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Ufumbuzi - Tiger algebra Suluhishaji

36662.05

36662.05

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (17510, 3, 1, 25)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 510$ , $r = 3 %$ , $n = 1$ , $t = 25$ .

$c i (17510, 3, 1, 25)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 510$ , $r = 3 %$ , $n = 1$ , $t = 25$ .

$P = 17510, r = 3 %, n = 1, t = 25$

$\frac{3}{100} = 0.03$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 510$ , $r = 3 %$ , $n = 1$ , $t = 25$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 510$ , $r = 3 %$ , $n = 1$ , $t = 25$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 510$ , $r = 3 %$ , $n = 1$ , $t = 25$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.03$

$n t = 25$

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

$r / n = 0.03, n t = 25, {(1 + r / n)}^{(n t)} = 2.0937779297$ .

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

${(1 + 0.03)}^{25} = 2.0937779297$

$r / n = 0.03, n t = 25, {(1 + r / n)}^{(n t)} = 2.0937779297$ .

$2.0937779297$

$r / n = 0.03, n t = 25, {(1 + r / n)}^{(n t)} = 2.0937779297$ .

3. Hesabu kiasi cha mwisho

$A = 36662.05$

$36662.05$

$17, 510 \cdot 2.0937779297 = 36662.05$ .

$36662.05$

$17, 510 \cdot 2.0937779297 = 36662.05$ .

$36662.05$

$17, 510 \cdot 2.0937779297 = 36662.05$ .

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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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