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Ufumbuzi - Tiger algebra Suluhishaji

21972.20

21972.20

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (19115, 2, 2, 7)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 115$ , $r = 2 %$ , $n = 2$ , $t = 7$ .

$c i (19115, 2, 2, 7)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 115$ , $r = 2 %$ , $n = 2$ , $t = 7$ .

$P = 19115, r = 2 %, n = 2, t = 7$

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 115$ , $r = 2 %$ , $n = 2$ , $t = 7$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 115$ , $r = 2 %$ , $n = 2$ , $t = 7$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 115$ , $r = 2 %$ , $n = 2$ , $t = 7$ .

2. Hesabu kipengele cha ukuaji

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

$n t = 14$

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

$r / n = 0.01, n t = 14, {(1 + r / n)}^{(n t)} = 1.1494742132$ .

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

${(1 + 0.01)}^{14} = 1.1494742132$

$r / n = 0.01, n t = 14, {(1 + r / n)}^{(n t)} = 1.1494742132$ .

$1.1494742132$

$r / n = 0.01, n t = 14, {(1 + r / n)}^{(n t)} = 1.1494742132$ .

3. Hesabu kiasi cha mwisho

$A = 21972.20$

$21972.20$

$19, 115 \cdot 1.1494742132 = 21972.20$ .

$21972.20$

$19, 115 \cdot 1.1494742132 = 21972.20$ .

$21972.20$

$19, 115 \cdot 1.1494742132 = 21972.20$ .

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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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