Free math worksheets

- Identify the highest point on the graphed path. - Read both coordinates of that point and match each coordinate to its axis label.

1. The maximum occurs at the highest point of the parabola. 2. The graph's highest point is \((3, 9)\), so the rocket reaches a maximum height of \(9\,\text{ft}\) at \(3\,\text{s}\).

The maximum height is \(9\,\text{ft}\), and it occurs at \(3\,\text{s}\).

- Identify the vertex directly from the quadratic's vertex form. - Interpret the vertex's x-coordinate using the stated unit for \(x\), and interpret its y-coordinate using the stated unit for \(P(x)\).

1. The model is in vertex form, so its vertex is \((6, 7200)\). 2. Because the coefficient of the squared term is negative, the vertex gives the maximum. 3. Since \(x=6\) represents \(6\) hundreds of units, the maximizing sales level is \(600\) units, and the maximum weekly profit is \(\$7200\).

Sell \(600\) units for a maximum weekly profit of \(\$7200\).

- Which variable represents time, and which represents distance? - How can you isolate the squared variable when the distance is given? - Compare both distances at the same time.

1. Evaluate Vehicle A's model: \(s_A(5) = 1.2 \cdot 5^2 = 1.2 \cdot 25 = 30\,\text{m}\). 2. Solve \(120 = 1.2t^2\): \(t^2 = 100\), so \(t = \pm 10\). Since time is nonnegative, \(t = 10\,\text{s}\). 3. Compare the distances at \(t = 10\): \(s_A(10) = 1.2 \cdot 10^2 = 120\,\text{m}\), while \(s_B(10) = 0.8 \cdot 10^2 + 2 \cdot 10 = 100\,\text{m}\). 4. Vehicle A is ahead because \(120\,\text{m} > 100\,\text{m}\).

a) \(30\,\text{m}\) b) \(10\,\text{s}\) c) Vehicle A; it has traveled \(120\,\text{m}\), compared with Vehicle B's \(100\,\text{m}\).

- What input represents the release point? - Which point of a downward-opening parabola gives the maximum value? - Substitute the specified horizontal distance into the model.

1. The release occurs at \(x = 0\): \(h(0) = 2\), so the release height is \(2\,\text{m}\). 2. The vertex occurs at \(x = -\frac{0.6}{2 \cdot (-0.05)} = 6\). Then \(h(6) = -0.05 \cdot 6^2 + 0.6 \cdot 6 + 2 = 3.8\). The maximum height is \(3.8\,\text{m}\). 3. Evaluate \(h(10) = -0.05 \cdot 10^2 + 0.6 \cdot 10 + 2 = 3\). The height is \(3\,\text{m}\).

a) \(2\,\text{m}\) b) \(3.8\,\text{m}\) c) \(3\,\text{m}\)

- Substitute each requested speed into the model. - What does equal output at two speeds suggest about the axis of symmetry? - Which formula gives the vertex of a quadratic in standard form?

1. Evaluate the model: \(g(40) = 0.002 \cdot 40^2 - 0.24 \cdot 40 + 10.7 = 4.3\), and \(g(80) = 0.002 \cdot 80^2 - 0.24 \cdot 80 + 10.7 = 4.3\). The two speeds have the same modeled fuel use. 2. Since the leading coefficient is positive, the minimum occurs at the vertex. Its speed is \(v = -\frac{-0.24}{2 \cdot 0.002} = 60\,\text{mph}\). 3. Evaluate \(g(60) = 0.002 \cdot 60^2 - 0.24 \cdot 60 + 10.7 = 3.5\). The minimum fuel use is \(3.5\) gallons per \(100\) miles.

a) \(4.3\) gallons per \(100\) miles at both \(40\,\text{mph}\) and \(80\,\text{mph}\) b) The minimum occurs at \(60\,\text{mph}\) and is \(3.5\) gallons per \(100\) miles.

- Substitute either measurement into \(s(v) = kv^2\). - Check the same value of \(k\) with the other measurement. - For part c, set the model equal to the given distance.

1. Use the first measurement: \(20 = k \cdot 20^2\), so \(k = \frac{20}{400} = 0.05\). 2. Check the second measurement: \(0.05 \cdot 40^2 = 80\), so both measurements fit the model \(s(v) = 0.05v^2\). 3. At \(60\,\text{mph}\), \(s(60) = 0.05 \cdot 60^2 = 180\). The predicted braking distance is \(180\,\text{ft}\). 4. Set \(0.05v^2 = 125\). Then \(v^2 = 2500\), so \(v = 50\) because speed is positive.

a) \(k = 0.05\), so \(s(v) = 0.05v^2\) b) \(180\,\text{ft}\) c) \(50\,\text{mph}\)

- Substitute each given horizontal distance into the function. - Where is the maximum of a downward-opening parabola? - What equation represents the points where the arch meets the ground?

1. Evaluate both heights: \(f(2) = -0.25 \cdot 2^2 + 2 \cdot 2 = 3\), and \(f(6) = -0.25 \cdot 6^2 + 2 \cdot 6 = 3\). Therefore, the arch is \(3\,\text{m}\) high at both locations. 2. The vertex occurs at \(x = -\frac{2}{2 \cdot (-0.25)} = 4\). Then \(f(4) = -0.25 \cdot 4^2 + 2 \cdot 4 = 4\), so the maximum height is \(4\,\text{m}\). 3. Set \(f(x) = 0\): \(-0.25x^2 + 2x = 0\). Factoring gives \(x(-0.25x + 2) = 0\), so the zeros are \(x = 0\) and \(x = 8\). The bases are \(8\,\text{m}\) apart.

a) \(f(2) = f(6) = 3\,\text{m}\) b) \(4\,\text{m}\) c) \(8\,\text{m}\), because the zeros are \(x = 0\) and \(x = 8\).

- Substitute each speed into the model. - Compare the factor by which the input changes with the factor by which the output changes. - If every output doubles, what happens to the coefficient?

1. Evaluate the model: \(s(25) = 0.04 \cdot 25^2 = 25\,\text{ft}\), and \(s(50) = 0.04 \cdot 50^2 = 100\,\text{ft}\). 2. The speed doubles from \(25\) to \(50\), while the braking distance changes from \(25\) to \(100\), which is multiplied by \(4\). This occurs because \((2v)^2 = 4v^2\). 3. Doubling every output doubles the coefficient: \(s_{\text{wet}}(v) = 2(0.04v^2) = 0.08v^2\).

a) \(25\,\text{ft}\) at \(25\,\text{mph}\); \(100\,\text{ft}\) at \(50\,\text{mph}\) b) Doubling the speed quadruples the braking distance. c) \(s_{\text{wet}}(v) = 0.08v^2\)

- Distinguish carefully between radius and diameter. - Substitute the known radius and mass into \(m(r) = kr^2\). - What scale factor for the radius has a square equal to \(4\)?

1. Substitute \(m = 1.2\) and \(r = 10\): \(1.2 = k \cdot 10^2 = 100k\). Therefore, \(k = 0.012\,\frac{\text{kg}}{\text{cm}^2}\). 2. A diameter of \(30\,\text{cm}\) gives a radius of \(15\,\text{cm}\). Then \(m(15) = 0.012 \cdot 15^2 = 2.7\,\text{kg}\). 3. Since mass is proportional to \(r^2\), multiplying the radius by \(2\) multiplies the mass by \(2^2 = 4\). The radius must be doubled.

a) \(k = 0.012\,\frac{\text{kg}}{\text{cm}^2}\), so \(m(r) = 0.012r^2\) b) \(2.7\,\text{kg}\) c) The radius must be doubled.

- Model each game as a pair of different players. - Make sure each matchup is counted only once. - Write the number of pairs in terms of \(n\). - Interpret both algebraic solutions in the context.

1. Let \(n\) be the number of players. Each game corresponds to an unordered pair of players, so the total number of games is \(\frac{n(n-1)}{2}\). 2. Set up the equation: \(\frac{n(n-1)}{2}=190\). 3. Rearrange: \(n^2-n-380=0\). 4. Apply the quadratic formula: \(n=\frac{1\pm\sqrt{(-1)^2-4\cdot1\cdot(-380)}}{2}=\frac{1\pm39}{2}\). 5. The solutions are \(n=20\) and \(n=-19\). Only the positive value is meaningful, so there are \(20\) players.

\(20\) players

- What does \(n\) represent in this model? - When an output value is given, what equation should you solve? - What restrictions does the context place on \(n\)?

1. Evaluate the model at \(n = 8\): \(d(8) = \frac{1}{2} \cdot 8^2 - 1.5 \cdot 8 = 32 - 12 = 20\). 2. Set the model equal to \(54\): \(54 = \frac{1}{2}n^2 - 1.5n\). Multiplying by \(2\) gives \(n^2 - 3n - 108 = 0\). 3. Factor: \((n - 12)(n + 9) = 0\), so \(n = 12\) or \(n = -9\). 4. A polygon cannot have a negative number of sides, so only \(n = 12\) is meaningful.

a) \(20\) diagonals b) \(12\) sides c) The solution \(n = -9\) is not meaningful because a polygon cannot have a negative number of sides.

