1. **Convert $ \frac{3 - 5i}{2 + 7i} $ into the form $ a + bi $.** 2. To convert a complex fraction into standard form $ a + bi $, multiply numerator and denominator by the conjugate of the denominator. 3. The conjugate of $ 2 + 7i $ is $ 2 - 7i $. 4. Multiply numerator and denominator: $$\frac{3 - 5i}{2 + 7i} \times \frac{2 - 7i}{2 - 7i} = \frac{(3 - 5i)(2 - 7i)}{(2 + 7i)(2 - 7i)}$$ 5. Expand numerator: $$3 \times 2 = 6$$ $$3 \times (-7i) = -21i$$ $$-5i \times 2 = -10i$$ $$-5i \times (-7i) = +35i^2$$ 6. Sum numerator terms: $$6 - 21i - 10i + 35i^2 = 6 - 31i + 35(-1) = 6 - 31i - 35 = -29 - 31i$$ 7. Denominator: $$(2)^2 - (7i)^2 = 4 - 49i^2 = 4 - 49(-1) = 4 + 49 = 53$$ 8. So the expression is: $$\frac{-29 - 31i}{53} = -\frac{29}{53} - \frac{31}{53}i$$ **Final answer:** $ -\frac{29}{53} - \frac{31}{53}i $ --- 2. **Find length of the median through vertex A of triangle ABC with points A(0,0,0), B(3,1,2), C(4,2,3).** 3. The median from vertex A goes to midpoint M of BC. 4. Find midpoint M: $$M = \left( \frac{3+4}{2}, \frac{1+2}{2}, \frac{2+3}{2} \right) = (3.5, 1.5, 2.5)$$ 5. Length of median AM is distance between A and M: $$AM = \sqrt{(3.5 - 0)^2 + (1.5 - 0)^2 + (2.5 - 0)^2} = \sqrt{12.25 + 2.25 + 6.25} = \sqrt{20.75}$$ 6. Simplify: $$\sqrt{20.75} \approx 4.555$$ **Final answer:** Approximately 4.555 units --- 3. **Find the rank of matrix $ A = \begin{bmatrix}4 & 5 & 27 \\ 2 & 1 & 5 \\ 3 & 4 & 7\end{bmatrix} $.** 4. Use row operations or determinant to find rank. 5. Calculate determinant: $$\det(A) = 4(1 \times 7 - 5 \times 4) - 5(2 \times 7 - 5 \times 3) + 27(2 \times 4 - 1 \times 3)$$ 6. Compute each term: $$4(7 - 20) = 4(-13) = -52$$ $$-5(14 - 15) = -5(-1) = 5$$ $$27(8 - 3) = 27(5) = 135$$ 7. Sum: $$-52 + 5 + 135 = 88$$ 8. Since determinant $ \neq 0 $, rank is 3. **Final answer:** Rank is 3 --- 4. **Find the projection of vector $ \mathbf{a} = 2\mathbf{i} + 3\mathbf{j} + 7\mathbf{k} $ on $ \mathbf{b} = 2\mathbf{i} + 7\mathbf{j} + \mathbf{k} $.** 5. Projection formula: $$\text{proj}_{\mathbf{b}} \mathbf{a} = \left( \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{b}\|^2} \right) \mathbf{b}$$ 6. Compute dot product: $$\mathbf{a} \cdot \mathbf{b} = 2 \times 2 + 3 \times 7 + 7 \times 1 = 4 + 21 + 7 = 32$$ 7. Compute $ \|\mathbf{b}\|^2 $: $$2^2 + 7^2 + 1^2 = 4 + 49 + 1 = 54$$ 8. Compute scalar multiplier: $$\frac{32}{54} = \frac{16}{27}$$ 9. Multiply $ \mathbf{b} $ by scalar: $$\text{proj}_{\mathbf{b}} \mathbf{a} = \frac{16}{27} (2\mathbf{i} + 7\mathbf{j} + \mathbf{k}) = \frac{32}{27} \mathbf{i} + \frac{112}{27} \mathbf{j} + \frac{16}{27} \mathbf{k}$$ **Final answer:** $ \frac{32}{27} \mathbf{i} + \frac{112}{27} \mathbf{j} + \frac{16}{27} \mathbf{k} $ --- 5. **Show that determinant of matrix $ \begin{bmatrix} b+c & c+b & a+b \\ a & b & c \\ 1 & 1 & 1 \end{bmatrix} = 0 $.** 6. Note that first row entries are symmetric sums. 7. Subtract second row from first row: $$R_1 \to R_1 - R_2 = \begin{bmatrix} (b+c)-a & (c+b)-b & (a+b)-c \\ a & b & c \\ 1 & 1 & 1 \end{bmatrix} = \begin{bmatrix} b+c - a & c & a + b - c \\ a & b & c \\ 1 & 1 & 1 \end{bmatrix}$$ 8. Notice the first row is now linear combination of second and third rows. 9. This implies rows are linearly dependent, so determinant is zero. **Final answer:** Determinant equals 0 --- 6. **Arithmetic series: first term $ a_1 = 5 $, last term $ a_n = 45 $, sum $ S_n = 500 $. Find number of terms $ n $ and common difference $ d $.