Economia/adm de Engenharia Mecânica
Por: Raimenson Silva • 17/5/2020 • Trabalho acadêmico • 956 Palavras (4 Páginas) • 147 Visualizações
ECONOMIA/ADM 4ºP
Mod01
05)
M= C(1+i.t), fazendo M1=M2 temos: C1(1+i1.t) = C2(1+i2.t) > 14000(1+4.10^-4t)= 15000(1+3.10^-4t) > 14000+5,6t= 15000+4,5t > t=909,09dias; 909,09/365= 2anos; 0,5252.12= 6meses; 0,3024.30=9dias
06)
07)
J=c.i.t > c=J/i.t > c=47650,2/(41.0,03) > c=R$:38740
08)
M=c+J > J=M-c > 83700-5400 > J=R$:297
09)
Observe que c2 será igual a M1; M= C(1+i.t) logo temos: c2=M/(1+i.t) > c2=35865,54/(1+0,012.12) > c2=31350,99 = M1; c1= M1/(1+i1.t1) > c1= 31350,99/(1+0,01.7) > c1=R$:29300
10)
Observe que: M=2c > M= c(1+i.t) > 2c= c +c.i.t > t=1/0,02 > t=50bimestre, 50/6=8anos >0,333.12=4meses; porém a resposta mais próxima é 4anos e 2meses.
11)
t1=t2= 238; M1=c1(1+i1.t) = 23000(1+5.10^-4.238) > M1=R$:25737; M1=M2; c2= M2/(1+i2.t) > c2= 25737/(1+4.10^-4.238) > c2=R$:23499,8
Mod02
01)
M1=c2; M1=c1(1+0,085)^2 > M1=77932,295; M2= 77932,295(1+0,028)^9
M2=99920,71
02)
M = C*(1 + i)^t > 3C = C*(1 + i)^t > 3 = (1 + 0.04)^t
t = log(3)/log(1.04) > t = 28,01meses ou t=2anos e 4meses
03)
M=c+J > 73500+16845 > M=90345; i=(M/c)^(1/t) -1 > i= (90345/73500)^(1/8)-1 > i=0,016; t=ln(M/c)/ln(1+i) > t=ln(90345/73500)/ln(1+0,016) > t= 13meses.
04)
M = C.(1 + i)^t ; t1=t2=t; M1=M2> C1.(1 + i1)^t= C2.(1 + i2)^t > 100000.(1+0,034)^t=150000.(1+0,0245)^t > t=ln(1,5)/ln(1,009) > t =45,254 > 42,254/12 > t= 3anos e 8mês.
05) i=(M/c)^1/t – 1; aplicando a formula mencionada temos: i1= 9.10^-3; i2= 11.10^-3; i3= 9.10^-3.
ECONOMIA/ADM 4ºP
Mod01
05)
M= C(1+i.t), fazendo M1=M2 temos: C1(1+i1.t) = C2(1+i2.t) > 14000(1+4.10^-4t)= 15000(1+3.10^-4t) > 14000+5,6t= 15000+4,5t > t=909,09dias; 909,09/365= 2anos; 0,5252.12= 6meses; 0,3024.30=9dias
06)
07)
J=c.i.t > c=J/i.t > c=47650,2/(41.0,03) > c=R$:38740
08)
M=c+J > J=M-c > 83700-5400 > J=R$:297
09)
Observe que c2 será igual a M1; M= C(1+i.t) logo temos: c2=M/(1+i.t) > c2=35865,54/(1+0,012.12) > c2=31350,99 = M1; c1= M1/(1+i1.t1) > c1= 31350,99/(1+0,01.7) > c1=R$:29300
10)
Observe que: M=2c > M= c(1+i.t) > 2c= c +c.i.t > t=1/0,02 > t=50bimestre, 50/6=8anos >0,333.12=4meses; porém a resposta mais próxima é 4anos e 2meses.
11)
t1=t2= 238; M1=c1(1+i1.t) = 23000(1+5.10^-4.238) > M1=R$:25737; M1=M2; c2= M2/(1+i2.t) > c2= 25737/(1+4.10^-4.238) > c2=R$:23499,8
Mod02
01)
M1=c2; M1=c1(1+0,085)^2 > M1=77932,295; M2= 77932,295(1+0,028)^9
M2=99920,71
02)
M = C*(1 + i)^t > 3C = C*(1 + i)^t > 3 = (1 + 0.04)^t
t = log(3)/log(1.04) > t = 28,01meses ou t=2anos e 4meses
03)
M=c+J > 73500+16845 > M=90345; i=(M/c)^(1/t) -1 > i= (90345/73500)^(1/8)-1 > i=0,016; t=ln(M/c)/ln(1+i) > t=ln(90345/73500)/ln(1+0,016) > t= 13meses.
04)
M = C.(1 + i)^t ; t1=t2=t; M1=M2> C1.(1 + i1)^t= C2.(1 + i2)^t > 100000.(1+0,034)^t=150000.(1+0,0245)^t > t=ln(1,5)/ln(1,009) > t =45,254 > 42,254/12 > t= 3anos e 8mês.
05) i=(M/c)^1/t – 1; aplicando a formula mencionada temos: i1= 9.10^-3; i2= 11.10^-3; i3= 9.10^-3.
ECONOMIA/ADM 4ºP
Mod01
05)
M= C(1+i.t), fazendo M1=M2 temos: C1(1+i1.t) = C2(1+i2.t) > 14000(1+4.10^-4t)= 15000(1+3.10^-4t) > 14000+5,6t= 15000+4,5t > t=909,09dias; 909,09/365= 2anos; 0,5252.12= 6meses; 0,3024.30=9dias
06)
07)
J=c.i.t > c=J/i.t > c=47650,2/(41.0,03) > c=R$:38740
08)
M=c+J > J=M-c > 83700-5400 > J=R$:297
09)
Observe que c2 será igual a M1; M= C(1+i.t) logo temos: c2=M/(1+i.t) > c2=35865,54/(1+0,012.12) > c2=31350,99 = M1; c1= M1/(1+i1.t1) > c1= 31350,99/(1+0,01.7) > c1=R$:29300
10)
Observe que: M=2c > M= c(1+i.t) > 2c= c +c.i.t > t=1/0,02 > t=50bimestre, 50/6=8anos >0,333.12=4meses; porém a resposta mais próxima é 4anos e 2meses.
11)
t1=t2= 238; M1=c1(1+i1.t) = 23000(1+5.10^-4.238) > M1=R$:25737; M1=M2; c2= M2/(1+i2.t) > c2= 25737/(1+4.10^-4.238) > c2=R$:23499,8
Mod02
01)
M1=c2; M1=c1(1+0,085)^2 > M1=77932,295; M2= 77932,295(1+0,028)^9
M2=99920,71
02)
M = C*(1 + i)^t > 3C = C*(1 + i)^t > 3 = (1 + 0.04)^t
t = log(3)/log(1.04) > t = 28,01meses ou t=2anos e 4meses
03)
M=c+J > 73500+16845 > M=90345; i=(M/c)^(1/t) -1 > i= (90345/73500)^(1/8)-1 > i=0,016; t=ln(M/c)/ln(1+i) > t=ln(90345/73500)/ln(1+0,016) > t= 13meses.
04)
M = C.(1 + i)^t ; t1=t2=t; M1=M2> C1.(1 + i1)^t= C2.(1 + i2)^t > 100000.(1+0,034)^t=150000.(1+0,0245)^t > t=ln(1,5)/ln(1,009) > t =45,254 > 42,254/12 > t= 3anos e 8mês.
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