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t³ - 3t² - 4t + 20 < 8
t³ - 3t² - 4t + 20 - 8 < 0
t³ - 3t² - 4t + 12 < 0
(t³ - 3t²) + (-4t + 12)
t²(t - 3) - 4(t - 3)
(t² - 4)(t - 3): Pero (t² - 4) = (t² - 2²) = (t + 2)(t - 2)
Entonces:
t³ - 3t² - 4t + 12 = (t + 2)(t - 2)(t - 3)
(t + 2)(t - 2)(t - 3) < 0
Analizamos
t + 2 < 0
t < - 2 (No nos sirve porque se sale del intervalo dado)
t - 2 < 0
t < 2 (Sirve esta dentro del intervalo dado)
t - 3 < 0
t < 3 (Sirve esta dentro del intervalo dado)
Los valores de que nos sirven estan dentro de 2 < t < 3
El intervalo es (2 , 3)
2)
Hallamos el area total de la ventana: (54 in)(24 in) = 1296 in²
Ahora la seccion que limpia la hoja seria un trapecio circular, para ello primero debemos encontrar la longitud de arco de:
Base menor = r(Ф)
Base Mayor = R(Ф)
Ф = 120°, debemos pasarlo a radianes
r = 5 in
R = 17 in + 5 in = 22 in
Regla de 3
π Radianes ============> 180°
X =================> 120°
X = [(120°)(π Radianes)]/[180°] = 2π/3 Radianes
Base menor = (5 in)(2π/3) = 10π/3
Base mayor = (22 in)(2π/3) = 44π/3
Area Trapecio Circular = [(Base Mayor + Base Menor)/2]*h
Donde h = 17 in
Area Trapecio Circular = [((44π/3) in+(10π/3) in)/2]*[17 in]
Area Trapecio Circular = [(18π)/2][17] [in²]
Area Trapecio Circular = (9π)(17) = 153π in²
Tomo π = 3.1416
Area Trapecio Circular = Area que limpia la hoja = 480.6648 in²
Para hallar el porcentaje aplico regla de 3
1296 in² =============> 100 %
480.6648 in² ==========> X
X = [(100 %)(480.6648 in²)]/[1296 in²]
X = [48066.48]/[1296] = 37.088 %
Rta: La hoja limpia el 37.088 % de la ventana