5. The factors of $x^2 + 7x + 6$ are: Step 1: Identify the quadratic expression. The expression is $x^2 + 7x + 6$. Step 2: Find two numbers that multiply to the constant term (6) and add up to the coefficient of the middle term (7). The numbers are 1 and 6, because $1 \times 6 = 6$ and $1 + 6 = 7$. Step 3: Write the factored form of the quadratic expression. $$x^2 + 7x + 6 = (x+1)(x+6)$$ The factors are $(x+1)$ and $(x+6)$. The correct option is A. $$\boxed{\text{A) } (x+1) \text{ and } (x+6)}$$ 6. The value of $\sqrt{6} \times \sqrt{27}$ is: Step 1: Combine the terms under a single square root. $$\sqrt{6} \times \sqrt{27} = \sqrt{6 \times 27}$$ Step 2: Multiply the numbers inside the square root. $$6 \times 27 = 162$$ So, the expression becomes $\sqrt{162}$. Step 3: Simplify the square root by finding perfect square factors of 162. $$162 = 81 \times 2$$ $$162 = 9^2 \times 2$$ Step 4: Extract the perfect square from the square root. $$\sqrt{162} = \sqrt{9^2 \times 2} = \sqrt{9^2} \times \sqrt{2} = 9\sqrt{2}$$ The value is $9\sqrt{2}$. The correct option is B. $$\boxed{\text{B) } 9\sqrt{2}}$$ 7. Observe the triangles, the correct statement among the following is: Step 1: Analyze the given triangles $\triangle ABC$ and $\triangle QPR$. In $\triangle ABC$: Side $AB = 6$ cm Side $BC = 4$ cm Angle $\angle B = 90^\circ$ (indicated by the square symbol) In $\triangle QPR$: Side $QP = 6$ cm Side $PR = 4$ cm Angle $\angle P = 90^\circ$ (indicated by the square symbol) Step 2: Compare corresponding sides and angles. We have: $AB = QP = 6$ cm $\angle B = \angle P = 90^\circ$ $BC = PR = 4$ cm Step 3: Apply the Side-Angle-Side (SAS) congruence criterion. Since two sides and the included angle of $\triangle ABC$ are equal to two sides and the included angle of $\triangle QPR$, the triangles are congruent. The correspondence of vertices is $A \leftrightarrow Q$, $B \leftrightarrow P$, $C \leftrightarrow R$. Step 4: Write the congruence statement. $$\triangle ABC \cong \triangle QPR$$ The correct option is A. $$\boxed{\text{A) } \triangle ABC \cong \triangle QPR}$$ 8. If one side of an isosceles triangle is 5 cm & its equal sides measure 3 cm each, then half of its perimeter is: Step 1: Identify the lengths of the sides of the isosceles triangle. The equal sides are 3 cm each. The third side is 5 cm. So, the side lengths are 3 cm, 3 cm, and 5 cm. Step 2: Calculate the perimeter of the triangle. Perimeter $P = \text{sum of all sides}$ $$P = 3 \text{ cm} + 3 \text{ cm} + 5 \text{ cm}$$ $$P = 11 \text{ cm}$$ Step 3: Calculate half of the perimeter. $$\text{Half of perimeter} = \frac{P}{2} = \frac{11 \text{ cm}}{2}$$ $$\text{Half of perimeter} = 5.5 \text{ cm}$$ The correct option is B. $$\boxed{\text{B) } 5.5 \text{ cm}}$$ 9. Two angles measure $(55^\circ + 3a)$ and $(115^\circ - 2a)$ if each angle is a supplement of the other, then calculate the value of "a". Step 1: Understand the definition of supplementary angles. If two angles are supplementary, their sum is $180^\circ$. Step 2: Set up the equation using the given angle measures. $$(55^\circ + 3a) + (115^\circ - 2a) = 180^\circ$$ Step 3: Combine like terms. $$55^\circ + 115^\circ + 3a - 2a = 180^\circ$$ $$170^\circ + a = 180^\circ$$ Step 4: Solve for $a$. $$a = 180^\circ - 170^\circ$$ $$a = 10^\circ$$ The value of $a$ is $10^\circ$. $$\boxed{a = 10^\circ}$$ 10. Write any two of Euclid's axioms. Here are two of Euclid's axioms: • Things which are equal to the same thing are equal to one another. • The whole is greater than the part.