1.1 Solve for x: 1.1.1 (7 - 3x)(x - 8 - x) = 0 1.1.2 3x^2 - 4x = \frac{1}{3} (correct to TWO decimal places) 1.1.3 -x^2 + 16 > 0 1.2 Solve for x and y if: x - y = 1 and x + 2xy + y^2 = 9 1.3 The diagram below shows an RCL circuit used for voltage magnification - NSC Technical Mathematics - Question 1 - 2023 - Paper 1

Rearranging gives:

[3x^2 - 4x - \frac{1}{3} = 0]

To eliminate the fraction, multiply through by 3:

[9x^2 - 12x - 1 = 0]

Using the quadratic formula: (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}):

[x = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 9 \cdot (-1)}}{2 \cdot 9}] [= \frac{12 \pm \sqrt{144 + 36}}{18} = \frac{12 \pm \sqrt{180}}{18} = \frac{12 \pm 13.42}{18}]

Thus, the two possible solutions in decimal places are approximately 1.41 and -0.08.

Rearranging gives:

[-x^2 > -16 \Rightarrow x^2 < 16]

Taking the square root of both sides (considering both positive and negative roots):

[x < 4 \text{ and } x > -4]

Therefore, the solution is (-4 < x < 4).

From the first equation:

[y = x - 1]

Substituting this value into the second equation:

[x + 2x(x - 1) + (x - 1)^2 = 9]

This simplifies to:

[x + 2x^2 - 2x + x^2 - 2x + 1 = 9] [3x^2 - 4x - 8 = 0]

Applying the quadratic formula gives:

[x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot (-8)}}{2 \cdot 3}] [= \frac{4 \pm \sqrt{16 + 96}}{6} = \frac{4 \pm \sqrt{112}}{6}]

Thus, both possible values for x are approximately 2.21 and -1.21. Substituting back gives values for y as approximately 1.21 and -2.21 respectively.

Starting from the resonant frequency formula:

[f_r = \frac{1}{2\pi\sqrt{LC}}]

We can rearrange it as follows:

1. Multiply both sides by (2\pi\sqrt{LC}): [f_r imes 2\pi\sqrt{LC} = 1]

2. Divide by (f_r): [2\pi\sqrt{LC} = \frac{1}{f_r}]

3. Next, square both sides: [(2\pi)^2(L)(C) = \left(\frac{1}{f_r}\right)^2]

4. Finally, isolate L: [L = \frac{1}{4\pi^2f_r^2C}]

Substituting in the values:

[L = \frac{1}{4\pi^2(1.59)^2(0.65 \times 10^{-6})}]

Calculating this step by step results in:

1. Calculate the denominator: [4\pi^2 \approx 39.478]
   [1.59^2 \approx 2.5281] [0.65 \times 10^{-6} = 0.00000065]

2. Final calculation: [L = \frac{1}{39.478 \times 2.5281 \times 0.00000065} \approx 15414.61 H]

To convert the decimal number 24 to binary:

• Start by dividing 24 by 2:
• 24 ÷ 2 = 12 remainder 0
• 12 ÷ 2 = 6 remainder 0
• 6 ÷ 2 = 3 remainder 0
• 3 ÷ 2 = 1 remainder 1
• 1 ÷ 2 = 0 remainder 1

Reading the remainders from bottom to top, we get: [24_{10} = 11000_2]

First convert (110_2) to decimal:

• 1 * 2^2 + 1 * 2^1 + 0 * 2^0 = 4 + 2 + 0 = 6.

Then perform the addition: [144 + 6 = 150].

Thus, the final answer is 150.