Question

1. marios $15,000 car depreciates in value at a rate of 19% per year. the value, v, after t years can be modeled by the function $v = 15,000(0.81)^t$.

which function is equivalent to the original function?

1. $v = 15,000(0.9)^{2t}$
2. $v = 15,000(0.9)^{2t}$
3. $v = 15,000(0.9)^{\frac{t}{5}}$
4. $v = 15,000(0.9)^{\frac{t}{2}}$

08 2017 16

1. if a population of 100 cells triples every hour, a function represents $p(t)$, the population

08 2017 14

1. anne invested $1000 in an account with a 1.3% annual interest rate. she made no deposits or withdrawals on the account for 2 years. if interest was compounded annually, which equation represents the balance in the account after the 2 years?
2. $a = 1000(1 - 0.013)^2$
3. $a = 1000(1 + 0.013)^2$
4. $a = 1000(1 - 1.3)^2$
5. $a = 1000(1 + 1.3)^2$

06 2017 12

Explanation:

Problem 6

Step1: Relate 0.81 to 0.9

Note that $0.81 = 0.9^2$, so substitute into the original function.
$V = 15,000(0.9^2)^t$

Step2: Simplify exponent using power rule

Use $(a^m)^n = a^{m \cdot n}$ to simplify the exponent.
$V = 15,000(0.9)^{2t}$

Step1: Identify compound interest formula

For annual compounding, the formula is $A = P(1 + r)^t$, where $P$ is principal, $r$ is annual interest rate, $t$ is time.

Step2: Substitute given values

$P=1000$, $r=0.013$, $t=2$.
$A = 1000(1 + 0.013)^2$

Step1: Identify exponential growth formula

Tripling population uses $P(t) = P_0(3)^t$, where $P_0=100$.

Step2: Substitute initial population

$P(t) = 100(3)^t$

Answer:

1. $V=15,000(0.9)^{2t}$

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Problem 9