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    • Posted by Cheshta Garg 1 week ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Guru Prasad 1 week ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Siya B 1 week ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Ruthvik Chowdary Kamma 1 week, 1 day ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Surya Tk 1 week, 1 day ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Topper Student 1 week, 1 day ago

    (-8k+5)-(5k-3)=(5k-3)-(2k+1) Solve this and you will get k

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    • Posted by Kartik Chauhan 1 week, 2 days ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Tanya Chauhan 19 hours 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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    • Posted by Tanishtha Varman 1 week, 2 days ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Shubhankar Nayak 1 week, 2 days ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Khushi Chaudhary 1 week 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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    • Posted by Cheshta Garg 1 week, 3 days ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Shubham Sinha 1 week, 3 days 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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    • Posted by Siri Priya 1 week, 3 days ago

    CBSE > Class 10 > Mathematics
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    • Posted by Ann ... 1 week, 3 days ago

    CBSE > Class 10 > Mathematics
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    • Posted by Aditya Jha 1 week, 3 days ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Sukhman Singh Bhullar 1 week, 3 days ago

    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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    • Posted by Anushka . 1 week, 3 days ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Laksh Heda 1 week, 3 days ago

    Df7

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    • Posted by Cheshta Garg 1 week, 3 days ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Poorvi Sharma 1 week, 4 days ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Kanaka Mahalakshmi Ganta 1 week, 1 day ago

    I want mind map of chapter 1,2,3

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    • Posted by Subhrajit Rout 1 week, 4 days ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Rishi Kumar 1 week, 4 days ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Sushant Singh 1 week, 4 days ago

    Solve. The question

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    • Posted by Aditya Shah 1 week, 4 days ago

    CBSE > Class 10 > Mathematics
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    • Posted by Risvanth Risvanth 1 week, 4 days ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Shayaan Shaikh 1 week, 4 days ago

    CBSE > Class 10 > Mathematics
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    • Posted by Priyanshi Kumre 1 week, 5 days ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Parfzfg Bunny 1 week, 5 days ago

    CBSE > Class 10 > Mathematics
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    • Posted by Lokesh Meena ❤️ 1 week, 5 days ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Sankalp Dinesh 1 week, 5 days ago

    What is the question bro??

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    • Posted by Kajal Sengar 1 week, 5 days ago

    CBSE > Class 10 > Mathematics
    • 2 answers

    Sankalp Dinesh 1 week, 5 days ago

    We know that, alpha + beta = -b/a alpha x beta = c/a Therefore, 1/alpha + 1/beta = beta + alpha/alpha x beta = -(-1)/1/-1/1 = 1/-1 = -1/1 So, -1 is the answer.

    1Thank You

    B Praveen 1 week, 5 days ago

    Geiub

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    • Posted by Abhi Bishnoi 1 week, 6 days ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    I Am Helper 1 week, 6 days ago

    Expand the left-hand side: [(x+2)^3 = x^3 + 3 \cdot x^2 \cdot 2 + 3 \cdot x \cdot 2^2 + 2^3] [= x^3 + 6x^2 + 12x + 8]Expand the right-hand side: [2x(x^2 - 1) = 2x \cdot x^2 - 2x \cdot 1] [= 2x^3 - 2x]Set the expanded forms equal: [x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x]Move all terms to one side to form a polynomial equation: [x^3 + 6x^2 + 12x + 8 - 2x^3 + 2x = 0] [ - x^3 + 6x^2 + 14x + 8 = 0]Simplify the equation: [x^3 - 6x^2 - 14x - 8 = 0]Solve the cubic equation (x^3 - 6x^2 - 14x - 8 = 0):

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    • Posted by Harman Kaur 1 week, 6 days ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Rakesh Kumar 1 week, 5 days ago

    3x is x and y is 5y and z is 17z

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    • Posted by Bikram Kumar Sarkar 2 weeks ago

    CBSE > Class 10 > Mathematics
    • 2 answers

    Yuvraj Shinde 1 week, 3 days ago

    2x+y=7 4x-3y+1=0

    0Thank You

    I Am Helper 1 week, 6 days ago

    Given that \(2 \tan A = 3\), we want to find the value of \(\sin A\) and \(\cos A\). 1. **Express \(\tan A\) in terms of a simpler fraction**: \[ \tan A = \frac{3}{2} \] 2. **Use the Pythagorean identity to find \(\sin A\) and \(\cos A\)**: We know that: \[ \sin^2 A + \cos^2 A = 1 \] and \[ \tan A = \frac{\sin A}{\cos A} \] Since \(\tan A = \frac{3}{2}\), we can set: \[ \sin A = 3k \quad \text{and} \quad \cos A = 2k \] where \(k\) is a common factor. 3. **Use the Pythagorean identity** to find \(k\): \[ \sin^2 A + \cos^2 A = 1 \] Substitute \(\sin A\) and \(\cos A\): \[ (3k)^2 + (2k)^2 = 1 \] Simplify the equation: \[ 9k^2 + 4k^2 = 1 \] \[ 13k^2 = 1 \] Solve for \(k\): \[ k^2 = \frac{1}{13} \] \[ k = \frac{1}{\sqrt{13}} \] 4. **Find \(\sin A\) and \(\cos A\)** using \(k\): \[ \sin A = 3k = 3 \times \frac{1}{\sqrt{13}} = \frac{3}{\sqrt{13}} = \frac{3 \sqrt{13}}{13} \] \[ \cos A = 2k = 2 \times \frac{1}{\sqrt{13}} = \frac{2}{\sqrt{13}} = \frac{2 \sqrt{13}}{13} \] Thus, the values of \(\sin A\) and \(\cos A\) are: \[ \sin A = \frac{3 \sqrt{13}}{13} \] \[ \cos A = \frac{2 \sqrt{13}}{13} \]

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    • Posted by Trupti R 2 weeks ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Khushi Yadav 2 weeks, 1 day ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Khushi Yadav 1 week, 5 days ago

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    • Posted by Srushti Biradar 2 weeks, 1 day ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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