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Ufumbuzi - Tiger algebra Suluhishaji

28783.17

28783.17

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (18697, 4, 1, 11)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 697$ , $r = 4 %$ , $n = 1$ , $t = 11$ .

$c i (18697, 4, 1, 11)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 697$ , $r = 4 %$ , $n = 1$ , $t = 11$ .

$P = 18697, r = 4 %, n = 1, t = 11$

$\frac{4}{100} = 0.04$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 697$ , $r = 4 %$ , $n = 1$ , $t = 11$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 697$ , $r = 4 %$ , $n = 1$ , $t = 11$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 697$ , $r = 4 %$ , $n = 1$ , $t = 11$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.04$

$n t = 11$

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

$r / n = 0.04, n t = 11, {(1 + r / n)}^{(n t)} = 1.5394540563$ .

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

${(1 + 0.04)}^{11} = 1.5394540563$

$r / n = 0.04, n t = 11, {(1 + r / n)}^{(n t)} = 1.5394540563$ .

$1.5394540563$

$r / n = 0.04, n t = 11, {(1 + r / n)}^{(n t)} = 1.5394540563$ .

3. Hesabu kiasi cha mwisho

$A = 28783.17$

$28783.17$

$18, 697 \cdot 1.5394540563 = 28783.17$ .

$28783.17$

$18, 697 \cdot 1.5394540563 = 28783.17$ .

$28783.17$

$18, 697 \cdot 1.5394540563 = 28783.17$ .

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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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