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Ufumbuzi - Tiger algebra Suluhishaji

29035.25

29035.25

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (16601, 15, 1, 4)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 16, 601$ , $r = 15 %$ , $n = 1$ , $t = 4$ .

$c i (16601, 15, 1, 4)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 16, 601$ , $r = 15 %$ , $n = 1$ , $t = 4$ .

$P = 16601, r = 15 %, n = 1, t = 4$

$\frac{15}{100} = 0.15$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 16, 601$ , $r = 15 %$ , $n = 1$ , $t = 4$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 16, 601$ , $r = 15 %$ , $n = 1$ , $t = 4$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 16, 601$ , $r = 15 %$ , $n = 1$ , $t = 4$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.15$

$n t = 4$

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

$r / n = 0.15, n t = 4, {(1 + r / n)}^{(n t)} = 1.74900625$ .

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

${(1 + 0.15)}^{4} = 1.74900625$

$r / n = 0.15, n t = 4, {(1 + r / n)}^{(n t)} = 1.74900625$ .

$1.74900625$

$r / n = 0.15, n t = 4, {(1 + r / n)}^{(n t)} = 1.74900625$ .

3. Hesabu kiasi cha mwisho

$A = 29035.25$

$29035.25$

$16, 601 \cdot 1.74900625 = 29035.25$ .

$29035.25$

$16, 601 \cdot 1.74900625 = 29035.25$ .

$29035.25$

$16, 601 \cdot 1.74900625 = 29035.25$ .

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Jifunze zaidi na Tiger

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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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