1. erica recently invested in gold that is growing in value 4% annually. she invested $1000 initially. find…

1. erica recently invested in gold that is growing in value 4% annually. she invested $1000 initially. find the value of her investment after 6 years.\n2. jericho bought a house in 2010 valued at $450,000. the house has appreciated by 5% each year from 2010 to 2019. find the value of jerichos house in 2019. (assume 2010 has an x - value of 0).\n3. jeffrey is buying a used bmw for $17,000. the value of the car depreciates 5% per year from the time he bought the car. the value of jeffreys car, v(t), can be represented by which of the following functions? select all that apply.\nv(t)=17,000(1 +.05)^t yes no\nv(t)=17,000(1 -.05)^t yes no\nv(t)=17,000(0.95)^t yes no\nv(t)=17,000(1.05)^t yes no\n4. write an equation that represents the amount of money in a savings account f(t), after t years, that started with $1000 and grows at a rate of 3%?\noptional reading & exercises: textbook lesson 11.2, check understanding & on your own 25 - 27, 29 - 33; textbook lesson 11.3, check understanding & on your own 1, 11a, 12a, 13a, 14ab, 15ab, 16ab, 17ab, 18ab

1. erica recently invested in gold that is growing in value 4% annually. she invested $1000 initially. find the value of her investment after 6 years.\n2. jericho bought a house in 2010 valued at $450,000. the house has appreciated by 5% each year from 2010 to 2019. find the value of jerichos house in 2019. (assume 2010 has an x - value of 0).\n3. jeffrey is buying a used bmw for $17,000. the value of the car depreciates 5% per year from the time he bought the car. the value of jeffreys car, v(t), can be represented by which of the following functions? select all that apply.\nv(t)=17,000(1 +.05)^t yes no\nv(t)=17,000(1 -.05)^t yes no\nv(t)=17,000(0.95)^t yes no\nv(t)=17,000(1.05)^t yes no\n4. write an equation that represents the amount of money in a savings account f(t), after t years, that started with $1000 and grows at a rate of 3%?\noptional reading & exercises: textbook lesson 11.2, check understanding & on your own 25 - 27, 29 - 33; textbook lesson 11.3, check understanding & on your own 1, 11a, 12a, 13a, 14ab, 15ab, 16ab, 17ab, 18ab

Answer

1.

Explanation:

Step1: Identify the compound - growth formula

The formula for compound growth is $A = P(1 + r)^t$, where $P$ is the principal amount, $r$ is the annual interest rate (as a decimal), and $t$ is the number of years. Here, $P=$4000$, $r = 0.04$, and $t = 6$.

Step2: Substitute the values into the formula

$A=4000\times(1 + 0.04)^6=4000\times(1.04)^6$.

Step3: Calculate $(1.04)^6$

$(1.04)^6=1.04\times1.04\times1.04\times1.04\times1.04\times1.04\approx1.265319$.

Step4: Calculate the final amount

$A = 4000\times1.265319=$5061.28$.

Answer:

$$5061.28$

2.

Explanation:

Step1: Identify the compound - growth formula

The formula for compound growth is $A = P(1 + r)^t$, where $P$ is the principal amount, $r$ is the annual interest rate (as a decimal), and $t$ is the number of years. Here, $P = 450000$, $r=0.05$, and $t = 2019 - 2010=9$.

Step2: Substitute the values into the formula

$A = 450000\times(1 + 0.05)^9=450000\times(1.05)^9$.

Step3: Calculate $(1.05)^9$

$(1.05)^9=1.05\times1.05\times\cdots\times1.05$ (9 times) $\approx1.551328$.

Step4: Calculate the final amount

$A=450000\times1.551328=$700197.6$.

Answer:

$$700197.6$

3.

Explanation:

The formula for depreciation is $V(t)=P(1 - r)^t$, where $P$ is the initial value, $r$ is the rate of depreciation (as a decimal), and $t$ is the number of years. Here, $P = 17000$ and $r=0.05$. So $V(t)=17000\times(1 - 0.05)^t=17000\times(0.95)^t$.

Answer:

$V(t)=17000(1 - 0.05)^t$: Yes $V(t)=17000(0.95)^t$: Yes $V(t)=17000(1 + 0.05)^t$: No $V(t)=17000(1.05)^t$: No

4.

Explanation:

Using the compound - growth formula $f(t)=P(1 + r)^t$, where $P$ is the initial amount, $r$ is the annual growth rate (as a decimal), and $t$ is the number of years. Here, $P = 1000$ and $r=0.03$.

Answer:

$f(t)=1000(1 + 0.03)^t=1000(1.03)^t$