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Ufumbuzi - Tiger algebra Suluhishaji

71679.40

71679.40

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (22032, 7, 4, 17)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 22, 032$ , $r = 7 %$ , $n = 4$ , $t = 17$ .

$c i (22032, 7, 4, 17)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 22, 032$ , $r = 7 %$ , $n = 4$ , $t = 17$ .

$P = 22032, r = 7 %, n = 4, t = 17$

$\frac{7}{100} = 0.07$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 22, 032$ , $r = 7 %$ , $n = 4$ , $t = 17$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 22, 032$ , $r = 7 %$ , $n = 4$ , $t = 17$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 22, 032$ , $r = 7 %$ , $n = 4$ , $t = 17$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.0175$

$n t = 68$

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

$r / n = 0.0175, n t = 68, {(1 + r / n)}^{(n t)} = 3.2534221343$ .

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

${(1 + 0.0175)}^{68} = 3.2534221343$

$r / n = 0.0175, n t = 68, {(1 + r / n)}^{(n t)} = 3.2534221343$ .

$3.2534221343$

$r / n = 0.0175, n t = 68, {(1 + r / n)}^{(n t)} = 3.2534221343$ .

3. Hesabu kiasi cha mwisho

$A = 71679.40$

$71679.40$

$22, 032 \cdot 3.2534221343 = 71679.40$ .

$71679.40$

$22, 032 \cdot 3.2534221343 = 71679.40$ .

$71679.40$

$22, 032 \cdot 3.2534221343 = 71679.40$ .

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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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