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Ufumbuzi - Tiger algebra Suluhishaji

50685.69

50685.69

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (22114, 5, 1, 17)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 22, 114$ , $r = 5 %$ , $n = 1$ , $t = 17$ .

$c i (22114, 5, 1, 17)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 22, 114$ , $r = 5 %$ , $n = 1$ , $t = 17$ .

$P = 22114, r = 5 %, n = 1, t = 17$

$\frac{5}{100} = 0.05$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 22, 114$ , $r = 5 %$ , $n = 1$ , $t = 17$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 22, 114$ , $r = 5 %$ , $n = 1$ , $t = 17$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 22, 114$ , $r = 5 %$ , $n = 1$ , $t = 17$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.05$

$n t = 17$

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

$r / n = 0.05, n t = 17, {(1 + r / n)}^{(n t)} = 2.2920183178$ .

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

${(1 + 0.05)}^{17} = 2.2920183178$

$r / n = 0.05, n t = 17, {(1 + r / n)}^{(n t)} = 2.2920183178$ .

$2.2920183178$

$r / n = 0.05, n t = 17, {(1 + r / n)}^{(n t)} = 2.2920183178$ .

3. Hesabu kiasi cha mwisho

$A = 50685.69$

$50685.69$

$22, 114 \cdot 2.2920183178 = 50685.69$ .

$50685.69$

$22, 114 \cdot 2.2920183178 = 50685.69$ .

$50685.69$

$22, 114 \cdot 2.2920183178 = 50685.69$ .

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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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