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Ufumbuzi - Tiger algebra Suluhishaji

23935.32

23935.32

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (21671, 5, 4, 2)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 671$ , $r = 5 %$ , $n = 4$ , $t = 2$ .

$c i (21671, 5, 4, 2)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 671$ , $r = 5 %$ , $n = 4$ , $t = 2$ .

$P = 21671, r = 5 %, n = 4, t = 2$

$\frac{5}{100} = 0.05$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 671$ , $r = 5 %$ , $n = 4$ , $t = 2$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 671$ , $r = 5 %$ , $n = 4$ , $t = 2$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 671$ , $r = 5 %$ , $n = 4$ , $t = 2$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.0125$

$n t = 8$

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

$r / n = 0.0125, n t = 8, {(1 + r / n)}^{(n t)} = 1.1044861012$ .

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

${(1 + 0.0125)}^{8} = 1.1044861012$

$r / n = 0.0125, n t = 8, {(1 + r / n)}^{(n t)} = 1.1044861012$ .

$1.1044861012$

$r / n = 0.0125, n t = 8, {(1 + r / n)}^{(n t)} = 1.1044861012$ .

3. Hesabu kiasi cha mwisho

$A = 23935.32$

$23935.32$

$21, 671 \cdot 1.1044861012 = 23935.32$ .

$23935.32$

$21, 671 \cdot 1.1044861012 = 23935.32$ .

$23935.32$

$21, 671 \cdot 1.1044861012 = 23935.32$ .

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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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