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Ufumbuzi - Tiger algebra Suluhishaji

21164.12

21164.12

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (19169, 2, 1, 5)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 169$ , $r = 2 %$ , $n = 1$ , $t = 5$ .

$c i (19169, 2, 1, 5)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 169$ , $r = 2 %$ , $n = 1$ , $t = 5$ .

$P = 19169, r = 2 %, n = 1, t = 5$

$\frac{2}{100} = 0.02$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 169$ , $r = 2 %$ , $n = 1$ , $t = 5$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 169$ , $r = 2 %$ , $n = 1$ , $t = 5$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 169$ , $r = 2 %$ , $n = 1$ , $t = 5$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.02$

$n t = 5$

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

$r / n = 0.02, n t = 5, {(1 + r / n)}^{(n t)} = 1.1040808032$ .

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

${(1 + 0.02)}^{5} = 1.1040808032$

$r / n = 0.02, n t = 5, {(1 + r / n)}^{(n t)} = 1.1040808032$ .

$1.1040808032$

$r / n = 0.02, n t = 5, {(1 + r / n)}^{(n t)} = 1.1040808032$ .

3. Hesabu kiasi cha mwisho

$A = 21164.12$

$21164.12$

$19, 169 \cdot 1.1040808032 = 21164.12$ .

$21164.12$

$19, 169 \cdot 1.1040808032 = 21164.12$ .

$21164.12$

$19, 169 \cdot 1.1040808032 = 21164.12$ .

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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

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