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Verify which of the following are solutions of the equation x + 2y = 6. (a) (2, 2) (b) (0, 3)

Mathematics

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  1. Define a linear polynomial. Give an example.

A linear polynomial is a polynomial of degree one. This means the highest power of the variable in the polynomial is 1.

Example: P(x)=3x+5P(x) = 3x + 5

  1. Find the value of (32)25(32)^{\frac{2}{5}}.

  1. Express the base 32 as a power of 2.

    32=2532 = 2^5

  2. Substitute this into the expression.

    (32)25=(25)25(32)^{\frac{2}{5}} = (2^5)^{\frac{2}{5}}

  3. Apply the exponent rule (am)n=am×n(a^m)^n = a^{m \times n}.

    (25)25=25×25(2^5)^{\frac{2}{5}} = 2^{5 \times \frac{2}{5}} =22= 2^2

  4. Calculate the final value.

    22=42^2 = 4

    The value is 4\boxed{4}.

    1. Represent the equation 5x=2y+25x = 2y + 2 in the form of ax+by+c=0ax + by + c = 0.

  5. Move all terms to one side of the equation to set it equal to zero.

    5x−2y−2=05x - 2y - 2 = 0

  6. Compare with the standard form ax+by+c=0ax + by + c = 0.

    Here, a=5a=5, b=−2b=-2, and c=−2c=-2.
    The equation in the required form is 5x−2y−2=0\boxed{5x - 2y - 2 = 0}.

    1. In the figure ABCD is a parallelogram if ∠DAB=75∘\angle DAB = 75^\circ and ∠DBC=60∘\angle DBC = 60^\circ, find ∠CDB\angle CDB and ∠ADB\angle ADB.

  7. Use the property of alternate interior angles for parallel lines.

    In a parallelogram ABCD, side AD∥BCAD \parallel BC.
    When BDBD is a transversal, ∠ADB=∠DBC\angle ADB = \angle DBC (alternate interior angles).
    Given ∠DBC=60∘\angle DBC = 60^\circ.
    Therefore, ∠ADB=60∘\angle ADB = 60^\circ.

  8. Use the property of adjacent angles in a parallelogram.

    Adjacent angles in a parallelogram are supplementary.

    ∠DAB+∠ABC=180∘\angle DAB + \angle ABC = 180^\circ

    Given ∠DAB=75∘\angle DAB = 75^\circ.

    75∘+∠ABC=180∘75^\circ + \angle ABC = 180^\circ ∠ABC=180∘−75∘\angle ABC = 180^\circ - 75^\circ ∠ABC=105∘\angle ABC = 105^\circ

  9. Find ∠ABD\angle ABD.

    We know that ∠ABC=∠ABD+∠DBC\angle ABC = \angle ABD + \angle DBC.

    105∘=∠ABD+60∘105^\circ = \angle ABD + 60^\circ ∠ABD=105∘−60∘\angle ABD = 105^\circ - 60^\circ ∠ABD=45∘\angle ABD = 45^\circ

  10. Use the property of alternate interior angles again.

    In a parallelogram ABCD, side AB∥DCAB \parallel DC.
    When BDBD is a transversal, ∠CDB=∠ABD\angle CDB = \angle ABD (alternate interior angles).
    Since ∠ABD=45∘\angle ABD = 45^\circ.
    Therefore, ∠CDB=45∘\angle CDB = 45^\circ.

    The values are ∠CDB=45∘\boxed{\angle CDB = 45^\circ} and ∠ADB=60∘\boxed{\angle ADB = 60^\circ}.

