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Ufumbuzi - Tiger algebra Suluhishaji

24016.92

24016.92

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (3218, 12, 4, 17)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 3, 218$ , $r = 12 %$ , $n = 4$ , $t = 17$ .

$c i (3218, 12, 4, 17)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 3, 218$ , $r = 12 %$ , $n = 4$ , $t = 17$ .

$P = 3218, r = 12 %, n = 4, t = 17$

$\frac{12}{100} = 0.12$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 3, 218$ , $r = 12 %$ , $n = 4$ , $t = 17$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 3, 218$ , $r = 12 %$ , $n = 4$ , $t = 17$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 3, 218$ , $r = 12 %$ , $n = 4$ , $t = 17$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.03$

$n t = 68$

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

$r / n = 0.03, n t = 68, {(1 + r / n)}^{(n t)} = 7.4633065436$ .

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

${(1 + 0.03)}^{68} = 7.4633065436$

$r / n = 0.03, n t = 68, {(1 + r / n)}^{(n t)} = 7.4633065436$ .

$7.4633065436$

$r / n = 0.03, n t = 68, {(1 + r / n)}^{(n t)} = 7.4633065436$ .

3. Hesabu kiasi cha mwisho

$A = 24016.92$

$24016.92$

$3, 218 \cdot 7.4633065436 = 24016.92$ .

$24016.92$

$3, 218 \cdot 7.4633065436 = 24016.92$ .

$24016.92$

$3, 218 \cdot 7.4633065436 = 24016.92$ .

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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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