Here are the solutions to questions 7, 8, and 9. Question 7: Given $P = 2i - 3j + k$, $Q = 3i - 4j - 3k$, and $R = 3P + 2Q$. Find $|R|$ correct to 2 decimal places. Step 1: Express vectors P and Q in component form. $$P = \begin{bmatrix} 2 \\ -3 \\ 1 \end{bmatrix}$$ $$Q = \begin{bmatrix} 3 \\ -4 \\ -3 \end{bmatrix}$$ Step 2: Calculate $3P$. $$3P = 3 \begin{bmatrix} 2 \\ -3 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \times 2 \\ 3 \times (-3) \\ 3 \times 1 \end{bmatrix} = \begin{bmatrix} 6 \\ -9 \\ 3 \end{bmatrix}$$ Step 3: Calculate $2Q$. $$2Q = 2 \begin{bmatrix} 3 \\ -4 \\ -3 \end{bmatrix} = \begin{bmatrix} 2 \times 3 \\ 2 \times (-4) \\ 2 \times (-3) \end{bmatrix} = \begin{bmatrix} 6 \\ -8 \\ -6 \end{bmatrix}$$ Step 4: Calculate vector R by adding $3P$ and $2Q$. $$R = 3P + 2Q = \begin{bmatrix} 6 \\ -9 \\ 3 \end{bmatrix} + \begin{bmatrix} 6 \\ -8 \\ -6 \end{bmatrix} = \begin{bmatrix} 6+6 \\ -9+(-8) \\ 3+(-6) \end{bmatrix} = \begin{bmatrix} 12 \\ -17 \\ -3 \end{bmatrix}$$ Step 5: Calculate the magnitude of R, $|R|$. The magnitude of a vector $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is given by $\sqrt{x^2 + y^2 + z^2}$. $$|R| = \sqrt{12^2 + (-17)^2 + (-3)^2}$$ $$|R| = \sqrt{144 + 289 + 9}$$ $$|R| = \sqrt{442}$$ Step 6: Calculate the numerical value and round to 2 decimal places. $$|R| \approx 21.023796...$$ Rounded to 2 decimal places: $$|R| \approx \boxed{\text{21.02}}$$ Question 8: Simplify: $\frac{2y^2 - 3xy - 2x^2}{4y^2 - x^2}$ Step 1: Factor the numerator $2y^2 - 3xy - 2x^2$. We can factor this quadratic expression by finding two terms that multiply to $2y^2 \times (-2x^2) = -4x^2y^2$ and add to $-3xy$. These terms are $-4xy$ and $xy$. $$2y^2 - 3xy - 2x^2 = 2y^2 - 4xy + xy - 2x^2$$ Factor by grouping: $$= 2y(y - 2x) + x(y - 2x)$$ $$= (2y + x)(y - 2x)$$ Step 2: Factor the denominator $4y^2 - x^2$. This is a difference of squares, $a^2 - b^2 = (a-b)(a+b)$, where $a = 2y$ and $b = x$. $$4y^2 - x^2 = (2y - x)(2y + x)$$ Step 3: Substitute the factored expressions back into the fraction and simplify. $$\frac{(2y + x)(y - 2x)}{(2y - x)(2y + x)}$$ Cancel out the common factor $(2y + x)$: $$= \frac{y - 2x}{2y - x}$$ The simplified expression is $\boxed{\frac{y - 2x}{2y - x}}$. Question 9: Find x if $3^{2x+3} + 1 = 28$. Step 1: Isolate the exponential term. $$3^{2x+3} = 28 - 1$$ $$3^{2x+3} = 27$$ Step 2: Express the right side of the equation with the same base as the left side. We know that $27 = 3^3$. $$3^{2x+3} = 3^3$$ Step 3: Equate the exponents. Since the bases are the same, the exponents must be equal. $$2x + 3 = 3$$ Step 4: Solve for x. $$2x = 3 - 3$$ $$2x = 0$$ $$x = \frac{0}{2}$$ $$x = \boxed{0}$$ 3 done, 2 left today. You're making progress.