- How many opponents does one team have? - Why does counting games team by team count each game twice? - What games must the new team play against the existing teams?

1. Each team has \(n - 1\) opponents and plays each opponent twice, giving \(2(n - 1)\) team-game counts per team. Multiplying by \(n\) counts each actual game twice, so \(S(n) = \frac{n \cdot 2(n - 1)}{2} = n(n - 1) = n^2 - n\). 2. Solve \(n^2 - n = 240\): \(n^2 - n - 240 = 0\). Factoring gives \((n - 16)(n + 15) = 0\), so \(n = 16\) or \(n = -15\). Only \(n = 16\) is meaningful. 3. The increase is \(S(n + 1) - S(n) = [(n + 1)^2 - (n + 1)] - (n^2 - n) = 2n\). Equivalently, the new team plays two games against each of the \(n\) existing teams.

a) \(S(n) = n(n - 1) = n^2 - n\) b) \(16\) teams c) \(2n\) additional games

- Consecutive odd integers differ by \(2\). - Express the second integer in terms of the first. - Remember that the product of two negative numbers is positive.

1. Let \(x\) be the smaller odd integer. The next odd integer is \(x+2\). 2. Set up the equation: \(x(x+2)=195\). 3. Rearrange: \(x^2+2x-195=0\). 4. Factor: \((x-13)(x+15)=0\), so \(x=13\) or \(x=-15\). 5. The corresponding pairs are \((13,15)\) and \((-15,-13)\).

\((-15,-13)\) and \((13,15)\)

- Relate length, width, and area for a rectangle. - Label the sides in terms of one variable. - The question asks for perimeter, not area.

1. Let \(x\) be the width in meters. Then the length is \(x+10\). 2. Use the area: \(x(x+10)=375\). 3. Rearrange: \(x^2+10x-375=0\). 4. Factor: \((x-15)(x+25)=0\), so \(x=15\) or \(x=-25\). 5. A width must be positive, so the width is \(15\,\text{m}\) and the length is \(25\,\text{m}\). 6. The perimeter is \(2(15+25)=80\,\text{m}\).

\(80\,\text{m}\)

- Use the Pythagorean theorem to relate the three side lengths. - Express the shorter leg in terms of the longer leg. - Use the two legs as the base and height when finding the area.

1. Let \(a\) be the longer leg in centimeters. The shorter leg is \(a-5\). 2. Apply the Pythagorean theorem: \(a^2+(a-5)^2=25^2\). 3. Expand and simplify: \(2a^2-10a-600=0\), so \(a^2-5a-300=0\). 4. Factor: \((a-20)(a+15)=0\), so \(a=20\) or \(a=-15\). 5. The positive side lengths are \(20\,\text{cm}\) and \(15\,\text{cm}\). 6. The area is \(\frac12\cdot20\cdot15=150\,\text{cm}^2\).

\(150\,\text{cm}^2\)

- What input represents the nozzle location? - How do you find the vertex of a quadratic model? - What value must \(h(x)\) have when the water reaches the surface? - Which root is meaningful as a distance from the nozzle?

1. At the nozzle, \(x = 0\): \(h(0) = 1.25\). The nozzle is \(1.25\,\text{m}\) above the water surface. 2. The vertex occurs at \(x = -\frac{1}{2 \cdot (-0.25)} = 2\). Then \(h(2) = -0.25 \cdot 2^2 + 2 + 1.25 = 2.25\). The highest point is \((2, 2.25)\). 3. Set \(h(x) = 0\): \(-0.25x^2 + x + 1.25 = 0\). Multiplying by \(-4\) gives \(x^2 - 4x - 5 = 0\), or \((x - 5)(x + 1) = 0\). 4. The solutions are \(x = 5\) and \(x = -1\). Only the positive distance is meaningful, so the water hits the surface \(5\,\text{m}\) from the nozzle.

a) \(1.25\,\text{m}\) b) \((2, 2.25)\) c) \(5\,\text{m}\)

- Consecutive numbers differ by \(1\). - Expand the square of the second number carefully. - Use the square-of-a-sum identity. - Select the solution that matches the stated number set.

1. Let the smaller number be \(n\). The next number is \(n+1\). 2. Set up the equation: \(n^2+(n+1)^2=313\). 3. Expand and simplify: \(2n^2+2n-312=0\), so \(n^2+n-156=0\). 4. Apply the quadratic formula: \(n=\frac{-1\pm\sqrt{1-4\cdot1\cdot(-156)}}{2}=\frac{-1\pm25}{2}\). 5. The solutions are \(n=12\) and \(n=-13\). Only \(12\) is a natural number in this context, so the numbers are \(12\) and \(13\).

\(12\) and \(13\)

- Express both rectangle side lengths in terms of the square side. - Write an equation using the new rectangle area. - Expand the product carefully. - Compare the original square area with \(84\,\text{cm}^2\).

1. Let \(x\) be the side length of the square. The rectangle has side lengths \(x+3\) and \(x-2\). 2. Set up the area equation: \((x+3)(x-2)=84\). 3. Expand and rearrange: \(x^2+x-90=0\). 4. Apply the quadratic formula: \(x=\frac{-1\pm\sqrt{1-4\cdot1\cdot(-90)}}{2}=\frac{-1\pm19}{2}\). 5. The solutions are \(x=9\) and \(x=-10\). Only \(x=9\,\text{cm}\) is a valid length. 6. The original area was \(9^2=81\,\text{cm}^2\), so the increase was \(84-81=3\,\text{cm}^2\).

The original side length was \(9\,\text{cm}\), and the area increased by \(3\,\text{cm}^2\).

- Write each square area in terms of its side length. - Express the larger side using the smaller side. - Interpret a negative algebraic solution in the measurement context.

1. Let \(s\) be the smaller side length in centimeters. The larger side length is \(s+10\). 2. The sum of the areas gives \(s^2+(s+10)^2=2500\). 3. Simplify: \(2s^2+20s-2400=0\), or \(s^2+10s-1200=0\). 4. Factor: \((s-30)(s+40)=0\), so \(s=30\) or \(s=-40\). 5. A side length cannot be negative, so the smaller tile has side \(30\,\text{cm}\), and the larger tile has side \(40\,\text{cm}\).

a) \(s^2+(s+10)^2=2500\) b) \(30\,\text{cm}\) and \(40\,\text{cm}\) c) The value \(s=-40\) is rejected because a physical length cannot be negative.

- What is the value of \(h_0\) at launch? - Substitute the known time and height to solve for the missing coefficient. - Once the model is complete, substitute the requested time.

1. The initial height is \(h_0 = 1.5\). Substitute \(t = 2\) and \(h(2) = 31.5\): \(31.5 = -5 \cdot 2^2 + 2v_0 + 1.5\). 2. Simplify: \(31.5 = -18.5 + 2v_0\), so \(50 = 2v_0\) and \(v_0 = 25\). 3. The height function is \(h(t) = -5t^2 + 25t + 1.5\). 4. Evaluate at \(t = 4\): \(h(4) = -5 \cdot 4^2 + 25 \cdot 4 + 1.5 = 21.5\).

a) \(h(t) = -5t^2 + 25t + 1.5\) b) \(21.5\,\text{m}\)

- What equation follows from being at the same height at \(t = 0\) and \(t = 0.8\)? - How does parabola symmetry locate the vertex between two equal-height times? - Evaluate the completed model at that time.

1. Since \(h_0 = 10\) and \(h(0.8) = 10\), substitute into the model: \(10 = -5 \cdot 0.8^2 + 0.8v_0 + 10\). 2. Simplify: \(0 = -3.2 + 0.8v_0\), so \(v_0 = 4\). Thus, \(h(t) = -5t^2 + 4t + 10\). 3. The equal heights at \(t = 0\) and \(t = 0.8\) are symmetric about the vertex, so the maximum occurs at \(t = 0.4\,\text{s}\). 4. Evaluate \(h(0.4) = -5 \cdot 0.4^2 + 4 \cdot 0.4 + 10 = 10.8\). The maximum height is \(10.8\,\text{m}\).

The maximum height is \(10.8\,\text{m}\).

- First find the maximum of the fully specified model. - How does the vertex-time formula determine Ball B's initial-velocity coefficient? - Evaluate both models at their vertex times and compare.

1. Ball A reaches its vertex at \(t = -\frac{20}{2 \cdot (-5)} = 2\). Its maximum height is \(h_A(2) = -5 \cdot 2^2 + 20 \cdot 2 + 2 = 22\,\text{m}\). 2. For Ball B, the vertex time satisfies \(1.5 = -\frac{v_0}{2 \cdot (-5)} = \frac{v_0}{10}\), so \(v_0 = 15\). 3. Ball B's model is \(h_B(t) = -5t^2 + 15t + 1\). Its maximum height is \(h_B(1.5) = -5 \cdot 1.5^2 + 15 \cdot 1.5 + 1 = 12.25\,\text{m}\). 4. Since \(22\,\text{m} > 12.25\,\text{m}\), Ball A reaches the greater maximum height.

