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Mathematical statement 2x² - 7x - 6

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4 parts
a

Quadratic expression\text{Quadratic expression}

b(i)

2x22x^2

b(iii)
-6

6.4 cm6.4 \text{ cm}

10 years10 \text{ years}

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14. Mathematical statement 2x2−7x−62x^2 - 7x - 6

Part a

Name given to this type of mathematical statement

This is a polynomial of degree 2.
The name is a quadratic expression or quadratic polynomial.
The name is Quadratic expression\boxed{\text{Quadratic expression}}.

Part b(i)

2x22x^2

Quadratic term

From the mathematical statement, write down the

The term containing x2x^2 is 2x22x^2.
The quadratic term is 2x2\boxed{2x^2}.

Part b(ii)-7

Coefficient of the linear term

The linear term is the term containing xx, which is −7x-7x. The coefficient is the number multiplying xx.
The coefficient of the linear term is −7\boxed{-7}.

Part b(iii)-6

Constant term

The constant term is the term without any variable.
The constant term is −6\boxed{-6}.

15. Volume of a cube is given as (2445×0.106) cm3(2445 \times 0.106) \text{ cm}^3. Use common logarithm tables to calculate the length of one side of the cube. Give your answer correct to 1 d.p.

Let VV be the volume of the cube and ss be the length of one side.
We are given V=2445×0.106 cm3V = 2445 \times 0.106 \text{ cm}^3.
The formula for the volume of a cube is V=s3V = s^3.
So, s3=2445×0.106s^3 = 2445 \times 0.106.
To find ss, we take the cube root: s=2445×0.1063s = \sqrt[3]{2445 \times 0.106}.

  1. Take the logarithm of both sides.

    log⁡s=log⁡((2445×0.106)13)\log s = \log \left( (2445 \times 0.106)^{\frac{1}{3}} \right) log⁡s=13log⁡(2445×0.106)\log s = \frac{1}{3} \log (2445 \times 0.106) log⁡s=13(log⁡2445+log⁡0.106)\log s = \frac{1}{3} (\log 2445 + \log 0.106)
  2. Find the logarithms of 2445 and 0.106 using logarithm tables.

    For log⁡2445\log 2445:
    Characteristic = 3 (since 2445 has 4 digits, 4−1=34-1=3).
    Mantissa for 24 under 4 is 0.3874. Mean difference for 5 is 9.
    So, log⁡2445=3.3874+0.0009=3.3883\log 2445 = 3.3874 + 0.0009 = 3.3883.

    For log⁡0.106\log 0.106:
    Characteristic = 1ˉ\bar{1} (since the first non-zero digit is in the first decimal place).
    Mantissa for 10 under 6 is 0.0253.
    So, log⁡0.106=1ˉ.0253\log 0.106 = \bar{1}.0253.

  3. Add the logarithms.

    log⁡(2445×0.106)=3.3883+1ˉ.0253\log (2445 \times 0.106) = 3.3883 + \bar{1}.0253 =(3+(−1))+(0.3883+0.0253)= (3 + (-1)) + (0.3883 + 0.0253) =2+0.4136= 2 + 0.4136 =2.4136= 2.4136
  4. Divide the sum by 3.

    log⁡s=13(2.4136)\log s = \frac{1}{3} (2.4136) log⁡s=0.804533...\log s = 0.804533...

    We will use 0.8045 for finding the antilog.

  5. Find the antilog of 0.8045.

    Antilog of 0.8045:
    Look up 0.80 under 4, which gives 6.368.
    The mean difference for 5 (in the row for 80) is 7.
    So, antilog 0.8045=6.368+0.007=6.3750.8045 = 6.368 + 0.007 = 6.375.
    Since the characteristic of log⁡s\log s is 0 (from 0.8045), the decimal point is after the first digit.
    So, s=6.375s = 6.375.

  6. Round the answer to 1 decimal place.

    s≈6.4s \approx 6.4 cm.
    The length of one side of the cube is 6.4 cm\boxed{6.4 \text{ cm}}.

    16. The population of a small village in Turkana grows at a rate modeled using indices. If the initial population PP triples every 5 years, use laws of indices to find how long it takes for the population to reach 9 times its original size.

    Let the initial population be P0P_0.
    The population triples every 5 years.
    After tt years, the population P(t)P(t) can be expressed as:

    P(t)=P0×(3)t5P(t) = P_0 \times (3)^{\frac{t}{5}}

    We want to find the time tt when the population reaches 9 times its original size, i.e., P(t)=9P0P(t) = 9P_0.

  7. Set up the equation.

    9P0=P0×(3)t59P_0 = P_0 \times (3)^{\frac{t}{5}}
  8. Divide both sides by P0P_0.

    9=3t59 = 3^{\frac{t}{5}}
  9. Express 9 as a power of 3.

    32=3t53^2 = 3^{\frac{t}{5}}
  10. Equate the exponents.

    Since the bases are the same, the exponents must be equal.

    2=t52 = \frac{t}{5}
  11. Solve for tt.

    t=2×5t = 2 \times 5 t=10t = 10

    It takes 10 years\boxed{10 \text{ years}} for the population to reach 9 times its original size.

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