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Ufumbuzi - Tiger algebra Suluhishaji

219733.30

219733.30

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (24306, 14, 4, 16)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 306$ , $r = 14 %$ , $n = 4$ , $t = 16$ .

$c i (24306, 14, 4, 16)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 306$ , $r = 14 %$ , $n = 4$ , $t = 16$ .

$P = 24306, r = 14 %, n = 4, t = 16$

$\frac{14}{100} = 0.14$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 306$ , $r = 14 %$ , $n = 4$ , $t = 16$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 306$ , $r = 14 %$ , $n = 4$ , $t = 16$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 306$ , $r = 14 %$ , $n = 4$ , $t = 16$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.035$

$n t = 64$

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

$r / n = 0.035, n t = 64, {(1 + r / n)}^{(n t)} = 9.0402905096$ .

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

${(1 + 0.035)}^{64} = 9.0402905096$

$r / n = 0.035, n t = 64, {(1 + r / n)}^{(n t)} = 9.0402905096$ .

$9.0402905096$

$r / n = 0.035, n t = 64, {(1 + r / n)}^{(n t)} = 9.0402905096$ .

3. Hesabu kiasi cha mwisho

$A = 219733.30$

$219733.30$

$24, 306 \cdot 9.0402905096 = 219733.30$ .

$219733.30$

$24, 306 \cdot 9.0402905096 = 219733.30$ .

$219733.30$

$24, 306 \cdot 9.0402905096 = 219733.30$ .

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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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