Ball A reaches the greater maximum height: \(22\,\text{m}\), compared with Ball B's \(12.25\,\text{m}\).

- What input represents the release point? - Where is the maximum of a downward-opening quadratic? - Evaluate the model at the obstacle's horizontal position. - What comparison determines whether the ball clears the obstacle?

1. At release, \(x = 0\): \(h(0) = 2\), so the release height is \(2\,\text{m}\). 2. The vertex occurs at \(x = -\frac{1}{2 \cdot (-0.1)} = 5\). Then \(h(5) = -0.1 \cdot 5^2 + 5 + 2 = 4.5\). The maximum height is \(4.5\,\text{m}\). 3. Evaluate the model at \(x = 9\): \(h(9) = -0.1 \cdot 9^2 + 9 + 2 = 2.9\). Since \(2.9\,\text{m} < 3\,\text{m}\), the ball does not clear the obstacle.

a) \(2\,\text{m}\) b) \(4.5\,\text{m}\) c) No. At \(x = 9\), the ball is \(2.9\,\text{m}\) high.

- Let an integer variable count the number of 10-cent price changes. - Revenue equals price times quantity sold. - The vertex lies between two allowable integer inputs, so compare both neighboring values.

1. Let \(x\) be the integer number of \(\$0.10\) price increases, where a negative value represents a decrease. The price is \(P(x) = 1.50 + 0.10x\), and the number sold is \(Q(x) = 200 - 10x\). 2. The revenue is \(R(x) = (1.50 + 0.10x)(200 - 10x) = -x^2 + 5x + 300\). 3. The continuous vertex occurs at \(x = -\frac{5}{2 \cdot (-1)} = 2.5\). Because \(x\) must be an integer, compare \(x = 2\) and \(x = 3\). 4. \(R(2) = 306\) and \(R(3) = 306\). The corresponding prices are \(P(2) = \$1.70\) and \(P(3) = \$1.80\). Both maximize the daily revenue.

The maximizing prices are \(\$1.70\) and \(\$1.80\). Each gives daily revenue of \(\$306\).

- Evaluate both models at the same speed. - Use the dry-pavement distance as the original amount when finding the percent increase. - For part b, first find the wet-pavement distance, then set the dry model equal to it.

1. At \(30\,\text{mph}\), \(s_d(30) = 0.04 \cdot 30^2 = 36\,\text{ft}\), and \(s_w(30) = 0.064 \cdot 30^2 = 57.6\,\text{ft}\). 2. The increase is \(57.6 - 36 = 21.6\,\text{ft}\). Relative to the dry distance, the percent increase is \(\frac{21.6}{36} \cdot 100\% = 60\%\). 3. The wet-pavement distance at \(50\,\text{mph}\) is \(s_w(50) = 0.064 \cdot 50^2 = 160\,\text{ft}\). 4. Set the dry-pavement model equal to \(160\): \(0.04v^2 = 160\). Then \(v^2 = 4000\), so \(v = \sqrt{4000} \approx 63.2\).

a) Dry: \(36\,\text{ft}\); wet: \(57.6\,\text{ft}\). The wet-pavement distance is \(60\%\) greater. b) Approximately \(63.2\,\text{mph}\)

- Evaluate each term of the model separately before adding. - Replace \(v\) with \(2v\) to compare the two parts of the model. - In part c, use only the quadratic braking-distance term.

1. Evaluate the model at \(65\): \(S(65) = 1.1 \cdot 65 + 0.04 \cdot 65^2 = 71.5 + 169 = 240.5\). The total stopping distance is \(240.5\,\text{ft}\). 2. If the speed changes from \(v\) to \(2v\), the reaction term becomes \(1.1(2v) = 2(1.1v)\), so it doubles. The braking term becomes \(0.04(2v)^2 = 4(0.04v^2)\), so it quadruples. 3. Set the braking term equal to \(144\): \(0.04v^2 = 144\). Then \(v^2 = 3600\), so \(v = 60\) because speed is positive.

a) \(240.5\,\text{ft}\) b) Doubling the speed doubles the reaction-distance term and quadruples the braking-distance term. c) \(60\,\text{mph}\)

- Where is the maximum of a downward-opening parabola? - What equation represents the moment the ball reaches the ground? - Which coefficient gives the height when \(x = 0\)?

1. The maximum occurs at the vertex. Its horizontal coordinate is \(x = -\frac{2}{2 \cdot (-0.5)} = 2\). 2. Evaluate the height: \(h(2) = -0.5 \cdot 2^2 + 2 \cdot 2 + 2.5 = 4.5\). The maximum height is \(4.5\,\text{m}\). 3. To find when the ball reaches the ground, solve \(-0.5x^2 + 2x + 2.5 = 0\). Multiplying by \(-2\) gives \(x^2 - 4x - 5 = 0\), which factors as \((x - 5)(x + 1) = 0\). The context-valid solution is \(x = 5\), so the horizontal distance is \(5\,\text{m}\). 4. Raising the initial height by \(1\,\text{m}\) increases the constant term from \(2.5\) to \(3.5\). The new function is \(h_{\text{new}}(x) = -0.5x^2 + 2x + 3.5\).

a) \(4.5\,\text{m}\) b) \(5\,\text{m}\) c) \(h_{\text{new}}(x) = -0.5x^2 + 2x + 3.5\)

- How many points determine a linear model? How many determine a quadratic model? - Substitute the data points into each model's general form. - Compare each week-\(5\) prediction with the measured height using absolute error.

1. For the linear model, \(h_{\text{lin}}(0) = 50\), so the y-intercept is \(50\). The slope through \((0, 50)\) and \((4, 170)\) is \(\frac{170 - 50}{4 - 0} = 30\). Thus, \(h_{\text{lin}}(t) = 30t + 50\). 2. Let \(h_{\text{quad}}(t) = at^2 + bt + c\). From \(h_{\text{quad}}(0) = 50\), \(c = 50\). 3. Using \((2, 90)\) gives \(4a + 2b = 40\), or \(2a + b = 20\). Using \((4, 170)\) gives \(16a + 4b = 120\), or \(4a + b = 30\). 4. Subtract the equations: \(2a = 10\), so \(a = 5\). Then \(2 \cdot 5 + b = 20\), so \(b = 10\). Therefore, \(h_{\text{quad}}(t) = 5t^2 + 10t + 50\). 5. At week \(5\), the linear model predicts \(h_{\text{lin}}(5) = 200\,\text{cm}\), with an absolute error of \(22.5\,\text{cm}\). The quadratic model predicts \(h_{\text{quad}}(5) = 225\,\text{cm}\), with an absolute error of \(2.5\,\text{cm}\). The quadratic model gives the better prediction.

a) \(h_{\text{lin}}(t) = 30t + 50\) \(h_{\text{quad}}(t) = 5t^2 + 10t + 50\) b) The quadratic model is better. Its prediction is \(225\,\text{cm}\), which is \(2.5\,\text{cm}\) from the measured value; the linear prediction is \(200\,\text{cm}\), which is \(22.5\,\text{cm}\) away.

- Use the point \((10, 80)\) to find the slope of the linear model. - Substitute the two nonzero data points into \(g(x) = ax^2 + bx\). - Compare whether each model has a constant or increasing rate of change.

1. For the linear model, substitute \((10, 80)\): \(80 = 10m\), so \(m = 8\). Therefore, \(f(x) = 8x\). 2. For the quadratic model, use \((5, 25)\): \(25a + 5b = 25\), so \(5a + b = 5\). 3. Use \((10, 80)\): \(100a + 10b = 80\), so \(10a + b = 8\). Subtracting the equations gives \(5a = 3\), so \(a = 0.6\). Then \(5 \cdot 0.6 + b = 5\), so \(b = 2\). Thus, \(g(x) = 0.6x^2 + 2x\). 4. In month \(12\), \(f(12) = 8 \cdot 12 = 96\), so the linear model predicts \(96{,}000\) users. Also, \(g(12) = 0.6 \cdot 12^2 + 2 \cdot 12 = 110.4\), so the quadratic model predicts \(110{,}400\) users. 5. A linear model has a constant growth rate, while this quadratic model has an increasing growth rate for \(x \geq 0\). Under the stated assumption, the quadratic model is more consistent.

a) \(f(x) = 8x\) b) \(g(x) = 0.6x^2 + 2x\) c) Linear model: \(96{,}000\) users Quadratic model: \(110{,}400\) users The quadratic model is more consistent with an increasing growth rate.

- What profit value represents breaking even? - Which feature of a downward-opening parabola gives the maximum profit? - Interpret the input and output when \(x = 0\).

