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Ufumbuzi - Tiger algebra Suluhishaji

21859.06

21859.06

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (20017, 9, 2, 1)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 20, 017$ , $r = 9 %$ , $n = 2$ , $t = 1$ .

$c i (20017, 9, 2, 1)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 20, 017$ , $r = 9 %$ , $n = 2$ , $t = 1$ .

$P = 20017, r = 9 %, n = 2, t = 1$

$\frac{9}{100} = 0.09$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 20, 017$ , $r = 9 %$ , $n = 2$ , $t = 1$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 20, 017$ , $r = 9 %$ , $n = 2$ , $t = 1$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 20, 017$ , $r = 9 %$ , $n = 2$ , $t = 1$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.045$

$n t = 2$

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

$r / n = 0.045, n t = 2, {(1 + r / n)}^{(n t)} = 1.092025$ .

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

${(1 + 0.045)}^{2} = 1.092025$

$r / n = 0.045, n t = 2, {(1 + r / n)}^{(n t)} = 1.092025$ .

$1.092025$

$r / n = 0.045, n t = 2, {(1 + r / n)}^{(n t)} = 1.092025$ .

3. Hesabu kiasi cha mwisho

$A = 21859.06$

$21859.06$

$20, 017 \cdot 1.092025 = 21859.06$ .

$21859.06$

$20, 017 \cdot 1.092025 = 21859.06$ .

$21859.06$

$20, 017 \cdot 1.092025 = 21859.06$ .

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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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