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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

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

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

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

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Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

I’m unable to analyze the image directly, but I can help you understand the table’s content based on your description. Answer: The table lists chemical reactions along with their rate constants

(3) BrO3- + 2H+ + Br- → HBrO2 + HOBr (2) HBrO2 + Br- + H+ → 2HOBr (4) 2HBrO2 → BrO3- + HOBr + H+ (5) BrO3- + HBrO2 + H+ → 2BrO2 (6) BrO2 + M_red + H+ → HBrO2 + M_ox (7) M_ox + Org → M_red + MA· + H+ (8) MA· → gBr- (10) MA· + O2 → ϕMA·

Answer: The image depicts a series of chemical reactions, likely part of a mechanism, possibly related to a bromate oscillating reaction or a similar redox process. Explanation: The reactions involve bromate

Math Question from Image

Answer: The image shows a series of chemical reactions, likely part of a mechanism for a chemical process. These reactions involve bromate (\( \text{BrO}_3^- \)), bromous acid (\( \text{HBrO}_2 \)), and

Math Question from Image

Answer: The image depicts a series of chemical reactions, likely part of a mechanism involving bromate (\( \text{BrO}_3^- \)) and organic compounds. The reactions suggest a redox process and radical formation.

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Answer: 1728 Explanation: Subtracting 3984 from 5712 gives 1728. You can verify by adding 3984 back to 1728, which should result in 5712. 100% (3 rated) Helpful Not Helpful

Answer: 4 quarts Explanation: There are 4 quarts in a gallon. This is a standard conversion in the U.S. customary system of measurement, where a gallon is divided into 4 equal