1.1 The picture below shows the curved flight path of an aircraft - NSC Technical Mathematics - Question 1 - 2019 - Paper 1

Setting $p(x) = 0$ gives us:
$2\left( x - \frac{2}{9} \right)\left( x + \frac{2}{9} \right) = 0$
Thus, we can solve for $x$:

1. $x - \frac{2}{9} = 0 \Rightarrow x = \frac{2}{9}$
2. $x + \frac{2}{9} = 0 \Rightarrow x = -\frac{2}{9}$
   Therefore, the solutions are $x = \frac{2}{9}$ and $x = -\frac{2}{9}$.

First, rewrite the equation:
$(3x-5)(x+2) + 13 = 0$
Expanding gives us:
$3x^2 + 6x - 5x - 10 + 13 = 0$
This simplifies to:
$3x^2 + x + 3 = 0$
Now, using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
Here, $a = 3$, $b = 1$, and $c = 3$.
Calculating the discriminant:
$b^2 - 4ac = 1^2 - 4(3)(3) = 1 - 36 = -35$
Since the discriminant is negative, we have complex solutions:
$x = \frac{-1 \pm i\sqrt{35}}{6}$

To solve the inequality, we can first find the critical values by setting $(4-x)(x+3) = 0$:\n1. $4 - x = 0 \Rightarrow x = 4$
2. $x + 3 = 0 \Rightarrow x = -3$
Thus, our critical points are $x = -3$ and $x = 4$.
Now, we test the intervals:

• For $x < -3$, choose $x = -4 \Rightarrow (4 - (-4))((-4) + 3) = (8)(-1) < 0$
• For $-3 < x < 4$, choose $x = 0 \Rightarrow (4 - 0)(0 + 3) = (4)(3) > 0$
• For $x > 4$, choose $x = 5 \Rightarrow (4 - 5)(5 + 3) = (-1)(8) < 0$
  The solution for the inequality $(4-x)(x+3) < 0$ is $x \in (-\infty, -3) \cup (4, \infty)$.

First, substitute $y=3x-8$ into the second equation:
$x^2 - x(3x - 8) + (3x - 8)^2 = 39$
Expanding yields:
$x^2 - 3x^2 + 8x + 9x^2 - 48x + 64 = 39$
Combining like terms gives:
$7x^2 - 40x + 25 = 39$
This simplifies to:
$7x^2 - 40x - 14 = 0$
Using the quadratic formula yields:
$x = \frac{-(-40) \pm \sqrt{(-40)^2 - 4 \times 7 \times (-14)}}{2 \times 7}$
Calculating the discriminant:
$1600 + 392 = 1992$
Now substituting gives:\n$x = \frac{40 \pm \sqrt{1992}}{14}$\nNow find $y = 3x - 8$ accordingly.

From the original formula $V = I \times Z$, we can isolate $I$:
$I = \frac{V}{Z}$

Using the expression for $I$:
$I = \frac{7i}{3 - i}$
Multiply the numerator and denominator by the conjugate of the denominator:
$I = \frac{7i(3 + i)}{(3 - i)(3 + i)} = \frac{21i + 7i^2}{9 + 1} = \frac{21i - 7}{10}$
Thus,
$I = -\frac{7}{10} + \frac{21}{10}i$

To simplify $101 \times 111_2$, first, convert $111_2$ to decimal.
$111_2 = 1 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0 = 4 + 2 + 1 = 7$\nNow multiply:
$101 \times 7 = 707$