    1. The surface area of a sphere is 100π cm2100\pi \text{ cm}^2. Find its radius.

  11. Write the formula for the surface area of a sphere.

    The surface area AA of a sphere with radius rr is given by:

    A=4πr2A = 4\pi r^2

  12. Substitute the given surface area into the formula.

    100π cm2=4πr2100\pi \text{ cm}^2 = 4\pi r^2

  13. Solve for r2r^2.

    Divide both sides by 4π4\pi:

    100π4π=r2\frac{100\pi}{4\pi} = r^2 25=r225 = r^2

  14. Solve for rr.

    r=25r = \sqrt{25} r=5 cmr = 5 \text{ cm}

    The radius of the sphere is 5 cm\boxed{5 \text{ cm}}.

    1. Verify whether x=45x = \frac{4}{5} is a zero of a polynomial p(x)=5x−πp(x) = 5x - \pi.

  15. Substitute the given value of xx into the polynomial.

    p(45)=5(45)−πp\left(\frac{4}{5}\right) = 5\left(\frac{4}{5}\right) - \pi

  16. Simplify the expression.

    p(45)=4−πp\left(\frac{4}{5}\right) = 4 - \pi

  17. Check if the result is zero.

    Since 4−π≠04 - \pi \neq 0 (as π≈3.14159\pi \approx 3.14159), x=45x = \frac{4}{5} is not a zero of the polynomial.
    No, x=45 is not a zero of p(x)=5x−π\boxed{\text{No, } x = \frac{4}{5} \text{ is not a zero of } p(x) = 5x - \pi}.

    1. Express 1.27‾1.\overline{27} in the form of pq\frac{p}{q}.

  18. Let xx be the given repeating decimal.

    x=1.272727...(1)x = 1.272727... \quad (1)

  19. Since two digits are repeating, multiply equation (1) by 102=10010^2 = 100.

    100x=127.272727...(2)100x = 127.272727... \quad (2)

  20. Subtract equation (1) from equation (2).

    100x−x=127.272727...−1.272727...100x - x = 127.272727... - 1.272727... 99x=12699x = 126

  21. Solve for xx.

    x=12699x = \frac{126}{99}

  22. Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 9.

    x=126÷999÷9x = \frac{126 \div 9}{99 \div 9} x=1411x = \frac{14}{11}

    The value of 1.27‾1.\overline{27} in the form pq\frac{p}{q} is 1411\boxed{\frac{14}{11}}.

    1. In the figure, O is the centre of the circle. If ∠BOC=100∘\angle BOC = 100^\circ, then find the value of xx.

  23. Identify the relationship between the angle at the center and the angle at the circumference.

    The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
    Here, arc BC subtends ∠BOC\angle BOC at the center O and ∠BDC\angle BDC (which is xx) at point D on the circumference.

  24. Apply the theorem.

    ∠BDC=12∠BOC\angle BDC = \frac{1}{2} \angle BOC

  25. Substitute the given value of ∠BOC\angle BOC.

    x=12×100∘x = \frac{1}{2} \times 100^\circ x=50∘x = 50^\circ

    The value of xx is 50∘\boxed{50^\circ}.

    1. Verify which of the following are solutions of the equation x+2y=6x + 2y = 6.

    (a) (2,2)(2, 2)

    (b) (0,3)(0, 3)

    To verify, substitute the xx and yy values from each option into the equation x+2y=6x + 2y = 6.

    *a) For the point (2,2)(2, 2):
    Substitute x=2x=2 and y=2y=2 into the equation:

    2+2(2)=62 + 2(2) = 6 2+4=62 + 4 = 6 6=66 = 6

    Since the left side equals the right side, (2,2)(2, 2) is a solution.

    *b) For the point (0,3)(0, 3):
    Substitute x=0x=0 and y=3y=3 into the equation:

    0+2(3)=60 + 2(3) = 6 0+6=60 + 6 = 6 6=66 = 6

    Since the left side equals the right side, (0,3)(0, 3) is a solution.

    Both (a) and (b) are solutions to the equation x+2y=6x + 2y = 6.

    Both (a) (2,2) and (b) (0,3) are solutions.\boxed{\text{Both (a) } (2, 2) \text{ and (b) } (0, 3) \text{ are solutions.}}
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