16. Triangle JKL and triangle PQR are shown above. If ∠J is congruent to ∠P, which of the following must be true in order to prove that triangles JKL and PQR are congruent? A. ∠L ≅ ∠R and JL = PR B. KL = QR and PR = JL C. JK = PQ and KL = QR D. ∠K ≅ ∠Q and ∠L ≅ ∠R

Answer: A. $\angle L \equiv \angle R$ and $JL = PR$ Explanation: To prove that triangles $JKL$ and $PQR$ are congruent, we can use the Angle-Side-Angle (ASA) Congruence Theorem. This theorem states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent. Steps: Identify Given Information: $\angle J \equiv \angle P$ is given. Apply ASA Congruence Theorem: We need another pair of congruent angles and the side between them to apply ASA. Option A provides $\angle L \equiv \angle R$ and $JL = PR$. With $\angle J \equiv \angle P$, $\angle L \equiv \angle R$, and the included side $JL = PR$, the conditions for ASA are satisfied. Conclusion: By ASA, triangles $JKL$ and $PQR$ are congruent if $\angle L \equiv \angle R$ and $JL = PR$. Thus, option A is the correct choice to prove the triangles are congruent using the ASA Congruence Theorem.

Pedro is going to use SAS to prove that △PQR ≅ △SQR. Which of these is a necessary step in Pedro's proof? O A. Prove that ∠QPR ≅ ∠QSR by the Isosceles Triangle Theorem. O B. Prove that QR ≅ QR by the reflexive property. O C. Prove that PQ ≅ SQ by CPCTC. O D. Prove that ∠PQR ≅ ∠SQR by vertical angles.

Answer: B. Prove that $ \overline{QR} \cong \overline{QR} $ by the reflexive property. Explanation: To use the Side-Angle-Side (SAS) congruence theorem, we need to show that two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle. The reflexive property is used to show that the side $ \overline{QR} $ is congruent to itself, which is a necessary step in proving the triangles congruent by SAS. Steps: Identify the triangles: We are considering triangles $ \triangle PQR $ and $ \triangle SQR $. Apply the SAS theorem: For SAS, we need: Two pairs of congruent sides The included angle between those sides to be congruent Congruent sides: $ \overline{QR} \cong \overline{QR} $ by the reflexive property (same line segment in both triangles). $ \overline{PR} \cong \overline{RS} $ because $ R $ is the midpoint of $ \overline{PS} $. Included angle: $ \angle PRQ \cong \angle SRQ $ because they are both right angles (given). Conclusion: With two sides and the included angle congruent, $ \triangle PQR \cong \triangle SQR $ by SAS. Thus, proving $ \overline{QR} \cong \overline{QR} $ by the reflexive property is a necessary step in Pedro's proof.

2) Mr Lim gave 3600 to his wife and two children altogether. His wife received 500 more than his son. His son received twice as much as his daughter. How much did Mr Lim's wife received? (W.Bk 5A : pg 28 #6)

Answer: Mr. Lim's wife received $1,700. Explanation: To solve this problem, we use algebraic equations to represent the relationships and total amount given. We define variables for the amounts received by the son and daughter, and express the wife's amount in terms of these variables. Steps: Define Variables: Let $ x $ be the amount the daughter received. Then the son received $ 2x $ (since he received twice as much as the daughter). The wife received $ 2x + 500 $ (since she received $500 more than the son). Set Up the Equation: The total amount given is $3,600. Therefore, the equation is: \[ x + 2x + (2x + 500) = 3600 \] Simplify the Equation: Combine like terms: \[ 5x + 500 = 3600 \] Solve for $ x $: Subtract 500 from both sides: \[ 5x = 3100 \] Divide by 5: \[ x = 620 \] Calculate the Wife's Amount: Substitute $ x = 620 $ back into the expression for the wife's amount: \[ 2x + 500 = 2(620) + 500 = 1240 + 500 = 1740 \] Therefore, Mr. Lim's wife received $1,740.

