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Ufumbuzi - Tiger algebra Suluhishaji

1709.75

1709.75

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (1294, 1, 1, 28)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 1, 294$ , $r = 1 %$ , $n = 1$ , $t = 28$ .

$c i (1294, 1, 1, 28)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 1, 294$ , $r = 1 %$ , $n = 1$ , $t = 28$ .

$P = 1294, r = 1 %, n = 1, t = 28$

$\frac{1}{100} = 0.01$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 1, 294$ , $r = 1 %$ , $n = 1$ , $t = 28$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 1, 294$ , $r = 1 %$ , $n = 1$ , $t = 28$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 1, 294$ , $r = 1 %$ , $n = 1$ , $t = 28$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.01$

$n t = 28$

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

$r / n = 0.01, n t = 28, {(1 + r / n)}^{(n t)} = 1.3212909669$ .

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

${(1 + 0.01)}^{28} = 1.3212909669$

$r / n = 0.01, n t = 28, {(1 + r / n)}^{(n t)} = 1.3212909669$ .

$1.3212909669$

$r / n = 0.01, n t = 28, {(1 + r / n)}^{(n t)} = 1.3212909669$ .

3. Hesabu kiasi cha mwisho

$A = 1709.75$

$1709.75$

$1, 294 \cdot 1.3212909669 = 1709.75$ .

$1709.75$

$1, 294 \cdot 1.3212909669 = 1709.75$ .

$1709.75$

$1, 294 \cdot 1.3212909669 = 1709.75$ .

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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

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