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Ufumbuzi - Tiger algebra Suluhishaji

2673.95

2673.95

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (1793, 2, 12, 20)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 1, 793$ , $r = 2 %$ , $n = 12$ , $t = 20$ .

$c i (1793, 2, 12, 20)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 1, 793$ , $r = 2 %$ , $n = 12$ , $t = 20$ .

$P = 1793, r = 2 %, n = 12, t = 20$

$\frac{2}{100} = 0.02$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 1, 793$ , $r = 2 %$ , $n = 12$ , $t = 20$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 1, 793$ , $r = 2 %$ , $n = 12$ , $t = 20$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 1, 793$ , $r = 2 %$ , $n = 12$ , $t = 20$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.0016666667$

$n t = 240$

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

$r / n = 0.0016666667, n t = 240, {(1 + r / n)}^{(n t)} = 1.4913280573$ .

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

${(1 + 0.0016666667)}^{240} = 1.4913280573$

$r / n = 0.0016666667, n t = 240, {(1 + r / n)}^{(n t)} = 1.4913280573$ .

$1.4913280573$

$r / n = 0.0016666667, n t = 240, {(1 + r / n)}^{(n t)} = 1.4913280573$ .

3. Hesabu kiasi cha mwisho

$A = 2673.95$

$2673.95$

$1, 793 \cdot 1.4913280573 = 2673.95$ .

$2673.95$

$1, 793 \cdot 1.4913280573 = 2673.95$ .

$2673.95$

$1, 793 \cdot 1.4913280573 = 2673.95$ .

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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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