1. A regular hexagon has 6 equal sides. The perimeter (P) of a hexagon is given by: P = 6 × side length Given that P = 24 cm, we solve for the side length: side length = 24 cm / 6 = 4 cm Thus, each side of the hexagon is 4 cm. Click here for more: https://www.tiwariacademy.com/ncert-solutions/class-11/Read more

    A regular hexagon has 6 equal sides. The perimeter (P) of a hexagon is given by:

    P = 6 × side length
    Given that P = 24 cm, we solve for the side length:
    side length = 24 cm / 6 = 4 cm
    Thus, each side of the hexagon is 4 cm.

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    https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-13/

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  2. The area (A) of a rectangle is given by the formula: A = l × b where: - l = 10 m (length), - b = 5 m (breadth). Substituting the values: A = 10 × 5 = 50 sq. m Thus, the area of the rectangle is 50 square meters. Click here for more: https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapRead more

    The area (A) of a rectangle is given by the formula:
    A = l × b
    where:
    – l = 10 m (length),
    – b = 5 m (breadth).

    Substituting the values:
    A = 10 × 5 = 50 sq. m

    Thus, the area of the rectangle is 50 square meters.

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    https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-13/

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  3. The perimeter (P) of a square is given by: P = 4 × side length Given that P = 36 cm, we solve for the side length: side length = 36 cm / 4 = 9 cm Thus, each side of the square will be 9 cm Click here for more: https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-6/

    The perimeter (P) of a square is given by:
    P = 4 × side length
    Given that P = 36 cm, we solve for the side length:
    side length = 36 cm / 4 = 9 cm

    Thus, each side of the square will be 9 cm

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    https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-6/

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  4. To find the interval for which the function y = x³ + 6x² + 6 is increasing, we will look at its derivative. Compute the derivative of the function: y'(x) = d/dx(x³ + 6x² + 6) = 3x² + 12x Determine where the derivative is positive, so the function is increasing: y'(x) > 0 3x² + 12x > 0 Factor tRead more

    To find the interval for which the function y = x³ + 6x² + 6 is increasing, we will look at its derivative.

    Compute the derivative of the function:
    y'(x) = d/dx(x³ + 6x² + 6) = 3x² + 12x

    Determine where the derivative is positive, so the function is increasing:
    y'(x) > 0
    3x² + 12x > 0

    Factor the expression:
    3x(x + 4) > 0
    This inequality holds when x 0. Thus, the function is increasing in the intervals (-∞, -4) and (0, ∞).

    Conclusion:
    The correct intervals where the function is increasing are (-∞, 0) U (4, ∞).

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    https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-6

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  5. We are given: cos⁻¹ α + cos⁻¹ β + cos⁻¹ γ = 3π Step 1: Using the property of inverse cosine The principal value range of cos⁻¹ x is [0, π]. For the sum of three inverse cosine terms to equal 3π, each term must be equal to π. This means: cos⁻¹ α = π, cos⁻¹ β = π, cos⁻¹ γ = π Step 2: Find the values oRead more

    We are given:
    cos⁻¹ α + cos⁻¹ β + cos⁻¹ γ = 3π

    Step 1: Using the property of inverse cosine
    The principal value range of cos⁻¹ x is [0, π].
    For the sum of three inverse cosine terms to equal 3π, each term must be equal to π. This means:
    cos⁻¹ α = π, cos⁻¹ β = π, cos⁻¹ γ = π

    Step 2: Find the values of α, β, and γ
    When cos⁻¹ α = π, then cos π = -1, so:
    α = -1, β = -1, γ = -1

    Step 3: Substitute into the expression
    We are given to calculate:
    α(β + γ) + β(γ + α) + γ(α + β)

    Replace α = -1, β = -1, and γ = -1:
    (-1)((-1) + (-1)) + (-1)((-1) + (-1)) + (-1)((-1) + (-1))

    Simplify each term:
    (-1)(-2) + (-1)(-2) + (-1)(-2) = 2 + 2 + 2 = 6

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    https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-2

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