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Ufumbuzi - Tiger algebra Suluhishaji

55947.23

55947.23

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (24039, 5, 4, 17)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 039$ , $r = 5 %$ , $n = 4$ , $t = 17$ .

$c i (24039, 5, 4, 17)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 039$ , $r = 5 %$ , $n = 4$ , $t = 17$ .

$P = 24039, r = 5 %, n = 4, t = 17$

$\frac{5}{100} = 0.05$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 039$ , $r = 5 %$ , $n = 4$ , $t = 17$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 039$ , $r = 5 %$ , $n = 4$ , $t = 17$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 039$ , $r = 5 %$ , $n = 4$ , $t = 17$ .

2. Hesabu kipengele cha ukuaji

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

$n t = 68$

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

$r / n = 0.0125, n t = 68, {(1 + r / n)}^{(n t)} = 2.3273525104$ .

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

${(1 + 0.0125)}^{68} = 2.3273525104$

$r / n = 0.0125, n t = 68, {(1 + r / n)}^{(n t)} = 2.3273525104$ .

$2.3273525104$

$r / n = 0.0125, n t = 68, {(1 + r / n)}^{(n t)} = 2.3273525104$ .

3. Hesabu kiasi cha mwisho

$A = 55947.23$

$55947.23$

$24, 039 \cdot 2.3273525104 = 55947.23$ .

$55947.23$

$24, 039 \cdot 2.3273525104 = 55947.23$ .

$55947.23$

$24, 039 \cdot 2.3273525104 = 55947.23$ .

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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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