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Ufumbuzi - Tiger algebra Suluhishaji

4216.61

4216.61

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (1065, 7, 2, 20)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 1, 065$ , $r = 7 %$ , $n = 2$ , $t = 20$ .

$c i (1065, 7, 2, 20)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 1, 065$ , $r = 7 %$ , $n = 2$ , $t = 20$ .

$P = 1065, r = 7 %, n = 2, t = 20$

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 1, 065$ , $r = 7 %$ , $n = 2$ , $t = 20$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 1, 065$ , $r = 7 %$ , $n = 2$ , $t = 20$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 1, 065$ , $r = 7 %$ , $n = 2$ , $t = 20$ .

2. Hesabu kipengele cha ukuaji

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

$n t = 40$

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

$r / n = 0.035, n t = 40, {(1 + r / n)}^{(n t)} = 3.9592597212$ .

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

${(1 + 0.035)}^{40} = 3.9592597212$

$r / n = 0.035, n t = 40, {(1 + r / n)}^{(n t)} = 3.9592597212$ .

$3.9592597212$

$r / n = 0.035, n t = 40, {(1 + r / n)}^{(n t)} = 3.9592597212$ .

3. Hesabu kiasi cha mwisho

$A = 4216.61$

$4216.61$

$1, 065 \cdot 3.9592597212 = 4216.61$ .

$4216.61$

$1, 065 \cdot 3.9592597212 = 4216.61$ .

$4216.61$

$1, 065 \cdot 3.9592597212 = 4216.61$ .

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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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