Given: $f(x) = x^2 - 3x - 10$ Solve for $x$ if: 1.1.1 $f(x) = 0$ 1.1.2 $f(x) < 0$ and represent the solution on a number line - NSC Technical Mathematics - Question 1 - 2022 - Paper 1

We analyze the quadratic inequality $x^2 - 3x - 10 < 0$. Based on the previous solutions, $x = -2$ and $x = 5$ are roots.

To determine the intervals where the quadratic is negative, we will test the intervals: $(-\infty, -2)$, $(-2, 5)$, and $(5, \infty)$.

• For $x < -2$, choose $x = -3$: $(-3)^2 - 3(-3) - 10 = 9 + 9 - 10 = 8 > 0$

• For $-2 < x < 5$, choose $x = 0$: $0^2 - 3(0) - 10 = -10 < 0$

• For $x > 5$, choose $x = 6$: $6^2 - 3(6) - 10 = 36 - 18 - 10 = 8 > 0$

Thus, the solution is in the interval: $-2 < x < 5$

On a number line, this is represented as the line segment between -2 and 5, not including the endpoints.

We start with the equation:

$2x^2 - 11 = -7x$

Rearranging gives:

$2x^2 + 7x - 11 = 0$

Using the quadratic formula with $a = 2$, $b = 7$, and $c = -11$:

$x = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 2 \cdot (-11)}}{2 \cdot 2}$ $x = \frac{-7 \pm \sqrt{49 + 88}}{4}$ $x = \frac{-7 \pm \sqrt{137}}{4}$

Calculating gives: $x \approx \frac{-7 \pm 11.70}{4}$

Thus, $x \approx 1.18 \, \text{or} \, x \approx -4.18$ (correct to TWO decimal places, round if necessary).

Starting with the equations: $y - x + 1 = 0 \implies y = x - 1$ $y + 7 = x^2 + 2x \implies y = x^2 + 2x - 7$

Setting the two expressions for $y$ equal gives: $x - 1 = x^2 + 2x - 7$

Rearranging leads to: $0 = x^2 + x - 6$

Factoring gives: $(x + 3)(x - 2) = 0$

Thus, $x = -3$ or $x = 2$. Now substituting back to find $y$:

For $x = -3$: $y = -3 - 1 = -4$

For $x = 2$: $y = 2 - 1 = 1$

So the solutions are:

• For $x = -3$, $y = -4$
• For $x = 2$, $y = 1$

Starting with the equation:

$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$

Taking the reciprocal gives:

$R_p = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}$

To simplify:

$R_p = \frac{R_1 R_2}{R_1 + R_2}$

Given: $R_1 = 40 \Omega \, \text{and} \, R_2 = 45 \Omega$

Substituting into the formula yields:

$R_p = \frac{40 \times 45}{40 + 45} = \frac{1800}{85} \approx 21.18 \Omega$

Converting both binary numbers to decimal:

$1101100_2 = 108_{10}$ $1100_2 = 12_{10}$

Adding them gives: $108 + 12 = 120$

Now convert $120_{10}$ back to binary: $120_{10} = 1111000_2$

Thus, the final answer in binary is:

$1111000_2$