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List prime numbers between 100 and 120 and find the product of the largest and smallest. Classify 1, 2, and sqrt(5) as rational or irrational. Calculate the fraction of journey covered.

Mathematics

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6. Prime numbers and their product

  1. List prime numbers between 100 and 120.

    A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.
    The prime numbers between 100 and 120 are:

    101,103,107,109,113101, 103, 107, 109, 113
  2. Identify the largest and smallest prime numbers.

    Smallest prime number = 101
    Largest prime number = 113

  3. Calculate the product of the largest and smallest prime numbers.

    Product = 101×113101 \times 113

    101×113=11413101 \times 113 = 11413

    The product of the largest and smallest prime numbers is 11413\boxed{11413}.

    7. Classification of numbers

  4. Define rational and irrational numbers.

    A rational number can be expressed as a fraction pq\frac{p}{q} where pp and qq are integers and q≠0q \neq 0.
    An irrational number cannot be expressed as a simple fraction; its decimal representation is non-terminating and non-repeating.

  5. Classify the given numbers.

    • 1: This can be written as 11\frac{1}{1}. Therefore, 1 is a rational number.
    • 2: This can be written as 21\frac{2}{1}. Therefore, 2 is a rational number.
    • 5\sqrt{5}: The square root of 5 is approximately 2.2360679..., which is a non-terminating and non-repeating decimal. Therefore, 5\sqrt{5} is an irrational number.

    8. Fraction of journey covered

  6. Determine the total time for the journey.

    Total time = 5 hours.

  7. Determine the fraction of the journey covered in 1 hour.

    Fraction covered in 1 hour = 1 hour5 hours=15\frac{1 \text{ hour}}{5 \text{ hours}} = \frac{1}{5}.

  8. Use reciprocal tables to find the value of 15\frac{1}{5}.

    From reciprocal tables, the reciprocal of 5 is 0.2.

    15=0.2\frac{1}{5} = 0.2

    The fraction of the journey covered in 1 hour is 0.2\boxed{0.2}.

    9. Evaluation using mathematical tables

  9. Evaluate 10.981\frac{1}{0.981} using reciprocal tables.

    From reciprocal tables, 10.981≈1.01936799...\frac{1}{0.981} \approx 1.01936799...
    Rounding to 5 decimal places for intermediate calculation: 1.01937

  10. Evaluate 19.81\frac{1}{9.81} using reciprocal tables.

    From reciprocal tables, 19.81≈0.101936799...\frac{1}{9.81} \approx 0.101936799...
    Rounding to 5 decimal places for intermediate calculation: 0.10194

  11. Add the results.

    1.01937+0.10194=1.121311.01937 + 0.10194 = 1.12131

    The value is approximately 1.1213\boxed{1.1213}.

    10. Laws of indices

  12. Rewrite the expression using fractional exponents.

    256813=(25681)13\sqrt[3]{\frac{256}{81}} = \left(\frac{256}{81}\right)^{\frac{1}{3}}
  13. Express the numbers as powers of their prime factors.

    256=28256 = 2^8
    81=3481 = 3^4

  14. Substitute these into the expression.

    (2834)13=(28)13(34)13=283343\left(\frac{2^8}{3^4}\right)^{\frac{1}{3}} = \frac{(2^8)^{\frac{1}{3}}}{(3^4)^{\frac{1}{3}}} = \frac{2^{\frac{8}{3}}}{3^{\frac{4}{3}}}
  15. Simplify the exponents.

    283=22+23=22⋅223=4⋅223=4432^{\frac{8}{3}} = 2^{2 + \frac{2}{3}} = 2^2 \cdot 2^{\frac{2}{3}} = 4 \cdot \sqrt[3]{2^2} = 4 \sqrt[3]{4} 343=31+13=31⋅313=3333^{\frac{4}{3}} = 3^{1 + \frac{1}{3}} = 3^1 \cdot 3^{\frac{1}{3}} = 3 \sqrt[3]{3}
  16. Combine the terms.

    443333\frac{4 \sqrt[3]{4}}{3 \sqrt[3]{3}}

    To rationalize the denominator, multiply the numerator and denominator by 323=93\sqrt[3]{3^2} = \sqrt[3]{9}.

    443333×9393=44×9333×93=43633273=43633×3=43639\frac{4 \sqrt[3]{4}}{3 \sqrt[3]{3}} \times \frac{\sqrt[3]{9}}{\sqrt[3]{9}} = \frac{4 \sqrt[3]{4 \times 9}}{3 \sqrt[3]{3 \times 9}} = \frac{4 \sqrt[3]{36}}{3 \sqrt[3]{27}} = \frac{4 \sqrt[3]{36}}{3 \times 3} = \frac{4 \sqrt[3]{36}}{9}

    The simplified expression is 43639\boxed{\frac{4 \sqrt[3]{36}}{9}}.

    11. Evaluation using logarithm tables
    Let x=0.724×20.37608x = \frac{0.724 \times 20.37}{608}.

  17. Take the logarithm base 10 of both sides.

    log⁡10x=log⁡10(0.724×20.37608)\log_{10} x = \log_{10} \left(\frac{0.724 \times 20.37}{608}\right)
  18. Apply logarithm rules.

    log⁡10x=log⁡10(0.724)+log⁡10(20.37)−log⁡10(608)\log_{10} x = \log_{10} (0.724) + \log_{10} (20.37) - \log_{10} (608)
  19. Find the logarithms of each number using tables.

    • log⁡10(0.724)\log_{10} (0.724): Characteristic is 1ˉ\bar{1}. Mantissa of 724 is 0.8597. So, log⁡10(0.724)=1ˉ.8597\log_{10} (0.724) = \bar{1}.8597.
    • log⁡10(20.37)\log_{10} (20.37): Characteristic is 1. Mantissa of 2037 is 0.3090. So, log⁡10(20.37)=1.3090\log_{10} (20.37) = 1.3090.
    • log⁡10(608)\log_{10} (608): Characteristic is 2. Mantissa of 608 is 0.7839. So, log⁡10(608)=2.7839\log_{10} (608) = 2.7839.
  20. Perform the addition and subtraction.

    log⁡10x=1ˉ.8597+1.3090−2.7839\log_{10} x = \bar{1}.8597 + 1.3090 - 2.7839 log⁡10x=(−1+0.8597)+(1+0.3090)−(2+0.7839)\log_{10} x = ( -1 + 0.8597 ) + ( 1 + 0.3090 ) - ( 2 + 0.7839 ) log⁡10x=−1+0.8597+1+0.3090−2−0.7839\log_{10} x = -1 + 0.8597 + 1 + 0.3090 - 2 - 0.7839 log⁡10x=(−1+1−2)+(0.8597+0.3090−0.7839)\log_{10} x = (-1 + 1 - 2) + (0.8597 + 0.3090 - 0.7839) log⁡10x=−2+(1.1687−0.7839)\log_{10} x = -2 + (1.1687 - 0.7839) log⁡10x=−2+0.3848\log_{10} x = -2 + 0.3848 log⁡10x=2ˉ.3848\log_{10} x = \bar{2}.3848
  21. Find the antilogarithm of 2ˉ.3848\bar{2}.3848.

    The mantissa is $0.3848

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