Solve for $x$: 1.1.1 $(3 – x)(x + 1) = 0$ 1.1.2 $2x^2 = 3x + 7$ (correct to TWO decimal places) 1.1.3 $x(x – 5) \leq 0$ Solve for $x$ and $y$ if: $y + x = 3$ and $x^2 + y^2 = 89$ The formula for calculating the electrical force between two charges is given by: $F = \frac{K Q_1 Q_2}{r^2}$ where $F$ = the force in N $K$ = Coulomb's constant in $Nm^2/C^2$ $r$ = distance in m $Q_1$ = charge in C $Q_2$ = charge in C 1.3.1 Make $r$ the subject of the formula - NSC Technical Mathematics - Question 1 - 2021 - Paper 1

Rearranging gives:

$2x^2 - 3x - 7 = 0$

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 2$, $b = -3$, and $c = -7$:

$x = \frac{3 \pm \sqrt{(-3)^2 - 4(2)(-7)}}{2(2)}$

Calculating yields:

$x = \frac{3 \pm \sqrt{9 + 56}}{4} = \frac{3 \pm \sqrt{65}}{4}$

The approximate values are $x \approx 2.766$ and $x \approx -0.766$, rounded to two decimal places.

To solve this inequality, we identify critical points by setting:

1. $x = 0$ (from $x = 0$)
2. $x = 5$ (from $x - 5 = 0$)

We perform a sign analysis:

• Choose test points from the intervals $(-\infty, 0)$, $(0, 5)$, and $(5, \infty)$.
• This shows that the solution satisfies the inequality in the interval:
  $[0, 5]$.

From $y + x = 3$, we express $y$ in terms of $x$:

$y = 3 - x$

Substituting into the second equation:

$x^2 + (3 - x)^2 = 89$

Expanding gives:

$x^2 + (9 - 6x + x^2) = 89 \Rightarrow 2x^2 - 6x - 80 = 0$

Dividing by 2 results in:

$x^2 - 3x - 40 = 0$ Using the quadratic formula here gives:

$x = \frac{3 \pm \sqrt{9 + 160}}{2} = \frac{3 \pm 13}{2}$

The solutions for $x$ yield $x = 8$ or $x = -5$. Substituting these back yields corresponding $y$ values:

• If $x = 8$, then $y = -5$.
• If $x = -5$, then $y = 8$.

Starting from:

$F = \frac{K Q_1 Q_2}{r^2}$

Rearranging to make $r$ the subject:

$r^2 = \frac{K Q_1 Q_2}{F}$

Taking the square root gives:

$r = \sqrt{\frac{K Q_1 Q_2}{F}}$

Substituting the values into the derived formula:

$r = \sqrt{\frac{(9 \times 10^9)(0.5 \times 10^{-6})(2.0 \times 10^{-6})}{2.25 \times 10^4}}$

Calculating this gives:

$r = \sqrt{\frac{9 \times 10^9 \times 1 \times 10^{-12}}{2.25 \times 10^4}} = \sqrt{\frac{9 \times 10^{-3}}{2.25}} \approx 2 m$

Adding the two binary numbers:

  1101
+ 1111
------
 11000

So, $1101_2 + 1111_2 = 11000_2$.