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Ufumbuzi - Tiger algebra Suluhishaji

60039.95

60039.95

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (23148, 10, 1, 10)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 23, 148$ , $r = 10 %$ , $n = 1$ , $t = 10$ .

$c i (23148, 10, 1, 10)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 23, 148$ , $r = 10 %$ , $n = 1$ , $t = 10$ .

$P = 23148, r = 10 %, n = 1, t = 10$

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 23, 148$ , $r = 10 %$ , $n = 1$ , $t = 10$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 23, 148$ , $r = 10 %$ , $n = 1$ , $t = 10$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 23, 148$ , $r = 10 %$ , $n = 1$ , $t = 10$ .

2. Hesabu kipengele cha ukuaji

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

$n t = 10$

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

$r / n = 0.1, n t = 10, {(1 + r / n)}^{(n t)} = 2.5937424601$ .

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

${(1 + 0.1)}^{10} = 2.5937424601$

$r / n = 0.1, n t = 10, {(1 + r / n)}^{(n t)} = 2.5937424601$ .

$2.5937424601$

$r / n = 0.1, n t = 10, {(1 + r / n)}^{(n t)} = 2.5937424601$ .

3. Hesabu kiasi cha mwisho

$A = 60039.95$

$60039.95$

$23, 148 \cdot 2.5937424601 = 60039.95$ .

$60039.95$

$23, 148 \cdot 2.5937424601 = 60039.95$ .

$60039.95$

$23, 148 \cdot 2.5937424601 = 60039.95$ .

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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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