6 years, 10 months ago
Posted by Rohan Alh 6 years, 10 months ago
- 1 answers
Yash Agarwal 6 years, 10 months ago
Posted by Abhishek Agrahari 6 years, 10 months ago
- 2 answers
Yogita Ingle 6 years, 10 months ago
an = 5 - 11n
If n = 1 ;
a1 = 5-(11×1)
a1 = -6
If n = 2 ;
a2 = 5-(11×2)
a2 = -17
Now , a1=-6 & a2=-17
Hence, common difference = d = a2-a1
= -17-(-6)
= -11
Posted by Ramkishor Koli 6 years, 10 months ago
- 1 answers
Posted by Uday Gupta 6 years, 10 months ago
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Posted by Pooja Rani 6 years, 10 months ago
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Posted by Riya Raj 6 years, 10 months ago
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Posted by Syam Prasad 6 years, 10 months ago
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Posted by Seema Bhansali 6 years, 10 months ago
- 1 answers
Gaurav Seth 6 years, 10 months ago
Secθ+tanθ=p ----------------------(1)
∵, sec²θ-tan²θ=1
or, (secθ+tanθ)(secθ-tanθ)=1
or, secθ-tanθ=1/p ----------------(2)
Adding (1) and (2) we get,
2secθ=p+1/p
or, secθ=(p²+1)/2p
∴, cosθ=1/secθ=2p/(p²+1)
∴, sinθ=√(1-cos²θ)
=√[1-{2p/(p²+1)}²]
=√[1-4p²/(p²+1)²]
=√[{(p²+1)²-4p²}/(p²+1)²]
=√[(p⁴+2p²+1-4p²)/(p²+1)²]
=√(p⁴-2p²+1)/(p²+1)
=√(p²-1)²/(p²+1)
=(p²-1)/(p²+1)
∴, cosecθ=1/sinθ=1/[(p²-1)/(p²+1)]=(p²+1)/(p²-1)
Posted by Riya Raj 6 years, 10 months ago
- 2 answers
Appu Enterprises 6 years, 10 months ago
Appu Enterprises 6 years, 10 months ago
Posted by Rishabh Vishwakarma 6 years, 10 months ago
- 4 answers
Posted by Kunik Duhan 6 years, 10 months ago
- 1 answers
Posted by Kulwant Singh 6 years, 10 months ago
- 1 answers
Posted by Ratan Lal Tailor 6 years, 10 months ago
- 1 answers
Yash Agarwal 6 years, 10 months ago
Posted by Neelam Gupta 6 years, 10 months ago
- 2 answers
Purvanshi Yadav 6 years, 10 months ago
Posted by Rina Singh 6 years, 10 months ago
- 2 answers
Gauri ❤ 6 years, 10 months ago
Posted by Vivek Thapliya 6 years, 10 months ago
- 2 answers
Posted by Mehak Dhalia 6 years, 10 months ago
- 3 answers
Sumit Kumar 5 years, 8 months ago
Yogita Ingle 6 years, 10 months ago
common difference (d) = common
So, 5 - 2x = 6 - x - 5
4 = x
Value of x = 4.
So, a = first term = 2x = 8
d = - 3.
a10 = a + 9d = 8 - 27 = -19
Tenth term is - 17.
Posted by Rahul Lathait 6 years, 10 months ago
- 2 answers
Posted by Bachitar Singh 6 years, 10 months ago
- 1 answers
Posted by Ashok Ray 6 years, 10 months ago
- 1 answers
Kartik Sharma 6 years, 10 months ago
Posted by Aditya Singh 6 years, 10 months ago
- 2 answers
Ram Kushwah 6 years, 10 months ago
let x is the angle then
{tex}\begin{array}{l}\sqrt3\mathrm{tanx}=3\\\mathrm{tanx}=\frac3{\sqrt3}=\sqrt3=\tan60^\circ\\\mathrm{so}\;\mathrm x=60^\circ\end{array}{/tex}
Posted by Ujjwal Shivhare 6 years, 10 months ago
- 1 answers
Posted by Mahi Choudhary 6 years, 10 months ago
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Posted by Tushat Sabat 6 years, 10 months ago
- 1 answers
Gaurav Seth 6 years, 10 months ago
Given quadratic polynomial f(x) = x2+px+45
Let α and β be the roots(zeros) of the given quadratic polynomial f(x) = x2+px+45
Also, given (α - β)2 = 144
We know that, (α - β)2 = (α + β)2 - 4αβ
Now, we have from the quadratic polynomial f(x) = x2+px+45, (α + β) = -p, αβ = 45
⇒144 = (-p)2 - 4(45)
⇒144 = p2 - 180
⇒144 + 180 = p2
⇒324 = p2
⇒p2 = 324
⇒p = + - 18
Posted by Vansh Saini 6 years, 10 months ago
- 1 answers
Gaurav Seth 6 years, 10 months ago
Let us assume, to the contrary, that √p is
rational.
So, we can find coprime integers a and b(b ≠ 0)
such that √p = a/b
=> √p b = a
=> pb2 = a2 ….(i) [Squaring both the sides]
=> a2 is divisible by p
=> a is divisible by p
So, we can write a = pc for some integer c.
Therefore, a2 = p2c2 ….[Squaring both the sides]
=> pb2 = p2c2 ….[From (i)]
=> b2 = pc2
=> b2 is divisible by p
=> b is divisible by p
=> p divides both a and b.
=> a and b have at least p as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction arises because we have
assumed that √p is rational.
Therefore, √p is irrational.
Posted by Sameer Ali 6 years, 10 months ago
- 2 answers
Ram Kushwah 6 years, 10 months ago
Least prime factor of a is 3 then 2 is not factor
as 3< 2
so a is odd number
Least prime factor of a is 5 then 2 is not factor
as 5< 2
so b is odd number
so a+b= odd number +odd number = even number
so a+b=2q ( q is any positive integer
hence least prime factor of a+b is 2
Gaurav Seth 6 years, 10 months ago
Given that, a is a positive integer and 3 is the least prime factor of a.
Also, b is a positive integer and 5 is the least prime factor of b.
Since least prime factor of a is 3, it means that a is an odd number.
So, if a is even then 2 is the least prime factor of a.
Similarly, b is also an odd number.
Now, it is known that sum of two odd numbers is an even number.
Hence, a + b will be an even number.
So, the least prime factor of a + b is 2
Posted by Vibhuti Narayan Srivastav 6 years, 10 months ago
- 1 answers
Gaurav Seth 6 years, 10 months ago
Consider the following numbers
72 and 124
First find the HCF of above numbers.
72 = 2 × 2 × 2 × 3 × 3
⇒ 2³ × 3²
124 = 2 × 2 × 31
⇒ 2² × 31
To find the HCF of these numbers, take the least power of each common factor and find the product.
Here 2 is the only common factor and least power of which is 2.
So, we have
HCF(72 and 124) = 4
Now we need to express HCF = 4 as a linear combination of 72 and 124.
That is,
4 = 72a + 124b, where a and b are integers.
Use hit and trial method.
Take a = -10 and b = 6
72(-10) + 124(6)
⇒ -720 + 744 = 24, which is not equal to 4.
So, take a = -12 and b =7
72(-12) + 124(7)
⇒ - 864 + 868 = 4
So, the required linear combination is
HCF(72 and 124) = 4 = 72(-12) + 124(7)
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