Ingiza equation au tatizo

Ingizo la kamera haitambuliki!

Ufumbuzi - Tiger algebra Suluhishaji

18324.10

18324.10

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (18142, 1, 4, 1)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 142$ , $r = 1 %$ , $n = 4$ , $t = 1$ .

$c i (18142, 1, 4, 1)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 142$ , $r = 1 %$ , $n = 4$ , $t = 1$ .

$P = 18142, r = 1 %, n = 4, t = 1$

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 142$ , $r = 1 %$ , $n = 4$ , $t = 1$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 142$ , $r = 1 %$ , $n = 4$ , $t = 1$ .

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

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 18, 142$ , $r = 1 %$ , $n = 4$ , $t = 1$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.0025$

$n t = 4$

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

$r / n = 0.0025, n t = 4, {(1 + r / n)}^{(n t)} = 1.0100375625$ .

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

${(1 + 0.0025)}^{4} = 1.0100375625$

$r / n = 0.0025, n t = 4, {(1 + r / n)}^{(n t)} = 1.0100375625$ .

$1.0100375625$

$r / n = 0.0025, n t = 4, {(1 + r / n)}^{(n t)} = 1.0100375625$ .

3. Hesabu kiasi cha mwisho

$A = 18324.10$

$18324.10$

$18, 142 \cdot 1.0100375625 = 18324.10$ .

$18324.10$

$18, 142 \cdot 1.0100375625 = 18324.10$ .

$18324.10$

$18, 142 \cdot 1.0100375625 = 18324.10$ .

Je, maelezo haya yalikuwa na manufaa?

Ndio Hapana

Tulifanyaje?

Tafadhali tuache maoni yako.

Kwa nini kujifunza hii

Jifunze zaidi na Tiger

Hebu tutatue hili hatua kwa hatua

Vigezo na mada

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

Viungo vinavyohusiana

Maswali ya hivi karibuni yaliyohusiana yalisuluhishwa