one year ago, grace put $200 she earned from tutoring into an investment account. today, the accounts value…

one year ago, grace put $200 she earned from tutoring into an investment account. today, the accounts value has increased to $212, and it is expected to continue growing each year. write an exponential equation in the form y = a(b)^x that can model the value of the investment account, y, x years after grace started the account. use whole numbers, decimals, or simplified fractions for the values of a and b. y = to the nearest dollar, what can grace expect the value of the account to be 6 years after starting the account if she makes no other deposits or withdrawals? $

one year ago, grace put $200 she earned from tutoring into an investment account. today, the accounts value has increased to $212, and it is expected to continue growing each year. write an exponential equation in the form y = a(b)^x that can model the value of the investment account, y, x years after grace started the account. use whole numbers, decimals, or simplified fractions for the values of a and b. y = to the nearest dollar, what can grace expect the value of the account to be 6 years after starting the account if she makes no other deposits or withdrawals? $

Answer

Explanation:

Step1: Encontrar el valor de a

El valor inicial del depósito es $200, entonces $a = 200$.

Step2: Encontrar el valor de b

El valor del depósito después de 1 año es $212. Sustituyendo $x = 1$, $y=212$ y $a = 200$ en $y=a(b)^x$, tenemos $212=200(b)^1$. Entonces $b=\frac{212}{200}=1.06$.

Step3: Escribir la ecuación exponencial

La ecuación exponencial es $y = 200(1.06)^x$.

Step4: Calcular el valor después de 6 años

Sustituir $x = 6$ en la ecuación $y = 200(1.06)^x$. $y=200\times(1.06)^6$. $(1.06)^6=1.06\times1.06\times1.06\times1.06\times1.06\times1.06\approx1.418519$. $y = 200\times1.418519\approx284$.

Answer:

$y = 200(1.06)^x$ $284$