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Ufumbuzi - Tiger algebra Suluhishaji

21349.16

21349.16

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (21137, 1, 4, 1)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 137$ , $r = 1 %$ , $n = 4$ , $t = 1$ .

$c i (21137, 1, 4, 1)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 137$ , $r = 1 %$ , $n = 4$ , $t = 1$ .

$P = 21137, r = 1 %, n = 4, t = 1$

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 137$ , $r = 1 %$ , $n = 4$ , $t = 1$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 137$ , $r = 1 %$ , $n = 4$ , $t = 1$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 137$ , $r = 1 %$ , $n = 4$ , $t = 1$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.0025$

$n t = 4$

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

$r / n = 0.0025, n t = 4, {(1 + r / n)}^{(n t)} = 1.0100375625$ .

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

${(1 + 0.0025)}^{4} = 1.0100375625$

$r / n = 0.0025, n t = 4, {(1 + r / n)}^{(n t)} = 1.0100375625$ .

$1.0100375625$

$r / n = 0.0025, n t = 4, {(1 + r / n)}^{(n t)} = 1.0100375625$ .

3. Hesabu kiasi cha mwisho

$A = 21349.16$

$21349.16$

$21, 137 \cdot 1.0100375625 = 21349.16$ .

$21349.16$

$21, 137 \cdot 1.0100375625 = 21349.16$ .

$21349.16$

$21, 137 \cdot 1.0100375625 = 21349.16$ .

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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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