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    • Posted by Lokesh Meena ❤️ 5 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Sankalp Dinesh 5 months ago

    What is the question bro??

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    • Posted by Kajal Sengar 5 months ago

    CBSE > Class 10 > Mathematics
    • 2 answers

    Sankalp Dinesh 5 months 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.

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    B Praveen 5 months ago

    Geiub

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    • Posted by Abhi Bishnoi 5 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    I Am Helper 5 months 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 5 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Rakesh Kumar 5 months ago

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

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    • Posted by Bikram Kumar Sarkar 5 months ago

    CBSE > Class 10 > Mathematics
    • 2 answers

    Yuvraj Shinde 5 months ago

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

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    I Am Helper 5 months 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 5 months ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Khushi Yadav 5 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Khushi Yadav 5 months ago

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    • Posted by Srushti Biradar 5 months ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Ku.Krishna Yadav 5 months, 1 week ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Ku.Krishna Yadav 5 months ago

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    Posted by Malvika Singh 5 months ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Beeraiah Bee 5 months, 1 week ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    Posted by Mannat Dadwal 5 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Aryan Garg 5 months, 1 week ago

    The given expression can be factored as: 13×17×19×23 + 13×23 = 13×(17×19×23 + 23) Since 17, 19, and 23 are all prime numbers, the factor (17×19×23 + 23) is a composite number. Therefore, the entire expression is also a composite number.

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    • Posted by Rakshith Hs 5 months, 1 week ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Sweety R 5 months, 1 week ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    H H 5 months, 1 week ago

    HCF 50 LCM 300

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    • Posted by Aachal 🤍 5 months, 1 week ago

    CBSE > Class 10 > Mathematics
    • 2 answers

    Rakshith Hs 5 months ago

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    Punit Patodi 5 months ago

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    • Posted by Vidhi Gupta 5 months, 1 week ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    Bcb

    • Posted by Shreeya Nayak 5 months, 1 week ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Ashish Ashish 5 months, 1 week ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    H H 5 months ago

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    • Posted by Nivritee Pal 5 months, 1 week ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Golu Kumar 5 months, 1 week ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Kritika Gupta 5 months ago

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    • Posted by Gouri Nandana 5 months, 1 week ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Naman Kumar Lakra 5 months ago

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    • Posted by Abhinav Kumar 5 months, 1 week ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Rakshith Hs 5 months ago

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    • Posted by Yashika Sain 5 months, 1 week ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Abhinav Kumar 5 months ago

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    • Posted by Kirti Rana 5 months, 1 week ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Zuha Yasmin 5 months, 1 week ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Zuha Yasmin 5 months, 1 week ago

    34

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    • Posted by Diya Debbarma 5 months, 1 week ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Aryan Garg 5 months, 1 week ago

    To find the value of α^4β^3 + α^3β^4, we first need to find the sum and product of the roots of the quadratic polynomial f(t) = t^2 - 4t + 3. The sum of the roots (α + β) = -(-4) = 4 The product of the roots (αβ) = 3 Now, we can use these values to find the desired expression: α^4β^3 + α^3β^4 = α^3β^3(α + β) = α^3β^3 * 4 (since α + β = 4) = 4 * (αβ)^3 = 4 * 3^3 = 4 * 27 = 108 Therefore, the value of α^4β^3 + α^3β^4 is 108.

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    • Posted by Faiza Asad 5 months, 1 week ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Durgesh Kavate 5 months, 2 weeks ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    Posted by Priyal Khandelwal 5 months, 1 week ago

    CBSE > Class 10 > Mathematics
    • 0 answers
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    • Posted by Pritika Limboo 5 months, 2 weeks ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Bhavya Sharma 5 months, 2 weeks ago

    Let x be the original average speed of the train in km/hr. We know the total distance traveled is 63 km + 72 km = 135 km. We can express the time taken for each part of the journey using the formula time = distance / speed. Time for the first 63 km: 63 km / x hours Time for the second 72 km: 72 km / (x + 6) hours (since the speed is 6 km/hr faster than the original speed) The total time taken is 3 hours, so we can set up an equation: 63 km / x hours + 72 km / (x + 6) hours = 3 hours

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