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Ufumbuzi - Tiger algebra Suluhishaji

25016.78

25016.78

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (14369, 4, 2, 14)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 14, 369$ , $r = 4 %$ , $n = 2$ , $t = 14$ .

$c i (14369, 4, 2, 14)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 14, 369$ , $r = 4 %$ , $n = 2$ , $t = 14$ .

$P = 14369, r = 4 %, n = 2, t = 14$

$\frac{4}{100} = 0.04$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 14, 369$ , $r = 4 %$ , $n = 2$ , $t = 14$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 14, 369$ , $r = 4 %$ , $n = 2$ , $t = 14$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 14, 369$ , $r = 4 %$ , $n = 2$ , $t = 14$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.02$

$n t = 28$

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

$r / n = 0.02, n t = 28, {(1 + r / n)}^{(n t)} = 1.7410242062$ .

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

${(1 + 0.02)}^{28} = 1.7410242062$

$r / n = 0.02, n t = 28, {(1 + r / n)}^{(n t)} = 1.7410242062$ .

$1.7410242062$

$r / n = 0.02, n t = 28, {(1 + r / n)}^{(n t)} = 1.7410242062$ .

3. Hesabu kiasi cha mwisho

$A = 25016.78$

$25016.78$

$14, 369 \cdot 1.7410242062 = 25016.78$ .

$25016.78$

$14, 369 \cdot 1.7410242062 = 25016.78$ .

$25016.78$

$14, 369 \cdot 1.7410242062 = 25016.78$ .

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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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