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Convert between numbers and words for large Indian numbering system values and perform polynomial division.

Mathematics

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Here are the solutions to the math problems.

Q1. Write in words / Write in numbers:

  1. 22, 05, 02, 038 Twenty-two crore five lakh two thousand thirty-eight.
  2. Thirteen crore two lakh eighty thousand nine hundred fifty-six 13,02,80,956

Q2. Divide:

  1. (25m4−15m3)÷5m(25m^4 - 15m^3) \div 5m
  1. Divide each term in the numerator by the denominator.

    25m4−15m35m=25m45m−15m35m\frac{25m^4 - 15m^3}{5m} = \frac{25m^4}{5m} - \frac{15m^3}{5m}
  2. Simplify each term using the exponent rule axay=ax−y\frac{a^x}{a^y} = a^{x-y}.

    25m45m=5m4−1=5m3\frac{25m^4}{5m} = 5m^{4-1} = 5m^3 15m35m=3m3−1=3m2\frac{15m^3}{5m} = 3m^{3-1} = 3m^2
  3. Combine the simplified terms.

    5m3−3m25m^3 - 3m^2

    The final answer is 5m3−3m2\boxed{5m^3 - 3m^2}.

    1. (y3+0y2+0y+216)÷(y+6)(y^3 + 0y^2 + 0y + 216) \div (y+6)
  4. Recognize the numerator as a sum of cubes, y3+216y^3 + 216.

    Note that 216=63216 = 6^3.
    So, the expression is y3+63y^3 + 6^3.

  5. Apply the sum of cubes formula

    a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a+b)(a^2 - ab + b^2).
    Here, a=ya=y and b=6b=6.

    y3+63=(y+6)(y2−(y)(6)+62)y^3 + 6^3 = (y+6)(y^2 - (y)(6) + 6^2) y3+63=(y+6)(y2−6y+36)y^3 + 6^3 = (y+6)(y^2 - 6y + 36)
  6. Substitute this back into the division problem.

    (y+6)(y2−6y+36)(y+6)\frac{(y+6)(y^2 - 6y + 36)}{(y+6)}
  7. Cancel out the common factor (y+6)(y+6) from the numerator and denominator.

    y2−6y+36y^2 - 6y + 36

    The final answer is y2−6y+36\boxed{y^2 - 6y + 36}.

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