1. **State whether each expression is True or False:** A) $3x^2 + 5x - 7 = 0$ is a quadratic equation in standard form, so it is **True**. B) $(x + 2)^2 = 2(x + 3)$ expands to $x^2 + 4x + 4 = 2x + 6$. Simplifying: $x^2 + 4x + 4 - 2x - 6 = 0 \Rightarrow x^2 + 2x - 2 = 0$. This is not an identity, so the original equality is **False**. C) $x^2 + 3x = (3 - 3x)^2$ expands right side to $9 - 18x + 9x^2$. So $x^2 + 3x = 9 - 18x + 9x^2$ rearranged: $0 = 9 - 18x + 9x^2 - x^2 - 3x = 9 - 21x + 8x^2$. Not an identity, so **False**. D) $(x + 3)(x - 1) = x^2 - 2x - 3$ expands left side to $x^2 - x + 3x - 3 = x^2 + 2x - 3$. Since $x^2 + 2x - 3 \neq x^2 - 2x - 3$, this is **False**. E) $x^2 + 3x + 1 = (x + 1)^2$ expands right side to $x^2 + 2x + 1$. Since $x^2 + 3x + 1 \neq x^2 + 2x + 1$, this is **False**. 4. **Solve the simultaneous equations:** Given: $$0.4x - 0.1y = -1$$ $$-4x - 0.5y = 2$$ Step 1: Multiply the first equation by 10 to clear decimals: $$4x - y = -10$$ Step 2: Write the system: $$4x - y = -10$$ $$-4x - 0.5y = 2$$ Step 3: Add the two equations: $$(4x - y) + (-4x - 0.5y) = -10 + 2$$ $$-1.5y = -8$$ Step 4: Solve for $y$: $$y = \frac{-8}{-1.5} = \frac{8}{1.5} = \frac{16}{3} \approx 5.33$$ Step 5: Substitute $y$ back into $4x - y = -10$: $$4x - \frac{16}{3} = -10$$ $$4x = -10 + \frac{16}{3} = -\frac{30}{3} + \frac{16}{3} = -\frac{14}{3}$$ $$x = -\frac{14}{3} \times \frac{1}{4} = -\frac{14}{12} = -\frac{7}{6} \approx -1.17$$ Step 6: Check which option matches approximately $x = -1.17$, $y = 5.33$. None exactly match, but closest is B) $(-2,4)$, so re-check calculations. Recalculate Step 3 carefully: Add equations: $4x - y = -10$ $-4x - 0.5y = 2$ Sum: $0x - 1.5y = -8$ $y = \frac{8}{1.5} = \frac{16}{3} \approx 5.33$ Substitute into first: $4x - 5.33 = -10$ $4x = -4.67$ $x = -1.167$ No exact match, so none of the options are correct. Possibly a typo in options. 5. **Domain of definition of** $$g(x) = \frac{x^2 + 3x - 12}{\sqrt{x - 2}}$$ Step 1: The denominator is $\sqrt{x - 2}$, so $x - 2 > 0 \Rightarrow x > 2$. Step 2: The numerator is a polynomial, defined for all real $x$. Step 3: Therefore, domain is $x \in (2, \infty)$. Answer: C) $x - (2, \infty)$. 6. **Determine if triangles are right-angled given side lengths:** Use Pythagoras theorem: For sides $a, b, c$ with $c$ largest, triangle is right-angled if $a^2 + b^2 = c^2$. Since side lengths are not provided, cannot determine YES or NO. Final answers: 1. A) True, B) False, C) False, D) False, E) False 4. No exact match; approximate solution $x \approx -1.17$, $y \approx 5.33$ 5. C) $x - (2, \infty)$ 6. Insufficient data