Free math worksheets

- At the same horizontal coordinate, distinguish the open point from the filled point. - Read the vertical coordinate of the point that is included in the function.

1. At \(t=2\), the open point is at \(72\), but an open point is not included in the function. 2. The filled point is at \(30\), so it gives the actual function value at the breakpoint.

\(30\)

- At \(t=4\), the second branch should still equal \(200\). - A \(10\%\) decrease leaves \(90\%\) of the previous value. - The exponent should count time elapsed since day \(4\).

At the moment the second branch begins, \(t=4\), the decay factor must have exponent \(0\) so the branch starts at \(200\). Only \(200(0.9)^{t-4}\) has that property. Also, a \(10\%\) decrease means a remaining factor of \(0.9\).

b) \(200(0.9)^{t-4}\)

- Compare each input time with the breakpoint \(t=4\). - After choosing the correct row, substitute the time only into that rule. - At the breakpoint, what is the value of the exponent \(t-4\)?

1. Since \(2<4\), use the first rule for \(Q(2)\). Since \(6\ge4\), use the second rule for \(Q(6)\). 2. \(Q(2)=500(1.2)^2=720\). 3. \(Q(6)=500(1.2)^4(0.9)^2=839.808\). 4. At \(t=4\), the second rule gives \(Q(4)=500(1.2)^4(0.9)^0=1036.8\). The exponent \(t-4\) is \(0\), so the new factor is \(1\) exactly at the breakpoint.

a) Use the first rule for \(Q(2)\) and the second rule for \(Q(6)\). b) \(Q(2)=720\); \(Q(6)=839.808\) c) \(Q(4)=1036.8\); at \(t=4\), \((0.9)^{t-4}=1\).

- Compare the two branch values at the breakpoint. - A growth factor greater than \(1\) indicates increase. - Convert each growth factor to a percent growth rate before comparing them.

a) At \(t=3\), the first branch gives \(100(1.10)^3\). The second branch approaches \(100(1.10)^3(1.04)^0=100(1.10)^3\), so the model is continuous. b) Both growth factors, \(1.10\) and \(1.04\), are greater than \(1\), so the quantity increases on both intervals. c) The first interval grows by \(10\%\) per time unit, while the second grows by \(4\%\). Therefore, the first interval has the larger percent growth.

a) Yes. b) Increasing on both intervals. c) The first interval; it grows by \(10\%\) per time unit instead of \(4\%\).

- Look for the time at which the graph changes from growth to decay. - Compare the marked values before the breakpoint to identify a doubling interval. - Compare the marked values after the breakpoint to identify a halving interval. - Continuity at the breakpoint requires the two branch values to agree there.

1. The graph shows \(P(0)=100\), a breakpoint at \(t=4\), and \(P(4)=200\). 2. From \(t=0\) to \(t=4\), the value doubles from \(100\) to \(200\), so the first branch is \(P(t)=100(2)^{t/4}\) for \(0\le t<4\). 3. After the breakpoint, the marked values fall from \(200\) at \(t=4\) to \(100\) at \(t=6\) and \(50\) at \(t=8\). The half-life is \(2\) time units, so the second branch is \(P(t)=200(0.5)^{(t-4)/2}\) for \(t\ge4\). 4. At \(t=4\), the first formula gives \(100(2)^1=200\), and the second gives \(200(0.5)^0=200\). The values agree, so the model is continuous at the breakpoint.

a) Initial value \(100\); breakpoint \(t=4\); breakpoint value \(200\) b) \(P(t)=100(2)^{t/4}\) for \(0\le t<4\), and \(P(t)=200(0.5)^{(t-4)/2}\) for \(t\ge4\) c) Both branches give \(200\) at \(t=4\), so the model is continuous there.

- Use the first weekly growth factor to find the value exactly at the change time. - For continuity, the second branch must begin from the value already reached at week \(5\). - In the second branch, count elapsed time from the breakpoint rather than from week \(0\).

