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Ufumbuzi - Tiger algebra Suluhishaji

75320.09

75320.09

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (20956, 13, 4, 10)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 20, 956$ , $r = 13 %$ , $n = 4$ , $t = 10$ .

$c i (20956, 13, 4, 10)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 20, 956$ , $r = 13 %$ , $n = 4$ , $t = 10$ .

$P = 20956, r = 13 %, n = 4, t = 10$

$\frac{13}{100} = 0.13$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 20, 956$ , $r = 13 %$ , $n = 4$ , $t = 10$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 20, 956$ , $r = 13 %$ , $n = 4$ , $t = 10$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 20, 956$ , $r = 13 %$ , $n = 4$ , $t = 10$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.0325$

$n t = 40$

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

$r / n = 0.0325, n t = 40, {(1 + r / n)}^{(n t)} = 3.5942014341$ .

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

${(1 + 0.0325)}^{40} = 3.5942014341$

$r / n = 0.0325, n t = 40, {(1 + r / n)}^{(n t)} = 3.5942014341$ .

$3.5942014341$

$r / n = 0.0325, n t = 40, {(1 + r / n)}^{(n t)} = 3.5942014341$ .

3. Hesabu kiasi cha mwisho

$A = 75320.09$

$75320.09$

$20, 956 \cdot 3.5942014341 = 75320.09$ .

$75320.09$

$20, 956 \cdot 3.5942014341 = 75320.09$ .

$75320.09$

$20, 956 \cdot 3.5942014341 = 75320.09$ .

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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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