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Ufumbuzi - Tiger algebra Suluhishaji

10121.48

10121.48

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (8295, 1, 1, 20)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 8, 295$ , $r = 1 %$ , $n = 1$ , $t = 20$ .

$c i (8295, 1, 1, 20)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 8, 295$ , $r = 1 %$ , $n = 1$ , $t = 20$ .

$P = 8295, r = 1 %, n = 1, t = 20$

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 8, 295$ , $r = 1 %$ , $n = 1$ , $t = 20$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 8, 295$ , $r = 1 %$ , $n = 1$ , $t = 20$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 8, 295$ , $r = 1 %$ , $n = 1$ , $t = 20$ .

2. Hesabu kipengele cha ukuaji

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

$n t = 20$

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

$r / n = 0.01, n t = 20, {(1 + r / n)}^{(n t)} = 1.2201900399$ .

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

${(1 + 0.01)}^{20} = 1.2201900399$

$r / n = 0.01, n t = 20, {(1 + r / n)}^{(n t)} = 1.2201900399$ .

$1.2201900399$

$r / n = 0.01, n t = 20, {(1 + r / n)}^{(n t)} = 1.2201900399$ .

3. Hesabu kiasi cha mwisho

$A = 10121.48$

$10121.48$

$8, 295 \cdot 1.2201900399 = 10121.48$ .

$10121.48$

$8, 295 \cdot 1.2201900399 = 10121.48$ .

$10121.48$

$8, 295 \cdot 1.2201900399 = 10121.48$ .

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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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