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Topper Student 4 months, 4 weeks ago

(-8k+5)-(5k-3)=(5k-3)-(2k+1) Solve this and you will get k
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Tanya Chauhan 4 months, 3 weeks ago

Very simple. LHS (Sin^2 A + cosec^2 A + 2 sin A cosec A)+( cos^2A+sec^2A +2cosA secA) (Sin^2A+cosec^2A+2)+ ( cos^2A+sec^2A+2) (Sin^2A +cos ^2A) +4+(cosec^2A+sec^2A) 1+4+(1+cot ^2A) +( 1+tan^A) (7+tan ^2A+cot^2A) = RHS
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Khushi Chaudhary 4 months, 4 weeks ago

Let the zeroes are alpha is 3 Bita is -3 Alpha+bita= 3+(-3) = 0 Alpha× Bita= (3) (-3) = -9 =K(x^2-(sum of zeroes)x +(product of zeroes) k is constant = (X^2-(0)x +(-9) Ans = x^2-0x-9
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Shubham Sinha 5 months ago

The given pair of equations are : - 3x + y = 1 ........... (1) (2k−1)x + (k−1)y = 2k + 1 .......... (2) Now rearranging eq1 and eq2 will get 3x + y − 1 = 0 ........... (3) (2k−1)x + (k−1)y − (2k+1) = 0 ........ (4) Now compare with a1 = 3, b1 = 1, c1 = −1 a2 = 2k−1, b2 = k−1, c2 = −(2k+1) Now we get a1a2 = 32k−1 , b1b2 = 1k−1 , c1c2 = −1−(2k+1) Now will take a1a2 = b1b2 ⇒32k−1 = 1k−1 ⇒3k−3 = 2k−1 ⇒3k−2k = −1+3 ⇒k = 2 Hence k = 2 is the value.
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Let √5=a/b is a rational number Where a and b are co-prime Squaring both sides (5)²=a²/b² 5=a²/b² b²=a²/5 -(1) .'. a² is divisible by 5, also a is divisible by 5 Let a/5=c a=5c Put this value of a in (1) b²=(5c)²/5 b²=25c²/5 b²=5c² c²=b²/5 b² is divisible by 5 also b is divisible by 5 .'. This is a contradication that √5 is a rational number .'.√5 is an irrational number
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I want mind map of chapter 1,2,3
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