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Ufumbuzi - Tiger algebra Suluhishaji

2023.62

2023.62

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (1017, 14, 4, 5)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 1, 017$ , $r = 14 %$ , $n = 4$ , $t = 5$ .

$c i (1017, 14, 4, 5)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 1, 017$ , $r = 14 %$ , $n = 4$ , $t = 5$ .

$P = 1017, r = 14 %, n = 4, t = 5$

$\frac{14}{100} = 0.14$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 1, 017$ , $r = 14 %$ , $n = 4$ , $t = 5$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 1, 017$ , $r = 14 %$ , $n = 4$ , $t = 5$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 1, 017$ , $r = 14 %$ , $n = 4$ , $t = 5$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.035$

$n t = 20$

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

$r / n = 0.035, n t = 20, {(1 + r / n)}^{(n t)} = 1.9897888635$ .

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

${(1 + 0.035)}^{20} = 1.9897888635$

$r / n = 0.035, n t = 20, {(1 + r / n)}^{(n t)} = 1.9897888635$ .

$1.9897888635$

$r / n = 0.035, n t = 20, {(1 + r / n)}^{(n t)} = 1.9897888635$ .

3. Hesabu kiasi cha mwisho

$A = 2023.62$

$2023.62$

$1, 017 \cdot 1.9897888635 = 2023.62$ .

$2023.62$

$1, 017 \cdot 1.9897888635 = 2023.62$ .

$2023.62$

$1, 017 \cdot 1.9897888635 = 2023.62$ .

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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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