Question

1. the value of a certain collectible card in dollars from january 2020 to may 2021 can be modeled by the function v(m) = 250m² - 3,500m + 18,000, where m is the approximate number of months since the start of 2020. over what period was the value of the card declining? \boxed{} < m < \boxed{} 18. which describes the end behavior of the absolute value function? f(x) = \frac{1}{2}|x - 6| + 2 a as x \to \infty, f(x) \to -\infty and as x \to -\infty, f(x) \to \infty. b as x \to \infty, f(x) \to \infty and as x \to -\infty, f(x) \to \infty. c as x \to \infty, f(x) \to \infty and as x \to -\infty, f(x) \to 6. d as x \to \infty, f(x) \to \infty and as x \to -\infty, f(x) \to 2. 19. the population of a town is modeled by the function f(x) = 465(1 + 0.003)^x, where x is the time in years. what type of change does this function represent? a linear growth b linear decay c exponential growth d exponential decay 20. what are the domain and the range of the function? f(x) = (x + 4)² + 3 a domain: x \geq 3 range: all real numbers b domain: all real numbers range: y \geq 3 c domain: x \geq -4 range: all real numbers d domain: all real numbers range: y \geq -4 21. vanessa purchased a used car on a payment plan. four months after purchasing the car, the balance was $1,200. seven months after purchasing the car, the balance was $975. write an equation that models the balance y after t months. y = \boxed{}t + \boxed{} 22. the equation y = 10x + 15 models the amount of money y, in dollars, that jack earns shoveling snow for x hours. how many hours does jack need to work to earn $55? \boxed{}

Explanation:

Question 17

Step1: Find vertex of parabola

For $V(m)=250m^2-3500m+18000$, vertex $m=-\frac{b}{2a}=-\frac{-3500}{2\times250}=7$

Step2: Identify decline interval

Parabola opens up ($a>0$), so declines before vertex. $m\geq0$, so $0

Step1: Analyze absolute value end behavior

For $f(x)=\frac{1}{2}|x-6|+2$, absolute value $|x-6|\to\infty$ as $x\to\infty$ or $x\to-\infty$. Thus $f(x)\to\infty$ in both directions.

Step1: Classify function type

The function $f(x)=465(1+0.003)^x$ follows the exponential growth form $f(x)=a(1+r)^x$ where $r=0.003>0$.

Answer:

$0 < m < 7$

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Question 18