We read your photo as:

Use logarithms to evaluate (58.72 x 406)/(0.845). A farmer fenced a rectangular plot of land. The length of the plot is 3 metres more than its width. Given that the total area of the plot is 180 m2. F

Mathematics

ScanSolve AI Answer

Answers

2 parts
a

w2+3w−180=0w^2 + 3w - 180 = 0

b

12 m12 \text{ m}

15 m15 \text{ m}

Step-by-step solution

13 steps

Here are the solutions to the questions from the image:

16. Use logarithms to evaluate 58.72×4060.845\frac{58.72 \times 406}{0.845}.

Step 1: Let N=58.72×4060.845N = \frac{58.72 \times 406}{0.845}.

Step 2: Take the logarithm of both sides.

log⁡N=log⁡(58.72)+log⁡(406)−log⁡(0.845)\log N = \log(58.72) + \log(406) - \log(0.845)

Step 3: Find the logarithms of each number using a logarithm table.

  • log⁡(58.72)=1.7690\log(58.72) = 1.7690
  • log⁡(406)=2.6085\log(406) = 2.6085
  • log⁡(0.845)=1ˉ.9269\log(0.845) = \bar{1}.9269

Step 4: Perform the addition for the numerator.

log⁡(58.72×406)=1.7690+2.6085=4.3775\log(58.72 \times 406) = 1.7690 + 2.6085 = 4.3775

Step 5: Perform the subtraction.

log⁡N=4.3775−1ˉ.9269\log N = 4.3775 - \bar{1}.9269 log⁡N=4.3775−(−1+0.9269)\log N = 4.3775 - (-1 + 0.9269) log⁡N=4.3775+1−0.9269\log N = 4.3775 + 1 - 0.9269 log⁡N=5.3775−0.9269\log N = 5.3775 - 0.9269 log⁡N=4.4506\log N = 4.4506

Step 6: Find the antilogarithm of 4.4506.

Antilog(0.4506)≈2.822\text{Antilog}(0.4506) \approx 2.822 N=2.822×104N = 2.822 \times 10^4 N=28220N = \boxed{28220}

17. A farmer fenced a rectangular plot of land. The length of the plot is 3 metres more than its width. Given that the total area of the plot is 180 m2^2.

Part a· 6 steps

Form quadratic equation to represent this situation.

  1. Let the width of the plot be ww metres.

  2. The length of the plot is 3 metres more than its width, so the length l=(w+3)l = (w + 3) metres.

  3. The area of a rectangle is given by length ×\times width.

    Area=l×w\text{Area} = l \times w
  4. Substitute the expressions for length and width, and the given area.

    180=(w+3)w180 = (w + 3)w
  5. Expand the equation.

    180=w2+3w180 = w^2 + 3w
  6. Rearrange into the standard quadratic form aw2+bw+c=0aw^2 + bw + c = 0.

    w2+3w−180=0\boxed{w^2 + 3w - 180 = 0}

Part b

12 m12 \text{ m}

Solve the equation by factorisation method to find the dimensions of the plot.

  1. Use the quadratic equation from part (a)

    w2+3w−180=0w^2 + 3w - 180 = 0.

  2. Find two numbers that multiply to -180 and add up to 3.

    These numbers are 15 and -12.

  3. Rewrite the middle term using these numbers.

    w2+15w−12w−180=0w^2 + 15w - 12w - 180 = 0
  4. Factor by grouping.

    w(w+15)−12(w+15)=0w(w + 15) - 12(w + 15) = 0 (w−12)(w+15)=0(w - 12)(w + 15) = 0
  5. Solve for ww.

    w−12=0orw+15=0w - 12 = 0 \quad \text{or} \quad w + 15 = 0 w=12orw=−15w = 12 \quad \text{or} \quad w = -15
  6. Since width cannot be negative, we take the positive value for ww.

    w=12 mw = 12 \text{ m}
  7. Calculate the length using l=w+3l = w + 3.

    l=12+3=15 ml = 12 + 3 = 15 \text{ m}

    The dimensions of the plot are:

    Width=12 m\text{Width} = \boxed{12 \text{ m}} Length=15 m\text{Length} = \boxed{15 \text{ m}}
Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp