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Ufumbuzi - Tiger algebra Suluhishaji

51119.98

51119.98

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (17271, 11, 4, 10)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 271$ , $r = 11 %$ , $n = 4$ , $t = 10$ .

$c i (17271, 11, 4, 10)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 271$ , $r = 11 %$ , $n = 4$ , $t = 10$ .

$P = 17271, r = 11 %, n = 4, t = 10$

$\frac{11}{100} = 0.11$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 271$ , $r = 11 %$ , $n = 4$ , $t = 10$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 271$ , $r = 11 %$ , $n = 4$ , $t = 10$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 271$ , $r = 11 %$ , $n = 4$ , $t = 10$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.0275$

$n t = 40$

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

$r / n = 0.0275, n t = 40, {(1 + r / n)}^{(n t)} = 2.9598739872$ .

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

${(1 + 0.0275)}^{40} = 2.9598739872$

$r / n = 0.0275, n t = 40, {(1 + r / n)}^{(n t)} = 2.9598739872$ .

$2.9598739872$

$r / n = 0.0275, n t = 40, {(1 + r / n)}^{(n t)} = 2.9598739872$ .

3. Hesabu kiasi cha mwisho

$A = 51119.98$

$51119.98$

$17, 271 \cdot 2.9598739872 = 51119.98$ .

$51119.98$

$17, 271 \cdot 2.9598739872 = 51119.98$ .

$51119.98$

$17, 271 \cdot 2.9598739872 = 51119.98$ .

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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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