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Evaluate (12.78 x 10⁻³)/9 x 10⁻¹ and find angles ACB and ADB.

Mathematics

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Answers

3 parts
a

1.42×10−21.42 \times 10^{-2}

b

1.42×10−21.42 \times 10^{-2}

c
0.014

25∘25^\circ

Step-by-step solution

3 steps

Here are the solutions to the remaining questions.

4. Evaluate (12.78×10−3)9×10−1\frac{(12.78 \times 10^{-3})}{9 \times 10^{-1}} expressing your answer in:

Step 1: Evaluate the expression.

12.78×10−39×10−1=(12.789)×(10−310−1)\frac{12.78 \times 10^{-3}}{9 \times 10^{-1}} = \left(\frac{12.78}{9}\right) \times \left(\frac{10^{-3}}{10^{-1}}\right) =1.42×10−3−(−1)= 1.42 \times 10^{-3 - (-1)} =1.42×10−3+1= 1.42 \times 10^{-3 + 1} =1.42×10−2= 1.42 \times 10^{-2}

Part a

Standard form

The number is already in standard form.

1.42×10−2\boxed{1.42 \times 10^{-2}}

Part b

1.42×10−21.42 \times 10^{-2}

Convert to 3 significant figures

The number 1.42×10−21.42 \times 10^{-2} has three significant figures (1, 4, 2).

1.42×10−2\boxed{1.42 \times 10^{-2}}

Part c0.014

In three decimal places

First, convert to decimal form:

1.42×10−2=0.01421.42 \times 10^{-2} = 0.0142

Now, round to three decimal places. The fourth decimal place is 2, so we round down.

0.014\boxed{0.014}

5. Find angle ACB and angle ADB

The image shows a circle with center O. The angle ∠AOB=50∘\angle AOB = 50^\circ is the central angle subtended by the arc AB. Angles ∠ACB\angle ACB and ∠ADB\angle ADB are angles subtended by the same arc AB at the circumference.

  1. Apply the circle theorem for angles subtended by the same arc.

    The angle subtended by an arc at the center is twice the angle subtended by the same arc at any point on the remaining part of the circle.
    Therefore, ∠ACB=12∠AOB\angle ACB = \frac{1}{2} \angle AOB.

  2. Calculate ∠ACB\angle ACB.

    ∠ACB=12×50∘\angle ACB = \frac{1}{2} \times 50^\circ ∠ACB=25∘\angle ACB = 25^\circ
  3. Calculate ∠ADB\angle ADB.

    Angles subtended by the same arc in the same segment are equal. Since ∠ADB\angle ADB is also subtended by arc AB in the same segment as ∠ACB\angle ACB, they are equal.

    ∠ADB=∠ACB\angle ADB = \angle ACB ∠ADB=25∘\angle ADB = 25^\circ

    The value of ∠ACB\angle ACB is 25∘\boxed{25^\circ} and the value of ∠ADB\angle ADB is 25∘\boxed{25^\circ}.

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