Free math worksheets

- Look only at points whose x-coordinate is \(1\). - Compare the meanings of an open point and a closed point. - A function value comes from the included point at that input.

1. At \(x=1\), the graph has an open point at \((1,2)\), so that point is not included in the function. 2. The graph has a closed point at \((1,-1)\), so this is the included output when \(x=1\). 3. Therefore, \(h(1)=-1\).

\(h(1)=-1\). The closed point gives the included value at \(x=1\); the open point is excluded.

- Check the inequality attached to each piece before substituting. - Which condition includes the boundary value \(x=3\)?

1. The input is \(x=3\), so use the piece whose condition includes equality at \(3\). 2. The condition \(x\ge3\) applies, so \(f(3)=7\).

\(f(3)=7\), using the \(x\ge3\) piece.

- Divide the trip into the three described distance intervals. - For a linear change, the value halfway through an interval is halfway between the endpoint values. - Check the speed at the start and end of the trip.

1. On \(0\le s\le0.2\), the speed rises linearly from \(0\) to \(40\). Therefore, halfway through that segment, at \(s=0.1\), the speed is \(20\,\text{mph}\). 2. On \(0.2\le s\le0.8\), the speed is constant at \(40\,\text{mph}\). 3. On \(0.8\le s\le1.0\), the speed falls linearly from \(40\) to \(0\). Therefore, at \(s=0.9\), the speed is \(20\,\text{mph}\).

<table><tr><td>Distance \(s\) in miles</td><td>\(0\)</td><td>\(0.1\)</td><td>\(0.2\)</td><td>\(0.5\)</td><td>\(0.8\)</td><td>\(0.9\)</td><td>\(1.0\)</td></tr><tr><td>Speed \(v\) in mph</td><td>\(0\)</td><td>\(20\)</td><td>\(40\)</td><td>\(40\)</td><td>\(40\)</td><td>\(20\)</td><td>\(0\)</td></tr></table>

- First determine the number of billed minutes. - Compare the costs at \(10\) minutes and the additional cost for each minute after that. - Decide whether the price changes continuously within a billed minute.

1. Let \(n\) be the number of billed minutes. Then \(C_S=1.00+0.15n\) and \(C_Z=0.25n\). 2. For \(5\), \(10\), and \(15\) minutes, Swift costs \(\$1.75\), \(\$2.50\), and \(\$3.25\). Zip costs \(\$1.25\), \(\$2.50\), and \(\$3.75\). 3. For whole-number ride times from \(1\) through \(12\): <table><tr><td>Minutes</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td><td>\(7\)</td><td>\(8\)</td><td>\(9\)</td><td>\(10\)</td><td>\(11\)</td><td>\(12\)</td></tr><tr><td>Swift</td><td>\(\$1.15\)</td><td>\(\$1.30\)</td><td>\(\$1.45\)</td><td>\(\$1.60\)</td><td>\(\$1.75\)</td><td>\(\$1.90\)</td><td>\(\$2.05\)</td><td>\(\$2.20\)</td><td>\(\$2.35\)</td><td>\(\$2.50\)</td><td>\(\$2.65\)</td><td>\(\$2.80\)</td></tr><tr><td>Zip</td><td>\(\$0.25\)</td><td>\(\$0.50\)</td><td>\(\$0.75\)</td><td>\(\$1.00\)</td><td>\(\$1.25\)</td><td>\(\$1.50\)</td><td>\(\$1.75\)</td><td>\(\$2.00\)</td><td>\(\$2.25\)</td><td>\(\$2.50\)</td><td>\(\$2.75\)</td><td>\(\$3.00\)</td></tr></table> 4. At \(10\) billed minutes, both cost \(\$2.50\). After that, Swift adds only \(\$0.15\) per minute while Zip adds \(\$0.25\), so the claim is true. 5. For an actual ride time between two minute marks, the cost remains constant until the next billing boundary and then jumps. A step graph, not straight segments between table points, represents this pricing.