1. At a break-even price, \(P(x) = 0\). Solve \(-2x^2 + 60x - 400 = 0\). Dividing by \(-2\) gives \(x^2 - 30x + 200 = 0\), which factors as \((x - 10)(x - 20) = 0\). The break-even prices are \(\$10\) and \(\$20\). 2. The maximum occurs at the vertex because the parabola opens downward. Its horizontal coordinate is \(x = -\frac{60}{2 \cdot (-2)} = 15\). Then \(P(15) = -2 \cdot 15^2 + 60 \cdot 15 - 400 = 50\). The profit is maximized at a price of \(\$15\), and the maximum modeled monthly profit is \(\$50\). 3. The value \(P(0) = -400\) means that the model predicts a \(\$400\) monthly loss if the shirts are priced at \(\$0\). The model alone does not show that the entire amount represents fixed costs.

a) \(\$10\) and \(\$20\) b) A price of \(\$15\) gives a maximum monthly profit of \(\$50\). c) The model predicts a monthly loss of \(\$400\) when the price is \(\$0\); the model does not establish that this amount consists entirely of fixed costs.

- Use the given sum to express one leg in terms of the other. - Apply the Pythagorean theorem. - Expand the square of a difference carefully. - Factor the resulting quadratic equation.

1. Let one leg be \(a\). Then the other leg is \(31-a\). 2. Apply the Pythagorean theorem: \(a^2+(31-a)^2=25^2\). 3. Expand and simplify: \(2a^2-62a+336=0\), so \(a^2-31a+168=0\). 4. Factor: \((a-7)(a-24)=0\), so \(a=7\) or \(a=24\). 5. The corresponding other leg is \(24\) or \(7\). Therefore, the legs are \(7\,\text{cm}\) and \(24\,\text{cm}\).

\(7\,\text{cm}\) and \(24\,\text{cm}\)

- Use the legs as the base and height in the triangle area formula. - Express the longer leg in terms of the shorter leg. - After finding the legs, apply the Pythagorean theorem.

1. Let the shorter leg be \(x\). The longer leg is \(x+7\). 2. Use the area formula: \(30=\frac12x(x+7)\). 3. Rearrange: \(x^2+7x-60=0\). 4. Factor: \((x-5)(x+12)=0\), so \(x=5\) or \(x=-12\). The valid leg lengths are \(5\,\text{cm}\) and \(12\,\text{cm}\). 5. Apply the Pythagorean theorem: \(c^2=5^2+12^2=169\), so \(c=13\,\text{cm}\).

a) \(5\,\text{cm}\) and \(12\,\text{cm}\) b) \(13\,\text{cm}\)

- What value of \(y\) represents ground level? - For part b, what height corresponds to the vehicle's top corners? - Because the vehicle is centered, its width spans equal distances on both sides of the y-axis.

1. At ground level, \(y = 0\). Solve \(0 = -0.2x^2 + 5\): \(0.2x^2 = 5\), so \(x^2 = 25\) and \(x = \pm 5\). The tunnel is \(5 - (-5) = 10\,\text{m}\) wide at the ground. 2. At the top corners of a \(4\,\text{m}\)-high vehicle, set \(y = 4\): \(4 = -0.2x^2 + 5\). Then \(0.2x^2 = 1\), so \(x^2 = 5\) and \(x = \pm\sqrt{5}\). 3. The greatest possible width is the distance between these two horizontal positions: \(\sqrt{5} - (-\sqrt{5}) = 2\sqrt{5} \approx 4.47\,\text{m}\).

a) \(10\,\text{m}\) b) \(2\sqrt{5}\,\text{m} \approx 4.47\,\text{m}\)

- Split the total distance equally on both sides of the y-axis. - Substitute a station's horizontal coordinate into the model. - At the water elevation, find both symmetric horizontal coordinates and calculate the full width.

1. Because the stations are at equal heights on opposite sides of the y-axis and are \(120\,\text{m}\) apart, their horizontal coordinates are \(x = -60\) and \(x = 60\). 2. Evaluate the model: \(h = 0.005 \cdot 60^2 = 18\). The stations are \(18\,\text{m}\) above the lowest point. 3. The water elevation is \(\frac{1}{2} \cdot 18 = 9\,\text{m}\). Set \(9 = 0.005x^2\), so \(x^2 = 1800\) and \(x = \pm\sqrt{1800} = \pm 30\sqrt{2}\). 4. The water surface extends from \(-30\sqrt{2}\) to \(30\sqrt{2}\), so its width is \(60\sqrt{2}\,\text{m} \approx 84.85\,\text{m}\).

a) \(18\,\text{m}\) b) \(60\sqrt{2}\,\text{m} \approx 84.85\,\text{m}\)

- Add the frame width on both sides of each photo dimension. - Write an equation for the outer rectangular area. - Simplify the quadratic equation before solving.

1. The outer dimensions are \(20+2x\) and \(30+2x\). 2. Use the total area: \((20+2x)(30+2x)=1064\). 3. Expand and rearrange: \(4x^2+100x-464=0\). 4. Divide by \(4\): \(x^2+25x-116=0\). 5. Apply the quadratic formula: \(x=\frac{-25\pm\sqrt{25^2-4\cdot1\cdot(-116)}}{2}=\frac{-25\pm33}{2}\). 6. The solutions are \(x=4\) and \(x=-29\). A frame width must be positive, so \(x=4\,\text{cm}\).

\(4\,\text{cm}\)

- Use both the perimeter and area formulas for a rectangle. - Express one side length in terms of the other. - Substitute into the area equation and factor.

1. Let the side lengths be \(a\) and \(b\). The perimeter gives \(2a+2b=60\), so \(a+b=30\) and \(b=30-a\). 2. Use the area: \(a(30-a)=221\). 3. Rearrange: \(a^2-30a+221=0\). 4. Factor: \((a-13)(a-17)=0\), so \(a=13\) or \(a=17\). 5. The two side lengths are \(13\,\text{m}\) and \(17\,\text{m}\).

\(13\,\text{m}\) by \(17\,\text{m}\)

- Subtract the border width from both sides of each dimension. - Explain why each center dimension decreases by \(2x\). - Check which algebraic solution fits the patio dimensions.

1. The center rectangle has dimensions \(15-2x\) and \(10-2x\). 2. Use its area: \((15-2x)(10-2x)=84\). 3. Expand and rearrange: \(4x^2-50x+66=0\), so \(2x^2-25x+33=0\). 4. Apply the quadratic formula: \(x=\frac{25\pm\sqrt{(-25)^2-4\cdot2\cdot33}}{2\cdot2}=\frac{25\pm19}{4}\). 5. The candidates are \(x=11\) and \(x=1.5\). Because the center width \(10-2x\) must be positive, \(0<x<5\). Therefore, \(x=1.5\,\text{ft}\).

\(1.5\,\text{ft}\)

- Use the two ground-level points as the zeros of the quadratic. - Substitute the additional point to determine the leading coefficient. - Where is the vertex relative to two symmetric zeros?

1. The arch has zeros at \(x = 0\) and \(x = 40\), so write \(h(x) = ax(x - 40)\). 2. Use the point \((10, 15)\): \(15 = a \cdot 10 \cdot (10 - 40) = -300a\). Thus, \(a = -0.05\). 3. The model is \(h(x) = -0.05x(x - 40) = -0.05x^2 + 2x\). 4. The vertex lies midway between the zeros at \(x = 20\). Evaluate \(h(20) = -0.05 \cdot 20^2 + 2 \cdot 20 = 20\). The maximum height is \(20\,\text{m}\).

a) \(h(x) = -0.05x^2 + 2x\) b) \(20\,\text{m}\)

- Use the perimeter to relate the two side lengths. - Express one side in terms of the other and write the area as a quadratic. - The vertex of the downward-opening area function gives the maximum.

1. The \(8\)-by-\(12\) bed has area \(8 \cdot 12 = 96\,\text{ft}^2\). 2. A square with perimeter \(40\,\text{ft}\) has side length \(40 \div 4 = 10\,\text{ft}\), so its area is \(10^2 = 100\,\text{ft}^2\). This is greater than \(96\,\text{ft}^2\). 3. Let the rectangle's side lengths be \(x\) and \(y\). From \(2x + 2y = 40\), \(y = 20 - x\). 4. The area is \(A(x) = x(20 - x) = -x^2 + 20x\). Its vertex occurs at \(x = -\frac{20}{2 \cdot (-1)} = 10\). 5. Then \(y = 10\), so the maximum-area rectangle is a \(10\,\text{ft}\) by \(10\,\text{ft}\) square with area \(100\,\text{ft}^2\).

a) \(96\,\text{ft}^2\) b) Yes. The square has area \(100\,\text{ft}^2\). c) \(10\,\text{ft}\) by \(10\,\text{ft}\), with maximum area \(100\,\text{ft}^2\)

- At an intersection, set the two function values equal. - Factor the resulting quadratic equation. - Use the vertex formula to find each maximum height.

