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Factorize completely: x2 - 2xy2 + y4 13. Solve for y given that y is acute and (3y - 50^) = (2y + 10^)

Mathematics

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12. Factorize completely: x2−2xy2+y4x^2 - 2xy^2 + y^4

  1. Recognize the pattern.

    The expression x2−2xy2+y4x^2 - 2xy^2 + y^4 is in the form of a perfect square trinomial, a2−2ab+b2=(a−b)2a^2 - 2ab + b^2 = (a-b)^2.

  2. Identify 'a' and 'b'.

    In this case, a=xa = x and b=y2b = y^2.

  3. Substitute 'a' and 'b' into the perfect square formula.

    x2−2x(y2)+(y2)2=(x−y2)2x^2 - 2x(y^2) + (y^2)^2 = (x - y^2)^2

    The completely factorized expression is:

    (x−y2)2\boxed{(x - y^2)^2}

    13. Solve for y given that y is acute and sin⁡(3y−50∘)=cos⁡(2y+10∘)\sin(3y - 50^\circ) = \cos(2y + 10^\circ)

  4. Use the trigonometric identity cos⁡θ=sin⁡(90∘−θ)\cos \theta = \sin(90^\circ - \theta).

    We can rewrite the right side of the equation:

    cos⁡(2y+10∘)=sin⁡(90∘−(2y+10∘))\cos(2y + 10^\circ) = \sin(90^\circ - (2y + 10^\circ)) cos⁡(2y+10∘)=sin⁡(90∘−2y−10∘)\cos(2y + 10^\circ) = \sin(90^\circ - 2y - 10^\circ) cos⁡(2y+10∘)=sin⁡(80∘−2y)\cos(2y + 10^\circ) = \sin(80^\circ - 2y)
  5. Set the arguments of the sine functions equal.

    Now the equation is sin⁡(3y−50∘)=sin⁡(80∘−2y)\sin(3y - 50^\circ) = \sin(80^\circ - 2y).
    For sin⁡A=sin⁡B\sin A = \sin B, we have two general cases:
    Case 1: A=B+360∘nA = B + 360^\circ n
    Case 2: A=180∘−B+360∘nA = 180^\circ - B + 360^\circ n
    Since y is acute, we will consider the primary solutions.

  6. Solve for y using Case 1.

    3y−50∘=80∘−2y3y - 50^\circ = 80^\circ - 2y 3y+2y=80∘+50∘3y + 2y = 80^\circ + 50^\circ 5y=130∘5y = 130^\circ y=130∘5y = \frac{130^\circ}{5} y=26∘y = 26^\circ

    This value is acute (0∘<26∘<90∘0^\circ < 26^\circ < 90^\circ).

  7. Solve for y using Case 2.

    3y−50∘=180∘−(80∘−2y)3y - 50^\circ = 180^\circ - (80^\circ - 2y) 3y−50∘=180∘−80∘+2y3y - 50^\circ = 180^\circ - 80^\circ + 2y 3y−50∘=100∘+2y3y - 50^\circ = 100^\circ + 2y 3y−2y=100∘+50∘3y - 2y = 100^\circ + 50^\circ y=150∘y = 150^\circ

    This value is not acute.

    Therefore, the acute value for y is:

    y=26∘\boxed{y = 26^\circ}

    14. A solid consists of a cone and a hemisphere. The common diameter of the cone and the hemisphere is 10 cm, and the slanting height of the cone is 12 cm. Calculate correct to two decimal places the surface area of the solid.

  8. Identify the dimensions.

    Common diameter = 10 cm.
    Radius (rr) = 10 cm2=5\frac{10 \text{ cm}}{2} = 5 cm.
    Slanting height of the cone (ll) = 12 cm.

  9. Determine the components of the total surface area.

    The solid is formed by joining a cone and a hemisphere at their circular bases. Therefore, the total surface area of the solid is the sum of the curved surface area of the cone and the curved surface area of the hemisphere.
    Total Surface Area = Curved Surface Area of Cone + Curved Surface Area of Hemisphere.

  10. Write down the formulas for each component.

    Curved Surface Area of Cone = πrl\pi r l
    Curved Surface Area of Hemisphere = 2πr22 \pi r^2

  11. Substitute the values into the formulas.

    Total Surface Area = π(5 cm)(12 cm)+2π(5 cm)2\pi (5 \text{ cm})(12 \text{ cm}) + 2 \pi (5 \text{ cm})^2
    Total Surface Area = 60π cm2+2π(25 cm2)60 \pi \text{ cm}^2 + 2 \pi (25 \text{ cm}^2)
    Total Surface Area = 60π cm2+50π cm260 \pi \text{ cm}^2 + 50 \pi \text{ cm}^2
    Total Surface Area = 110π cm2110 \pi \text{ cm}^2

  12. Calculate the numerical value and round to two decimal places.

    Using π≈3.14159265...\pi \approx 3.14159265...
    Total Surface Area = 110×3.14159265110 \times 3.14159265
    Total Surface Area ≈345.5751915\approx 345.5751915 cm2^2

    Rounding to two decimal places:
    Total Surface Area ≈345.58\approx 345.58 cm2^2

    The surface area of the solid is:

    345.58 cm2\boxed{345.58 \text{ cm}^2}
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