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Ufumbuzi - Tiger algebra Suluhishaji

11744.07

11744.07

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (11172, 1, 4, 5)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 11, 172$ , $r = 1 %$ , $n = 4$ , $t = 5$ .

$c i (11172, 1, 4, 5)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 11, 172$ , $r = 1 %$ , $n = 4$ , $t = 5$ .

$P = 11172, r = 1 %, n = 4, t = 5$

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 11, 172$ , $r = 1 %$ , $n = 4$ , $t = 5$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 11, 172$ , $r = 1 %$ , $n = 4$ , $t = 5$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 11, 172$ , $r = 1 %$ , $n = 4$ , $t = 5$ .

2. Hesabu kipengele cha ukuaji

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

$n t = 20$

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

$r / n = 0.0025, n t = 20, {(1 + r / n)}^{(n t)} = 1.0512055033$ .

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

${(1 + 0.0025)}^{20} = 1.0512055033$

$r / n = 0.0025, n t = 20, {(1 + r / n)}^{(n t)} = 1.0512055033$ .

$1.0512055033$

$r / n = 0.0025, n t = 20, {(1 + r / n)}^{(n t)} = 1.0512055033$ .

3. Hesabu kiasi cha mwisho

$A = 11744.07$

$11744.07$

$11, 172 \cdot 1.0512055033 = 11744.07$ .

$11744.07$

$11, 172 \cdot 1.0512055033 = 11744.07$ .

$11744.07$

$11, 172 \cdot 1.0512055033 = 11744.07$ .

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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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