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Ufumbuzi - Tiger algebra Suluhishaji

338377.02

338377.02

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (17163, 15, 12, 20)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 163$ , $r = 15 %$ , $n = 12$ , $t = 20$ .

$c i (17163, 15, 12, 20)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 163$ , $r = 15 %$ , $n = 12$ , $t = 20$ .

$P = 17163, r = 15 %, n = 12, t = 20$

$\frac{15}{100} = 0.15$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 163$ , $r = 15 %$ , $n = 12$ , $t = 20$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 163$ , $r = 15 %$ , $n = 12$ , $t = 20$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 163$ , $r = 15 %$ , $n = 12$ , $t = 20$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.0125$

$n t = 240$

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

$r / n = 0.0125, n t = 240, {(1 + r / n)}^{(n t)} = 19.7154935184$ .

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

${(1 + 0.0125)}^{240} = 19.7154935184$

$r / n = 0.0125, n t = 240, {(1 + r / n)}^{(n t)} = 19.7154935184$ .

$19.7154935184$

$r / n = 0.0125, n t = 240, {(1 + r / n)}^{(n t)} = 19.7154935184$ .

3. Hesabu kiasi cha mwisho

$A = 338377.02$

$338377.02$

$17, 163 \cdot 19.7154935184 = 338377.02$ .

$338377.02$

$17, 163 \cdot 19.7154935184 = 338377.02$ .

$338377.02$

$17, 163 \cdot 19.7154935184 = 338377.02$ .

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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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