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Ufumbuzi - Tiger algebra Suluhishaji

20705.37

20705.37

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (14787, 2, 1, 17)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 14, 787$ , $r = 2 %$ , $n = 1$ , $t = 17$ .

$c i (14787, 2, 1, 17)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 14, 787$ , $r = 2 %$ , $n = 1$ , $t = 17$ .

$P = 14787, r = 2 %, n = 1, t = 17$

$\frac{2}{100} = 0.02$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 14, 787$ , $r = 2 %$ , $n = 1$ , $t = 17$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 14, 787$ , $r = 2 %$ , $n = 1$ , $t = 17$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 14, 787$ , $r = 2 %$ , $n = 1$ , $t = 17$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.02$

$n t = 17$

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

$r / n = 0.02, n t = 17, {(1 + r / n)}^{(n t)} = 1.4002414192$ .

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

${(1 + 0.02)}^{17} = 1.4002414192$

$r / n = 0.02, n t = 17, {(1 + r / n)}^{(n t)} = 1.4002414192$ .

$1.4002414192$

$r / n = 0.02, n t = 17, {(1 + r / n)}^{(n t)} = 1.4002414192$ .

3. Hesabu kiasi cha mwisho

$A = 20705.37$

$20705.37$

$14, 787 \cdot 1.4002414192 = 20705.37$ .

$20705.37$

$14, 787 \cdot 1.4002414192 = 20705.37$ .

$20705.37$

$14, 787 \cdot 1.4002414192 = 20705.37$ .

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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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