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Ufumbuzi - Tiger algebra Suluhishaji

111992.69

111992.69

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (10366, 12, 1, 21)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 10, 366$ , $r = 12 %$ , $n = 1$ , $t = 21$ .

$c i (10366, 12, 1, 21)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 10, 366$ , $r = 12 %$ , $n = 1$ , $t = 21$ .

$P = 10366, r = 12 %, n = 1, t = 21$

$\frac{12}{100} = 0.12$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 10, 366$ , $r = 12 %$ , $n = 1$ , $t = 21$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 10, 366$ , $r = 12 %$ , $n = 1$ , $t = 21$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 10, 366$ , $r = 12 %$ , $n = 1$ , $t = 21$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.12$

$n t = 21$

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

$r / n = 0.12, n t = 21, {(1 + r / n)}^{(n t)} = 10.8038482645$ .

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

${(1 + 0.12)}^{21} = 10.8038482645$

$r / n = 0.12, n t = 21, {(1 + r / n)}^{(n t)} = 10.8038482645$ .

$10.8038482645$

$r / n = 0.12, n t = 21, {(1 + r / n)}^{(n t)} = 10.8038482645$ .

3. Hesabu kiasi cha mwisho

$A = 111992.69$

$111992.69$

$10, 366 \cdot 10.8038482645 = 111992.69$ .

$111992.69$

$10, 366 \cdot 10.8038482645 = 111992.69$ .

$111992.69$

$10, 366 \cdot 10.8038482645 = 111992.69$ .

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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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