** 7. Sum formula: $$S_n = \frac{n}{2} (a_1 + a_n)$$ 8. Substitute known values: $$500 = \frac{n}{2} (5 + 45) = \frac{n}{2} \times 50 = 25n$$ 9. Solve for $ n $: $$n = \frac{500}{25} = 20$$ 10. Use formula for last term: $$a_n = a_1 + (n-1)d$$ 11. Substitute values: $$45 = 5 + (20 - 1)d = 5 + 19d$$ 12. Solve for $ d $: $$19d = 40 \Rightarrow d = \frac{40}{19}$$ **Final answer:** Number of terms $ n = 20 $, common difference $ d = \frac{40}{19} $ --- 7. **Find sum $ \sum_{k=1}^5 \frac{2k + 1}{3k - 1} $.** 8. Calculate each term: $ k=1: \frac{3}{2} = 1.5 $ $ k=2: \frac{5}{5} = 1 $ $ k=3: \frac{7}{8} = 0.875 $ $ k=4: \frac{9}{11} \approx 0.8182 $ $ k=5: \frac{11}{14} \approx 0.7857 $ 9. Sum: $$1.5 + 1 + 0.875 + 0.8182 + 0.7857 = 4.9789$$ **Final answer:** Approximately 4.979 --- 8. **Number of different arrangements of letters in "SHOPKEEPER" taken all together.** 9. Count letters and their frequencies: S(1), H(1), O(1), P(2), K(1), E(3), R(1) 10. Total letters = 10 11. Number of arrangements: $$\frac{10!}{2! \times 3!}$$ 12. Calculate factorials: $$10! = 3628800$$ $$2! = 2$$ $$3! = 6$$ 13. Compute: $$\frac{3628800}{2 \times 6} = \frac{3628800}{12} = 302400$$ **Final answer:** 302400 arrangements --- 9. **Find coefficient of $ x^3 $ in expansion of $ (x + \frac{3}{x})^7 $.** 10. General term: $$T_{k+1} = \binom{7}{k} x^{7-k} \left( \frac{3}{x} \right)^k = \binom{7}{k} 3^k x^{7-k-k} = \binom{7}{k} 3^k x^{7 - 2k}$$ 11. We want exponent of $ x $ to be 3: $$7 - 2k = 3 \Rightarrow 2k = 4 \Rightarrow k = 2$$ 12. Coefficient: $$\binom{7}{2} 3^2 = 21 \times 9 = 189$$ **Final answer:** 189 --- 10. **Find inverse function $ g^{-1}(x) $ for $ g(x) = 7x + 4 $.** 11. Let $ y = 7x + 4 $. 12. Solve for $ x $: $$y = 7x + 4 \Rightarrow 7x = y - 4 \Rightarrow x = \frac{y - 4}{7}$$ 13. Replace $ y $ by $ x $ for inverse: $$g^{-1}(x) = \frac{x - 4}{7}$$ **Final answer:** $ g^{-1}(x) = \frac{x - 4}{7} $ --- 11. **Prove $ \frac{2 \cot \theta}{1 + \cot^2 \theta} = \sin 2\theta $.** 12. Recall identity: $$1 + \cot^2 \theta = \csc^2 \theta$$ 13. Rewrite left side: $$\frac{2 \cot \theta}{1 + \cot^2 \theta} = \frac{2 \cot \theta}{\csc^2 \theta} = 2 \cot \theta \sin^2 \theta$$ 14. Express $ \cot \theta = \frac{\cos \theta}{\sin \theta} $: $$2 \times \frac{\cos \theta}{\sin \theta} \times \sin^2 \theta = 2 \cos \theta \sin \theta$$ 15. Right side is: $$\sin 2\theta = 2 \sin \theta \cos \theta$$ 16. Both sides equal, identity proven. **Final answer:** Identity holds true --- 12. **Solve triangle with sides 5, 12, 13.** 13. Recognize 5-12-13 is a right triangle (since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$). 14. Hypotenuse = 13, legs = 5 and 12. 15. Angles: $$\theta = \arcsin \left( \frac{5}{13} \right) \approx 22.62^\circ$$ $$\phi = \arcsin \left( \frac{12}{13} \right) \approx 67.38^\circ$$ 16. Right angle is $90^\circ$. **Final answer:** Angles approximately $22.62^\circ, 67.38^\circ, 90^\circ$ --- 13. **Find general solution of $ \cos x = \frac{1}{\sqrt{2}} $.** 14. Recall $ \cos x = \frac{1}{\sqrt{2}} = \cos \frac{\pi}{4} $. 15. General solution for $ \cos x = \cos \alpha $ is: $$x = 2n\pi \pm \alpha, \quad n \in \mathbb{Z}$$ 16. Substitute $ \alpha = \frac{\pi}{4} $: $$x = 2n\pi \pm \frac{\pi}{4}$$ **Final answer:** $ x = 2n\pi \pm \frac{\pi}{4}, n \in \mathbb{Z} $