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AlgebraQuestion and Answers: Page 302

Question Number 66769 Answers: 1 Comments: 2

simplify ((x+4)/(x−4))−((5x+20)/(x^2 −16))

$${simplify} \\ $$$$\frac{{x}+\mathrm{4}}{{x}−\mathrm{4}}−\frac{\mathrm{5}{x}+\mathrm{20}}{{x}^{\mathrm{2}} −\mathrm{16}} \\ $$

Question Number 66760 Answers: 2 Comments: 2

By writing your answer in the form a^y simplify (3^(5x) ×5^(2x) ×3^(−x) ÷5^(−2x) )^(1/4)

$${By}\:{writing}\:{your}\:{answer}\:{in}\:{the} \\ $$$${form}\:{a}^{{y}} \:{simplify} \\ $$$$\left(\mathrm{3}^{\mathrm{5}{x}} ×\mathrm{5}^{\mathrm{2}{x}} ×\mathrm{3}^{−{x}} \boldsymbol{\div}\mathrm{5}^{−\mathrm{2}{x}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$

Question Number 66743 Answers: 0 Comments: 4

Each month a store owner can spend at most $100,000 on PC′s and laptops. A PC costs the store owner $1000 and a laptop costs him $1500. Each PC is sold for a profit of $400 while a laptop is sold for a profit of $700. The store owner estimates that at least 15 PC′s but no more than 80 are sold each month. He also estimates that the number of laptops sold is at most half the PC′s. How many PC′s and how many laptops should be sold in order to maximize the profit?

$$\mathrm{Each}\:\mathrm{month}\:\mathrm{a}\:\mathrm{store}\:\mathrm{owner}\:\mathrm{can}\:\mathrm{spend}\:\mathrm{at} \\ $$$$\mathrm{most}\:\$\mathrm{100},\mathrm{000}\:\mathrm{on}\:\mathrm{PC}'\mathrm{s}\:\mathrm{and}\:\mathrm{laptops}.\:\mathrm{A} \\ $$$$\mathrm{PC}\:\mathrm{costs}\:\mathrm{the}\:\mathrm{store}\:\mathrm{owner}\:\$\mathrm{1000}\:\mathrm{and}\:\mathrm{a} \\ $$$$\mathrm{laptop}\:\mathrm{costs}\:\mathrm{him}\:\$\mathrm{1500}.\:\mathrm{Each}\:\mathrm{PC}\:\mathrm{is}\:\mathrm{sold} \\ $$$$\mathrm{for}\:\mathrm{a}\:\mathrm{profit}\:\mathrm{of}\:\$\mathrm{400}\:\mathrm{while}\:\mathrm{a}\:\mathrm{laptop}\:\mathrm{is}\:\mathrm{sold} \\ $$$$\mathrm{for}\:\mathrm{a}\:\mathrm{profit}\:\mathrm{of}\:\$\mathrm{700}.\:\mathrm{The}\:\mathrm{store}\:\mathrm{owner}\:\mathrm{estimates} \\ $$$$\mathrm{that}\:\mathrm{at}\:\mathrm{least}\:\mathrm{15}\:\mathrm{PC}'\mathrm{s}\:\mathrm{but}\:\mathrm{no}\:\mathrm{more}\:\mathrm{than} \\ $$$$\mathrm{80}\:\mathrm{are}\:\mathrm{sold}\:\mathrm{each}\:\mathrm{month}.\:\mathrm{He}\:\mathrm{also}\:\mathrm{estimates} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{laptops}\:\mathrm{sold}\:\mathrm{is}\:\mathrm{at}\:\mathrm{most} \\ $$$$\mathrm{half}\:\mathrm{the}\:\mathrm{PC}'\mathrm{s}.\:\mathrm{How}\:\mathrm{many}\:\mathrm{PC}'\mathrm{s}\:\mathrm{and}\:\mathrm{how} \\ $$$$\mathrm{many}\:\mathrm{laptops}\:\mathrm{should}\:\mathrm{be}\:\mathrm{sold}\:\mathrm{in}\:\mathrm{order}\:\mathrm{to} \\ $$$$\mathrm{maximize}\:\mathrm{the}\:\mathrm{profit}? \\ $$

Question Number 66715 Answers: 0 Comments: 4

if f(x)=((ln (x+(√(1+x^2 ))))/(√(1+x^2 ))) f^(−1) (x)=?

$${if}\:{f}\left({x}\right)=\frac{\mathrm{ln}\:\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=? \\ $$

Question Number 66619 Answers: 1 Comments: 0

solve for x,y∈R ((√(1+x^2 ))/(ln (x+(√(1+x^2 )))))=((√(1+y^2 ))/(ln (y+(√(1+y^2 )))))

$${solve}\:{for}\:{x},{y}\in{R} \\ $$$$\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{ln}\:\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}=\frac{\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }}{\mathrm{ln}\:\left({y}+\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\right)} \\ $$

Question Number 66546 Answers: 0 Comments: 6

Question Number 66508 Answers: 0 Comments: 3

Question Number 66382 Answers: 0 Comments: 14

Give me any Quintic, i shall solve it. For sure! At^5 +Bt^4 +Ct^3 +Dt^2 +Et+F=0 wont even assume A=1, or B=0. but if A+C+E=B+D+F then my formula dont work but then obviously t=−1 is a root!

$${Give}\:{me}\:{any}\:{Quintic},\:{i}\:{shall}\:{solve} \\ $$$${it}.\:{For}\:{sure}! \\ $$$${At}^{\mathrm{5}} +{Bt}^{\mathrm{4}} +{Ct}^{\mathrm{3}} +{Dt}^{\mathrm{2}} +{Et}+{F}=\mathrm{0} \\ $$$${wont}\:{even}\:{assume}\:{A}=\mathrm{1},\:{or}\:{B}=\mathrm{0}. \\ $$$${but}\:{if}\:{A}+{C}+{E}={B}+{D}+{F}\: \\ $$$${then}\:{my}\:{formula}\:{dont}\:{work} \\ $$$${but}\:{then}\:{obviously}\:{t}=−\mathrm{1}\:{is}\:{a}\:{root}! \\ $$

Question Number 66381 Answers: 0 Comments: 0