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Ufumbuzi - Tiger algebra Suluhishaji

234952.38

234952.38

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (21570, 12, 12, 20)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 570$ , $r = 12 %$ , $n = 12$ , $t = 20$ .

$c i (21570, 12, 12, 20)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 570$ , $r = 12 %$ , $n = 12$ , $t = 20$ .

$P = 21570, r = 12 %, n = 12, t = 20$

$\frac{12}{100} = 0.12$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 570$ , $r = 12 %$ , $n = 12$ , $t = 20$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 570$ , $r = 12 %$ , $n = 12$ , $t = 20$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 570$ , $r = 12 %$ , $n = 12$ , $t = 20$ .

2. Hesabu kipengele cha ukuaji

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

$n t = 240$

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

$r / n = 0.01, n t = 240, {(1 + r / n)}^{(n t)} = 10.8925536539$ .

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

${(1 + 0.01)}^{240} = 10.8925536539$

$r / n = 0.01, n t = 240, {(1 + r / n)}^{(n t)} = 10.8925536539$ .

$10.8925536539$

$r / n = 0.01, n t = 240, {(1 + r / n)}^{(n t)} = 10.8925536539$ .

3. Hesabu kiasi cha mwisho

$A = 234952.38$

$234952.38$

$21, 570 \cdot 10.8925536539 = 234952.38$ .

$234952.38$

$21, 570 \cdot 10.8925536539 = 234952.38$ .

$234952.38$

$21, 570 \cdot 10.8925536539 = 234952.38$ .

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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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