1.1 Los op vir x: 1.1.1 (7 - 3x)(x - 8 - x) = 0 1.1.2 3x² - 4x = \frac{1}{3} (korrek tot TWEE desimale plekke) 1.1.3 -x² + 16 > 0 1.2 Los op vir x en y indien: x = y - 1 en x + 2xy + y² = 9 1.3 Die diagram hieronder toon 'n RCL-stroombaan wat vir spanningsverhoging gebruik word - NSC Technical Mathematics - Question 1 - 2023 - Paper 1

First, rearrange the equation:

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

Multiply through by 3 to eliminate the fraction:

$9x^2 - 12x - 1 = 0$

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we find:

$x = \frac{12 \pm \sqrt{144 + 36}}{18} = \frac{12 \pm \sqrt{180}}{18}$

Calculating further leads to:

$x = \frac{12 \pm 6 \sqrt{5}}{18} = \frac{2 \pm \sqrt{5}}{3}$

The approximate values give us:

$x \approx 1.41 \text{ or } x \approx -0.08$

Rearranging gives:

$x^2 < 16$

Therefore,

$-4 < x < 4$.

This indicates that the solution for x must be in the range of $(-4, 4)$.

From the first equation, solving for y:

$y = x + 1$.

Substituting into the second equation gives:

$x + 2x(x+1) + (x+1)^2 = 9$

This simplifies to:

$x + 2x^2 + 2x + x^2 + 2x + 1 = 9$

Combine like terms:

$3x^2 + 5x - 8 = 0$

Using the quadratic formula:

$x = \frac{-5 \pm \sqrt{25 + 96}}{2 \cdot 3} = \frac{-5 \pm 11}{6}$

Thus,

$x \approx 1 ext{ or } x \approx -2.67$

Then substitute these x-values back into $y = x + 1$ to find the corresponding y-values:

Starting with:

$f_r = \frac{1}{2\pi \sqrt{LC}}$

Rearranging to make L the subject:

$\sqrt{LC} = \frac{1}{2\pi f_r}$

By squaring both sides:

$LC = \left(\frac{1}{2\pi f_r}\right)^2$

Thus,

$L = \frac{1}{C(2\pi f_r)^2}$

Substituting the values gives:

$L = \frac{1}{0.65 \times 10^{-6} \times (2\pi \cdot 1.59)^2}$

Calculating this results in:

$L \approx 15414.61 H$.

The binary representation of 24 is:

$24 = 11000_2$.

Calculating this simply:

$144 + 110 = 254$.