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Ufumbuzi - Tiger algebra Suluhishaji

11906.73

11906.73

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (11333, 5, 2, 1)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 11, 333$ , $r = 5 %$ , $n = 2$ , $t = 1$ .

$c i (11333, 5, 2, 1)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 11, 333$ , $r = 5 %$ , $n = 2$ , $t = 1$ .

$P = 11333, r = 5 %, n = 2, t = 1$

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 11, 333$ , $r = 5 %$ , $n = 2$ , $t = 1$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 11, 333$ , $r = 5 %$ , $n = 2$ , $t = 1$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 11, 333$ , $r = 5 %$ , $n = 2$ , $t = 1$ .

2. Hesabu kipengele cha ukuaji

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

$n t = 2$

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

$r / n = 0.025, n t = 2, {(1 + r / n)}^{(n t)} = 1.050625$ .

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

${(1 + 0.025)}^{2} = 1.050625$

$r / n = 0.025, n t = 2, {(1 + r / n)}^{(n t)} = 1.050625$ .

$1.050625$

$r / n = 0.025, n t = 2, {(1 + r / n)}^{(n t)} = 1.050625$ .

3. Hesabu kiasi cha mwisho

$A = 11906.73$

$11906.73$

$11, 333 \cdot 1.050625 = 11906.73$ .

$11906.73$

$11, 333 \cdot 1.050625 = 11906.73$ .

$11906.73$

$11, 333 \cdot 1.050625 = 11906.73$ .

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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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