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Ufumbuzi - Tiger algebra Suluhishaji

13065.35

13065.35

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (12432, 5, 4, 1)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 12, 432$ , $r = 5 %$ , $n = 4$ , $t = 1$ .

$c i (12432, 5, 4, 1)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 12, 432$ , $r = 5 %$ , $n = 4$ , $t = 1$ .

$P = 12432, r = 5 %, n = 4, t = 1$

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 12, 432$ , $r = 5 %$ , $n = 4$ , $t = 1$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 12, 432$ , $r = 5 %$ , $n = 4$ , $t = 1$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 12, 432$ , $r = 5 %$ , $n = 4$ , $t = 1$ .

2. Hesabu kipengele cha ukuaji

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

$n t = 4$

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

$r / n = 0.0125, n t = 4, {(1 + r / n)}^{(n t)} = 1.0509453369$ .

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

${(1 + 0.0125)}^{4} = 1.0509453369$

$r / n = 0.0125, n t = 4, {(1 + r / n)}^{(n t)} = 1.0509453369$ .

$1.0509453369$

$r / n = 0.0125, n t = 4, {(1 + r / n)}^{(n t)} = 1.0509453369$ .

3. Hesabu kiasi cha mwisho

$A = 13065.35$

$13065.35$

$12, 432 \cdot 1.0509453369 = 13065.35$ .

$13065.35$

$12, 432 \cdot 1.0509453369 = 13065.35$ .

$13065.35$

$12, 432 \cdot 1.0509453369 = 13065.35$ .

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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

Vigezo na mada

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