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Ufumbuzi - Tiger algebra Suluhishaji

19368.72

19368.72

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (17568, 10, 2, 1)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 568$ , $r = 10 %$ , $n = 2$ , $t = 1$ .

$c i (17568, 10, 2, 1)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 568$ , $r = 10 %$ , $n = 2$ , $t = 1$ .

$P = 17568, r = 10 %, n = 2, t = 1$

$\frac{10}{100} = 0.1$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 568$ , $r = 10 %$ , $n = 2$ , $t = 1$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 568$ , $r = 10 %$ , $n = 2$ , $t = 1$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 17, 568$ , $r = 10 %$ , $n = 2$ , $t = 1$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.05$

$n t = 2$

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

$r / n = 0.05, n t = 2, {(1 + r / n)}^{(n t)} = 1.1025$ .

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

${(1 + 0.05)}^{2} = 1.1025$

$r / n = 0.05, n t = 2, {(1 + r / n)}^{(n t)} = 1.1025$ .

$1.1025$

$r / n = 0.05, n t = 2, {(1 + r / n)}^{(n t)} = 1.1025$ .

3. Hesabu kiasi cha mwisho

$A = 19368.72$

$19368.72$

$17, 568 \cdot 1.1025 = 19368.72$ .

$19368.72$

$17, 568 \cdot 1.1025 = 19368.72$ .

$19368.72$

$17, 568 \cdot 1.1025 = 19368.72$ .

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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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