a) Swift: \(\$1.75\), \(\$2.50\), \(\$3.25\); Zip: \(\$1.25\), \(\$2.50\), \(\$3.75\) b) <table><tr><td>Minutes</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td><td>\(7\)</td><td>\(8\)</td><td>\(9\)</td><td>\(10\)</td><td>\(11\)</td><td>\(12\)</td></tr><tr><td>Swift</td><td>\(\$1.15\)</td><td>\(\$1.30\)</td><td>\(\$1.45\)</td><td>\(\$1.60\)</td><td>\(\$1.75\)</td><td>\(\$1.90\)</td><td>\(\$2.05\)</td><td>\(\$2.20\)</td><td>\(\$2.35\)</td><td>\(\$2.50\)</td><td>\(\$2.65\)</td><td>\(\$2.80\)</td></tr><tr><td>Zip</td><td>\(\$0.25\)</td><td>\(\$0.50\)</td><td>\(\$0.75\)</td><td>\(\$1.00\)</td><td>\(\$1.25\)</td><td>\(\$1.50\)</td><td>\(\$1.75\)</td><td>\(\$2.00\)</td><td>\(\$2.25\)</td><td>\(\$2.50\)</td><td>\(\$2.75\)</td><td>\(\$3.00\)</td></tr></table> c) The claim is true. d) No. The graph should be a step graph because each started minute is billed as a whole minute.

- After the free period, round each partial paid hour up to a full billed hour. - Identify the jumps at the billing boundaries and the horizontal section at the daily maximum. - Recall the graph and constant-ratio tests for proportional relationships.

1. For \(45\) minutes, the first paid hour has begun, so the fee is \(\$2.50\). For \(2\) hours, \(90\) paid minutes have elapsed, so two started hours cost \(\$5.00\). For \(6\) hours and \(10\) hours, the daily maximum applies, so each costs \(\$15.00\). 2. The graph is a step graph. It is \(\$0\) through \(0.5\) hour, then increases in \(\$2.50\) steps. At exactly \(5.5\) hours, the price is \(\$12.50\); for times greater than \(5.5\) hours, the maximum price is \(\$15.00\). 3. The graph is not a line through the origin, and doubling the parking time does not consistently double the price. Therefore, the relation is not proportional.

a) \(45\) min: \(\$2.50\); \(2\) h: \(\$5.00\); \(6\) h: \(\$15.00\); \(10\) h: \(\$15.00\) b) A step graph that begins at \(\$0\), rises by \(\$2.50\) per started paid hour, and is capped at \(\$15.00\) c) It is not a line through the origin, and it does not have a constant ratio of price to time.

- Choose the formula whose condition includes the given input. - A constant formula produces a horizontal graph. - Compare the value approached by the left piece with \(f(0)\).

1. For \(x=-3\), use the first piece: \(f(-3)=-2\). For \(x=0\), use the second piece: \(f(0)=0-2=-2\). For \(x=3\), \(f(3)=3-2=1\). 2. For \(-4\le x<0\), the graph is the horizontal segment \(y=-2\). For \(0\le x\le4\), the graph is the line \(y=x-2\). 3. The left-hand piece stays at \(-2\) as \(x\) approaches \(0\), and the second piece gives \(f(0)=-2\). Therefore, both pieces meet at \((0, -2)\) with no gap.

a) \(f(-3)=-2\), \(f(0)=-2\), and \(f(3)=1\) b) \(y=-2\) for \(-4\le x<0\), and \(y=x-2\) for \(0\le x\le4\) c) Yes. The pieces meet at \((0, -2)\).

- Read the horizontal step directly above each usage value. - At a jump, a filled point is included and an open point is excluded. - Pay special attention to exact threshold values.