1. For the first \(5\) weeks, \(S(t)=120(1.08)^t\). Thus, \(S(5)=120(1.08)^5\approx176.32\,\text{cm}^2\). 2. To keep the model continuous, use the week-5 value as the starting value of the second branch. 3. Therefore, \(S(t)=120(1.08)^t\) for \(0\le t\le5\), and \(S(t)=120(1.08)^5(1.03)^{t-5}\) for \(t>5\). 4. At week \(9\), \(S(9)=120(1.08)^5(1.03)^4\approx198.45\,\text{cm}^2\).

a) About \(176.32\,\text{cm}^2\) b) \(S(t)=120(1.08)^t\) for \(0\le t\le5\), and \(S(t)=120(1.08)^5(1.03)^{t-5}\) for \(t>5\) c) About \(198.45\,\text{cm}^2\)

- Treat the instant when the filter turns on separately from the growth that happens before it. - The second branch should begin from the rate that remains immediately after the filter acts. - In the second branch, the exponent measures elapsed time since the filter was activated. - Solve the threshold equation using the branch that applies after the filter is active.

1. The first rule reaches \(400(1.5)^3=1350\) requests per minute just before the filter is activated. The filter blocks half, so \(R(3)=675\). 2. Before the filter, \(R(t)=400(1.5)^t\) for \(0\le t<3\). After the filter, time in the new regime is measured from \(t=3\), so \(R(t)=675(0.9)^{t-3}\) for \(t\ge3\). 3. After the filter, solve \(675(0.9)^{t-3}=500\). 4. Then \(t-3=\frac{\ln(500/675)}{\ln(0.9)}\approx2.848\), so \(t\approx5.85\) hours. 5. The rate equals \(500\) at that time and is below \(500\) for later times.

a) Just before: \(1350\) requests per minute; immediately after: \(675\) requests per minute b) \(R(t)=400(1.5)^t\) for \(0\le t<3\), and \(R(t)=675(0.9)^{t-3}\) for \(t\ge3\) c) The rate reaches \(500\) at about \(5.85\) hours after monitoring begins and is below \(500\) for later times.

- First determine what remains of \(S_k(t)\) as the time-dependent factor approaches zero. - Before day \(50\), decide which campaign is actually active. - After day \(50\), measure Campaign 2's elapsed time from its own start. - The combined model adds contributions only from campaigns that have begun.

1. As \(t\to\infty\), \((20-t)e^{-0.1t}\to0\), so \(S_k(t)\to10k\). 2. Before \(t=50\), only Campaign 1 contributes, so \(G(t)=S_{100}(t)=100(20-t)e^{-0.1t}+1000\) for \(0\le t<50\). 3. From \(t=50\) onward, add Campaign 1 and the time-shifted second campaign: \(G(t)=S_{100}(t)+S_{150}(t-50)\). 4. Thus, for \(t\ge50\), \(G(t)=100(20-t)e^{-0.1t}+150(70-t)e^{-0.1(t-50)}+2500\). 5. At \(t=60\), \(G(60)=-4000e^{-6}+1500e^{-1}+2500\approx3041.90\), which rounds to \(3042\) visits. 6. The long-term combined value is \(10(100)+10(150)=2500\) visits.

a) \(10k\) b) \(G(t)=100(20-t)e^{-0.1t}+1000\) for \(0\le t<50\), and \(G(t)=100(20-t)e^{-0.1t}+150(70-t)e^{-0.1(t-50)}+2500\) for \(t\ge50\) c) About \(3042\) visits d) \(2500\) visits

- Read the marked rate-change point against the horizontal axis before writing the second branch. - The second branch of \(B\) must start from the value already reached at the breakpoint. - The crossover after the breakpoint comes from setting that second branch equal to \(A\). - For the final comparison, distinguish the accumulated value at the breakpoint from the growth factor used afterward.

