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Ufumbuzi - Tiger algebra Suluhishaji

140092.75

140092.75

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (12930, 10, 1, 25)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 12, 930$ , $r = 10 %$ , $n = 1$ , $t = 25$ .

$c i (12930, 10, 1, 25)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 12, 930$ , $r = 10 %$ , $n = 1$ , $t = 25$ .

$P = 12930, r = 10 %, n = 1, t = 25$

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 12, 930$ , $r = 10 %$ , $n = 1$ , $t = 25$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 12, 930$ , $r = 10 %$ , $n = 1$ , $t = 25$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 12, 930$ , $r = 10 %$ , $n = 1$ , $t = 25$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.1$

$n t = 25$

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

$r / n = 0.1, n t = 25, {(1 + r / n)}^{(n t)} = 10.8347059434$ .

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

${(1 + 0.1)}^{25} = 10.8347059434$

$r / n = 0.1, n t = 25, {(1 + r / n)}^{(n t)} = 10.8347059434$ .

$10.8347059434$

$r / n = 0.1, n t = 25, {(1 + r / n)}^{(n t)} = 10.8347059434$ .

3. Hesabu kiasi cha mwisho

$A = 140092.75$

$140092.75$

$12, 930 \cdot 10.8347059434 = 140092.75$ .

$140092.75$

$12, 930 \cdot 10.8347059434 = 140092.75$ .

$140092.75$

$12, 930 \cdot 10.8347059434 = 140092.75$ .

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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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