1. At \(0.3\,\text{GB}\), the graph is at \(\$0\). At exactly \(0.5\,\text{GB}\), the filled point is also at \(\$0\), so the base price still applies. The value \(1.6\,\text{GB}\) lies in the interval \((1.5, 2.0]\), where the graph is at \(\$15\). 2. The filled point at \(v=1.0\,\text{GB}\) shows that the \(\$5\) price level includes exactly \(1.0\,\text{GB}\). The next price level, \(\$10\), applies only when usage exceeds \(1.0\,\text{GB}\).

a) \(0.3\,\text{GB}\): \(\$0\); \(0.5\,\text{GB}\): \(\$0\); \(1.6\,\text{GB}\): \(\$15\) b) At exactly \(1.0\,\text{GB}\), the added cost is \(\$5\); the \(\$10\) level begins above \(1.0\,\text{GB}\).

- Move vertically from each input to the graph and read the output. - At a jump, use the closed endpoint rather than the open endpoint. - Compare each input with the two nearest integers.

1. Since \(2.4\in[2, 3)\), the graph has height \(2\), so \(g(2.4)=2\). 2. Since \(0.5\in[0, 1)\), the graph has height \(0\), so \(g(0.5)=0\). 3. Since \(-1.3\in[-2, -1)\), the graph has height \(-2\), so \(g(-1.3)=-2\). 4. The function assigns each real number the greatest integer less than or equal to it. It is the greatest integer function, also called the floor function: \(g(x)=\lfloor x\rfloor\).

a) \(g(2.4)=2\), \(g(0.5)=0\), and \(g(-1.3)=-2\) b) It rounds each number down to the greatest integer less than or equal to it. It is the greatest integer, or floor, function.

- Compare each input with the breakpoint \(3\) before evaluating. - The equality sign tells you which rule applies when \(x=3\). - Use only one branch for each input.

1. Since \(2.9<3\), use \(2x+1\): \(p(2.9)=2(2.9)+1=6.8\). 2. Since equality belongs to \(x\ge3\), use \(x^2-4\): \(p(3)=3^2-4=5\). 3. Since \(3.1>3\), use \(x^2-4\): \(p(3.1)=3.1^2-4=5.61\).

\(p(2.9)=6.8\) \(p(3)=5\) \(p(3.1)=5.61\) At \(x=3\), use the \(x\ge3\) branch.

- Graph each rule only on the interval stated for it. - Which inequality includes equality at \(x=1\)? - Compare the endpoint markers as well as the shapes of the pieces.

1. For \(x\le1\), the graph follows \(y=-x\) and includes \(x=1\), so \((1,-1)\) must be closed. 2. For \(x>1\), the graph is the horizontal line \(y=2\) and excludes \(x=1\), so \((1,2)\) must be open. 3. Only panel b) has both pieces and both endpoint markers correct.

b). The point \((1,-1)\) is closed and the point \((1,2)\) is open.

- Read the leftmost and rightmost included x-values for the domain. - Read the lowest and highest included y-values for the range. - Zeros are x-values where an included piece meets the x-axis. - At \(x=0\), distinguish the closed point from the open point.

1. The graph covers all x-values from \(-4\) through \(4\), so the domain is \([-4,4]\). 2. The included y-values run from \(-2\) through \(3\), so the range is \([-2,3]\). 3. The two included x-axis crossings are \((-2,0)\) and \((1,0)\), so the zeros are \(-2\) and \(1\). 4. At \(x=0\), the closed point is \((0,2)\), so the y-intercept is \((0,2)\). The open point at \((0,-1)\) is not part of the function.

Domain: \([-4,4]\) Range: \([-2,3]\) Zeros: \(x=-2\) and \(x=1\) y-intercept: \((0,2)\)

- Compare which endpoint of each horizontal step is included. - Floor returns the greatest integer less than or equal to the input. - Ceiling returns the least integer greater than or equal to the input. - Locate \(1.4\) between \(1\) and \(2\).

1. Panel a) includes the left endpoint of each step and excludes the right endpoint, so it represents \(y=\lfloor x\rfloor\). 2. Since \(1\le1.4<2\), \(\lfloor1.4\rfloor=1\). 3. Panel b) excludes the left endpoint of each step and includes the right endpoint, so it represents \(y=\lceil x\rceil\). 4. Since \(1<1.4\le2\), \(\lceil1.4\rceil=2\).

a) \(y=\lfloor x\rfloor\), and \(\lfloor1.4\rfloor=1\) b) \(y=\lceil x\rceil\), and \(\lceil1.4\rceil=2\)

- For each function, compare what the left piece approaches at the boundary with the included boundary value. - Equality at \(x=2\) belongs to the second piece in both functions. - Matching boundary values indicate the pieces meet; different values indicate a jump.