1. The graph marks the rate change at \(t=4\). 2. The single-rate model is \(A(t)=100(1.12)^t\). For \(B\), use \(B(t)=100(1.05)^t\) for \(0\le t\le4\), and \(B(t)=100(1.05)^4(1.25)^{t-4}\) for \(t>4\). 3. For the crossover after year \(4\), solve \(100(1.12)^t=100(1.05)^4(1.25)^{t-4}\). 4. Taking logarithms gives \(t\ln(1.12)=4\ln(1.05)+(t-4)\ln(1.25)\), so \(t=\frac{4(\ln(1.25)-\ln(1.05))}{\ln(1.25)-\ln(1.12)}\approx6.351\). 5. At year \(10\), \(A(10)=100(1.12)^{10}\approx310.58\), while \(B(10)=100(1.05)^4(1.25)^6\approx463.68\). 6. Although \(A\) is larger at the rate-change time, afterward \(B\) has the larger annual growth factor, so it eventually catches and then exceeds \(A\).

a) Year \(4\) b) \(A(t)=100(1.12)^t\); \(B(t)=100(1.05)^t\) for \(0\le t\le4\), and \(B(t)=100(1.05)^4(1.25)^{t-4}\) for \(t>4\). The crossover is \(t=\frac{4(\ln(1.25)-\ln(1.05))}{\ln(1.25)-\ln(1.12)}\approx6.351\,\text{years}\). c) \(B\) is larger at year \(10\): \(A(10)\approx310.58\) and \(B(10)\approx463.68\). After the rate change, \(B\) has the larger yearly growth factor.

- Identify when the factor \(0.8\) first begins to act. - At a later time, count only the time spent in the second regime. - Compare what each proposed second branch would do immediately after the breakpoint. - At day \(8\), compare how many decay factors the two models apply.

1. The factor \(0.8\) begins acting only after day \(6\). At time \(t\), the number of days spent in the decay regime is \(t-6\), not \(t\). Jordan's exponent incorrectly applies six extra days of decay. 2. The correct model is \(P(t)=50(1.1)^t\) for \(0\le t\le6\), and \(P(t)=50(1.1)^6(0.8)^{t-6}\) for \(t>6\). 3. At day \(8\), the correct model gives \(50(1.1)^6(0.8)^2\approx56.69\). 4. The proposed model gives \(50(1.1)^6(0.8)^8\approx14.86\). 5. The incorrect model has multiplied by six additional factors of \(0.8\), so its prediction is only \((0.8)^6\approx0.262\) times the correct prediction.

a) The decay exponent must count time since day \(6\), so it should be \(t-6\), not \(t\). b) \(P(t)=50(1.1)^t\) for \(0\le t\le6\), and \(P(t)=50(1.1)^6(0.8)^{t-6}\) for \(t>6\) c) Correct: about \(56.69\); proposed: about \(14.86\). The proposed model applies six extra daily decay factors.

- Treat each change time as a new elapsed-time origin for the factor that begins there. - At day \(4\), distinguish the value just before the intervention from the value immediately after it. - At day \(10\), use the value already reached by the second branch as the starting value of the third branch. - Continuity depends on whether the branch values agree at the breakpoint, not merely on whether both branches are exponential.

1. Before the first breakpoint, \(Q(t)=500(1.10)^t\) for \(0\le t<4\). 2. Just before day \(4\), the value is \(500(1.10)^4\). The intervention leaves \(80\%\) of that value, so the second branch begins at \(0.8\cdot500(1.10)^4\). 3. Therefore, \(Q(t)=0.8\cdot500(1.10)^4(0.95)^{t-4}\) for \(4\le t<10\). 4. At day \(10\), there is no jump. The second branch reaches \(0.8\cdot500(1.10)^4(0.95)^6\), so the third branch is \(Q(t)=0.8\cdot500(1.10)^4(0.95)^6(1.03)^{t-10}\) for \(t\ge10\). 5. At \(t=4\), the left-hand branch approaches \(500(1.10)^4\), while the actual branch value is \(80\%\) of that amount. The model is discontinuous there. 6. At \(t=10\), the second-branch value and third-branch starting value are identical, so the model is continuous there. 7. \(Q(12)=0.8\cdot500(1.10)^4(0.95)^6(1.03)^2\approx456.7\).

a) \(Q(t)=500(1.10)^t\) for \(0\le t<4\); \(Q(t)=0.8\cdot500(1.10)^4(0.95)^{t-4}\) for \(4\le t<10\); \(Q(t)=0.8\cdot500(1.10)^4(0.95)^6(1.03)^{t-10}\) for \(t\ge10\) b) Discontinuous at \(t=4\) because of the instantaneous \(20\%\) drop; continuous at \(t=10\) because the third branch starts from the second branch's day-10 value. c) \(Q(12)\approx456.7\)