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Ufumbuzi - Tiger algebra Suluhishaji

22060.99

22060.99

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (20169, 9, 12, 1)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 20, 169$ , $r = 9 %$ , $n = 12$ , $t = 1$ .

$c i (20169, 9, 12, 1)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 20, 169$ , $r = 9 %$ , $n = 12$ , $t = 1$ .

$P = 20169, r = 9 %, n = 12, t = 1$

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 20, 169$ , $r = 9 %$ , $n = 12$ , $t = 1$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 20, 169$ , $r = 9 %$ , $n = 12$ , $t = 1$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 20, 169$ , $r = 9 %$ , $n = 12$ , $t = 1$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.0075$

$n t = 12$

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

$r / n = 0.0075, n t = 12, {(1 + r / n)}^{(n t)} = 1.0938068977$ .

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

${(1 + 0.0075)}^{12} = 1.0938068977$

$r / n = 0.0075, n t = 12, {(1 + r / n)}^{(n t)} = 1.0938068977$ .

$1.0938068977$

$r / n = 0.0075, n t = 12, {(1 + r / n)}^{(n t)} = 1.0938068977$ .

3. Hesabu kiasi cha mwisho

$A = 22060.99$

$22060.99$

$20, 169 \cdot 1.0938068977 = 22060.99$ .

$22060.99$

$20, 169 \cdot 1.0938068977 = 22060.99$ .

$22060.99$

$20, 169 \cdot 1.0938068977 = 22060.99$ .

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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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