1. For \(f\), the left-hand piece approaches \(2(2)+1=5\), and \(f(2)=5\). The two pieces meet continuously at \(x=2\). 2. For \(g\), the left-hand piece approaches \(2+2=4\), while \(g(2)=2-1=1\). The two values differ, so \(g\) has a jump at \(x=2\).

\(f\): left-hand value \(5\), \(f(2)=5\); continuous at \(x=2\). \(g\): left-hand value \(4\), \(g(2)=1\); jump at \(x=2\).

- Determine which inequality includes equality at \(x=0\). - Evaluate each formula at the boundary only to locate its endpoint. - Compare the required inclusion with the open and closed markers shown.

1. The first rule includes \(x=0\), and \(q(0)=1\), so \((0,1)\) must be a closed point. 2. The second rule excludes \(x=0\), and its boundary point would be \((0,2)\), so \((0,2)\) must be an open point. 3. Jisoo's graph reverses those endpoint markers.

The markers are reversed. \((0,1)\) should be closed, and \((0,2)\) should be open.

- Determine how many complete or partial \(30\)-minute blocks are included in the total time. - A “started \(30\)-minute block” means that any amount of time past a block boundary counts as a new full block. - Write an inequality to determine when the uncapped charge reaches or exceeds the daily maximum.

1. For \(145\) minutes, \(145 \div 30 \approx 4.83\), so \(5\) blocks have been started. The charge is \(\$2.50 + 4\cdot \$1.50 = \$8.50\). 2. Let \(n\) be the number of started blocks. The uncapped charge reaches the daily maximum when \(2.50 + 1.50(n-1) \ge 12.00\). This gives \(n \ge 7.33\), so the cap first applies during the eighth block. 3. The eighth block begins when the usage time is greater than \(7\cdot 30=210\) minutes. With whole-minute recording, the first such duration is \(211\) minutes. 4. The graph is an increasing step function with jumps every \(30\) minutes. It is constant at \(12.00\) for \(x>210\).

a) \(\$8.50\) b) \(211\) minutes c) An increasing step function, or piecewise-constant function, capped at \(\$12.00\).

- Divide each piece of the total-cost function by the number of guests. - Evaluate the correct formula on each side of the breakpoint at \(80\). - Solve the equation in the interval where its solution belongs. - For \(x\ge100\), compare \(\frac{150}{x}\) with \(\frac{150}{100}\).

1. The average cost is the total cost divided by \(x\): \(k(x)=25+\frac{600}{x}\) for \(1\le x<80\), \(k(x)=25+\frac{300}{x}\) for \(80\le x<100\), and \(k(x)=25+\frac{150}{x}\) for \(x\ge100\). 2. \(k(79)=25+\frac{600}{79}\approx32.59\), and \(k(80)=25+\frac{300}{80}=28.75\). Therefore, \(k(79)-k(80)=\frac{600}{79}-\frac{300}{80}\approx3.84\), so the decrease is about \(\$3.84\) per guest. 3. In the first interval, solve \(25+\frac{600}{x}=35\). Then \(\frac{600}{x}=10\), so \(x=60\), which satisfies \(x<80\). 4. For \(x\ge100\), \(\frac{150}{x}\le\frac{150}{100}=1.5\). Therefore, \(k(x)=25+\frac{150}{x}\le26.5<27\).

a) \(k(x)=25+\frac{600}{x}\) for \(1\le x<80\); \(k(x)=25+\frac{300}{x}\) for \(80\le x<100\); \(k(x)=25+\frac{150}{x}\) for \(x\ge100\) b) About \(\$3.84\) per guest c) \(60\) guests d) For \(x\ge100\), \(k(x)\le26.5<27\).

