Free math worksheets

- Look for the output that corresponds to one unit of input. - The constant tells how much \(y\) changes for each \(1\) unit of \(x\). - Because the relationship is proportional, the graph value at \(x=1\) is the multiplier in \(y=kx\).

1. At \(x=1\), the graph has \(y=3\). 2. Therefore the unit rate and constant of proportionality are \(k=3\).

\(k=3\)

- Start with the given time on the horizontal axis. - Move to the plotted relationship, then read the matching value on the vertical axis. - Use the axis label to attach the correct unit to the value you read.

1. Locate \(t=3\) on the x-axis. 2. Read the y-value where the graph is above that input. 3. The graph gives \(d=6\), so the distance is \(6\,\text{miles}\).

\(6\,\text{miles}\)

- Locate \(1\) on the horizontal axis. - Read the corresponding cost from the line. - Use the axis labels to state the unit rate with units.

1. At \(x=1\), the graph has \(C=3\). 2. Therefore, the unit rate is \(\$3\) per pound.

The unit rate is \(\$3\) per pound.

- A proportional relationship has a graph that passes through the origin. - Choose a point on the line that lies at a grid intersection. - Divide cost by weight to find the unit rate \(k\).

1. A directly proportional relationship has the form \(y = kx\). 2. The point \((4, 3)\) lies on the graph, so \(4\) pounds cost \(\$3\). 3. The constant of proportionality is \(k = \frac{y}{x} = \frac{3}{4} = 0.75\). 4. Therefore, the equation is \(y = 0.75x\).

\(y = 0.75x\)

- Look for the graph value that corresponds to one unit of time. - Use the axis labels to determine the units of the constant. - Connect the constant rate to the standard equation for a proportional relationship.

1. At \(t=1\), the graph has \(d=2.5\), so the constant of proportionality is \(2.5\) miles per hour. 2. The proportional relationship is \(d=2.5t\).

a) \(2.5\) miles per hour; Maya walks \(2.5\) miles each hour. b) \(d=2.5t\)

- Locate the input value of one on the horizontal axis. - Read the matching output from the line. - Interpret both coordinates using the axis units.

1. At \(t=1\), the graph gives \(V=6\), so the point is \((1,6)\). 2. The point means that after \(1\,\text{minute}\), \(6\,\text{gallons}\) have been added; equivalently, the constant of proportionality is \(6\,\text{gallons per minute}\).

a) \((1,6)\) b) It means the tank receives \(6\,\text{gallons}\) in \(1\,\text{minute}\), so the rate is \(6\,\text{gallons per minute}\).

- Match each coordinate with the variable on its axis. - Think about what an x-coordinate of \(1\) means in context. - Interpret the origin using the real quantities, not only the graph rule.

1. Point \(P=(4,10)\) means \(4\,\text{pounds}\) of apples cost \(\$10\). 2. The point \((1,2.5)\) means \(1\,\text{pound}\) costs \(\$2.50\); this is the unit rate. 3. The point \((0,0)\) means buying \(0\,\text{pounds}\) costs \(\$0\), which is required for this proportional relationship.

a) \(4\,\text{pounds}\) of apples cost \(\$10\). b) \(1\,\text{pound}\) costs \(\$2.50\). c) \(0\,\text{pounds}\) costs \(\$0\), so the graph passes through the origin.

- Inspect where the segment already drawn on the graph ends. - Identify the later marked point that the completed graph must reach. - Decide whether every time between those two x-values has a meaningful volume.

1. The unfinished graph already contains the segment ending at \((2,8)\) and shows the later endpoint \((5,20)\). 2. Therefore, the missing segment must run from \((2,8)\) to \((5,20)\). 3. Filling occurs continuously, so every time between \(2\) and \(5\) minutes has a corresponding volume. The graph needs the entire connecting segment, not only an isolated point at \((5,20)\).

a) Add the segment from \((2,8)\) to \((5,20)\). b) Use a connected line segment because time and water volume vary continuously during filling.

- Compare how much distance each graph shows after the same amount of time. - On a proportional graph, the y-value at an x-value of \(1\) has a special interpretation. - Use the difference between the two unit rates for part b).

1. Both graphs pass through the origin, so each speed is the slope \(\frac{d}{t}\). 2. Graph p passes through \((1,4)\), so Cyclist P travels at \(4\) miles per hour. 3. Graph q passes through \((1,6)\), so Cyclist Q travels at \(6\) miles per hour. 4. Cyclist Q is faster by \(6-4=2\) miles per hour. 5. In a proportional graph \(d=kt\), the point \((1,k)\) has y-coordinate equal to the unit rate. Here those y-coordinates are \(4\) and \(6\).

a) Cyclist P: \(4\) miles per hour; Cyclist Q: \(6\) miles per hour b) Cyclist Q is faster by \(2\) miles per hour. c) The points \((1,4)\) and \((1,6)\) show the unit rates because the y-coordinate at \(t=1\) equals \(k\).

