henry begins a savings account with $500. the savings account accumulates 2.5% annual interest based on the…

henry begins a savings account with $500. the savings account accumulates 2.5% annual interest based on the equation $v_{h}=500(1.025)^{t}$, where $v_{h}$ is the value of the account after $t$ years. andres begins a savings account three years later with the same initial amount of money at the same interest rate. which equations represent the value of andress account after $t$ years?\n$v_{a}=500(1.025)^{t - 3}$ and $v_{a}=484.30(1.025)^{t}$\n$v_{a}=500(1.025)^{t+3}$ and $v_{a}=538.45(1.025)^{t}$\n$v_{a}=500(1.025)^{t - 3}$ and $v_{a}=538.45(1.025)^{t}$\n$v_{a}=500(1.025)^{t+3}$ and $v_{a}=484.30(1.025)^{t}$

henry begins a savings account with $500. the savings account accumulates 2.5% annual interest based on the equation $v_{h}=500(1.025)^{t}$, where $v_{h}$ is the value of the account after $t$ years. andres begins a savings account three years later with the same initial amount of money at the same interest rate. which equations represent the value of andress account after $t$ years?\n$v_{a}=500(1.025)^{t - 3}$ and $v_{a}=484.30(1.025)^{t}$\n$v_{a}=500(1.025)^{t+3}$ and $v_{a}=538.45(1.025)^{t}$\n$v_{a}=500(1.025)^{t - 3}$ and $v_{a}=538.45(1.025)^{t}$\n$v_{a}=500(1.025)^{t+3}$ and $v_{a}=484.30(1.025)^{t}$

Answer

Explanation:

Step1: Analyze Andres' time - delay

Andres starts 3 years later. So when calculating the value of his account after $t$ years, the number of years of interest - accumulation is $t - 3$ compared to Henry's account. The formula for compound - interest is $V=P(1 + r)^n$, where $P$ is the principal amount, $r$ is the interest rate per period, and $n$ is the number of periods. Here, $P = 500$, $r=0.025$, and the number of years of interest - accumulation for Andres is $t - 3$. So the first formula for the value of Andres' account is $V_A=500(1.025)^{t - 3}$.

Step2: Simplify the formula

We know that $500(1.025)^{t - 3}=500\times(1.025)^{- 3}\times(1.025)^{t}$. Calculate $500\times(1.025)^{-3}=500\div(1.025)^{3}=500\div1.076890625\approx464.30$. So another form of the formula is $V_A = 464.30(1.025)^{t}$.

Answer:

$V_A = 500(1.025)^{t - 3}$ and $V_A=464.30(1.025)^{t}$