Free math worksheets

- Read the rectangle's width from the horizontal axis. - Read its height from the vertical axis. - Multiply the rate by the elapsed time and check how the units combine.

1. Read the rectangle's width from the time axis: \(4\,\text{s}\). 2. Read its height from the rate axis: \(3\,\frac{\text{L}}{\text{s}}\). 3. The accumulated amount is \(4\,\text{s}\cdot 3\,\frac{\text{L}}{\text{s}}=12\,\text{L}\).

\(12\,\text{L}\)

- Read where the graph changes height and split the time interval there. - Treat each constant-rate piece as a rectangle. - Add the areas of the rectangles and keep the resulting quantity in gallons.

1. From \(t=0\) to \(t=2\), the rectangular area is \(2\,\text{min}\cdot 2\,\frac{\text{gal}}{\text{min}}=4\,\text{gal}\). 2. From \(t=2\) to \(t=3\), the rectangular area is \(1\,\text{min}\cdot 5\,\frac{\text{gal}}{\text{min}}=5\,\text{gal}\). 3. Add the accumulated amounts: \(4\,\text{gal}+5\,\text{gal}=9\,\text{gal}\).

\(9\,\text{gal}\)

- Match each time interval with the speed at its left endpoint. - Every rectangle has the same width here. - Multiply each speed by its interval width, then add the resulting distances.

1. The three intervals are \([0,1]\), \([1,2]\), and \([2,3]\), each with width \(1\,\text{s}\). 2. The left-endpoint speeds are \(2\), \(3\), and \(5\,\frac{\text{m}}{\text{s}}\). 3. The estimate is \(1\cdot 2+1\cdot 3+1\cdot 5=10\,\text{m}\).

\(10\,\text{m}\)

- Read the height of each shaded rectangle from the graph. - Every rectangle has width \(1\) minute. - Think about whether a decreasing graph lies above or below a rectangle whose height is chosen at the right edge.

a) The right-endpoint rectangle heights are \(5\), \(4\), \(3\), and \(2\,\frac{\text{L}}{\text{min}}\). With width \(1\,\text{min}\), the estimate is \(5+4+3+2=14\,\text{L}\). b) Because the rate decreases throughout the interval, each right-endpoint height is below the graph over most of its rectangle, while each left-endpoint height is above the graph over most of its rectangle. Therefore the right-endpoint estimate is smaller.

a) \(14\,\text{L}\) b) For a decreasing rate, right-endpoint rectangles lie below the graph while left-endpoint rectangles lie above it, so the right-endpoint estimate is smaller.

- Find the width of each time interval before multiplying by its rate. - The intervals do not all have the same width. - Add the accumulated amounts from the three intervals.

1. On \([0,1]\), the accumulated amount is \(1\cdot 4=4\,\text{L}\). 2. On \([1,2.5]\), the interval width is \(1.5\) minutes, so the amount is \(1.5\cdot 2=3\,\text{L}\). 3. On \([2.5,4]\), the interval width is also \(1.5\) minutes, so the amount is \(1.5\cdot 5=7.5\,\text{L}\). 4. The total is \(4+3+7.5=14.5\,\text{L}\).

\(14.5\,\text{L}\)

- Split the region under the graph where its shape changes. - Identify the familiar geometric shape in each section. - Because velocity is nonnegative, each geometric area contributes positively to distance traveled.

1. From \(0\) to \(2\) seconds, the region is a triangle with area \(\frac{1}{2}\cdot 2\cdot 4=4\,\text{m}\). 2. From \(2\) to \(5\) seconds, the region is a rectangle with area \(3\cdot 4=12\,\text{m}\). 3. From \(5\) to \(6\) seconds, the region is a triangle with area \(\frac{1}{2}\cdot 1\cdot 4=2\,\text{m}\). 4. The total distance is \(4+12+2=18\,\text{m}\).

\(18\,\text{m}\)

- Read the two endpoint heights from the vertical axis. - Read the horizontal width of the interval from the time axis. - Use the trapezoid area formula and check the resulting units.

1. Read the endpoint rates from the graph: \(2\,\frac{\text{L}}{\text{min}}\) and \(6\,\frac{\text{L}}{\text{min}}\). 2. The time interval has width \(4\,\text{min}\), so the trapezoid area is \(\frac{2+6}{2}\cdot 4=16\). 3. Multiplying \(\frac{\text{L}}{\text{min}}\) by minutes gives liters, so the volume added is \(16\,\text{L}\).

\(16\,\text{L}\)

- Split the graph at the times where the velocity is \(0\). - Areas above and below the time axis have different signs when finding displacement. - For total distance, use the magnitude of each geometric area.

1. From \(t=0\) to \(t=4\), the graph forms a triangle above the time axis with area \(\frac{1}{2}\cdot4\cdot4=8\,\text{m}\). 2. From \(t=4\) to \(t=6\), the graph forms a triangle below the time axis with area magnitude \(\frac{1}{2}\cdot2\cdot2=2\,\text{m}\). 3. Displacement uses signed area, so the displacement is \(8-2=6\,\text{m}\). 4. Distance uses the magnitudes of both areas, so the total distance is \(8+2=10\,\text{m}\). 5. The negative velocity indicates motion in the opposite direction, which subtracts from displacement but still adds to distance traveled.

a) \(6\,\text{m}\) b) \(10\,\text{m}\) c) The below-axis area represents motion in the opposite direction, so it is negative for displacement but positive when counting total distance.

- Find the signed change contributed by each time interval. - Keep negative-rate intervals negative when combining the changes. - After finding the net change, combine it with the initial amount.