- Inspect which allowed whole-number game counts are already represented on the graph. - Use the equation only for the allowed inputs that are missing from the graph. - Think about whether a point between two whole-number game counts would represent a possible purchase.

1. The graph already shows the allowed inputs \(g=0\), \(1\), \(3\), and \(5\), so the missing whole-number inputs are \(2\) and \(4\). 2. Using \(C=3g\), the missing points are \((2,6)\) and \((4,12)\). 3. The points should not be connected because the number of games is discrete; values such as \(2.5\) games are not possible in this situation.

a) \((2,6)\) and \((4,12)\) b) No. Keep the points isolated because the number of games can only be a whole number.

- Read a convenient point on the line and compare its distance coordinate with its time coordinate. - Use the rate you found from the graph to build the whole-hour table. - Think about whether Sofia also travels during parts of an hour. - Convert \(150\) minutes to hours and locate that time directly on the graph before checking by calculation.

1. The graph contains points such as \((1,12)\) and \((2,24)\). The ratio of distance to time is \(12\), so Sofia's speed is \(12\,\text{mi/h}\). 2. Use the rate \(12\,\text{mi/h}\) to complete the table. <table><tr><td>Time</td><td>\(1\,\text{h}\)</td><td>\(2\,\text{h}\)</td><td>\(3\,\text{h}\)</td><td>\(4\,\text{h}\)</td><td>\(5\,\text{h}\)</td></tr><tr><td>Distance</td><td>\(12\,\text{mi}\)</td><td>\(24\,\text{mi}\)</td><td>\(36\,\text{mi}\)</td><td>\(48\,\text{mi}\)</td><td>\(60\,\text{mi}\)</td></tr></table> 3. A connected line makes sense because time and distance vary continuously during the ride, including between whole-hour times. 4. Since \(150\) minutes is \(2.5\) hours, the graph gives \(30\) miles at \(t=2.5\). The calculation \(12\cdot2.5=30\) verifies the graph reading.

a) \(12\,\text{mi/h}\); for example, the graph contains \((1,12)\). b) <table><tr><td>Time</td><td>\(1\,\text{h}\)</td><td>\(2\,\text{h}\)</td><td>\(3\,\text{h}\)</td><td>\(4\,\text{h}\)</td><td>\(5\,\text{h}\)</td></tr><tr><td>Distance</td><td>\(12\,\text{mi}\)</td><td>\(24\,\text{mi}\)</td><td>\(36\,\text{mi}\)</td><td>\(48\,\text{mi}\)</td><td>\(60\,\text{mi}\)</td></tr></table> c) A connected line makes sense because Sofia travels continuously between the listed times. d) \(30\,\text{mi}\); the graph shows \((2.5,30)\), and \(12\cdot2.5=30\).

- Choose a plotted point with nonzero coordinates and use it to determine ink per page. - Check where the plotted pattern begins and whether all plotted points align. - Decide whether the number of fully printed pages can take non-whole-number values. - For the cartridge question, first read the page count paired with \(40\,\text{mL}\) from the graph, then check it using your rate.

1. One convenient plotted point is \((400,10)\). The unit rate is \(10\div400=0.025\,\text{mL}\) per page. 2. The plotted points include the origin and lie on one straight-line pattern. Those features are consistent with a proportional relationship. The points are not connected because page count is discrete. 3. On the graph, the plotted point with ink use \(40\,\text{mL}\) has page count \(1600\). 4. The calculation \(40\div0.025=1600\) verifies the graph reading.

a) \(0.025\,\text{mL}\) per page; for example, using \((400,10)\), \(10\div400=0.025\). b) The plotted points include the origin and lie on one straight-line pattern. They remain isolated because the number of fully printed pages is discrete. c) \(1600\) pages. The graph shows \((1600,40)\), and \(40\div0.025=1600\).

- Find a feature of the table that stays constant from column to column. - Use one or more table points to compare with the graphs. - Check both the rate and whether the graph has the required proportional form.

1. The table has constant ratio \(\frac{y}{x}=1.5\). 2. Graph a) represents \(y=1.5x\) and contains the table points, including \((4,6)\). 3. Graph b) has a larger constant of proportionality, and graph c) does not pass through the origin.

Graph a) represents the table.

- Start with the given cost on the vertical axis and locate the plotted point at that height. - Use the plotted point for one ticket to identify the cost per ticket. - Remember that ticket counts are whole numbers, so the graph uses isolated points.

1. The plotted point with \(C=35\) has \(n=5\). 2. The points show a unit cost of \(7\) dollars per ticket, so \(C=7n\) for whole-number ticket counts. 3. Substituting \(C=35\) gives \(35=7n\), so \(n=5\).

a) \(5\) tickets b) \(C=7n\) for whole-number \(n\), and \(35=7\cdot5\).