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 $2y - x = 7$ - NSC Technical Mathematics - Question 1 - 2024 - Paper 1

Here, since $p$ can be any non-positive value, we rewrite the equation as:

$x^2 - x - 12 = p \\leq 0$

The roots of the equation $x^2 - x - 12 = 0$ are $x = 4$ and $x = -3$. Therefore, the regions in which the quadratic is less than or equal to zero lies between the roots:

Thus, $x \in [-3, 4]$

For $p = -5$, we solve the equation:

$x^2 - x - 12 = -5$

which simplifies to:

$x^2 - x - 7 = 0$

Applying the quadratic formula:

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-7)}}{2(1)} = \frac{1 \pm \sqrt{1 + 28}}{2} = \frac{1 \pm \sqrt{29}}{2}$

Calculating the square root gives us two values:

$x = \frac{1 + 5.385}{2} \approx 3.19$ and $x = \frac{1 - 5.385}{2} \approx -2.19$

So the solutions correct to TWO decimal places are:

$x \approx 3.19 ext{ and } x \approx -2.19$

Starting from the equation:

$2y - x = 7$

We isolate $x$ to make it the subject:

Now, substitute $x = 2y - 7$ into the second equation:

$x^2 + xy - 21 = y^2$

Replacing $x$ gives:

$(2y - 7)^2 + (2y - 7)y - 21 = y^2$

Expanding this yields:

Starting from:

$T = \frac{12.5D}{D + 4d}$

Multiply both sides by $(D + 4d)$ to clear the fraction:

$T(D + 4d) = 12.5D$

Expanding gives:

$TD + 4Td = 12.5D$

Rearranging for $d$:

Substituting the values into the rearranged equation:

$d = \frac{12.5(32) - 10(32)}{4(10)}$

Calculating step by step:

$= \frac{400 - 320}{40} = \frac{80}{40} = 2$

Thus, the depth of the cutter is $d = 2 ext{ cm}$.

Calculating within the parentheses first:

$111110 + 38 = 111148$

Now multiply by 2:

$2(111148) = 222296$

Thus, the final answer is:

$222296$