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Ufumbuzi - Tiger algebra Suluhishaji

294009.94

294009.94

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (18505, 10, 4, 28)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 505$ , $r = 10 %$ , $n = 4$ , $t = 28$ .

$c i (18505, 10, 4, 28)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 505$ , $r = 10 %$ , $n = 4$ , $t = 28$ .

$P = 18505, r = 10 %, n = 4, t = 28$

$\frac{10}{100} = 0.1$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 505$ , $r = 10 %$ , $n = 4$ , $t = 28$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 505$ , $r = 10 %$ , $n = 4$ , $t = 28$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 505$ , $r = 10 %$ , $n = 4$ , $t = 28$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.025$

$n t = 112$

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

$r / n = 0.025, n t = 112, {(1 + r / n)}^{(n t)} = 15.8881350934$ .

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

${(1 + 0.025)}^{112} = 15.8881350934$

$r / n = 0.025, n t = 112, {(1 + r / n)}^{(n t)} = 15.8881350934$ .

$15.8881350934$

$r / n = 0.025, n t = 112, {(1 + r / n)}^{(n t)} = 15.8881350934$ .

3. Hesabu kiasi cha mwisho

$A = 294009.94$

$294009.94$

$18, 505 \cdot 15.8881350934 = 294009.94$ .

$294009.94$

$18, 505 \cdot 15.8881350934 = 294009.94$ .

$294009.94$

$18, 505 \cdot 15.8881350934 = 294009.94$ .

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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

Vigezo na mada

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