1.1 Given: $x^2 - x - 12 = p$ Solve for $x$ if: 1.1.1 $p = 0$ 1.1.2 $p \leq 0$ 1.1.3 $p = -5$ (correct to TWO decimal places) 1.2 Given: $2y - x = 7$ and $x^2 + xy = 21 - y^2$ 1.2.1 Make $x$ the subject of the equation $2y - x = 7$ 1.2.2 Hence, or otherwise solve for $x$ and $y$ - NSC Technical Mathematics - Question 1 - 2024 - Paper 1

Since $p \leq 0$, we set the equation as follows:

$x^2 - x - 12 \leq 0$
The roots of the equation from the previous part are $x = 4$ and $x = -3$.
The inequality will hold between the roots:

$x \in [-3, 4]$

Substituting $p = -5$ gives:

$x^2 - x - 12 = -5$
This simplifies to:

$x^2 - x - 7 = 0$
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we have:

$x = \frac{1 \pm \sqrt{(1)^2 - 4(1)(-7)}}{2(1)} = \frac{1 \pm \sqrt{1 + 28}}{2} = \frac{1 \pm \sqrt{29}}{2}$
Calculating the approximate values results in:

$x \approx 3.19 \quad \text{or} \quad x \approx -2.19$

To make $x$ the subject, rearranging gives:

$x = 2y - 7$

Substituting $x = 2y - 7$ into the second equation $x^2 + xy = 21 - y^2$ gives:

$(2y - 7)^2 + (2y - 7)y = 21 - y^2$
Expanding and simplifying leads to a quadratic equation in terms of $y$. Solving it will yield corresponding values for $y$ which can then be substituted back to find $x$.

We start from the equation:

$T = \frac{12.5D}{D + 4d}$
Multiplying both sides by $(D + 4d)$ gives:

$T(D + 4d) = 12.5D$
Rearranging yields:

$TD + 4Td = 12.5D$
Now, isolating $d$:

$4Td = 12.5D - TD \Rightarrow \ d = \frac{12.5D - TD}{4T}$

Substituting $T = 10$ and $D = 32$ into $d = \frac{12.5D - TD}{4T}$:

$d = \frac{12.5(32) - 10(32)}{4(10)} = \frac{400 - 320}{40} = \frac{80}{40} = 2$
Thus, the depth of the cutter is $d = 2$ cm.

Calculating the expression:

$111110 + 38 = 111148$
Then, multiply by 2:

$2 \times 111148 = 222296$
Thus, the final answer is $222296$.