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Ufumbuzi - Tiger algebra Suluhishaji

25070.20

25070.20

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (24575, 1, 2, 2)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 575$ , $r = 1 %$ , $n = 2$ , $t = 2$ .

$c i (24575, 1, 2, 2)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 575$ , $r = 1 %$ , $n = 2$ , $t = 2$ .

$P = 24575, r = 1 %, n = 2, t = 2$

$\frac{1}{100} = 0.01$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 575$ , $r = 1 %$ , $n = 2$ , $t = 2$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 575$ , $r = 1 %$ , $n = 2$ , $t = 2$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 575$ , $r = 1 %$ , $n = 2$ , $t = 2$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.005$

$n t = 4$

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

$r / n = 0.005, n t = 4, {(1 + r / n)}^{(n t)} = 1.0201505006$ .

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

${(1 + 0.005)}^{4} = 1.0201505006$

$r / n = 0.005, n t = 4, {(1 + r / n)}^{(n t)} = 1.0201505006$ .

$1.0201505006$

$r / n = 0.005, n t = 4, {(1 + r / n)}^{(n t)} = 1.0201505006$ .

3. Hesabu kiasi cha mwisho

$A = 25070.20$

$25070.20$

$24, 575 \cdot 1.0201505006 = 25070.20$ .

$25070.20$

$24, 575 \cdot 1.0201505006 = 25070.20$ .

$25070.20$

$24, 575 \cdot 1.0201505006 = 25070.20$ .

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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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