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Ufumbuzi - Tiger algebra Suluhishaji

29013.09

29013.09

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (24498, 1, 1, 17)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 498$ , $r = 1 %$ , $n = 1$ , $t = 17$ .

$c i (24498, 1, 1, 17)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 498$ , $r = 1 %$ , $n = 1$ , $t = 17$ .

$P = 24498, r = 1 %, n = 1, t = 17$

$\frac{1}{100} = 0.01$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 498$ , $r = 1 %$ , $n = 1$ , $t = 17$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 498$ , $r = 1 %$ , $n = 1$ , $t = 17$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 24, 498$ , $r = 1 %$ , $n = 1$ , $t = 17$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.01$

$n t = 17$

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

$r / n = 0.01, n t = 17, {(1 + r / n)}^{(n t)} = 1.1843044314$ .

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

${(1 + 0.01)}^{17} = 1.1843044314$

$r / n = 0.01, n t = 17, {(1 + r / n)}^{(n t)} = 1.1843044314$ .

$1.1843044314$

$r / n = 0.01, n t = 17, {(1 + r / n)}^{(n t)} = 1.1843044314$ .

3. Hesabu kiasi cha mwisho

$A = 29013.09$

$29013.09$

$24, 498 \cdot 1.1843044314 = 29013.09$ .

$29013.09$

$24, 498 \cdot 1.1843044314 = 29013.09$ .

$29013.09$

$24, 498 \cdot 1.1843044314 = 29013.09$ .

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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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