Question

question in the attachment​

Answer 1

Answer:

\Large A = P \left(1+\frac{r}{100}\right)^{n} => 7986 = 6000 \left(1+\frac{10}{100}\right)^{n} \)

=> \( \Large \frac{7986}{6000}=\frac{11}{10}^{n} => \frac{1331}{1000}= \left(\frac{11}{10}\right)^{n} => n = 3 years \)

Answer 2

Answer:

Explanation:   s 5s t      

E s