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Ufumbuzi - Tiger algebra Suluhishaji

19971.00

19971.00

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (19020, 5, 1, 1)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 020$ , $r = 5 %$ , $n = 1$ , $t = 1$ .

$c i (19020, 5, 1, 1)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 020$ , $r = 5 %$ , $n = 1$ , $t = 1$ .

$P = 19020, r = 5 %, n = 1, t = 1$

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 020$ , $r = 5 %$ , $n = 1$ , $t = 1$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 020$ , $r = 5 %$ , $n = 1$ , $t = 1$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 020$ , $r = 5 %$ , $n = 1$ , $t = 1$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.05$

$n t = 1$

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

$r / n = 0.05, n t = 1, {(1 + r / n)}^{(n t)} = 1.05$ .

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

${(1 + 0.05)}^{1} = 1.05$

$r / n = 0.05, n t = 1, {(1 + r / n)}^{(n t)} = 1.05$ .

$1.05$

$r / n = 0.05, n t = 1, {(1 + r / n)}^{(n t)} = 1.05$ .

3. Hesabu kiasi cha mwisho

$A = 19971.00$

$19971.00$

$19, 020 \cdot 1.05 = 19971.00$ .

$19971.00$

$19, 020 \cdot 1.05 = 19971.00$ .

$19971.00$

$19, 020 \cdot 1.05 = 19971.00$ .

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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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