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Ufumbuzi - Tiger algebra Suluhishaji

46820.10

46820.10

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (18320, 5, 2, 19)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 320$ , $r = 5 %$ , $n = 2$ , $t = 19$ .

$c i (18320, 5, 2, 19)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 320$ , $r = 5 %$ , $n = 2$ , $t = 19$ .

$P = 18320, r = 5 %, n = 2, t = 19$

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 320$ , $r = 5 %$ , $n = 2$ , $t = 19$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 320$ , $r = 5 %$ , $n = 2$ , $t = 19$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 320$ , $r = 5 %$ , $n = 2$ , $t = 19$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.025$

$n t = 38$

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

$r / n = 0.025, n t = 38, {(1 + r / n)}^{(n t)} = 2.5556824161$ .

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

${(1 + 0.025)}^{38} = 2.5556824161$

$r / n = 0.025, n t = 38, {(1 + r / n)}^{(n t)} = 2.5556824161$ .

$2.5556824161$

$r / n = 0.025, n t = 38, {(1 + r / n)}^{(n t)} = 2.5556824161$ .

3. Hesabu kiasi cha mwisho

$A = 46820.10$

$46820.10$

$18, 320 \cdot 2.5556824161 = 46820.10$ .

$46820.10$

$18, 320 \cdot 2.5556824161 = 46820.10$ .

$46820.10$

$18, 320 \cdot 2.5556824161 = 46820.10$ .

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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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