The question asks for the best description of the change in bacteria population each hour and the value of P(3) rounded to the nearest whole number. This is an exponential growth function, distinguished by its multiplicative rate rather than additive changes seen in linear functions. The function P(t) = $500(1.12)^t$ shows multiplication by 1.12 each hour, representing a 12% increase. To find P(3), calculate $1.12^3$ = 1.12 * 1.12 = 1.2544, then 1.2544 * 1.12 = 1.404928, and 500 * 1.404928 ≈ 702.464, rounding to 702. A common error is confusing multiplication by 1.12 with adding 12% or a fixed amount, leading to incorrect values like in choices A or D. As a test-taking strategy, identify the base in exponential functions to confirm growth or decay and compute powers step-by-step for accuracy.

The question asks for the domain of the function f(x)=√(2x-5) in real numbers. This is a square root (radical) function, distinguished by its requirement that the radicand be non-negative, unlike unrestricted domains in polynomials or exponentials. To find the domain, solve the inequality 2x-5 ≥ 0, which gives 2x ≥ 5, so x ≥ 5/2. Thus, the domain is all real numbers from 5/2 to infinity, including 5/2 where f(5/2)=0. A common error is mistakenly allowing x < 5/2, leading to imaginary values, or excluding the boundary point. Another error confuses it with ranges of other functions like rationals. As a test-taking strategy, for radical functions, always set the inside expression ≥ 0 and solve for x to determine the domain.

The question asks for the value not in the domain of h(x)=(x+3)/(x-2) and the value of h(5). This is a rational function, characterized by restrictions where the denominator is zero, unlike polynomials or exponentials with full domains. The domain excludes x=2, as it makes the denominator zero, causing undefined behavior. Now compute h(5)=(5+3)/(5-2)=8/3. Common errors include mistaking x=-3 as excluded (it makes numerator zero but is defined as 0) or miscalculating the fraction. Another error is confusing domain with range in rational functions. As a test-taking strategy, set the denominator to zero to find exclusions and plug in values directly for evaluation.

The question asks which equation could be k(x) with a vertical asymptote at x=1, horizontal at y=0, and passing through (2,3). This describes a rational function like a hyperbola, identified by asymptotes from denominator factors, unlike linear or exponential without vertical lines. For y=3/(x-1), vertical asymptote at x=1, horizontal at y=0; at (2,3): 3/(2-1)=3, it fits. Choices like y=3/(x+1) have asymptote at x=-1, y=3(x-1) is linear without asymptotes, and $y=3^x$ approaches y=0 but no vertical. A key error is ignoring the asymptote positions or not substituting the point. As a test-taking strategy, verify asymptotes first, then plug in points to confirm rational functions.

The question asks for the value(s) of x where m(x)=|2x-7| equals 5, choosing the option listing all solutions. This is an absolute value function, forming a V-shape with linear pieces, unlike smooth curves in quadratics or radicals. Solve |2x-7|=5 by considering cases: 2x-7=5 gives 2x=12, x=6; 2x-7=-5 gives 2x=2, x=1. Both x=1 and x=6 are solutions, as m(1)=|2-7|=5 and m(6)=|12-7|=5. A common error is solving only one case or including extraneous solutions like x=-1. As a test-taking strategy, always solve absolute value equations by splitting into positive and negative branches and verify both.

The question asks which equation best matches a graph starting at (1,0), increasing with downward curve through (2,1) and (5,2), shown for x≥1. This describes a square root function, identified by its concave-down shape and restricted domain, unlike absolute value's linear pieces or quadratic's symmetry. For y=√(x-1), at x=1: √0=0; x=2: √1=1; x=5: √4=2, matching points and domain. Choices like y=|x-1| give y=4 at x=5, not 2; $y=(x-1)^2$ gives y=16 at x=5; y=1/(x-1) decreases after x=2. A common error is selecting without checking all points or ignoring the curve's concavity. As a test-taking strategy, verify domain, points, and curvature to distinguish radical functions from others.

The question requires an equation for a rational function with vertical asymptote x=2, horizontal y=1, passing through (3,2). Rational functions have asymptotes based on degrees and factors. The form y = 1 + 1/(x - 2) has vertical at x=2 (denom zero), horizontal at 1 (constant term), and at x=3: 1 + 1/1 = 2. It fits perfectly. Errors include no constant like 1/(x-2) (horizontal 0) or wrong asymptote like 1 + 2/(x+2) (vertical x=-2). Linear-like (x-2)/1 has no asymptotes. Test points and asymptotes in choices to match.

The question asks for the interpretation of 0.86 in V(t) = 18000 * $(0.86)^t$ for car value. This is exponential decay, as base 0.86 < 1, unlike growth >1. The factor 0.86 means value multiplies by 0.86 yearly, a 14% drop (1 - 0.86 = 0.14). It implies decreasing value over time. Errors include thinking 86% drop (confusing with 0.14) or growth if misreading base. Linear would subtract constant, not multiply. In exponential models, interpret base as retention rate for decay.

The question asks for the type of function modeling the bacterial culture and its implication for growth over time. This is an exponential function of the form P(t) = 250 * $(1.12)^t$. To identify it, recognize the base raised to the power of t, where the base 1.12 indicates growth since it is greater than 1. The factor 1.12 means the population multiplies by 1.12 each hour, implying a 12% growth rate per hour, so the culture size increases exponentially. A common error is confusing it with linear growth, but linear would add a constant, not multiply. Another mistake is thinking it decays if overlooking that 1.12 > 1. When encountering growth models, check if the change is additive (linear) or multiplicative (exponential) to distinguish types.

The question requires simplifying q(x) = (x² - 9)/(x - 3) for x ≠ 3. This rational simplifies by factoring numerator (x-3)(x+3), canceling (x-3), yielding x+3. It's linear for x ≠ 3, but original has hole at x=3. Direct division also gives x+3. Errors include not canceling, leaving as (x-3)/(x+3), or wrong factoring. Denominator as x+3 is incorrect. For rationals, factor and cancel common terms, noting restrictions.