1. At an intersection, the heights are equal: \(-0.5x^2 + 2x + 1 = -x^2 + 3x + 1\). 2. Simplify: \(0.5x^2 - x = 0\). Factoring gives \(x(0.5x - 1) = 0\), so \(x = 0\) or \(x = 2\). The nonstarting intersection has \(x = 2\). 3. Evaluate either function: \(f(2) = -0.5 \cdot 2^2 + 2 \cdot 2 + 1 = 3\). The second intersection is \((2, 3)\). 4. For \(f\), the vertex occurs at \(x = -\frac{2}{2 \cdot (-0.5)} = 2\), and \(f(2) = 3\). Its vertex is \((2, 3)\), so its maximum height is \(3\,\text{m}\). 5. For \(g\), the vertex occurs at \(x = -\frac{3}{2 \cdot (-1)} = 1.5\), and \(g(1.5) = 3.25\). Its vertex is \((1.5, 3.25)\), so its maximum height is \(3.25\,\text{m}\). Therefore, the stream modeled by \(g\) reaches the greater maximum height.

a) \((2, 3)\) b) The stream modeled by \(g\). Its vertex is \((1.5, 3.25)\), compared with \((2, 3)\) for \(f\), so its maximum height is \(3.25\,\text{m}\) rather than \(3\,\text{m}\).

- Evaluate the function at \(x = 0\) to interpret the constant term. - Use the vertex formula to find the maximum. - Setting the function equal to \(10\) gives two distances; which one is on the descending side of the vertex?

1. Since \(h(0) = 6\), the ball is released from a height of \(6\,\text{ft}\). 2. The vertex occurs at \(x = -\frac{\frac{16}{15}}{2 \cdot \left(-\frac{4}{75}\right)} = 10\). Evaluate the height: \(h(10) = -\frac{4}{75} \cdot 10^2 + \frac{16}{15} \cdot 10 + 6 = \frac{34}{3}\). The maximum height is \(\frac{34}{3}\,\text{ft} \approx 11.33\,\text{ft}\). 3. Set the height equal to the hoop height: \(-\frac{4}{75}x^2 + \frac{16}{15}x + 6 = 10\). 4. Multiplying by \(75\) and simplifying gives \(x^2 - 20x + 75 = 0\), which factors as \((x - 5)(x - 15) = 0\). The ball is at hoop height at \(x = 5\) on the way up and at \(x = 15\) on the way down. 5. Therefore, the hoop must be \(15\,\text{ft}\) horizontally from the release point.

a) The ball is released from a height of \(6\,\text{ft}\). b) \(\frac{34}{3}\,\text{ft} \approx 11.33\,\text{ft}\) c) \(15\,\text{ft}\)

- Isolate \(t^2\) before taking the physically meaningful positive square root. - Keep the two calculations unrounded until the final requested hundredths. - For the factor comparison, compare the heights inside the square root rather than relying only on rounded times.

1. Solve the model for positive time: \(t=\sqrt{\frac{2h}{g}}\). 2. For \(h_1=45\,\text{m}\), \(t_1=\sqrt{\frac{2\cdot45}{9.81}}\,\text{s}\approx3.03\,\text{s}\). 3. For \(h_2=180\,\text{m}\), \(t_2=\sqrt{\frac{2\cdot180}{9.81}}\,\text{s}\approx6.06\,\text{s}\). 4. Exactly, \(\frac{t_2}{t_1}=\sqrt{\frac{180}{45}}=\sqrt{4}=2\). Multiplying the height by \(4\) doubles the fall time.

a) \(t_1\approx3.03\,\text{s}\); \(t_2\approx6.06\,\text{s}\) b) The fall time is multiplied by \(2\).

- Express the planned duration using total acreage and daily rate. - Determine the number of days spent at the increased rate. - The acreage remains fixed even though the completion time changes. - Select the positive solution for a physical rate.

1. Let \(x\) be the planned rate in acres per day and \(t\) the planned duration. Then \(xt=220\), so \(t=\frac{220}{x}\). 2. The actual duration is \(t-3\) days. After the first two days, the higher rate is used for \(t-5\) days. 3. Model the total acreage: \(2x+(t-5)(x+10)=220\). 4. Substitute \(t=\frac{220}{x}\): \(2x+\left(\frac{220}{x}-5\right)(x+10)=220\). 5. Simplify: \(-3x+\frac{2200}{x}-50=0\). Multiplying by \(x\) gives \(3x^2+50x-2200=0\). 6. Factor: \((3x+110)(x-20)=0\). The candidates are \(x=20\) and \(x=-\frac{110}{3}\). 7. The positive rate is \(20\) acres per day.

\(20\) acres per day

- Express the planned rate as total brochures divided by planned time. - Increase that rate by \(10\). - Use the reduced time in the actual production equation. - Report the actual time, not the planned time.

1. Let \(x\) be the originally planned time in hours. The planned rate is \(\frac{600}{x}\) brochures per hour. 2. The new rate is \(\frac{600}{x}+10\), and the actual time is \(x-3\). 3. Use rate times time equals total production: \((x-3)\left(\frac{600}{x}+10\right)=600\). 4. Simplifying gives \(x^2-3x-180=0\). 5. Factor: \((x-15)(x+12)=0\), so \(x=15\) or \(x=-12\). 6. The planned time was \(15\) hours, so the actual time was \(15-3=12\) hours.

\(12\) hours

- Use \(\text{time}=\frac{\text{distance}}{\text{speed}}\). - Convert the time savings to hours. - Write a time expression for each trip. - Reject a negative speed.

1. Let \(v\) be the outbound speed in miles per hour. The return speed is \(v+5\). 2. The outbound time is \(\frac{30}{v}\) hours, and the return time is \(\frac{30}{v+5}\) hours. 3. Since \(30\) minutes is \(\frac12\) hour, write \(\frac{30}{v}-\frac{30}{v+5}=\frac12\). 4. Multiply by \(2v(v+5)\): \(60(v+5)-60v=v(v+5)\). 5. Simplify: \(v^2+5v-300=0\). 6. Factor: \((v-15)(v+20)=0\), so \(v=15\) or \(v=-20\). 7. Speed must be positive, so the outbound speed was \(15\,\text{mph}\).

\(15\,\text{mph}\)

- Add the wind speed for the tailwind flight and subtract it for the headwind flight. - Use distance divided by ground speed for each flight time. - Express “one hour longer” as a difference of the two times. - Check the solution against the domain.

1. Let \(v\) be the airplane's airspeed in still air, in miles per hour. Its ground speeds are \(v+50\) with the tailwind and \(v-50\) with the headwind. 2. The return flight takes one hour longer, so \(\frac{600}{v-50}-\frac{600}{v+50}=1\). The domain requires \(v>50\). 3. Multiply by \((v-50)(v+50)\): \(600(v+50)-600(v-50)=v^2-2500\). 4. Simplify: \(60{,}000=v^2-2500\), so \(v^2=62{,}500\). 5. Therefore, \(v=\pm250\). Only \(v=250\) satisfies the physical context and the domain.

\(250\,\text{mph}\)

- Use \(\text{time}=\frac{\text{distance}}{\text{speed}}\) for each traveler. - Determine how the two travel times differ. - The car's speed is \(15\,\text{mph}\) greater than the bicycle's speed. - Reject a negative speed.

1. Let \(v\) be the bicycle's average speed in miles per hour. The car's average speed is \(v+15\). 2. The bicycle's travel time is \(\frac{30}{v}\) hours, and the car's travel time is \(\frac{30}{v+15}\) hours. 3. Because the car leaves one hour later but arrives at the same time, \(\frac{30}{v}-\frac{30}{v+15}=1\). 4. Multiply by \(v(v+15)\): \(30(v+15)-30v=v(v+15)\). 5. Simplify: \(v^2+15v-450=0\). 6. Factor: \((v-15)(v+30)=0\), so \(v=15\) or \(v=-30\). 7. Speed must be positive, so the bicycle's average speed was \(15\,\text{mph}\).

\(15\,\text{mph}\)

- Express each person's hourly work rate as the reciprocal of the time needed alone. - Add the two individual rates. - The combined rate is one job divided by \(2\) hours. - Check each algebraic solution against the context.

1. Let \(x\) be the number of hours Marcus would need alone. Elena would need \(x+3\) hours. 2. Marcus completes \(\frac{1}{x}\) of the lawn per hour, and Elena completes \(\frac{1}{x+3}\) per hour. Together, they complete \(\frac12\) of the lawn per hour. 3. Write the work-rate equation: \(\frac{1}{x}+\frac{1}{x+3}=\frac12\). 4. Multiply by \(2x(x+3)\): \(2(x+3)+2x=x(x+3)\). 5. Simplify: \(x^2-x-6=0\). 6. Factor: \((x-3)(x+2)=0\), so \(x=3\) or \(x=-2\). 7. A work time cannot be negative, so \(x=-2\) is not meaningful. Marcus would take \(3\) hours alone.