- Find where the expression inside the absolute value equals \(0\). - On one side of the breakpoint, keep the expression unchanged; on the other side, negate it. - Solve each linear equation and check that each solution belongs to its corresponding interval.

1. The expression inside the absolute value changes sign where \(2x-5=0\), so the breakpoint is \(x=2.5\). 2. If \(x\ge 2.5\), then \(2x-5\ge 0\), so \(f(x)=2x-5-3=2x-8\). 3. If \(x<2.5\), then \(2x-5<0\), so \(f(x)=-(2x-5)-3=-2x+2\). 4. On the first interval, \(2x-8=0\) gives \(x=4\), which satisfies \(x\ge 2.5\). 5. On the second interval, \(-2x+2=0\) gives \(x=1\), which satisfies \(x<2.5\).

\(f(x)=\begin{cases}-2x+2, &x<2.5\\2x-8, &x\ge 2.5\end{cases}\) The zeros are \(x=1\) and \(x=4\).

- Find where the expression inside the absolute value changes sign. - Simplify the formula separately for \(x<3\) and \(x\ge 3\). - If the variable cancels completely, the resulting function is constant on that interval.

1. The expression inside the absolute value equals zero at \(x=3\). 2. If \(x\ge 3\), then \(|x-3|=x-3\), so \(g(x)=x-3+x=2x-3\). 3. If \(x<3\), then \(|x-3|=-(x-3)=-x+3\), so \(g(x)=-x+3+x=3\). 4. Therefore, \(g(x)=\begin{cases}3, &x<3\\2x-3, &x\ge 3\end{cases}\). For \(x<3\), every function value equals \(3\), so the function is constant on that interval.

\(g(x)=\begin{cases}3, &x<3\\2x-3, &x\ge 3\end{cases}\) For \(x<3\), the function is constant with value \(3\).

- Find the horizontal and vertical changes for each section. - Use rise divided by run for each slope. - Make sure the second formula gives the correct height at the joining point. - Choose the correct piece before evaluating or solving.

1. The first slope is \(m_1=\frac{30}{150}=0.2\). 2. The second section rises \(130-30=100\) meters over \(400-150=250\) meters, so \(m_2=\frac{100}{250}=0.4\). 3. On the first section, \(h(x)=0.2x\) for \(0\le x\le150\). On the second, use the point \((150, 30)\): \(h(x)=0.4(x-150)+30=0.4x-30\) for \(150<x\le400\). 4. Since \(250>150\), \(h(250)=0.4\cdot 250-30=70\). The height is \(70\,\text{m}\). 5. Since \(90>30\), use the second piece: \(90=0.4x-30\). Then \(x=300\,\text{m}\).

a) \(m_1=0.2\); \(m_2=0.4\) b) \(h(x)=\begin{cases}0.2x, &0\le x\le150\\0.4x-30, &150<x\le400\end{cases}\) c) \(70\,\text{m}\) d) \(300\,\text{m}\)

- Find the slope and equation of each straight section. - Choose the appropriate formula for each table entry. - To find a missing input, set the appropriate piece equal to the given height. - Average slope uses the total rise and total horizontal run.

1. The first slope is \(\frac{15}{50}=0.3\), so \(h(x)=0.3x\) for \(0\le x\le50\). 2. The second slope is \(\frac{75-15}{150-50}=0.6\). Through \((50, 15)\), its equation is \(h(x)=0.6(x-50)+15=0.6x-15\) for \(50<x\le150\). 3. For \(x=20\), \(h_1=0.3\cdot 20=6\). 4. For \(h=9\), use the first piece: \(0.3x_2=9\), so \(x_2=30\). 5. For \(x=100\), \(h_3=0.6\cdot 100-15=45\). 6. For \(h=60\), use the second piece: \(0.6x_4-15=60\), so \(x_4=125\). 7. The average slope from start to finish is \(\frac{75}{150}=0.5\), equivalent to a \(50\%\) grade.

a) \(h(x)=\begin{cases}0.3x, &0\le x\le50\\0.6x-15, &50<x\le150\end{cases}\) b) \(h_1=6\); \(x_2=30\); \(h_3=45\); \(x_4=125\) c) \(0.5\), or \(50\%\)

- Recover each line equation from two vertices on the same segment. - Keep the given interval convention aligned with the segment formulas. - For \(f(x+2)\), transform both the formulas and the interval boundaries.

