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Ufumbuzi - Tiger algebra Suluhishaji

10479.62

10479.62

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (9971, 1, 1, 5)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 9, 971$ , $r = 1 %$ , $n = 1$ , $t = 5$ .

$c i (9971, 1, 1, 5)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 9, 971$ , $r = 1 %$ , $n = 1$ , $t = 5$ .

$P = 9971, r = 1 %, n = 1, t = 5$

$\frac{1}{100} = 0.01$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 9, 971$ , $r = 1 %$ , $n = 1$ , $t = 5$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 9, 971$ , $r = 1 %$ , $n = 1$ , $t = 5$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 9, 971$ , $r = 1 %$ , $n = 1$ , $t = 5$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.01$

$n t = 5$

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

$r / n = 0.01, n t = 5, {(1 + r / n)}^{(n t)} = 1.0510100501$ .

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

${(1 + 0.01)}^{5} = 1.0510100501$

$r / n = 0.01, n t = 5, {(1 + r / n)}^{(n t)} = 1.0510100501$ .

$1.0510100501$

$r / n = 0.01, n t = 5, {(1 + r / n)}^{(n t)} = 1.0510100501$ .

3. Hesabu kiasi cha mwisho

$A = 10479.62$

$10479.62$

$9, 971 \cdot 1.0510100501 = 10479.62$ .

$10479.62$

$9, 971 \cdot 1.0510100501 = 10479.62$ .

$10479.62$

$9, 971 \cdot 1.0510100501 = 10479.62$ .

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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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