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Ufumbuzi - Tiger algebra Suluhishaji

23737.50

23737.50

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (21505, 10, 4, 1)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 505$ , $r = 10 %$ , $n = 4$ , $t = 1$ .

$c i (21505, 10, 4, 1)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 505$ , $r = 10 %$ , $n = 4$ , $t = 1$ .

$P = 21505, r = 10 %, n = 4, t = 1$

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 505$ , $r = 10 %$ , $n = 4$ , $t = 1$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 505$ , $r = 10 %$ , $n = 4$ , $t = 1$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 21, 505$ , $r = 10 %$ , $n = 4$ , $t = 1$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.025$

$n t = 4$

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

$r / n = 0.025, n t = 4, {(1 + r / n)}^{(n t)} = 1.1038128906$ .

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

${(1 + 0.025)}^{4} = 1.1038128906$

$r / n = 0.025, n t = 4, {(1 + r / n)}^{(n t)} = 1.1038128906$ .

$1.1038128906$

$r / n = 0.025, n t = 4, {(1 + r / n)}^{(n t)} = 1.1038128906$ .

3. Hesabu kiasi cha mwisho

$A = 23737.50$

$23737.50$

$21, 505 \cdot 1.1038128906 = 23737.50$ .

$23737.50$

$21, 505 \cdot 1.1038128906 = 23737.50$ .

$23737.50$

$21, 505 \cdot 1.1038128906 = 23737.50$ .

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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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