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Ufumbuzi - Tiger algebra Suluhishaji

29004.35

29004.35

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (17037, 6, 2, 9)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 037$ , $r = 6 %$ , $n = 2$ , $t = 9$ .

$c i (17037, 6, 2, 9)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 037$ , $r = 6 %$ , $n = 2$ , $t = 9$ .

$P = 17037, r = 6 %, n = 2, t = 9$

$\frac{6}{100} = 0.06$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 037$ , $r = 6 %$ , $n = 2$ , $t = 9$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 037$ , $r = 6 %$ , $n = 2$ , $t = 9$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 037$ , $r = 6 %$ , $n = 2$ , $t = 9$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.03$

$n t = 18$

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

$r / n = 0.03, n t = 18, {(1 + r / n)}^{(n t)} = 1.7024330612$ .

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

${(1 + 0.03)}^{18} = 1.7024330612$

$r / n = 0.03, n t = 18, {(1 + r / n)}^{(n t)} = 1.7024330612$ .

$1.7024330612$

$r / n = 0.03, n t = 18, {(1 + r / n)}^{(n t)} = 1.7024330612$ .

3. Hesabu kiasi cha mwisho

$A = 29004.35$

$29004.35$

$17, 037 \cdot 1.7024330612 = 29004.35$ .

$29004.35$

$17, 037 \cdot 1.7024330612 = 29004.35$ .

$29004.35$

$17, 037 \cdot 1.7024330612 = 29004.35$ .

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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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