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Ufumbuzi - Tiger algebra Suluhishaji

11289.45

11289.45

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (4691, 10, 2, 9)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 4, 691$ , $r = 10 %$ , $n = 2$ , $t = 9$ .

$c i (4691, 10, 2, 9)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 4, 691$ , $r = 10 %$ , $n = 2$ , $t = 9$ .

$P = 4691, r = 10 %, n = 2, t = 9$

$\frac{10}{100} = 0.1$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 4, 691$ , $r = 10 %$ , $n = 2$ , $t = 9$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 4, 691$ , $r = 10 %$ , $n = 2$ , $t = 9$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 4, 691$ , $r = 10 %$ , $n = 2$ , $t = 9$ .

2. Hesabu kipengele cha ukuaji

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

$n t = 18$

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

$r / n = 0.05, n t = 18, {(1 + r / n)}^{(n t)} = 2.4066192337$ .

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

${(1 + 0.05)}^{18} = 2.4066192337$

$r / n = 0.05, n t = 18, {(1 + r / n)}^{(n t)} = 2.4066192337$ .

$2.4066192337$

$r / n = 0.05, n t = 18, {(1 + r / n)}^{(n t)} = 2.4066192337$ .

3. Hesabu kiasi cha mwisho

$A = 11289.45$

$11289.45$

$4, 691 \cdot 2.4066192337 = 11289.45$ .

$11289.45$

$4, 691 \cdot 2.4066192337 = 11289.45$ .

$11289.45$

$4, 691 \cdot 2.4066192337 = 11289.45$ .

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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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