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Ufumbuzi - Tiger algebra Suluhishaji

18629.28

18629.28

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (14531, 5, 4, 5)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 14, 531$ , $r = 5 %$ , $n = 4$ , $t = 5$ .

$c i (14531, 5, 4, 5)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 14, 531$ , $r = 5 %$ , $n = 4$ , $t = 5$ .

$P = 14531, r = 5 %, n = 4, t = 5$

$\frac{5}{100} = 0.05$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 14, 531$ , $r = 5 %$ , $n = 4$ , $t = 5$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 14, 531$ , $r = 5 %$ , $n = 4$ , $t = 5$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 14, 531$ , $r = 5 %$ , $n = 4$ , $t = 5$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.0125$

$n t = 20$

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

$r / n = 0.0125, n t = 20, {(1 + r / n)}^{(n t)} = 1.2820372317$ .

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

${(1 + 0.0125)}^{20} = 1.2820372317$

$r / n = 0.0125, n t = 20, {(1 + r / n)}^{(n t)} = 1.2820372317$ .

$1.2820372317$

$r / n = 0.0125, n t = 20, {(1 + r / n)}^{(n t)} = 1.2820372317$ .

3. Hesabu kiasi cha mwisho

$A = 18629.28$

$18629.28$

$14, 531 \cdot 1.2820372317 = 18629.28$ .

$18629.28$

$14, 531 \cdot 1.2820372317 = 18629.28$ .

$18629.28$

$14, 531 \cdot 1.2820372317 = 18629.28$ .

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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

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