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Ufumbuzi - Tiger algebra Suluhishaji

12508.88

12508.88

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (11101, 12, 12, 1)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 11, 101$ , $r = 12 %$ , $n = 12$ , $t = 1$ .

$c i (11101, 12, 12, 1)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 11, 101$ , $r = 12 %$ , $n = 12$ , $t = 1$ .

$P = 11101, r = 12 %, n = 12, t = 1$

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 11, 101$ , $r = 12 %$ , $n = 12$ , $t = 1$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 11, 101$ , $r = 12 %$ , $n = 12$ , $t = 1$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 11, 101$ , $r = 12 %$ , $n = 12$ , $t = 1$ .

2. Hesabu kipengele cha ukuaji

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

$n t = 12$

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

$r / n = 0.01, n t = 12, {(1 + r / n)}^{(n t)} = 1.1268250301$ .

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

${(1 + 0.01)}^{12} = 1.1268250301$

$r / n = 0.01, n t = 12, {(1 + r / n)}^{(n t)} = 1.1268250301$ .

$1.1268250301$

$r / n = 0.01, n t = 12, {(1 + r / n)}^{(n t)} = 1.1268250301$ .

3. Hesabu kiasi cha mwisho

$A = 12508.88$

$12508.88$

$11, 101 \cdot 1.1268250301 = 12508.88$ .

$12508.88$

$11, 101 \cdot 1.1268250301 = 12508.88$ .

$12508.88$

$11, 101 \cdot 1.1268250301 = 12508.88$ .

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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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