Summary

This document contains a variety of mathematical formulae and problems requiring transposition, along with practice questions. Suitable for secondary-level math students.

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S1A6 Transposing Formulae (In-Class Activity) Option 1: questions 1-12, Option 2: questions 10-15 1 Solve for the variable shown in ; red. State the restrictions, if any. (Ex: π‘₯ = ; π‘š β‰  0)...

S1A6 Transposing Formulae (In-Class Activity) Option 1: questions 1-12, Option 2: questions 10-15 1 Solve for the variable shown in ; red. State the restrictions, if any. (Ex: π‘₯ = ; π‘š β‰  0) π‘š 𝑣 1) 𝐢 = 2Ο€π‘Ÿ;π‘Ÿ 2) 𝑠 = π‘Ÿ ;π‘Ÿ 3) 𝑙 = π‘ƒπ‘Ÿπ‘‘;𝑃 2 1 4) 𝐴 = 2π‘Ž + 4π‘Žβ„Ž;β„Ž 5) 𝐴 = 2 β„Ž(π‘Ž + 𝑏);β„Ž 6) 𝑝 = 2(𝑙 + 𝑀);𝑀 π‘₯+𝑦 𝑛 π‘£βˆ’π‘’ 7) π‘š = 2 ;𝑦 8) 𝑆 = 2 (π‘Ž + 𝑙);π‘Ž 9) π‘Ž = 𝑑 ;𝑑 2 2 π‘Ž 𝑓𝑔 10) 𝑣 =𝑒 + 2π‘Žπ‘ ;𝑠 11) 𝑆 = π‘Žβˆ’π‘Ÿ ;π‘Ÿ 12) 𝐹 = 𝑓+π‘”βˆ’π‘‘ ;𝑑 Challenge: (these may require factoring) 13) π‘Ž = 180(π‘›βˆ’2) 𝑛 ;𝑛 14) π‘Ÿ = π‘Žπ‘ π‘Ž+𝑏 ;π‘Ž 15) 𝐢 = 𝐾 ( );𝑅 π‘…π‘Ÿ π‘…βˆ’π‘Ÿ ________________________________________________________ Additional Practice: (optional) π‘š 1) 𝐹 = π‘šπ‘Ž;π‘Ž 2) 𝑑 = 𝑣 ;𝑣 3) 𝐴 = 𝑃 + π‘ƒπ‘Ÿπ‘‘;𝑑 2 𝑛 4) 𝑠 = 𝑣𝑑 + 16𝑑 ;𝑣 5) 𝑆 = 2 (π‘Ž + 𝑙);𝑛 6) 𝐴 = 𝑃(𝑙 + π‘Ÿπ‘‘);π‘Ÿ π‘£βˆ’π‘’ 5 π‘₯+𝑦+𝑧 7) π‘Ž = 𝑑 ;𝑣 8) 𝐢 = 9 (𝐹 βˆ’ 32);𝐹 9) π‘š = 3 ;𝑦 𝑛 π‘Žβˆ’π‘Ÿπ‘™ 10) 𝑠 = 2 (π‘Ž + 𝑙);𝑙 11) 𝑙 = π‘Ž + (𝑛 βˆ’ 1)𝑑;𝑛 12) 𝑆 = 1βˆ’π‘Ÿ ;𝑙 Challenge: (these may require factoring) π‘Ÿ 𝑓𝑔 3𝑛+2 13) 𝑆 = 1βˆ’π‘Ÿ ;π‘Ÿ 14) 𝐹 = 𝑓+π‘”βˆ’π‘‘ ;𝑓 15) 𝐾 = 𝑛+1 ;𝑛 Answers: Additional practice: 𝐹 1) π‘Ÿ = 𝐢 1) π‘Ž = π‘š ;π‘š β‰  0 2Ο€ π‘š 2) π‘Ÿ = 𝑣 ;𝑠 β‰  0 2) 𝑣 = 𝑑 ;𝑑 β‰  0 𝑠 π΄βˆ’π‘ƒ 3) 𝑃 = 𝑙 ; π‘Ÿ β‰  0, 𝑑≠0 3) 𝑃 = π‘ƒπ‘Ÿ ; π‘Ÿ β‰  0, 𝑃≠0 π‘Ÿπ‘‘ 2 2 π‘ βˆ’16𝑑 4) β„Ž = π΄βˆ’2π‘Ž ;π‘Ž β‰ 0 4) 𝑣= 𝑑 ;𝑑 β‰  0 4π‘Ž 2𝑠 2π‘Ž 5) β„Ž = 2𝐴 ;π‘Ž + 𝑏≠0 5) 𝑛 = π‘Ž+𝑙 ; π‘Ž+𝑏 ; π‘Ž + 𝑙≠0 π‘Ž+𝑏 π΄βˆ’π‘™π‘ƒ 6) 𝑀 = π‘βˆ’2𝑙 6) π‘Ÿ = 𝑃𝑑 ; 𝑃 β‰  0, 𝑑≠0 2 7) 𝑦 = 2π‘š βˆ’ π‘₯ 7) 𝑣 = π‘Žπ‘‘ + 𝑒 9 8) π‘Ž = 2π‘†βˆ’π‘›π‘™ ;𝑛 β‰  0 8) 𝐹 = 5 𝐢 + 32 𝑛 π‘£βˆ’π‘’ 9) 𝑦 = 3π‘š βˆ’ π‘₯ βˆ’ 𝑧 9) 𝑑 = π‘Ž ;π‘Ž β‰  0 2𝑠 2 2 10) 𝑙 = 𝑛 βˆ’ π‘Ž; 𝑛 β‰  0 𝑣 βˆ’π‘’ 10) 𝑠 = 2π‘Ž ;π‘Ž β‰  0 π‘™βˆ’π‘Ž+𝑑 11) 𝑛 = 𝑑 ; 𝑑 β‰  0 π‘†π‘Žβˆ’π‘Ž 11) π‘Ÿ = 𝑆 ;𝑆 β‰  0 π‘†βˆ’π‘†π‘Ÿβˆ’π‘Ž 12) 𝑙 = βˆ’π‘Ÿ ; π‘Ÿ β‰  0 𝐹𝑓+πΉπ‘”βˆ’π‘“π‘” 12) 𝑑 = 𝐹 ;𝐹 β‰  0 𝑆 13) π‘Ÿ = 𝑆+1 ; 𝑆 β‰ βˆ’ 1 360 13) 𝑛 = βˆ’ π‘Žβˆ’180 ; π‘Ž β‰  180 πΉπ‘‘βˆ’πΉπ‘” 14) 𝑓 = πΉβˆ’π‘” ; 𝐹 β‰  𝑔 or βˆ’π‘Ÿπ‘ 14) π‘Ž = π‘Ÿβˆ’π‘ ;𝑏 β‰  π‘Ÿ πΉπ‘”βˆ’πΉπ‘‘ 𝑓 = π‘”βˆ’πΉ ;𝐹 β‰  𝑔 πΆπ‘Ÿ βˆ’πΆπ‘Ÿ 15) 𝑅 = πΆβˆ’πΎπ‘Ÿ π‘œπ‘Ÿ πΎπ‘Ÿβˆ’πΆ ;𝐢 β‰  πΎπ‘Ÿ 2βˆ’πΎ 15) 𝑛 = πΎβˆ’3 ; 𝐾 β‰  3

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