Jak z podanego poniżej wzoru wyznaczyć literę r ?
P = πr(r+l)
P = πr² + πlr
πr² + πlr - P = 0
Δ = b² - 4ac
Δ = π²l² - 4πP
r = -b + √Δ/2a
r = -l + √π²l² - 4πP/2 x π
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Litterarum radices amarae sunt, fructus iucundiores
Pozdrawiam :)
P = πr²+πrl
πr²+πlr-P = 0
a = π, b = πl, c = -P
Δ = b²-4ac = (πl)²-4·π·(-P) = π²l²+4πP = π(πl²+4P)
√Δ = √[π (πl²+4P)]
r = (-b+√Δ)/2a = [-πl+√((πl²+4P))]/2π = -l/2 + √((π(l²+4P)/4π²)) = -l/2 + √(((l²+4P)/2π))
r = -l/2 + √((l²+4P)/2π))
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P = πr(r+l)
P = πr² + πlr
πr² + πlr - P = 0
Δ = b² - 4ac
Δ = π²l² - 4πP
r = -b + √Δ/2a
r = -l + √π²l² - 4πP/2 x π
----------------------------------------------------------------------------------------------------
Litterarum radices amarae sunt, fructus iucundiores
Pozdrawiam :)
P = πr(r+l)
P = πr²+πrl
πr²+πlr-P = 0
a = π, b = πl, c = -P
Δ = b²-4ac = (πl)²-4·π·(-P) = π²l²+4πP = π(πl²+4P)
√Δ = √[π (πl²+4P)]
r = (-b+√Δ)/2a = [-πl+√((πl²+4P))]/2π = -l/2 + √((π(l²+4P)/4π²)) = -l/2 + √(((l²+4P)/2π))
r = -l/2 + √((l²+4P)/2π))