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Ufumbuzi - Tiger algebra Suluhishaji

25824.35

25824.35

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (24813, 1, 4, 4)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 813$ , $r = 1 %$ , $n = 4$ , $t = 4$ .

$c i (24813, 1, 4, 4)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 813$ , $r = 1 %$ , $n = 4$ , $t = 4$ .

$P = 24813, r = 1 %, n = 4, t = 4$

$\frac{1}{100} = 0.01$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 813$ , $r = 1 %$ , $n = 4$ , $t = 4$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 813$ , $r = 1 %$ , $n = 4$ , $t = 4$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 813$ , $r = 1 %$ , $n = 4$ , $t = 4$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.0025$

$n t = 16$

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

$r / n = 0.0025, n t = 16, {(1 + r / n)}^{(n t)} = 1.0407588215$ .

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

${(1 + 0.0025)}^{16} = 1.0407588215$

$r / n = 0.0025, n t = 16, {(1 + r / n)}^{(n t)} = 1.0407588215$ .

$1.0407588215$

$r / n = 0.0025, n t = 16, {(1 + r / n)}^{(n t)} = 1.0407588215$ .

3. Hesabu kiasi cha mwisho

$A = 25824.35$

$25824.35$

$24, 813 \cdot 1.0407588215 = 25824.35$ .

$25824.35$

$24, 813 \cdot 1.0407588215 = 25824.35$ .

$25824.35$

$24, 813 \cdot 1.0407588215 = 25824.35$ .

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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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