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Ufumbuzi - Tiger algebra Suluhishaji

23891.18

23891.18

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (21227, 12, 4, 1)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 227$ , $r = 12 %$ , $n = 4$ , $t = 1$ .

$c i (21227, 12, 4, 1)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 227$ , $r = 12 %$ , $n = 4$ , $t = 1$ .

$P = 21227, r = 12 %, n = 4, t = 1$

$\frac{12}{100} = 0.12$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 227$ , $r = 12 %$ , $n = 4$ , $t = 1$ .

$A = P {(1 + r / n)}^{(n t)}$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 227$ , $r = 12 %$ , $n = 4$ , $t = 1$ .

$N e x t c o m p u t e r / n, n t, a n d {(1 + r / n)}^{(n t)} .$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 227$ , $r = 12 %$ , $n = 4$ , $t = 1$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.03$

$n t = 4$

${(1 + r / n)}^{(n t)}$

$r / n = 0.03, n t = 4, {(1 + r / n)}^{(n t)} = 1.12550881$ .

${(1 + r / n)}^{(n t)}$

${(1 + 0.03)}^{4} = 1.12550881$

$r / n = 0.03, n t = 4, {(1 + r / n)}^{(n t)} = 1.12550881$ .

$1.12550881$

$r / n = 0.03, n t = 4, {(1 + r / n)}^{(n t)} = 1.12550881$ .

3. Hesabu kiasi cha mwisho

$A = 23891.18$

$23891.18$

$21, 227 \cdot 1.12550881 = 23891.18$ .

$23891.18$

$21, 227 \cdot 1.12550881 = 23891.18$ .

$23891.18$

$21, 227 \cdot 1.12550881 = 23891.18$ .

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Tiger algebra Suluhishaji

Ufumbuzi - Tiger algebra Suluhishaji

Maelezo kwa hatua

1. Weka fomula ya riba changamano

2. Hesabu kipengele cha ukuaji

3. Hesabu kiasi cha mwisho

Kwa nini kujifunza hii

Vigezo na mada

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