1. The first segment goes through \((-3,-1)\) and \((-1,2)\), so its slope is \(\frac{3}{2}\) and its rule is \(\frac{3}{2}x+\frac{7}{2}\). 2. The second segment goes through \((-1,2)\) and \((2,0)\), so its slope is \(-\frac{2}{3}\) and its rule is \(-\frac{2}{3}x+\frac{4}{3}\). 3. The third segment goes through \((2,0)\) and \((4,1)\), so its slope is \(\frac{1}{2}\) and its rule is \(\frac{1}{2}x-1\). 4. Therefore, \(f\) uses those three rules on \([-3,-1)\), \([-1,2)\), and \([2,4]\), respectively. 5. For \(h(x)=f(x+2)\), each breakpoint shifts left \(2\) units, so the intervals become \([-5,-3)\), \([-3,0)\), and \([0,2]\). 6. Substituting \(x+2\) into the three branch formulas gives \(\frac{3}{2}x+\frac{13}{2}\), \(-\frac{2}{3}x\), and \(\frac{1}{2}x\). The new domain is \([-5,2]\).

a) \(f(x)=\begin{cases}\frac{3}{2}x+\frac{7}{2}&-3\le x<-1\\-\frac{2}{3}x+\frac{4}{3}&-1\le x<2\\\frac{1}{2}x-1&2\le x\le4\end{cases}\) b) \(h(x)=\begin{cases}\frac{3}{2}x+\frac{13}{2}&-5\le x<-3\\-\frac{2}{3}x&-3\le x<0\\\frac{1}{2}x&0\le x\le2\end{cases}\), domain \([-5,2]\). Every breakpoint moves left \(2\) units.

- Determine the slope of each segment from two visible points. - Use one point to recover each line equation. - Let the open and closed endpoint markers determine the inequalities.

1. The left segment goes through \((-4,5)\) and \((-1,-1)\). Its slope is \(-2\), so its rule is \(h(x)=-2x-3\). 2. The left endpoint is closed and the point at \(x=-1\) is open, so this rule applies for \(-4\le x<-1\). 3. The right segment goes through \((-1,1)\) and \((3,5)\). Its slope is \(1\), so its rule is \(h(x)=x+2\). 4. Both endpoints of the right segment are closed, so this rule applies for \(-1\le x\le3\).

\(h(x)=\begin{cases}-2x-3&\text{if }-4\le x<-1\\x+2&\text{if }-1\le x\le3\end{cases}\)

- “Each started pound” means round the weight up before calculating Plan A. - Treat Plan A as a step function, one pound-interval at a time. - Compare the plans with an inequality on each interval. - Check strict inequality carefully at interval endpoints.

1. At \(x=0.5\), Plan A costs \(2\cdot 1=\$2.00\), and Plan B costs \(1.5\cdot 0.5+1=\$1.75\). 2. At \(x=1.0\), Plan A costs \(2\cdot 1=\$2.00\), and Plan B costs \(1.5\cdot 1.0+1=\$2.50\). 3. At \(x=2.2\), Plan A costs \(2\cdot 3=\$6.00\), and Plan B costs \(1.5\cdot 2.2+1=\$4.30\). 4. On \((0, 1]\), Plan A costs \(\$2.00\). Solve \(2<1.5x+1\): \(x>\frac{2}{3}\). This gives \(\frac{2}{3}<x\le1\). 5. On \((1, 2]\), Plan A costs \(\$4.00\). The inequality \(4<1.5x+1\) requires \(x>2\), so there is no solution in that interval. 6. On every later interval \((k-1, k]\), the condition is \(x>\frac{4k-2}{3}\), and this threshold is greater than or equal to \(k\) for \(k\ge2\). Therefore, there are no additional solutions.

a) At \(0.5\) lb: A \(\$2.00\), B \(\$1.75\) At \(1.0\) lb: A \(\$2.00\), B \(\$2.50\) At \(2.2\) lb: A \(\$6.00\), B \(\$4.30\) b) Plan A costs less for \(\frac{2}{3}<x\le1\) lb.

