Math Question from Image

Answer: The feasible region for the number of tickets that can be given away is defined by the inequalities $ x + y \geq 20 $ and $ 10x + 20y \leq 300 $. Explanation: This problem involves linear inequalities and graphing a feasible region. The goal is to determine how many tickets of each kind (costing $10 and $20) can be given away while satisfying the conditions of giving away at least 20 tickets and not exceeding a total cost of $300. Steps: Define Variables: Let $ x $ be the number of $10 tickets. Let $ y $ be the number of $20 tickets. Set Up Inequalities: The total number of tickets must be at least 20: \[ x + y \geq 20 \] The total cost of the tickets must not exceed $300: \[ 10x + 20y \leq 300 \] Simplify the Cost Inequality: Divide the entire inequality by 10 to simplify: \[ x + 2y \leq 30 \] Graph the Inequalities: Plot the line $ x + y = 20 $. This line represents the boundary where exactly 20 tickets are given away. Plot the line $ x + 2y = 30 $. This line represents the boundary where the total cost is exactly $300. Determine the Feasible Region: The feasible region is the area on the graph where both conditions are satisfied. It is the intersection of the regions defined by the inequalities $ x + y \geq 20 $ and $ x + 2y \leq 30 $. Find Intersections: Solve the system of equations to find intersection points: $ x + y = 20 $ $ x + 2y = 30 $ Subtract the first equation from the second: \[ (x + 2y) - (x + y) = 30 - 20 \] \[ y = 10 \] Substitute $ y = 10 $ into $ x + y = 20 $: \[ x + 10 = 20 \Rightarrow x = 10 \] Intersection point: $ (10, 10) $. Check Boundary Points: $ x = 0 $: $ y = 20 $ (from $ x + y = 20 $) $ y = 0 $: $ x = 30 $ (from $ x + 2y = 30 $) Conclusion: The feasible region is bounded by the points $ (0, 20) $, $ (10, 10) $, and $ (30, 0) $. The solution set includes all integer points within and on the boundary of this region, representing combinations of tickets that satisfy both constraints.

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

Q: P Q S R 22° 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 supplementary angles. In a straight line, the sum of angles is 180°. Given that $\angle PQR$ is 75° and $\angle PQS$ is 22°, we can find $\angle SQR$ by subtracting the sum of these two angles from 180°. Steps: Identify the known angles: $\angle PQR = 75^\circ$ $\angle PQS = 22^\circ$ Use the supplementary angle theorem: \[ \angle PQR + \angle PQS + \angle SQR = 180^\circ \] 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: \[ \angle SQR = 180^\circ - 97^\circ = 83^\circ \] The correct calculation should be: \[ \angle SQR = 180^\circ - 75^\circ - 22^\circ = 83^\circ \] Therefore, the measure of $\angle SQR$ is 83°.

Q: 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 angles and the included side to be congruent. Option A provides $\angle L \equiv \angle R$ and $JL = PR$. Verify ASA Conditions: $\angle J \equiv \angle P$ (Given) $\angle L \equiv \angle R$ (Option A) $JL = PR$ (Option A) Conclusion: With $\angle J \equiv \angle P$, $\angle L \equiv \angle R$, and $JL = PR$, the ASA conditions are satisfied, proving triangles $JKL$ and $PQR$ are congruent. Thus, the correct choice is A, as it fulfills the requirements of the ASA Congruence Theorem.

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

Answer: B. Prove that $ \overline{QR} \cong \overline{QR} $ by the reflexive property. Explanation: To prove that $\triangle PQR \cong \triangle SQR$ using the Side-Angle-Side (SAS) congruence theorem, we need to establish that two sides and the included angle of one triangle are congruent to two sides and the included angle of the other triangle. The reflexive property is used to show that the side $\overline{QR}$ is congruent to itself, which is a necessary step in the proof. Steps: Identify the triangles and the goal: We want to prove $\triangle PQR \cong \triangle SQR$. Use the SAS Congruence Theorem: This theorem states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent. Establish congruence of sides: $\overline{QR} \cong \overline{QR}$ by the reflexive property. This is the side common to both triangles. Identify the included angle: $\angle QRP = \angle QRS$ because they are both right angles. Prove congruence of the other sides: Since $\overline{PR} \cong \overline{RS}$ (as indicated by the diagram showing equal segments), this satisfies the second side requirement. Conclude using SAS: With $\overline{PR} \cong \overline{RS}$, $\angle QRP = \angle QRS$, and $\overline{QR} \cong \overline{QR}$, we can conclude $\triangle PQR \cong \triangle SQR$ by SAS. This step-by-step approach confirms that the reflexive property is crucial for proving the congruence of the triangles using SAS.

Q: 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: $1,700 Explanation: To solve this problem, we use algebraic equations to represent the relationships between the amounts received by Mr. Lim's wife, son, and daughter. We set up equations based on the given conditions and solve for the unknowns. Steps: Define Variables: Let $ x $ be the amount the daughter received. Then the son received $ 2x $ (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. \[ x + 2x + (2x + 500) = 3600 \] Simplify and Solve the Equation: \[ x + 2x + 2x + 500 = 3600 \] \[ 5x + 500 = 3600 \] \[ 5x = 3600 - 500 \] \[ 5x = 3100 \] \[ x = \frac{3100}{5} = 620 \] Calculate the Amounts: Daughter: $ x = 620 $ Son: $ 2x = 1240 $ Wife: $ 2x + 500 = 1240 + 500 = 1740 $ Verify the Total: Total = $ 620 + 1240 + 1740 = 3600 $ Thus, Mr. Lim's wife received $1,740.

Q: Which of the following is equivalent to the expression below? 7^8.27 A. 7^8 7^2/10 7^7/100 B. 7^6 7^2/10 7^7/100 C. 7^8 * 7^27/100 D. 7^8 + 7^2/10 + 7^7/100

Answer: C. $ 7^6 \times 7^{27/100} $ Explanation: The expression $ 7^{6.27} $ can be rewritten using the properties of exponents. Specifically, we use the rule that $ a^{m+n} = a^m \times a^n $. Here, we split the exponent $ 6.27 $ into $ 6 + 0.27 $. Steps: Identify the Expression: \[ 7^{6.27} \] Split the Exponent: $ 6.27 = 6 + 0.27 $ Apply the Exponent Rule: \[ 7^{6.27} = 7^{6 + 0.27} = 7^6 \times 7^{0.27} \] Convert Decimal to Fraction: $ 0.27 = \frac{27}{100} $ Rewrite the Expression: \[ 7^{6.27} = 7^6 \times 7^{27/100} \] Thus, the equivalent expression is $ 7^6 \times 7^{27/100} $, which corresponds to option C.

Answer: 53°

Explanation: The problem involves the concept of supplementary angles. In a straight line, the sum of angles is 180°. Given that $\angle PQR$ is 75° and $\angle PQS$ is 22°, we can find $\angle SQR$ by subtracting the sum of these two angles from 180°.

Steps: