S1A6 Transposing Formulae PDF
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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 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 β 0) π π£ 1) πΆ = 2Οπ;π 2) π = π ;π 3) π = πππ‘;π 2 1 4) π΄ = 2π + 4πβ;β 5) π΄ = 2 β(π + π);β 6) π = 2(π + π€);π€ π₯+π¦ π π£βπ’ 7) π = 2 ;π¦ 8) π = 2 (π + π);π 9) π = π‘ ;π‘ 2 2 π ππ 10) π£ =π’ + 2ππ ;π 11) π = πβπ ;π 12) πΉ = π+πβπ ;π Challenge: (these may require factoring) 13) π = 180(πβ2) π ;π 14) π = ππ π+π ;π 15) πΆ = πΎ ( );π π π π βπ ________________________________________________________ Additional Practice: (optional) π 1) πΉ = ππ;π 2) π = π£ ;π£ 3) π΄ = π + πππ‘;π‘ 2 π 4) π = π£π‘ + 16π‘ ;π£ 5) π = 2 (π + π);π 6) π΄ = π(π + ππ‘);π π£βπ’ 5 π₯+π¦+π§ 7) π = π‘ ;π£ 8) πΆ = 9 (πΉ β 32);πΉ 9) π = 3 ;π¦ π πβππ 10) π = 2 (π + π);π 11) π = π + (π β 1)π;π 12) π = 1βπ ;π Challenge: (these may require factoring) π ππ 3π+2 13) π = 1βπ ;π 14) πΉ = π+πβπ ;π 15) πΎ = π+1 ;π Answers: Additional practice: πΉ 1) π = πΆ 1) π = π ;π β 0 2Ο π 2) π = π£ ;π β 0 2) π£ = π ;π β 0 π π΄βπ 3) π = π ; π β 0, π‘β 0 3) π = ππ ; π β 0, πβ 0 ππ‘ 2 2 π β16π‘ 4) β = π΄β2π ;π β 0 4) π£= π‘ ;π‘ β 0 4π 2π 2π 5) β = 2π΄ ;π + πβ 0 5) π = π+π ; π+π ; π + πβ 0 π+π π΄βππ 6) π€ = πβ2π 6) π = ππ‘ ; π β 0, π‘β 0 2 7) π¦ = 2π β π₯ 7) π£ = ππ‘ + π’ 9 8) π = 2πβππ ;π β 0 8) πΉ = 5 πΆ + 32 π π£βπ’ 9) π¦ = 3π β π₯ β π§ 9) π‘ = π ;π β 0 2π 2 2 10) π = π β π; π β 0 π£ βπ’ 10) π = 2π ;π β 0 πβπ+π 11) π = π ; π β 0 ππβπ 11) π = π ;π β 0 πβππβπ 12) π = βπ ; π β 0 πΉπ+πΉπβππ 12) π = πΉ ;πΉ β 0 π 13) π = π+1 ; π β β 1 360 13) π = β πβ180 ; π β 180 πΉπβπΉπ 14) π = πΉβπ ; πΉ β π or βππ 14) π = πβπ ;π β π πΉπβπΉπ π = πβπΉ ;πΉ β π πΆπ βπΆπ 15) π = πΆβπΎπ ππ πΎπβπΆ ;πΆ β πΎπ 2βπΎ 15) π = πΎβ3 ; πΎ β 3