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Ufumbuzi - Tiger algebra Suluhishaji

30555.08

30555.08

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (18740, 13, 1, 4)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 740$ , $r = 13 %$ , $n = 1$ , $t = 4$ .

$c i (18740, 13, 1, 4)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 740$ , $r = 13 %$ , $n = 1$ , $t = 4$ .

$P = 18740, r = 13 %, n = 1, t = 4$

$\frac{13}{100} = 0.13$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 740$ , $r = 13 %$ , $n = 1$ , $t = 4$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 740$ , $r = 13 %$ , $n = 1$ , $t = 4$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 740$ , $r = 13 %$ , $n = 1$ , $t = 4$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.13$

$n t = 4$

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

$r / n = 0.13, n t = 4, {(1 + r / n)}^{(n t)} = 1.63047361$ .

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

${(1 + 0.13)}^{4} = 1.63047361$

$r / n = 0.13, n t = 4, {(1 + r / n)}^{(n t)} = 1.63047361$ .

$1.63047361$

$r / n = 0.13, n t = 4, {(1 + r / n)}^{(n t)} = 1.63047361$ .

3. Hesabu kiasi cha mwisho

$A = 30555.08$

$30555.08$

$18, 740 \cdot 1.63047361 = 30555.08$ .

$30555.08$

$18, 740 \cdot 1.63047361 = 30555.08$ .

$30555.08$

$18, 740 \cdot 1.63047361 = 30555.08$ .

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Jifunze zaidi na Tiger

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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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