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Ufumbuzi - Tiger algebra Suluhishaji

21865.56

21865.56

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (18099, 1, 1, 19)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 099$ , $r = 1 %$ , $n = 1$ , $t = 19$ .

$c i (18099, 1, 1, 19)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 099$ , $r = 1 %$ , $n = 1$ , $t = 19$ .

$P = 18099, r = 1 %, n = 1, t = 19$

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 099$ , $r = 1 %$ , $n = 1$ , $t = 19$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 099$ , $r = 1 %$ , $n = 1$ , $t = 19$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 099$ , $r = 1 %$ , $n = 1$ , $t = 19$ .

2. Hesabu kipengele cha ukuaji

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

$n t = 19$

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

$r / n = 0.01, n t = 19, {(1 + r / n)}^{(n t)} = 1.2081089504$ .

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

${(1 + 0.01)}^{19} = 1.2081089504$

$r / n = 0.01, n t = 19, {(1 + r / n)}^{(n t)} = 1.2081089504$ .

$1.2081089504$

$r / n = 0.01, n t = 19, {(1 + r / n)}^{(n t)} = 1.2081089504$ .

3. Hesabu kiasi cha mwisho

$A = 21865.56$

$21865.56$

$18, 099 \cdot 1.2081089504 = 21865.56$ .

$21865.56$

$18, 099 \cdot 1.2081089504 = 21865.56$ .

$21865.56$

$18, 099 \cdot 1.2081089504 = 21865.56$ .

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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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