Vereenvoudig die volgende sonder om 'n sakrekenaar te gebruik: 3.1.1 $$ \frac{8 \cdot x^{3} \cdot y^{2}}{16 \cdot x^{4} \cdot y^{4}} $$ (laat die antwoord met positiwe eksponente) 3.1.2 $$ \frac{\sqrt{48} + \sqrt{12}}{27} $$ 3.2 Indien $log_{5} m = m$, bepaal die volgende in terme van $m$: 3.2.1 $log 25$ 3.2.2 $log 2$ 3.3 Los op vir $x$: $$ log_{2}(x + 3) - 3 = - log_{2}(x - 4) $$ 3.4 Gegee die komplekse getalle: $$ z_{1} = -1 + 3i \ en \ z_{2} = \sqrt{2} \ ext{cis} 135^{\circ} $$ 3.4.1 Skryf die toegedoven van $z_{1}$ neer - NSC Technical Mathematics - Question 3 - 2022 - Paper 1

To simplify the expression $\frac{\sqrt{48} + \sqrt{12}}{27}$, we start by simplifying the square roots:

$\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3} \quad \text{and} \quad \sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}$.

Thus, we have:

$\sqrt{48} + \sqrt{12} = 4\sqrt{3} + 2\sqrt{3} = 6\sqrt{3}$.

Now we can substitute back into the original expression:

$\frac{6\sqrt{3}}{27} = \frac{2\sqrt{3}}{9}$.

Final answer: $\frac{2\sqrt{3}}{9}$.

Given that $log_{5} m = m$, we can use the property of logarithms to find $log 25$:

$log 25 = log(5^{2}) = 2 \cdot log 5 = 2m$.

Thus, the answer is: $2m$.

Using the same property as before:

$log 2 = log(\frac{10}{5}) = log 10 - log 5 = 1 - m$.

Therefore, the answer is: $1 - m$.

To solve for $x$ in the equation:

$log_{2}(x + 3) - 3 = - log_{2}(x - 4)$,

first, we can move the logs to one side:

$log_{2}(x + 3) + log_{2}(x - 4) = 3$.

Using log properties:

$log_{2}((x + 3)(x - 4)) = 3$.

This exponentiates as:

$(x + 3)(x - 4) = 2^{3} = 8$.

Expanding gives:

$x^{2} - x - 12 = 0$.

Factoring this quadratic gives:

$(x + 3)(x - 4) = 0$,

thus, $x = -3$ or $x = 4$. However,

only $x = 5$ satisfies the original logarithm's domain condition of $x + 3 > 0$ and $x - 4 > 0$.

Final solutions: $x = 5 ext{ or } x = 4$.

For the complex number:

$z_{1} = -1 + 3i$,

the conjugate $\overline{z_{1}}$ is:

$\overline{z_{1}} = -1 - 3i$.

To express $z_{2} = \sqrt{2} \text{cis} 135^{\circ}$ in standard form:

Recall $\text{cis} \theta = cos \theta + i sin \theta$:

$z_{2} = \sqrt{2} (cos 135^{\circ} + i sin 135^{\circ})$.

Calculating, we find:

$cos 135^{\circ} = -\frac{1}{\sqrt{2}} \quad \text{and} \quad sin 135^{\circ} = \frac{1}{\sqrt{2}}$

Thus: $z_{2} = \sqrt{2}( -\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} ) = -1 + i$.

Given: $z_{1} = z_{2}$,

i.e.: $-1 + 3i = -1 + i$.

This simplifies to: $3i = i$,

yielding: $2i = 0$, confirming that $z_{1}$ does not equal $z_{2}$. Thus:

The equality does not hold true.

To solve for $x$ and $y$ in the equation: $x + yi - (1 - i) = 4 + 5i$,

we rearrange it: $(x - 1) + (y + 1)i = 4 + 5i$

Equating real parts and imaginary parts gives: $x - 1 = 4 \quad \text{and} \quad y + 1 = 5$.

Thus, from $x - 1 = 4$ we find: $x = 5$, and from $y + 1 = 5$: $y = 4$.

Final answers are: $x = 5 \text{ and } y = 4$.