The question asks for the equation modeling pages p printed after t minutes, with 18/min and 6 already at t=0. The model is linear because pages add constantly at 18 per minute after the initial, emphasizing additive growth. The equation is p=18t +6, where 6 is the intercept and 18 the slope. At t=1, p=24, fitting the pattern. A common error is using exponential like $6(18)^t$, misapplying multiplicative change. In production models, use linear for constant rate additive increases from a starting point.

The question asks for the annual percent increase in water usage from $W(t)=12000(1.03)^t$. This is exponential growth with multiplicative factor 1.03 per year. The 1.03 means 3% increase annually. Emphasize multiplicative over additive. A common error is misreading as 0.03% or adding 103%. Identify the base minus 1 for percent in exponential models.

This question asks us to evaluate an exponential function at t = 3. Given A = $800(1.05)^t$, we substitute t = 3 to get A = $800(1.05)^3$. First calculate $(1.05)^3$ = 1.05 × 1.05 × 1.05 = 1.157625. Then multiply: 800 × 1.157625 = 926.10. Common errors include calculating 800 + 3(0.05×800) = 920 (treating it as simple interest instead of compound) or just adding 5% three times. The key is recognizing that exponential growth compounds, meaning each year's growth is calculated on the new total, not the original amount.

When you encounter problems comparing linear and exponential functions, you're looking for intersection points where the two functions have equal values. This requires setting the equations equal and solving for the variable.

To find when the balances are equal, set $$F(t) = G(t)$$: $$2{,}000 + 150t = 2{,}000(1.04)^t$$

This equation can't be solved algebraically, so you need to test values or use a graphing approach. Since Plan F grows by a constant $150 per year while Plan G grows by 4% annually, the exponential plan will eventually overtake the linear one.

Testing strategic values:

• At $$t = 25$$: $$F(25) = 5{,}750$$ and $$G(25) ≈ 5{,}332$$
• At $$t = 30$$: $$F(30) = 6{,}500$$ and $$G(30) ≈ 6{,}493$$
• At $$t = 31$$: $$F(31) = 6{,}650$$ and $$G(31) ≈ 6{,}752$$

The crossover occurs between years 30 and 31, making the answer approximately 31 years.

Choice A (25 years) is too early—the linear function still exceeds the exponential at this point. Choice B (28 years) is also premature; you haven't reached the intersection yet. Choice D (never equal) misunderstands exponential growth—while the exponential function starts slower, it eventually surpasses any linear function.

Study tip: For linear vs. exponential comparisons, remember that exponential functions always eventually overtake linear ones, but it may take longer than you initially expect. Always test values around your estimate to pinpoint the intersection.

The question asks which model represents the bank balance B after m months with 5% compounded monthly. This is exponential with monthly multiplicative factor $1 + 0.05/12 ≈ 1.0041667$. Thus $B=1000(1.0041667)^m$. Emphasize compounding multiplication over linear addition. A key error is using annual rate without adjusting. Divide annual rate by 12 for monthly exponential.

The question models phone battery percent B losing 12 points per hour after 100% at noon, and predicts at t=5. This is linear decay with constant subtraction of 12 per hour, emphasizing additive change over multiplicative. The equation is B=100-12t; at t=5, B=100-60=40%. Verify: at t=1, 88%; t=2, 76%, linear. Errors include exponential decay (A) or positive slope (C). Distinguish: constant differences mean linear, ratios mean exponential.

This question interprets the parameter 1.08 in the rabbit population model $R=300(1.08)^t$, where t is years. The model is exponential growth because population multiplies by 1.08 annually, showing multiplicative change. The 1.08 means 8% growth per year (1 + 0.08). Initial is 300, not 1.08 (A); it's not additive (B) or decay (D). Error: confusing growth factor with additive rate. In exponentials, factors over 1 mean growth, under 1 decay, unlike linear addition.

The question asks for b in the exponential model y=a $b^x$ passing through (0,5) and (1,15). The model is exponential because it uses a growth factor b for multiplicative change, unlike linear's additive slope. At x=0, y=5=a. At x=1, 15=a b=5b, so b=3. This means y triples each unit x multiplicatively. A key error is setting b=1/3 for decay. For points on exponentials, solve for a at x=0, then b using the ratio of y-values.

The question requires finding the exponential equation that fits a curve starting at 300 when x=0 and decaying to 75 when x=2. This is an exponential decay model, identified by constant multiplicative ratios, unlike linear models with additive differences. The general form is y = 300 × $r^x$; using x=2, 300 × $r^2$ = 75 gives $r^2$ = 0.25, so r = 0.5, yielding y = $300(0.5)^x$. A key error is confusing it with linear, like choice D, or miscalculating the base as 0.25 instead of 0.5. Another mistake might be using x/2 in the exponent, which doesn't fit the points. Emphasize checking ratios for exponential fits, and verify by plugging in given points.

The question asks for the predicted number of tickets sold after 6 hours using the linear model T(h) = 95h + 240. This is a linear model, characterized by constant additive increases of 95 tickets per hour, in contrast to exponential models where changes are multiplicative. Calculate by substituting h = 6: 95 × 6 = 570, plus 240 equals 810 tickets. Common errors include omitting the +240 to get 570 or adding incorrectly to reach 900 or 1110. Remember, linear growth adds a fixed amount each time, not a percentage. For test-taking, double-check substitutions in linear equations to ensure all terms are accounted for.