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Ufumbuzi - Tiger algebra Suluhishaji

41458.14

41458.14

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (18664, 3, 1, 27)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 664$ , $r = 3 %$ , $n = 1$ , $t = 27$ .

$c i (18664, 3, 1, 27)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 664$ , $r = 3 %$ , $n = 1$ , $t = 27$ .

$P = 18664, r = 3 %, n = 1, t = 27$

$\frac{3}{100} = 0.03$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 664$ , $r = 3 %$ , $n = 1$ , $t = 27$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 664$ , $r = 3 %$ , $n = 1$ , $t = 27$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 664$ , $r = 3 %$ , $n = 1$ , $t = 27$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.03$

$n t = 27$

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

$r / n = 0.03, n t = 27, {(1 + r / n)}^{(n t)} = 2.2212890056$ .

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

${(1 + 0.03)}^{27} = 2.2212890056$

$r / n = 0.03, n t = 27, {(1 + r / n)}^{(n t)} = 2.2212890056$ .

$2.2212890056$

$r / n = 0.03, n t = 27, {(1 + r / n)}^{(n t)} = 2.2212890056$ .

3. Hesabu kiasi cha mwisho

$A = 41458.14$

$41458.14$

$18, 664 \cdot 2.2212890056 = 41458.14$ .

$41458.14$

$18, 664 \cdot 2.2212890056 = 41458.14$ .

$41458.14$

$18, 664 \cdot 2.2212890056 = 41458.14$ .

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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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