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Ufumbuzi - Tiger algebra Suluhishaji

19234.95

19234.95

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (17090, 3, 1, 4)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 090$ , $r = 3 %$ , $n = 1$ , $t = 4$ .

$c i (17090, 3, 1, 4)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 090$ , $r = 3 %$ , $n = 1$ , $t = 4$ .

$P = 17090, r = 3 %, n = 1, t = 4$

$\frac{3}{100} = 0.03$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 090$ , $r = 3 %$ , $n = 1$ , $t = 4$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 090$ , $r = 3 %$ , $n = 1$ , $t = 4$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 090$ , $r = 3 %$ , $n = 1$ , $t = 4$ .

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

$A = 19234.95$

$19234.95$

$17, 090 \cdot 1.12550881 = 19234.95$ .

$19234.95$

$17, 090 \cdot 1.12550881 = 19234.95$ .

$19234.95$

$17, 090 \cdot 1.12550881 = 19234.95$ .

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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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