Marcus would take \(3\) hours alone. The solution \(x=-2\) is not meaningful because a work time cannot be negative.

- Add the current speed when traveling downstream and subtract it when traveling upstream. - Write each travel time as distance divided by speed. - Express the three-hour difference with an equation. - Check the solution against the physical domain.

1. Let \(c\) be the current's speed in miles per hour. The downstream speed is \(12+c\), and the upstream speed is \(12-c\). The domain requires \(0<c<12\). 2. The upstream trip takes \(3\) hours longer, so \(\frac{48}{12-c}-\frac{48}{12+c}=3\). 3. Multiply by \((12-c)(12+c)\): \(48(12+c)-48(12-c)=3(144-c^2)\). 4. Simplify: \(96c=432-3c^2\), so \(c^2+32c-144=0\). 5. Factor: \((c-4)(c+36)=0\), so \(c=4\) or \(c=-36\). 6. Only \(c=4\) satisfies the domain and represents a current speed.

\(4\,\text{mph}\)

- Express the new price and the number of tickets sold using the same variable. - Revenue equals ticket price times tickets sold. - Factor the resulting quadratic equation. - Verify both possible price increases in the original revenue model.

1. Let \(x\) be the price increase in dollars. The new ticket price is \(40+x\), and the number of tickets sold is \(500-10x\). 2. Set revenue equal to the target: \((40+x)(500-10x)=20{,}210\). 3. Expand: \(20{,}000+100x-10x^2=20{,}210\). 4. Rearrange and divide by \(-10\): \(x^2-10x+21=0\). 5. Factor: \((x-3)(x-7)=0\), so \(x=3\) or \(x=7\). 6. Both values are meaningful: \(43\cdot470=20{,}210\) and \(47\cdot430=20{,}210\).

The ticket price could be increased by \(\$3\) or by \(\$7\).

- Write a growth factor for each year. - Remember that the second year begins with the first year's ending balance. - A one-percentage-point increase changes \(p\%\) to \((p+1)\%\). - Check whether each algebraic solution is meaningful as an interest rate.

1. The two annual growth factors are \(1+\frac{p}{100}\) and \(1+\frac{p+1}{100}\). 2. Model the final balance: \(2000\left(1+\frac{p}{100}\right)\left(1+\frac{p+1}{100}\right)=2226\). 3. Rewrite the factors and simplify: \(\frac{(p+100)(p+101)}{5}=2226\), so \((p+100)(p+101)=11{,}130\). 4. Expand and rearrange: \(p^2+201p-1030=0\). 5. Factor: \((p-5)(p+206)=0\), so \(p=5\) or \(p=-206\). 6. The value \(p=-206\) would represent a loss greater than \(100\%\) in the first year, which is not meaningful for this investment model. Therefore, \(p=5\).

\(5\%\)

- Use the formula \(\frac{n(n-3)}{2}\) for the number of diagonals. - Translate “\(12\) more than the number of vertices” into an equation. - Factor the resulting quadratic. - Check which solution can represent a polygon.

1. Let \(n\) be the number of vertices. The number of diagonals is \(\frac{n(n-3)}{2}\). 2. Translate the condition into an equation: \(\frac{n(n-3)}{2}=n+12\). 3. Multiply by \(2\) and rearrange: \(n^2-5n-24=0\). 4. Factor: \((n-8)(n+3)=0\), so \(n=8\) or \(n=-3\). 5. A polygon must have a positive whole number of vertices, so \(n=8\).

The polygon has \(8\) vertices.

- Represent the two vertex counts with \(n\) and \(n+1\). - Apply the diagonal formula to each polygon. - Add the two expressions and set the sum equal to \(14\). - Check which solution gives valid polygons.

1. Let \(n\) be the number of vertices in the smaller polygon. The larger polygon has \(n+1\) vertices. 2. Use the diagonal formula for both polygons: \(\frac{n(n-3)}{2}+\frac{(n+1)(n-2)}{2}=14\). 3. Combine and simplify: \(2n^2-4n-2=28\). 4. Divide by \(2\) and rearrange: \(n^2-2n-15=0\). 5. Factor: \((n-5)(n+3)=0\), so \(n=5\) or \(n=-3\). 6. The smaller polygon must have at least \(3\) vertices, so \(n=5\). The polygons have \(5\) and \(6\) vertices.

The polygons have \(5\) vertices and \(6\) vertices.

- Use the formula \(\frac{n(n-3)}{2}\) for the number of diagonals. - Translate “\(4\) more than twice the number of vertices” into an equation. - Factor the quadratic equation. - Interpret the valid vertex count as a polygon name.

1. Let \(n\) be the number of vertices. The number of diagonals is \(\frac{n(n-3)}{2}\). 2. Translate the condition into an equation: \(\frac{n(n-3)}{2}=2n+4\). 3. Multiply by \(2\) and rearrange: \(n^2-7n-8=0\). 4. Factor: \((n-8)(n+1)=0\), so \(n=8\) or \(n=-1\). 5. A polygon must have a positive whole number of vertices, so \(n=8\). A polygon with \(8\) sides is an octagon.

The polygon is an octagon.

- Write the diagonal formula for a polygon with \(n\) vertices. - Write it again for a polygon with \(2n\) vertices. - Set the difference between the two expressions equal to \(18\). - Check which solution is a valid vertex count.

1. Let \(n\) be the original number of vertices. The original number of diagonals is \(D_1=\frac{n(n-3)}{2}\). 2. After the number of vertices is doubled, the new number of diagonals is \(D_2=\frac{2n(2n-3)}{2}=n(2n-3)\). 3. The increase is \(18\), so \(n(2n-3)-\frac{n(n-3)}{2}=18\). 4. Multiply by \(2\) and simplify: \(4n^2-6n-(n^2-3n)=36\), so \(n^2-n-12=0\). 5. Factor: \((n-4)(n+3)=0\), so \(n=4\) or \(n=-3\). 6. The valid vertex count is \(n=4\).

The original polygon had \(4\) vertices; it was a quadrilateral.

- Find the target lawn area from the given percentage. - Subtract \(2x\) from each outside dimension. - Set the area of the inner rectangle equal to the target lawn area. - Check each solution against half of the park's smaller dimension.

1. The original area is \(120\cdot80=9600\,\text{ft}^2\). 2. The remaining lawn must have area \(0.625\cdot9600=6000\,\text{ft}^2\). 3. The lawn dimensions are \(120-2x\) and \(80-2x\), so \((120-2x)(80-2x)=6000\). 4. Expand and simplify: \(4x^2-400x+3600=0\), so \(x^2-100x+900=0\). 5. Factor: \((x-10)(x-90)=0\), so \(x=10\) or \(x=90\). 6. The path width must satisfy \(0<x<40\), so \(x=90\) is impossible. Therefore, \(x=10\,\text{ft}\).

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

- Determine the base dimensions after the corner squares are removed. - The cutout side length becomes the box height. - Use the volume formula for a rectangular prism. - Check which algebraic solution can represent the original sheet.

1. Let \(a\) be the original side length in inches. 2. Cutting \(2\)-inch squares from both ends of each side leaves a square base with side length \(a-4\). The box height is \(2\,\text{in}\). 3. Use the volume: \(2(a-4)^2=72\). 4. Divide by \(2\): \((a-4)^2=36\). 5. Therefore, \(a-4=6\) or \(a-4=-6\), giving \(a=10\) or \(a=-2\). 6. A side length must be positive and greater than \(4\), so \(a=10\,\text{in}\).

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

- Express the length in terms of the width. - Add the walkway width twice to each bed dimension. - Use the area of the entire outside rectangle. - Reject a negative dimension.

1. Let \(x\) be the flower bed's width in feet. Its length is \(x+5\). 2. Because the walkway is \(2\,\text{ft}\) wide on every side, the outside dimensions are \(x+4\) and \(x+9\). 3. Use the total area: \((x+4)(x+9)=150\). 4. Expand and rearrange: \(x^2+13x-114=0\). 5. Factor: \((x-6)(x+19)=0\), so \(x=6\) or \(x=-19\). 6. A width must be positive, so the flower bed is \(6\,\text{ft}\) wide and \(11\,\text{ft}\) long.

The flower bed is \(6\,\text{ft}\) wide and \(11\,\text{ft}\) long.

- Determine the distance remaining after the delay. - Convert the delay to hours. - Compare the planned and actual travel times for the remaining distance. - Reject a negative speed.

1. Let \(v\) be the original speed in miles per hour. The increased speed is \(v+15\). 2. The driver must save \(15\) minutes, or \(\frac14\) hour, over the remaining \(45\) miles. 3. Write the time-savings equation: \(\frac{45}{v}-\frac{45}{v+15}=\frac14\). 4. Multiply by \(4v(v+15)\): \(180(v+15)-180v=v(v+15)\). 5. Simplify: \(v^2+15v-2700=0\). 6. Factor: \((v-45)(v+60)=0\), so \(v=45\) or \(v=-60\). 7. Speed must be positive, so the original speed was \(45\,\text{mph}\).