- First recover the line rule on each visible segment before applying either transformation. - For an input transformation, transform the interval conditions as well as the formulas. - In \(f(2x)\), ask which values of \(x\) make \(2x\) land in each original interval. - In \(f(x-1)\), shift every breakpoint consistently with the input shift.

1. The four graph segments have slopes \(1,-1,1,-1\). Using the visible vertices gives \(f(x)=x+2\) on \([-2,0)\), \(f(x)=-x+2\) on \([0,2)\), \(f(x)=x-2\) on \([2,4)\), and \(f(x)=-x+6\) on \([4,6]\). 2. For \(k(x)=f(2x)-1\), transform each input interval by requiring \(2x\) to lie in the corresponding interval for \(f\). The breakpoints become \(-1,0,1,2,3\). 3. Substituting \(2x\) into each branch and subtracting \(1\) gives \(2x+1\), \(-2x+1\), \(2x-3\), and \(-2x+5\), respectively. Thus, the domain of \(k\) is \([-1,3]\). 4. For \(p(x)=0.5f(x-1)+1\), require \(x-1\) to lie in each original interval. The breakpoints become \(-1,1,3,5,7\). 5. Substituting \(x-1\) into each branch, multiplying by \(0.5\), and adding \(1\) gives \(0.5x+1.5\), \(-0.5x+2.5\), \(0.5x-0.5\), and \(-0.5x+4.5\), respectively. Thus, the domain of \(p\) is \([-1,7]\).

a) \(f(x)=\begin{cases}x+2&-2\le x<0\\-x+2&0\le x<2\\x-2&2\le x<4\\-x+6&4\le x\le6\end{cases}\) b) \(k(x)=\begin{cases}2x+1&-1\le x<0\\-2x+1&0\le x<1\\2x-3&1\le x<2\\-2x+5&2\le x\le3\end{cases}\), domain \([-1,3]\) c) \(p(x)=\begin{cases}0.5x+1.5&-1\le x<1\\-0.5x+2.5&1\le x<3\\0.5x-0.5&3\le x<5\\-0.5x+4.5&5\le x\le7\end{cases}\), domain \([-1,7]\)

- Find the horizontal cross-sectional area below and above the step. - Treat the lower and upper height intervals separately. - When the water passes \(h=5\,\text{cm}\), include the full volume of the lower section. - Use the physical minimum and maximum fill heights to determine the domain and range.

1. For \(0\le h\le5\), the horizontal cross section measures \(20\,\text{cm}\times10\,\text{cm}\), so its area is \(200\,\text{cm}^2\). Thus, \(V(h)=200h\). 2. For \(5<h\le10\), the lower section already contains \(200\cdot5=1000\,\text{cm}^3\). The upper cross section measures \(10\,\text{cm}\times10\,\text{cm}\), so each additional centimeter of height adds \(100\,\text{cm}^3\). Therefore, \(V(h)=1000+100(h-5)=100h+500\). 3. The piecewise function is \(V(h)=\begin{cases}200h,&0\le h\le5\\100h+500,&5<h\le10\end{cases}\). 4. \(V(4)=200\cdot4=800\,\text{cm}^3\). Also, \(V(8)=100\cdot8+500=1300\,\text{cm}^3\). 5. The possible fill heights are from \(0\) to \(10\,\text{cm}\), and the volume increases from \(0\) to \(1500\,\text{cm}^3\). The domain is \([0,10]\), and the range is \([0,1500]\).

a) \(V(h)=\begin{cases}200h,&0\le h\le5\\100h+500,&5<h\le10\end{cases}\) b) \(V(4)=800\,\text{cm}^3\); \(V(8)=1300\,\text{cm}^3\) c) Domain: \([0,10]\); range: \([0,1500]\)