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Ufumbuzi - Tiger algebra Suluhishaji

37729.35

37729.35

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (19179, 4, 4, 17)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 179$ , $r = 4 %$ , $n = 4$ , $t = 17$ .

$c i (19179, 4, 4, 17)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 179$ , $r = 4 %$ , $n = 4$ , $t = 17$ .

$P = 19179, r = 4 %, n = 4, t = 17$

$\frac{4}{100} = 0.04$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 179$ , $r = 4 %$ , $n = 4$ , $t = 17$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 179$ , $r = 4 %$ , $n = 4$ , $t = 17$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 179$ , $r = 4 %$ , $n = 4$ , $t = 17$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.01$

$n t = 68$

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

$r / n = 0.01, n t = 68, {(1 + r / n)}^{(n t)} = 1.9672222021$ .

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

${(1 + 0.01)}^{68} = 1.9672222021$

$r / n = 0.01, n t = 68, {(1 + r / n)}^{(n t)} = 1.9672222021$ .

$1.9672222021$

$r / n = 0.01, n t = 68, {(1 + r / n)}^{(n t)} = 1.9672222021$ .

3. Hesabu kiasi cha mwisho

$A = 37729.35$

$37729.35$

$19, 179 \cdot 1.9672222021 = 37729.35$ .

$37729.35$

$19, 179 \cdot 1.9672222021 = 37729.35$ .

$37729.35$

$19, 179 \cdot 1.9672222021 = 37729.35$ .

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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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