1. On \([0,2]\), the change is \(2\cdot3=6\,\text{L}\). 2. On \([2,5]\), the change is \(3\cdot(-2)=-6\,\text{L}\). 3. On \([5,6]\), the change is \(1\cdot1=1\,\text{L}\). 4. The net accumulated change is \(6-6+1=1\,\text{L}\). 5. Add the net change to the initial amount: \(20+1=21\,\text{L}\).

a) \(1\,\text{L}\) b) \(21\,\text{L}\)

- Convert each constant-rate interval into a rectangular accumulated amount. - Express the unknown interval’s contribution in terms of \(r\). - The two interval contributions must add to the stated total.

1. The first interval contributes \(2\cdot 3=6\) units. 2. The second interval has width \(3\) hours, so it contributes \(3r\) units. 3. Use the total: \(6+3r=18\). 4. Then \(3r=12\), so \(r=4\,\frac{\text{units}}{\text{h}}\).

\(r=4\,\frac{\text{units}}{\text{h}}\)

- Read each rectangle's width and height from the axes and grid. - Add rectangle areas, not just rectangle heights. - Compare the top of each left-endpoint rectangle with the increasing curve over the same subinterval.

1. In panel a), the two rectangles each have width \(1\) and heights \(1\) and \(2\). The estimate is \(1(1+2)=3\). 2. In panel b), the four rectangles each have width \(0.5\) and heights \(1\), \(1.25\), \(2\), and \(3.25\). The estimate is \(0.5(1+1.25+2+3.25)=3.75\). 3. The graph is increasing, so each left-endpoint rectangle lies below the curve over its interval. The refined partition uses later, larger left-endpoint heights on some subintervals, so its lower estimate is larger.

a) \(3\) b) \(3.75\) c) Because the rate is increasing, both are lower estimates, and the finer left-endpoint partition gives the larger one.

- Read the four measured rate values from the marked points. - Decide which endpoint gives the smaller rectangle height on each interval when the rate is increasing. - Use the other endpoint heights for the upper bound.

1. Read the marked rates as \(1\), \(2\), \(4\), and \(7\) units per minute. 2. Because the rate is increasing, left-endpoint rectangles give a lower bound: \(1+2+4=7\) units. 3. Right-endpoint rectangles give an upper bound: \(2+4+7=13\) units. 4. Thus the accumulated amount lies between \(7\) and \(13\) units under the stated monotonicity assumption.

Lower bound: \(7\) units Upper bound: \(13\) units

- Equal starting and ending rates do not determine what happens between those times. - Break each region under the rate graph into familiar geometric shapes. - Compare the two total areas rather than only the endpoint heights.

1. For machine a), the rate is constantly \(2\) units per minute, so the rectangular area is \(4\cdot 2=8\) units. 2. For machine b), view the region as the same \(4\)-by-\(2\) rectangle plus a triangle above it with base \(4\) minutes and height \(2\) units per minute. 3. Machine b) accumulates \(8+\frac{1}{2}\cdot 4\cdot 2=12\) units. 4. Therefore, machine b) accumulates \(12-8=4\) more units, even though the two machines have the same initial and final rates.

Machine a) accumulates \(8\) units. Machine b) accumulates \(12\) units. Machine b) accumulates \(4\) more units.

- Relate the sign of the rate to whether the accumulated amount is rising or falling. - A zero rate adds no new amount over time. - To compare \(A(2)\) and \(A(5)\), account for the signed area added between those inputs.

1. From \(0\) to \(2\), the rate is positive, so \(A\) increases. 2. From \(2\) to \(3\), the rate is \(0\), so \(A\) is constant. 3. From \(3\) to \(5\), the rate is negative, so \(A\) decreases. 4. From \(5\) to \(6\), the rate is positive, so \(A\) increases again. 5. The accumulated amount at \(x=2\) is \(2\cdot2=4\) units. From \(2\) to \(5\), the zero-rate interval adds nothing and the negative rectangle contributes \(-1\cdot2=-2\) units, so \(A(5)=2\). Therefore, \(A(2)>A(5)\).

a) \(A\) increases on \((0,2)\) and \((5,6)\), is constant on \((2,3)\), and decreases on \((3,5)\). b) \(A(2)>A(5)\); specifically, \(A(2)=4\) and \(A(5)=2\).

- Read the marked power values from the graph before calculating any areas. - For each trapezoid, average its two endpoint heights and multiply by its width. - Keep the four-trapezoid and two-trapezoid calculations separate. - Compare which visible peaks are represented by each partition.

1. Read the power values \(1\), \(3\), \(2\), \(5\), and \(3\,\text{kW}\) at \(t=0,1,2,3,4\). 2. For part a), the four trapezoid areas are \(\frac{1+3}{2}(1)=2\), \(\frac{3+2}{2}(1)=2.5\), \(\frac{2+5}{2}(1)=3.5\), and \(\frac{5+3}{2}(1)=4\). Their total is \(12\,\text{kWh}\). 3. For part b), the two trapezoids have width \(2\) hours: \(2\cdot\frac{1+2}{2}+2\cdot\frac{2+3}{2}=8\,\text{kWh}\). 4. The coarse estimate skips the higher intermediate power values at \(t=1\) and \(t=3\), so its straight segments cut below substantial parts of the displayed model. Power in kilowatts multiplied by time in hours gives kilowatt-hours.

a) \(12\,\text{kWh}\) b) \(8\,\text{kWh}\) c) The coarse estimate misses the intermediate peaks, so it is smaller. The accumulated energy unit is kilowatt-hours.