SOLUTION: USF Applied Mathematics Maximum & Minimum Value Extraneous Root Worksheet

Good Day! How are you? Please see the attachments. If you have questions and concerns please feel free to approach me. I am looking forward on your approval. Thank you very much. Have a nice day ahead ^_^ πŸ™‚ :)1. a.) 8𝑒 7π‘Ž+15 + 35 = 838𝑒 7π‘Ž+15 = 48𝑒 7π‘Ž+15 = 6ln (𝑒 7π‘Ž+15 ) = ln (6)7π‘Ž + 15 = ln (6)𝒂=π₯𝐧(πŸ”)βˆ’πŸπŸ“πŸ•β‰… βˆ’πŸ. πŸ–πŸ—b.) 𝑒 2π‘₯ + 4𝑒 π‘₯ βˆ’ 221 = 0𝐿𝑒𝑑 𝑧 = 𝑒 π‘₯∴ 𝑧 2 + 4𝑧 βˆ’ 221 = 0πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘›π‘”,(𝑧 + 17)(𝑧 βˆ’ 13) = 0∴ 𝑧 + 17 = 0|𝑧 βˆ’ 13 = 0𝑧 = βˆ’17𝑧 = 13𝑒 π‘₯ = βˆ’17𝑒 π‘₯ = 13π‘₯ = ln (βˆ’17)𝒙 = π₯𝐧 (πŸπŸ‘) β‰… 𝟐. πŸ“πŸ”(extraneous root)2. β„Ž(π‘₯) = βˆ’5π‘₯ 2 + 13π‘₯ βˆ’ 44β„Ž(π‘₯) 𝑖𝑠 𝑖𝑛 π‘‘β„Žπ‘’ π‘“π‘œπ‘Ÿπ‘š π‘œπ‘“ π‘Žπ‘₯ 2 + 𝑏π‘₯ + 𝑐 ∴ π‘‘β„Žπ‘’ π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘› 𝑖𝑠 π‘žπ‘’π‘Žπ‘‘π‘Ÿπ‘Žπ‘‘π‘–π‘ π‘€β„Žπ‘’π‘Ÿπ‘’:π‘Ž = βˆ’5, 𝑏 = βˆ’13, 𝑐 = βˆ’44𝐼𝑓 π‘Ž
𝐼𝑓 π‘Ž > 0 , π‘”π‘Ÿπ‘Žπ‘β„Ž 𝑖𝑠 π‘šπ‘–π‘›π‘–π‘šπ‘’π‘š (π‘œπ‘π‘’π‘›π‘  π‘’π‘π‘€π‘Žπ‘Ÿπ‘‘)𝑆𝑖𝑛𝑐𝑒 (π‘Ž = βˆ’5)
π‘€π‘Žπ‘₯π‘–π‘šπ‘’π‘š π‘£π‘Žπ‘™π‘’π‘’ 𝑖𝑠 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ β„Ž 𝑖𝑛 π‘‘β„Žπ‘’ π‘£π‘’π‘Ÿπ‘‘π‘’π‘₯ (β„Ž, π‘˜)π‘πΉπ‘œπ‘Ÿπ‘šπ‘’π‘™π‘Ž: β„Ž = βˆ’ 2π‘Žβ„Ž=βˆ’βˆ’(βˆ’13)2(βˆ’5)13= βˆ’ 10 = βˆ’πŸ. πŸ‘ β†’ π’Žπ’‚π’™π’Šπ’Žπ’–π’Ž π’—π’‚π’π’–π’†π‘΄π’Šπ’π’Šπ’Žπ’–π’Ž 𝒗𝒂𝒍𝒖𝒆: 𝑆𝑖𝑛𝑐𝑒 π‘‘β„Žπ‘’ π‘”π‘Ÿπ‘Žπ‘β„Ž 𝑖𝑠 π‘Ž π‘žπ‘’π‘Žπ‘‘π‘Ÿπ‘Žπ‘‘π‘–π‘ π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘› π‘œπ‘π‘’π‘›π‘–π‘›π‘” π‘‘π‘œπ‘€π‘›π‘€π‘Žπ‘Ÿπ‘‘ ,π‘‘β„Žπ‘’π‘Ÿπ‘’ 𝑖𝑠 π‘›π‘œ π‘‘π‘’π‘‘π‘’π‘Ÿπ‘šπ‘–π‘›π‘Žπ‘π‘™π‘’ π‘šπ‘–π‘›π‘–π‘šπ‘’π‘š π‘£π‘Žπ‘™π‘’π‘’.3. βˆ’4π‘₯ + 3𝑦 + 7𝑧 = 25 β†’ eq. 12π‘₯ βˆ’ 𝑦 + 6𝑧 = 17 β†’eq. 2βˆ’8π‘₯ βˆ’ 5𝑦 + 3𝑧 = βˆ’5 β†’ eq. 3βž” eq. 1 + 2* eq.2βˆ’4π‘₯ + 3𝑦 + 7𝑧 = 254π‘₯ βˆ’2y +12z =34𝑦 + 19𝑧 = 59 β†’ eq. 4βž” 4*eq. 2+eq.38π‘₯ βˆ’ 4𝑦 + 24𝑧 = 68βˆ’8π‘₯ βˆ’ 5𝑦 + 3𝑧 = βˆ’5βˆ’9𝑦 + 27𝑧 = 63 β†’ eq. 5βž” 9*eq.4+eq.59𝑦 + 171𝑧 = 531βˆ’9𝑦 + 27𝑧 = 63198z= 594βˆ΄π’›=πŸ‘Substitute the value of z either to eq. 4 or eq. 5 to get y.@eq.4: 𝑦 = 59 βˆ’ 19𝑧 = 59 βˆ’ 19(3) = 2@eq.5: 𝑦 =63βˆ’27π‘§βˆ’963βˆ’27(3)=βˆ’9=2βˆ΄π’š=𝟐Substitute the value of y and z to either eq. 1, eq.2 or eq. 3 to get x.@ eq.1: π‘₯ =@ eq.2: π‘₯ =@eq.3: π‘₯ =25βˆ’3π‘¦βˆ’7π‘§βˆ’417+π‘¦βˆ’6𝑧=2βˆ’5+5π‘¦βˆ’3π‘§βˆ’8=25βˆ’3(2)βˆ’7(3)βˆ’417+(2)βˆ’6(3)=1= 2 = 0.51= = 0.522βˆ’5+5(2)βˆ’3(3)1βˆ’8= 2 = 0.5∴ 𝒙 = 𝟎. πŸ“π‘­π’Šπ’π’‚π’ π‘¨π’π’”π’˜π’†π’“: 𝒙 = 𝟎. πŸ“, π’š = 𝟐, 𝒛 = πŸ‘14. [7 log(π‘₯ 2 + 8) + 2 log(π‘₯ + 11)] βˆ’ [5log(3π‘₯ βˆ’ 1) + log(π‘₯ 2 + 4π‘₯ βˆ’ 77)]1=[log(π‘₯ 2 + 8)7 + log(π‘₯ + 11)2 ] βˆ’ [log(3π‘₯ βˆ’ 1)5 + log(π‘₯ 2 + 4π‘₯ βˆ’ 77)]1= log [(π‘₯ 2 + 8)7 (π‘₯ + 11)2 ] βˆ’ log[(3π‘₯ βˆ’ 1)5 (π‘₯ 2 + 4π‘₯ βˆ’ 77)]1(π‘₯ 2 + 8)7 (π‘₯ + 11)2)= log ((3π‘₯ βˆ’ 1)5 (π‘₯ 2 + 4π‘₯ βˆ’ 77)1(π‘₯ 2 + 8)7 (π‘₯ + 11)2)= log ((3π‘₯ βˆ’ 1)5 (π‘₯ βˆ’ 7)(π‘₯ + 11)𝟏(π’™πŸ + πŸ–)πŸ• (𝒙 + 𝟏𝟏))= π₯𝐨𝐠 ((πŸ‘π’™ βˆ’ 𝟏)πŸ“ (𝒙 βˆ’ πŸ•)5.Mode of TransportationCarBicycleBusTrainWalkingNumber of Children140601507080Children’s Mode of Transportation16%28%Car14%Bicycle12%BusTrain30%Walking6. i.) Arrange the data in ascending order then look for the median by inspection.401st462nd543rd554th625th656th697th728th789th8410th85 8991 9799thththth11 1213 1415thMEDIAN∴ π‘΄π’†π’…π’Šπ’‚π’ = πŸ•πŸii.) There is no mode since value only appears once and there is no value thatappears more than any other.π‘…π‘Žπ‘›π‘”π‘’ = π»π‘–π‘”β„Ž…

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