\(45\,\text{mph}\)

- Subtract the current speed when traveling upstream. - Write the lake and upstream travel times. - Use the one-hour difference to form an equation. - For the downstream trip, add the current speed.

1. For part a), let \(v\) be the boat's speed in still water, in miles per hour. The upstream ground speed is \(v-2\), so the domain requires \(v>2\). 2. The upstream trip takes one hour longer: \(\frac{40}{v-2}=\frac{40}{v}+1\). 3. Multiply by \(v(v-2)\): \(40v=40(v-2)+v(v-2)\). 4. Simplify: \(v^2-2v-80=0\). 5. Factor: \((v-10)(v+8)=0\), so \(v=10\) or \(v=-8\). 6. Only \(v=10\) satisfies the domain, so the still-water speed is \(10\,\text{mph}\). 7. For part b), the downstream speed is \(10+2=12\,\text{mph}\). 8. The downstream time is \(\frac{40}{12}=\frac{10}{3}\) hours, or \(3\) hours \(20\) minutes.

a) \(10\,\text{mph}\) b) \(3\) hours \(20\) minutes

- Use the perimeter to find the sum of the two side lengths. - Express one side length in terms of the other. - Use the area to form a quadratic equation. - Factor the resulting quadratic equation.

1. Let one side length be \(x\) inches. The perimeter gives a total side-length sum of \(13\), so the other side is \(13-x\). 2. Use the area: \(x(13-x)=40\). 3. Rearrange: \(x^2-13x+40=0\). 4. Factor: \((x-5)(x-8)=0\), so \(x=5\) or \(x=8\). 5. The corresponding other side is \(8\) or \(5\), so the rectangle's side lengths are \(5\,\text{in}\) and \(8\,\text{in}\).

\(5\,\text{in}\) and \(8\,\text{in}\)

- Use the identity for \((u+v)^2\) to find the product. - Express one number in terms of the other using their sum. - Substitute into the sum-of-squares condition. - Factor the resulting quadratic equation.

1. For part a), use \((u+v)^2=u^2+2uv+v^2\). 2. Substitute the given values: \((-2)^2=52+2uv\), so \(4=52+2uv\) and \(uv=-24\). 3. For part b), the sum condition gives \(v=-2-u\). 4. Substitute into the square-sum condition: \(u^2+(-2-u)^2=52\). 5. Expand and simplify: \(2u^2+4u-48=0\), so \(u^2+2u-24=0\). 6. Factor: \((u-4)(u+6)=0\), so \(u=4\) or \(u=-6\). 7. The corresponding values of \(v\) are \(-6\) and \(4\), so the two numbers are \(4\) and \(-6\).

a) \(uv=-24\) b) The two numbers are \(4\) and \(-6\).

- Represent the total number of bulbs as rows times bulbs per row. - Write one equation for each arrangement. - Substitute to eliminate the number of bulbs per row. - Reject a negative row count.

1. Let \(r\) be the original number of rows and \(b\) the number of bulbs per row. Then \(rb=400\). 2. The revised array satisfies \((r-4)(b+5)=360\). 3. Since \(b=\frac{400}{r}\), substitute: \((r-4)\left(\frac{400}{r}+5\right)=360\). 4. Multiply by \(r\) and simplify: \(r^2+4r-320=0\). 5. Factor: \((r-16)(r+20)=0\), so \(r=16\) or \(r=-20\). 6. A row count must be positive, so the original plan used \(16\) rows.

\(16\) rows

- Express each pumping rate as volume divided by time. - Relate the two individual times. - Set the difference between the rates equal to \(5\,\text{m}^3/\text{h}\). - Reject a negative time.

1. Let \(t\) be the number of hours the faster pump needs. The slower pump needs \(t+2\) hours. 2. Their pumping rates are \(\frac{120}{t}\) and \(\frac{120}{t+2}\) cubic meters per hour. 3. The rate difference is \(5\): \(\frac{120}{t}-\frac{120}{t+2}=5\). 4. Divide by \(5\) and clear the denominators: \(24(t+2)-24t=t(t+2)\). 5. Simplify: \(t^2+2t-48=0\). 6. Factor: \((t-6)(t+8)=0\), so \(t=6\) or \(t=-8\). 7. Time must be positive, so the faster pump takes \(6\) hours and the slower pump takes \(8\) hours.

The slower pump takes \(8\) hours, and the faster pump takes \(6\) hours.

- Express the original price as total budget divided by number of tablets. - Reduce that price by \(\$50\) and increase the quantity by \(8\). - Set the new total cost equal to the original budget. - Reject a negative quantity.

1. Let \(x\) be the original number of tablets. Measured in dollars, the original price per tablet is \(\frac{12{,}000}{x}\). 2. After the discount, the price in dollars is \(\frac{12{,}000}{x}-50\), and the school buys \(x+8\) tablets. 3. Use the unchanged total cost: \(\left(\frac{12{,}000}{x}-50\right)(x+8)=12{,}000\). 4. Clear the denominator and simplify: \(x^2+8x-1920=0\). 5. Factor: \((x-40)(x+48)=0\), so \(x=40\) or \(x=-48\). 6. The number of tablets must be positive, so \(x=40\). 7. The original price was \(\frac{12{,}000}{40}=300\), so it was \(\$300\) per tablet.

The school originally planned to buy \(40\) tablets at \(\$300\) each.

- Express average time per mile as total time divided by distance. - Convert the seconds to hours. - Subtract the two time-per-mile expressions. - Reject a negative distance.

1. Driver A's average time per mile is \(\frac{2}{x}\) hours, and driver B's is \(\frac{2}{x+10}\) hours. 2. Convert \(36\) seconds to hours: \(\frac{36}{3600}=\frac{1}{100}\) hour. 3. Write the time-per-mile difference: \(\frac{2}{x}-\frac{2}{x+10}=\frac{1}{100}\). 4. Multiply by \(100x(x+10)\): \(200(x+10)-200x=x(x+10)\). 5. Simplify: \(x^2+10x-2000=0\). 6. Factor: \((x-40)(x+50)=0\), so \(x=40\) or \(x=-50\). 7. Distance must be positive, so driver A's route is \(40\) miles.

\(40\) miles

- Express the length in terms of the width. - Add the path width on both sides of each dimension. - Use the area of the outside rectangle. - Reject a negative dimension.

1. Let \(x\) be the garden's width in feet. Its length is \(2x\). 2. The path adds \(4\) feet to each overall dimension, so the outside dimensions are \(x+4\) and \(2x+4\). 3. Use the total area: \((x+4)(2x+4)=160\). 4. Expand and simplify: \(x^2+6x-72=0\). 5. Factor: \((x-6)(x+12)=0\), so \(x=6\) or \(x=-12\). 6. Width must be positive, so the garden is \(6\,\text{ft}\) wide and \(12\,\text{ft}\) long.

The garden is \(6\,\text{ft}\) wide and \(12\,\text{ft}\) long.

- Express each person's daily work rate as the reciprocal of the individual completion time. - Multiply each rate by the number of days that person works. - The two completed portions add to one whole project. - Reject a negative time.

1. Let \(x\) be the number of days the designer would need alone. The assistant would need \(x+10\) days. 2. The designer completes \(\frac{4}{x}\) of the project, and the assistant completes \(\frac{9}{x+10}\). 3. Their completed fractions sum to one: \(\frac{4}{x}+\frac{9}{x+10}=1\). 4. Multiply by \(x(x+10)\): \(4(x+10)+9x=x(x+10)\). 5. Simplify: \(x^2-3x-40=0\). 6. Factor: \((x-8)(x+5)=0\), so \(x=8\) or \(x=-5\). 7. Time must be positive, so the designer needs \(8\) days and the assistant needs \(18\) days.

The designer needs \(8\) days, and the assistant needs \(18\) days.

- Add the two hourly work rates. - Express the apprentice's time using the landscaper's time. - For part b), multiply the landscaper's hourly rate by \(6\) hours.

1. Let \(g\) be the number of hours the landscaper would need alone. The apprentice would need \(g+9\) hours. 2. Add their work rates: \(\frac{1}{g}+\frac{1}{g+9}=\frac16\). 3. Multiply by \(6g(g+9)\): \(6(g+9)+6g=g(g+9)\). 4. Simplify: \(g^2-3g-54=0\). 5. Factor: \((g-9)(g+6)=0\), so \(g=9\) or \(g=-6\). 6. The positive solution gives \(9\) hours for the landscaper and \(18\) hours for the apprentice. 7. During \(6\) hours, the landscaper completes \(6\left(\frac19\right)=\frac23\) of the job.

a) The landscaper needs \(9\) hours, and the apprentice needs \(18\) hours. b) The landscaper completes \(\frac23\) of the job.

