Solve the inequality −7 < 5x − 2x − 3. Represent the solution on a real number line. If g(x) = x2 + 2 and h(x) = x + 2, find g(2) and h⁻¹(x).

$$-7 < (5x - 2x) - 3$$ $$-7 < 3x - 3$$

$$-7 + 3 < 3x - 3 + 3$$ $$-4 < 3x$$

$$\frac{-4}{3} < \frac{3x}{3}$$ $$-\frac{4}{3} < x$$

The solution is $x > -\frac{4}{3}$.

$$\boxed{x > -\frac{4}{3}}$$

b) Represent the solution of (a) on a real number line.

To represent $x > -\frac{4}{3}$ on a real number line, draw a number line. Place an open circle at the point $-\frac{4}{3}$ (which is approximately $-1.33$). Draw an arrow extending to the right from this open circle, indicating that all numbers greater than $-\frac{4}{3}$ are part of the solution.

:
The functions $g$ and $h$ are defined on $\mathbb{R}$ as follows:
$g(x) = x^2 + 2$
$h(x) = x + 2$

a) Find $g(2)$.

$$g(2) = (2)^2 + 2$$ $$g(2) = 4 + 2$$ $$g(2) = 6$$ $$\boxed{g(2) = 6}$$

b) Find $h^{-1}(x)$.

$$y = x - 2$$

So, the inverse function is $h^{-1}(x) = x - 2$.

$$\boxed{h^{-1}(x) = x - 2}$$

c) Find $hg(x)$. (This means $h(g(x))$)

$$h(x^2 + 2) = (x^2 + 2) + 2$$ $$h(x^2 + 2) = x^2 + 4$$ $$\boxed{h(g(x)) = x^2 + 4}$$

:
Given $y$ varies directly as $x$ and that $y = 10$ when $x = \frac{1}{2}$.

a) Find the relation between $x$ and $y$.

$$y = kx$$

where $k$ is the constant of proportionality.

$$10 \times 2 = k$$ $$k = 20$$

$$y = 20x$$ $$\boxed{y = 20x}$$

b) Hence find the value of $x$ when $y = 6$.

$$x = \frac{6}{20}$$ $$x = \frac{3}{10}$$ $$\boxed{x = \frac{3}{10}}$$