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Dimas [21]
2 years ago
6

Pls help on my maths question

Mathematics
1 answer:
dimulka [17.4K]2 years ago
8 0

Answer:

Somewhat illegible but looks like the equation should be:

S = u t + 1/2 a t^2

100 = 20 + 1/2 * a * 4    where t = 2

a* 4 = 160

a = 40

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Colton is going to invest $59,000 and leave it in an account for 10 years. Assuming
siniylev [52]

Answer:

4.66

Step-by-step explanation:

A=P(1+

n

r

​

)

nt

Compound interest formula

A=94000\hspace{35px}P=59000\hspace{35px}t=10\hspace{35px}n=365

A=94000P=59000t=10n=365

Given values

94000=

94000=

\,\,59000\left(1+\frac{r}{365}\right)^{365(10)}

59000(1+

365

r

​

)

365(10)

Plug in values

94000=

94000=

\,\,59000\left(1+\frac{r}{365}\right)^{3650}

59000(1+

365

r

​

)

3650

Multiply

\frac{94000}{59000}=

59000

94000

​

=

\,\,\frac{59000\left(1+\frac{r}{365}\right)^{3650}}{59000}

59000

59000(1+

365

r

​

)

3650

​

Divide by 59000

1.593220339=

1.593220339=

\,\,\left(1+\frac{r}{365}\right)^{3650}

(1+

365

r

​

)

3650

\left(1.593220339\right)^{1/3650}=

(1.593220339)

1/3650

=

\,\,\left[\left(1+\frac{r}{365}\right)^{3650}\right]^{1/3650}

[(1+

365

r

​

)

3650

]

1/3650

Raise both sides to 1/3650 power

1.000127613=

1.000127613=

\,\,1+\frac{r}{365}

1+

365

r

​

-1\phantom{=}

−1=

\,\,-1

−1

Subtract 1

0.000127613=

0.000127613=

\,\,\frac{r}{365}

365

r

​

365\left(0.0001276\right)=

365(0.0001276)=

\,\,\left(\frac{r}{365}\right)365

(

365

r

​

)365

Multiply by 365

0.046578745=

0.046578745=

\,\,r

r

4.6578745\%=

4.6578745%=

\,\,r

r

Convert to percent (multiply by 100)

r\approx

r≈

\,\,4.66\%

4.66%

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