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Ufumbuzi - Tiger algebra Suluhishaji

12010.27

12010.27

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (4279, 7, 2, 15)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 4, 279$ , $r = 7 %$ , $n = 2$ , $t = 15$ .

$c i (4279, 7, 2, 15)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 4, 279$ , $r = 7 %$ , $n = 2$ , $t = 15$ .

$P = 4279, r = 7 %, n = 2, t = 15$

$\frac{7}{100} = 0.07$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 4, 279$ , $r = 7 %$ , $n = 2$ , $t = 15$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 4, 279$ , $r = 7 %$ , $n = 2$ , $t = 15$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 4, 279$ , $r = 7 %$ , $n = 2$ , $t = 15$ .

2. Hesabu kipengele cha ukuaji

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

$n t = 30$

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

$r / n = 0.035, n t = 30, {(1 + r / n)}^{(n t)} = 2.8067937047$ .

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

${(1 + 0.035)}^{30} = 2.8067937047$

$r / n = 0.035, n t = 30, {(1 + r / n)}^{(n t)} = 2.8067937047$ .

$2.8067937047$

$r / n = 0.035, n t = 30, {(1 + r / n)}^{(n t)} = 2.8067937047$ .

3. Hesabu kiasi cha mwisho

$A = 12010.27$

$12010.27$

$4, 279 \cdot 2.8067937047 = 12010.27$ .

$12010.27$

$4, 279 \cdot 2.8067937047 = 12010.27$ .

$12010.27$

$4, 279 \cdot 2.8067937047 = 12010.27$ .

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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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