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Ufumbuzi - Tiger algebra Suluhishaji

137616.11

137616.11

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (11212, 12, 12, 21)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 11, 212$ , $r = 12 %$ , $n = 12$ , $t = 21$ .

$c i (11212, 12, 12, 21)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 11, 212$ , $r = 12 %$ , $n = 12$ , $t = 21$ .

$P = 11212, r = 12 %, n = 12, t = 21$

$\frac{12}{100} = 0.12$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 11, 212$ , $r = 12 %$ , $n = 12$ , $t = 21$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 11, 212$ , $r = 12 %$ , $n = 12$ , $t = 21$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 11, 212$ , $r = 12 %$ , $n = 12$ , $t = 21$ .

2. Hesabu kipengele cha ukuaji

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

$n t = 252$

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

$r / n = 0.01, n t = 252, {(1 + r / n)}^{(n t)} = 12.2740020992$ .

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

${(1 + 0.01)}^{252} = 12.2740020992$

$r / n = 0.01, n t = 252, {(1 + r / n)}^{(n t)} = 12.2740020992$ .

$12.2740020992$

$r / n = 0.01, n t = 252, {(1 + r / n)}^{(n t)} = 12.2740020992$ .

3. Hesabu kiasi cha mwisho

$A = 137616.11$

$137616.11$

$11, 212 \cdot 12.2740020992 = 137616.11$ .

$137616.11$

$11, 212 \cdot 12.2740020992 = 137616.11$ .

$137616.11$

$11, 212 \cdot 12.2740020992 = 137616.11$ .

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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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