1. **Problem 1: Determine the truth of each statement for any nonzero rational number $a$.** a. The product $4a$ is negative. - Since $a$ can be positive or negative, $4a$ can be positive or negative. - Therefore, the statement is **false**. b. The product of $a$ and its opposite (additive inverse) is positive. - The opposite of $a$ is $-a$. - Their product is $a \times (-a) = -a^2$. - Since $a^2$ is positive, $-a^2$ is negative. - So the product is negative, not positive. - The statement is **false**. c. The product $a \times a$ is negative. - $a \times a = a^2$ which is always positive. - The statement is **false**. d. Twice the number $a$ is positive. - Twice $a$ is $2a$. - Since $a$ can be positive or negative, $2a$ can be positive or negative. - The statement is **false**. 2. **Problem 2: Calculate the expressions $A$ and $B$ showing all steps.** - $A = 1 - 12 \div 2 + 4 \times (-5)$ 1. Calculate division: $12 \div 2 = 6$ 2. Calculate multiplication: $4 \times (-5) = -20$ 3. Substitute: $A = 1 - 6 - 20$ 4. Calculate subtraction: $1 - 6 = -5$ 5. Calculate: $-5 - 20 = -25$ - So, $A = -25$. - $B = \frac{-3 + 4 - 7}{-6 + 3 \times 4}$ 1. Calculate numerator: $-3 + 4 = 1$, then $1 - 7 = -6$ 2. Calculate denominator: $3 \times 4 = 12$, then $-6 + 12 = 6$ 3. Substitute: $B = \frac{-6}{6} = -1$ 3. **Problem 3: Calculate and simplify rational expressions $E$, $F$, $G$, and $H$ with all steps.** - $E = \left(-\frac{7}{50}\right) + \frac{2}{25} \times \frac{4}{3}$ 1. Multiply fractions: $\frac{2}{25} \times \frac{4}{3} = \frac{8}{75}$ 2. Find common denominator for $-\frac{7}{50}$ and $\frac{8}{75}$: - LCD of 50 and 75 is 150. - Convert: $-\frac{7}{50} = -\frac{21}{150}$, $\frac{8}{75} = \frac{16}{150}$ 3. Add: $-\frac{21}{150} + \frac{16}{150} = -\frac{5}{150} = -\frac{1}{30}$ - So, $E = -\frac{1}{30}$. - $F = \frac{3}{16} - \frac{1}{6} \div 2$ 1. Division: $\frac{1}{6} \div 2 = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}$ 2. Subtract: $\frac{3}{16} - \frac{1}{12}$ 3. Find LCD of 16 and 12 is 48. - Convert: $\frac{3}{16} = \frac{9}{48}$, $\frac{1}{12} = \frac{4}{48}$ 4. Subtract: $\frac{9}{48} - \frac{4}{48} = \frac{5}{48}$ - So, $F = \frac{5}{48}$. - $G = -2 \left( \frac{3}{4} - \frac{2}{-5} \right) + \frac{3}{2}$ 1. Simplify inside parentheses: $\frac{2}{-5} = -\frac{2}{5}$ 2. Calculate: $\frac{3}{4} - (-\frac{2}{5}) = \frac{3}{4} + \frac{2}{5}$ 3. Find LCD of 4 and 5 is 20. - Convert: $\frac{3}{4} = \frac{15}{20}$, $\frac{2}{5} = \frac{8}{20}$ 4. Add: $\frac{15}{20} + \frac{8}{20} = \frac{23}{20}$ 5. Multiply by $-2$: $-2 \times \frac{23}{20} = -\frac{46}{20} = -\frac{23}{10}$ 6. Add $\frac{3}{2} = \frac{15}{10}$ 7. Sum: $-\frac{23}{10} + \frac{15}{10} = -\frac{8}{10} = -\frac{4}{5}$ - So, $G = -\frac{4}{5}$. - $H = \frac{7}{-8} + \frac{5}{4} - 1$ 1. Rewrite: $\frac{7}{-8} = -\frac{7}{8}$ 2. Find LCD of 8 and 4 is 8. - Convert: $-\frac{7}{8}$ stays the same, $\frac{5}{4} = \frac{10}{8}$ 3. Add: $-\frac{7}{8} + \frac{10}{8} = \frac{3}{8}$ 4. Subtract 1: $\frac{3}{8} - 1 = \frac{3}{8} - \frac{8}{8} = -\frac{5}{8}$ - So, $H = -\frac{5}{8}$. 4. **Problem 4: Geometry with points A, B symmetric about Ox, points C, D on Oy, with $OA = OC$ and $OB = OD$.** 1. Construct the figure with acute angle $xOy$, points $A$ and $B$ symmetric about $Ox$, points $C$ and $D$ on $Oy$ such that $OA = OC$ and $OB = OD$. 2. Prove triangles $OBC$ and $ODA$ are similar. - Since $A$ and $B$ are symmetric about $Ox$, $OA = OB$ and angles at $O$ are equal. - Given $OA = OC$ and $OB = OD$, triangles $OBC$ and $ODA$ have two sides equal and included angle equal. - By SAS similarity criterion, $\triangle OBC \sim \triangle ODA$. 3. Let $I$ be the intersection of segments $AD$ and $BC$. - Prove triangles $ICD$ and $IAB$ are similar. - Using properties of symmetric points and equal lengths, angles at $I$ are equal. - By angle-angle similarity, $\triangle ICD \sim \triangle IAB$. **Final answers:** - Problem 1: All statements a, b, c, d are false. - Problem 2: $A = -25$, $B = -1$. - Problem 3: $E = -\frac{1}{30}$, $F = \frac{5}{48}$, $G = -\frac{4}{5}$, $H = -\frac{5}{8}$. - Problem 4: Triangles $OBC$ and $ODA$ are similar; triangles $ICD$ and $IAB$ are similar.