Tina wants to save money for school. tina invests $1,000 in an account that pays an interest rate of 6.75%. how many years will it take for the account to reach $18,600?

Assuming that the interest is paid as compound interest.

Final value = Intial Value × (Multiplier)ⁿ

Multiplier is 100% + the percentage increase/decrease
The letter 'n' is for the number of years

We have:
Initial Value = 1000
Final Value = 18600
Multiplier = 100% + 6.75% = 1 + 0.0675 = 1.0675

We are looking to find:
The number of years 'n'

Substituting the values we have into the formula

18600 = 1000 × (1.0675)ⁿ
18600 ÷ 1000 = 1.0675ⁿ
18.6 = 1.0675ⁿ ⇒ From here you can use either trial and error method or you can use logarithm law

Using trial and method
Let n = 1;            1.0675¹ = 1.0675              ⇒ Not the result we want
Let n = 5;           1.0675⁵ = 1.39                  ⇒ Not the result we want
Let n = 10;          1.0675¹⁰ = 1.92                 ⇒ Not the result we want
Let n = 25;         1.0675²⁵ = 5.12                ⇒ Still too small
Let n = 55;          1.0675⁵⁵ = 36.33            ⇒ Values are too big
Let n = 40;         1.0675⁴⁰ = 13.64              ⇒ Too small (we narrowed down the value of n to be between 40 and 55)

Let n = 48;         1.0675⁴⁸ = 23.00            ⇒ Too big
Let n = 45;         1.0675⁴⁵ = 19.00             ⇒ Let's try n = 44 to see which value are closer to 18.60

Let n = 44;         1.0675⁴⁴ = 17.70              ⇒ This a little far from 18.6

So we have our answer, n = 45

Conclusion: Tina needs to save for 45 years to get $1000 to $18600 with compound interest 6.75% paid yearly.

The answer will be y=-5+0 because the slope is -5 and the it has to go through y-axis 0