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Ufumbuzi - Tiger algebra Suluhishaji

102022.31

102022.31

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (19253, 6, 4, 28)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 253$ , $r = 6 %$ , $n = 4$ , $t = 28$ .

$c i (19253, 6, 4, 28)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 253$ , $r = 6 %$ , $n = 4$ , $t = 28$ .

$P = 19253, r = 6 %, n = 4, t = 28$

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 253$ , $r = 6 %$ , $n = 4$ , $t = 28$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 253$ , $r = 6 %$ , $n = 4$ , $t = 28$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 19, 253$ , $r = 6 %$ , $n = 4$ , $t = 28$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.015$

$n t = 112$

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

$r / n = 0.015, n t = 112, {(1 + r / n)}^{(n t)} = 5.2990343153$ .

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

${(1 + 0.015)}^{112} = 5.2990343153$

$r / n = 0.015, n t = 112, {(1 + r / n)}^{(n t)} = 5.2990343153$ .

$5.2990343153$

$r / n = 0.015, n t = 112, {(1 + r / n)}^{(n t)} = 5.2990343153$ .

3. Hesabu kiasi cha mwisho

$A = 102022.31$

$102022.31$

$19, 253 \cdot 5.2990343153 = 102022.31$ .

$102022.31$

$19, 253 \cdot 5.2990343153 = 102022.31$ .

$102022.31$

$19, 253 \cdot 5.2990343153 = 102022.31$ .

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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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