Photo AI

3.1 Simplify the following WITHOUT using a calculator: 3.1.1 $$\frac{3^1 \cdot 3^{2-2}}{9^{1}}$$ 3.1.2 $$\left(\sqrt{5 + 4}\right) - \sqrt{-45}$$ 3.1.3 $$\log_{3} 8 + \log_{10} 25$$ 3.2 Solve for $x$: $$\log_{x} + \log_{(x-6)} = \log 25$$ 3.3 In the RLC circuit, the impedance of the two impedances connected in series are: $z_{1} = \frac{4}{2} \text{ cis } 225^{\circ}$ and $z_{2} = 3 - 4i$ 3.3.1 Express $z_{1}$ in rectangular form - NSC Technical Mathematics - Question 3 - 2022 - Paper 1

Question icon

Question 3

3.1-Simplify-the-following-WITHOUT-using-a-calculator:--3.1.1---$$\frac{3^1-\cdot-3^{2-2}}{9^{1}}$$----3.1.2---$$\left(\sqrt{5-+-4}\right)---\sqrt{-45}$$----3.1.3---$$\log_{3}-8-+-\log_{10}-25$$----3.2-Solve-for-$x$:---$$\log_{x}-+-\log_{(x-6)}-=-\log-25$$----3.3-In-the-RLC-circuit,-the-impedance-of-the-two-impedances-connected-in-series-are:---$z_{1}-=-\frac{4}{2}-\text{-cis-}-225^{\circ}$-and-$z_{2}-=-3---4i$----3.3.1-Express-$z_{1}$-in-rectangular-form-NSC Technical Mathematics-Question 3-2022-Paper 1.png

3.1 Simplify the following WITHOUT using a calculator: 3.1.1 $$\frac{3^1 \cdot 3^{2-2}}{9^{1}}$$ 3.1.2 $$\left(\sqrt{5 + 4}\right) - \sqrt{-45}$$ 3.1.3 ... show full transcript

Worked Solution & Example Answer:3.1 Simplify the following WITHOUT using a calculator: 3.1.1 $$\frac{3^1 \cdot 3^{2-2}}{9^{1}}$$ 3.1.2 $$\left(\sqrt{5 + 4}\right) - \sqrt{-45}$$ 3.1.3 $$\log_{3} 8 + \log_{10} 25$$ 3.2 Solve for $x$: $$\log_{x} + \log_{(x-6)} = \log 25$$ 3.3 In the RLC circuit, the impedance of the two impedances connected in series are: $z_{1} = \frac{4}{2} \text{ cis } 225^{\circ}$ and $z_{2} = 3 - 4i$ 3.3.1 Express $z_{1}$ in rectangular form - NSC Technical Mathematics - Question 3 - 2022 - Paper 1

Step 1

3.1.1 $$\frac{3^1 \cdot 3^{2-2}}{9^{1}}$$

96%

114 rated

Answer

To simplify this expression, we can express the base of the denominator in terms of the base of the numerator.

  1. Notice that 91=329^{1} = 3^{2}.
  2. Rewrite the expression:
    31⋅32−232=31⋅3032\frac{3^{1} \cdot 3^{2-2}}{3^{2}} = \frac{3^{1} \cdot 3^{0}}{3^{2}}
  3. Simplifying gives:
    131=3−1=13\frac{1}{3^{1}} = 3^{-1} = \frac{1}{3}.

Step 2

3.1.2 $$\left(\sqrt{5 + 4}\right) - \sqrt{-45}$$

99%

104 rated

Answer

  1. Calculate 5+4\sqrt{5 + 4}:
    9=3.\sqrt{9} = 3.
  2. Next, simplify −45\sqrt{-45}:
    −45=45⋅i=35i.\sqrt{-45} = \sqrt{45} \cdot i = 3\sqrt{5} i.
  3. The final expression becomes:
    3−35i.3 - 3\sqrt{5} i.

Step 3

3.1.3 $$\log_{3} 8 + \log_{10} 25$$

96%

101 rated

Answer

  1. Using the change of base formula for logarithms:
    log⁡38=log⁡108log⁡103\log_{3} 8 = \frac{\log_{10} 8}{\log_{10} 3}
  2. Calculating:
    • log⁡108=3log⁡102\log_{10} 8 = 3\log_{10} 2
    • log⁡1025=2log⁡105\log_{10} 25 = 2\log_{10} 5
  3. Therefore:
    log⁡38+log⁡1025=3log⁡102log⁡103+2log⁡105.\log_{3} 8 + \log_{10} 25 = \frac{3\log_{10} 2}{\log_{10} 3} + 2\log_{10} 5.

Step 4

3.2 Solve for $x$: $$\log_{x} + \log_{(x-6)} = \log 25$$

98%

120 rated

Answer

  1. Combine the logarithms using the product property:
    log⁡x(x−6)=log⁡25\log_{x(x-6)} = \log 25
  2. This gives:
    x(x−6)=25x(x - 6) = 25
  3. Rearranging leads to the quadratic equation:
    x2−6x−25=0.x^2 - 6x - 25 = 0.
  4. Solving via the quadratic formula:
    x=6±62+4×252=8 or −2.x = \frac{6 \pm \sqrt{6^2 + 4 \times 25}}{2} = 8 \text{ or } -2.
  5. Thus, the valid solution is x=8x = 8.

Step 5

3.3.1 Express $z_{1}$ in rectangular form.

97%

117 rated

Answer

  1. Start with:
    z1=42 cis 225∘=2(cos⁡225∘+isin⁡225∘).z_{1} = \frac{4}{2} \text{ cis } 225^{\circ} = 2(\cos 225^{\circ} + i\sin 225^{\circ}).
  2. Knowing that cos⁡225∘=−22\cos 225^{\circ} = -\frac{\sqrt{2}}{2} and sin⁡225∘=−22\sin 225^{\circ} = -\frac{\sqrt{2}}{2}, we get:
    z1=2(−22−22i)=−2−2i.z_{1} = 2\left(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i\right) = -\sqrt{2} - \sqrt{2} i.

Step 6

3.3.2 Hence, determine $(z_{1} + z_{2})$, the total impedance of the circuit.

97%

121 rated

Answer

  1. We have:
    z2=3−4i.z_{2} = 3 - 4i.
  2. Therefore:
    z1+z2=(−2+3)+(−2−4)i=(3−2)−(4+2)i.z_{1} + z_{2} = (-\sqrt{2} + 3) + (-\sqrt{2} - 4)i = (3 - \sqrt{2}) - (4 + \sqrt{2})i.
  3. This gives us the total impedance as Total Impedance=(3−2)−(4+2)i.\text{Total Impedance} = (3 - \sqrt{2}) - (4 + \sqrt{2})i.

Step 7

3.4 Determine (showing ALL working) the numerical values of $p$ and $q$ if: $$-p + qi = 4i^{-2} - 2(7 + 3i)$$

96%

114 rated

Answer

  1. Begin by simplifying:
    4i−2=4⋅(−1)=−4.4i^{-2} = 4 \cdot (-1) = -4.
  2. Now evaluate:
    −2(7+3i)=−14−6i.-2(7 + 3i) = -14 - 6i.
  3. Combine these results:
    −4−14−6i=−18−6i.-4 - 14 - 6i = -18 - 6i.
  4. This yields:
    −p=−18  ⟹  p=18-p = -18 \implies p = 18
    q=−6.q = -6.

Join the NSC students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;