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QUESTION 6 [EXPONENTS] Calculate the values of the following: 6.1 2/60, 6.2 53 x 54, 6.3 85 + 83, 6.4 Show by means of calculations that 25 is bigger than 52.

Mathematics

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QUESTION 6 [EXPONENTS]

6.1.

  1. Evaluate the term with the exponent.

    Any non-zero number raised to the power of 0 is 1.

    60=16^0 = 1
  2. Substitute this value back into the expression and simplify.

    21=2\frac{2}{1} = 2

    The value is 2\boxed{2}.

    6.2.

  3. Apply the rule for multiplying exponents with the same base

    am×an=am+na^m \times a^n = a^{m+n}.

    53×54=53+45^3 \times 5^4 = 5^{3+4}
  4. Add the exponents and calculate the final value.

    57=781255^7 = 78125

    The value is 78125\boxed{78125}.

    6.3.

  5. Apply the rule for dividing exponents with the same base

    am÷an=am−na^m \div a^n = a^{m-n}.

    85÷83=85−38^5 \div 8^3 = 8^{5-3}
  6. Subtract the exponents and calculate the final value.

    82=648^2 = 64

    The value is 64\boxed{64}.

    6.4.

  7. Calculate the value of 252^5.

    25=2×2×2×2×2=322^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32
  8. Calculate the value of 525^2.

    52=5×5=255^2 = 5 \times 5 = 25
  9. Compare the two values.

    32>2532 > 25

    Therefore, 252^5 is bigger than 525^2.

    QUESTION 7 [NUMERIC AND GEOMETRIC PATTERNS]

    7.1.1.

  10. Find the common difference between consecutive terms.

    −5−(−2)=−3-5 - (-2) = -3 −8−(−5)=−3-8 - (-5) = -3 −11−(−8)=−3-11 - (-8) = -3
  11. Add the common difference to the last given term to find the next term.

    −11+(−3)=−14-11 + (-3) = -14

    The next term is −14\boxed{-14}.

    7.1.2.

  12. Identify the first term (aa) and the common difference (dd).

    a=−2a = -2 d=−3d = -3
  13. Use the general formula for an arithmetic sequence, Tn=a+(n−1)dT_n = a + (n-1)d.

    Tn=−2+(n−1)(−3)T_n = -2 + (n-1)(-3)
  14. Simplify the expression to find the general rule.

    Tn=−2−3n+3T_n = -2 - 3n + 3 Tn=1−3nT_n = 1 - 3n

    The general rule is Tn=1−3n\boxed{T_n = 1 - 3n}.

    7.1.3.

  15. Use the general rule Tn=1−3nT_n = 1 - 3n found in 7.1.2.

  16. Substitute n=50n = 50 into the rule to find the 50th term.

    T50=1−3(50)T_{50} = 1 - 3(50)
  17. Perform the multiplication and subtraction.

    T50=1−150T_{50} = 1 - 150 T50=−149T_{50} = -149

    The 50th term is −149\boxed{-149}.

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