Milena fez uma aplicação de $2300 a juros compostos de 2% a.m qual sera o montante que ela irá poder retirar em :
A: 3 meses
B:6 meses
C:9 meses
D: 12 meses
E: 15 meses ?
Respostas
Juros compostos
M= C * (1+i)ⁿ
M= ?
C= 2.300
i= 2%/100>> 0,02
n= 3
M= 2.300 * (1+0,02)³
M= 2.300 * (1,02)³
M= 2.300 * 1,061208
M= 2.440,77
____________________
M= C * (1+i)ⁿ
M= ?
C= 2.300
i= 2%/100>> 0,02
n= 6
M= 2.300 * (1+0,02)^6
M= 2.300 * (1,02)^6
M= 2.300 * 1,126162
M= 2.590,17
____________________
M= C * (1+i)ⁿ
M= ?
C= 2.300
i= 2%/100>> 0,02
n= 9
M= 2.300 * (1+0,03)^9
M= 2.300 * (1,02)^9
M= 2.300 * 1,195092
M= 2.748,71
____________________
M= C * (1+i)ⁿ
M= ?
C= 2.300
i= 2%/100>> 0,02
n= 12
M= 2.300 * (1+0,02)¹²
M= 2.300 * (1,02)¹²
M= 2.300 * 1,268241
M= 2.916,95
____________________
M= C * (1+i)ⁿ
M= ?
C= 2.300
i= 2%/100>> 0,02
n= 15
M= 2.300 * (1+0,02)^15
M= 2.300 * (1,02)^15
M= 2.300 * 1,345868
M= 3.095,49
____________________
Boa formatura !!!
Olá.
A: 3 meses = M = 2.440,77
m = c(1+i)t
m = 2300 (1+0,02)³
m = 2300 (1,02)³
m = 2300 . 1,061208
m = 2.440,77
B: 6 meses = M = 2.590,17
m = c(1+i)t
m = 2300(1+0,02)⁶
m = 2300(1,02)⁶
m = 2300 . 1,12616242
m = 2.590,17
C: 9 meses = M = 2.748,71
m = c(1+i)t
m = 2300(1+0,02)⁹
m = 2300(1,02)⁹
m = 2300 . 1,19509257
m = 2.748,71
D: 12 meses = M = 2.916,95
m = c(1+i)t
m = 2300(1+0,02)¹²
m = 2300(1,02)¹²
m = 2300 . 1,26824179
m = 2.916,95
E: 15 meses = M = 3.095,49
m = c(1+i)t
m = 2300 (1+0,02)¹⁵
m = 2300 (1,02)¹⁵
m = 2300 . 1,34586834
m = 3.095,49