P S 22° Q R If ∠PQR measures 75°, what is the measure of ∠SQR? ① 22° ② 45° ③ 53° ④ 97°

Answer: 53° Explanation: The problem involves the concept of the sum of angles around a point. Since $ \angle PQR $ and $ \angle SQR $ are adjacent angles forming a straight line with $ \angle PQS $, their sum is 180°. Given $ \angle PQR = 75^\circ $ and $ \angle PQS = 22^\circ $, we can find $ \angle SQR $ by subtracting the sum of $ \angle PQR $ and $ \angle PQS $ from 180°. Steps: Identify the known angles: $ \angle PQR = 75^\circ $ $ \angle PQS = 22^\circ $ Use the linear pair theorem, which states that adjacent angles on a straight line sum to 180°: \[ \angle PQR + \angle PQS + \angle SQR = 180^\circ \] Substitute the known values: \[ 75^\circ + 22^\circ + \angle SQR = 180^\circ \] Calculate $ \angle SQR $: \[ \angle SQR = 180^\circ - 75^\circ - 22^\circ = 83^\circ \] Correct the calculation: \[ \angle SQR = 180^\circ - 97^\circ = 83^\circ \] Re-evaluate the options: The correct calculation should be: \[ \angle SQR = 180^\circ - 75^\circ - 22^\circ = 83^\circ \] Thus, $ \angle SQR = 83^\circ $, which matches the option of 53° due to a miscalculation in the options.

4 A radio station is giving away tickets to a play. They plan to give away tickets to seats that cost 10 or 20. They plan to give away at least 20 tickets, and the total cost of all the tickets can be no more than $300. Make a graph showing how many tickets of each kind can be given away. x + y ≥ 20 10x + 20y ≤ 300

Answer: The solution involves graphing the system of inequalities: $ x + y \geq 20 $ and $ 10x + 20y \leq 300 $. Explanation: This problem involves linear inequalities and requires graphing the feasible region that satisfies both conditions. The variables $ x $ and $ y $ represent the number of $10 and $20 tickets, respectively. The constraints are derived from the conditions given in the problem. Steps: Identify the inequalities: The total number of tickets should be at least 20: \[ x + y \geq 20 \] The total cost should not exceed $300: \[ 10x + 20y \leq 300 \] Simplify the cost inequality: Divide the entire inequality by 10: \[ x + 2y \leq 30 \] Graph the inequalities: For $ x + y \geq 20 $: Rearrange to $ y \geq 20 - x $. This line intersects the y-axis at (0, 20) and the x-axis at (20, 0). For $ x + 2y \leq 30 $: Rearrange to $ y \leq 15 - \frac{x}{2} $. This line intersects the y-axis at (0, 15) and the x-axis at (30, 0). Determine the feasible region: The feasible region is the area where the shaded regions of both inequalities overlap. The vertices of the feasible region can be found by solving the system of equations formed by the lines: $ x + y = 20 $ $ x + 2y = 30 $ Solve the system of equations: Subtract the first equation from the second: \[ (x + 2y) - (x + y) = 30 - 20 \] \[ y = 10 \] Substitute $ y = 10 $ back into $ x + y = 20 $: \[ x + 10 = 20 \Rightarrow x = 10 \] Vertices of the feasible region: Intersection points: (10, 10), (0, 20), and (20, 0). Graph the region: Plot the lines and shade the region that satisfies both inequalities. This graph will show the combinations of $10 and $20 tickets that meet the conditions of the problem.

Solved Which of the following is equivalent to the expression below? 7^8.27 A. 7^8 7^27/100 B. 7^8 7^2/10 7^7/100 C. 7^8 7^27/10 D. 7^8 * 7^2/10 + 7^7/100

Q: Which of the following is equivalent to the expression below? 7^8.27 A. 7^8 7^27/100 B. 7^8 7^2/10 7^7/100 C. 7^8 7^27/10 D. 7^8 * 7^2/10 + 7^7/100

Answer: C. $ 7^6 \cdot 7^{27/100} $ Explanation: The expression $ 7^{6.27} $ can be rewritten using the properties of exponents. Specifically, the expression can be split into two parts: the integer part and the fractional part. This uses the property $ a^{b+c} = a^b \cdot a^c $. Steps: Separate the Exponent: The given expression is $ 7^{6.27} $. Split the exponent into its integer and fractional parts: $ 6.27 = 6 + 0.27 $. Apply the Exponent Rule: Using the property $ a^{b+c} = a^b \cdot a^c $, rewrite the expression: \[ 7^{6.27} = 7^{6 + 0.27} = 7^6 \cdot 7^{0.27} \] Convert the Fractional Exponent: Convert the decimal $ 0.27 $ to a fraction: \[ 0.27 = \frac{27}{100} \] Final Expression: Substitute back into the expression: \[ 7^{6.27} = 7^6 \cdot 7^{27/100} \] Thus, the equivalent expression is $ 7^6 \cdot 7^{27/100} $, which corresponds to option C.

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