- Express each work rate as the reciprocal of its completion time. - Write each individual time in terms of the shared time. - Set the shared rate equal to the sum of the individual rates. - Use the positive square root for a time.

1. Let \(x\) be the number of hours the two mowers need together. Their individual times are \(x+16\) and \(x+9\). 2. The combined rate equals the sum of the individual rates: \(\frac{1}{x}=\frac{1}{x+16}+\frac{1}{x+9}\). 3. Multiply by \(x(x+16)(x+9)\): \((x+16)(x+9)=x(x+9)+x(x+16)\). 4. Expand and simplify: \(x^2+25x+144=2x^2+25x\), so \(x^2=144\). 5. Since time is positive, \(x=12\) hours. 6. The individual times are \(12+16=28\) hours and \(12+9=21\) hours.

Together, the mowers take \(12\) hours. Mower 1 takes \(28\) hours alone, and mower 2 takes \(21\) hours alone.

- Add the wind speed for the tailwind flight and subtract it for the headwind flight. - Write a travel-time expression for each flight. - Set their sum equal to the total time. - Check the result against the requirement that the airplane can fly into the wind.

1. Let \(v\) be the airspeed in still air, in miles per hour. The domain requires \(v>30\). 2. The outbound and return ground speeds are \(v+30\) and \(v-30\). 3. The total-time equation is \(\frac{180}{v+30}+\frac{180}{v-30}=2.5\). 4. Multiply by \((v+30)(v-30)\) and simplify: \(360v=2.5(v^2-900)\). 5. Multiply by \(2\) and rearrange: \(v^2-144v-900=0\). 6. Factor: \((v-150)(v+6)=0\), so \(v=150\) or \(v=-6\). 7. Only \(v=150\) satisfies the domain, so the airspeed is \(150\,\text{mph}\).

\(150\,\text{mph}\)

- Convert both time differences to hours. - Add the starting advantage and arrival advantage to obtain the travel-time difference. - Use distance divided by speed for each train. - Reject a negative speed.

1. Let \(v\) be the freight train's speed in miles per hour. The express train's speed is \(v+20\). 2. Their travel times are \(\frac{300}{v}\) and \(\frac{300}{v+20}\) hours. 3. The freight train's travel time exceeds the express train's by the \(30\)-minute head start plus the \(45\)-minute earlier arrival: \(\frac12+\frac34=\frac54\) hours. 4. Write the equation \(\frac{300}{v}-\frac{300}{v+20}=\frac54\). 5. Clear the denominators and simplify: \(v^2+20v-4800=0\). 6. Factor: \((v-60)(v+80)=0\), so \(v=60\) or \(v=-80\). 7. Speed must be positive, so the freight train travels at \(60\,\text{mph}\).

\(60\,\text{mph}\)

- Use the vertex of the quadratic to find the minimum. - Convert each function value from hundreds of dollars to dollars. - Subtract the earlier cost from the later cost. - Compare the two first differences.

1. The parabola opens upward. Its vertex occurs at \(x=-\frac{b}{2a}=\frac{12}{1}=12\). 2. Evaluate \(K(12)=0.5\cdot 12^2-12\cdot 12+120=48\). Because the units are hundreds of dollars, the minimum cost is \(\$4800\). 3. Evaluate \(K(15)=52.5\) and \(K(14)=50\). The increase is \(2.5\) hundreds of dollars, or \(\$250\). 4. Evaluate \(K(19)=72.5\) and \(K(18)=66\). The increase is \(6.5\) hundreds of dollars, or \(\$650\). 5. The one-unit cost increase is larger farther to the right of the minimum. The cost graph becomes steeper as production increases in that region.

a) Minimum at \(x=12\); minimum cost \(\$4800\) b) \(\$250\) c) The increase from \(18\) to \(19\) is \(\$650\). Costs rise at an increasing rate to the right of the minimum.

- Substitute each speed directly into the model. - Find an increase by subtracting the earlier output from the later output. - Set the model equal to the given distance and solve the quadratic equation. - Reject solutions that are not meaningful speeds.

1. Evaluate \(s(30)=0.02\cdot 30^2+0.5\cdot 30=18+15=33\) feet. 2. Evaluate \(s(60)=0.02\cdot 60^2+0.5\cdot 60=72+30=102\) feet. 3. Compute \(s(50)=75\) and \(s(55)=88\), so the increase is \(13\) feet. 4. Set \(s(v)=75\): \(0.02v^2+0.5v-75=0\). Multiplying by \(50\) gives \(v^2+25v-3750=0\). 5. The quadratic formula gives \(v=\frac{-25\pm125}{2}\), so \(v=50\) or \(v=-75\). A negative speed is not meaningful in this context, so the answer is \(50\) mph.

a) \(33\) ft at \(30\) mph; \(102\) ft at \(60\) mph b) \(13\) ft c) \(50\) mph

- Raising the starting position adds the same amount to every height. - A vertical shift does not change the horizontal location of the maximum. - Track how \(f(x)+c\) changes the coordinates of a point.

1. Raising the nozzle by \(2.5\) feet adds \(2.5\) to every height: \(h_{\mathrm{new}}(d)=h(d)+2.5=-0.5d^2+2d+3.5\). 2. This is a vertical shift up \(2.5\) units. 3. A vertical shift does not change the horizontal coordinate of the vertex. The vertical coordinate increases by \(2.5\), so \((2, 3)\) moves to \((2, 5.5)\).

a) \(h_{\mathrm{new}}(d)=-0.5d^2+2d+3.5\) b) Shift the graph up \(2.5\) units. c) The new vertex is \((2, 5.5)\).

- Find the positive zero of the quadratic in terms of \(k\). - Translate “does not travel past the edge” into an inequality. - The vertex lies halfway between the two zeros of a parabola.

1. The water returns to ground level at the positive zero of \(f_k\). Solve \(0.8x-\frac{k}{100}x^2=0\): \(x\left(0.8-\frac{k}{100}x\right)=0\). The positive zero is \(x=\frac{80}{k}\). 2. The water must land no farther than \(16\,\text{ft}\) from the nozzle, so \(\frac{80}{k}\le16\). Because \(k>0\), this gives \(k\ge5\). The smallest possible value is \(k=5\). 3. For \(k=5\), the zeros are \(0\) and \(16\), so the vertex lies halfway between them at \(x=8\). 4. Evaluate the function at the vertex: \(f_5(8)=0.8\cdot 8-0.05\cdot 8^2=6.4-3.2=3.2\).

a) \(k=5\) b) \(3.2\,\text{ft}\)

- Use the given horizontal range to write an equation involving \(c\). - Substitute the obstacle's horizontal position into the function. - Compare the resulting height with the obstacle's height.

1. Returning to the ground at \(x=50\) means \(h_c(50)=0\). Thus, \(50c-0.01\cdot 50^2=0\). 2. Simplify: \(50c-25=0\), so \(c=0.5\). 3. Evaluate the height at the obstacle: \(h_{0.5}(25)=0.5\cdot 25-0.01\cdot 25^2=12.5-6.25=6.25\). 4. Since \(6.25<7\), the projectile does not clear the obstacle.

a) \(c=0.5\) b) No. Its height at \(x=25\) is \(6.25\,\text{ft}\), which is less than \(7\,\text{ft}\).

- Choose an exact point on each graph. - Substitute the point's speed and force into \(F = av^2\). - For “percent greater,” use the racing-position force as the reference value.

1. For the upright position, the graph includes the point \((10, 40)\). Substitute into \(F = av^2\): \(40 = a \cdot 10^2\), so \(a = 0.4\). 2. For the racing position, the graph includes the point \((10, 25)\). Then \(25 = a \cdot 10^2\), so \(a = 0.25\). 3. At \(10\,\text{m/s}\), the difference is \(40 - 25 = 15\,\text{N}\). Relative to the racing-position value, the percent increase is \(\frac{15}{25} \cdot 100\% = 60\%\).

a) Upright position: \(a = 0.4\) Racing position: \(a = 0.25\) b) \(60\%\) greater

- Compare the graph's y-intercept, opening direction, and vertex with each option. - Set the correct function equal to the hoop height and choose the solution after the vertex. - For each shot, compare the vertex height and the positive zero.

1. The graph starts at \((0, 6)\), opens downward, and has vertex \((10, 11)\). Function (1) has \(f(0) = 6\), opens downward, and has vertex at \(x = -\frac{1}{2 \cdot (-0.05)} = 10\), with \(f(10) = 11\). Therefore, function (1) matches the graph. 2. Set the first function equal to the hoop height: \(-0.05x^2 + x + 6 = 10\). This is equivalent to \(x^2 - 20x + 80 = 0\), so \(x = 10 \pm 2\sqrt{5}\). The descending solution is \(x = 10 + 2\sqrt{5} \approx 14.47\,\text{ft}\). 3. The first ball has maximum height \(11\,\text{ft}\). Its positive ground-level intercept solves \(-0.05x^2 + x + 6 = 0\), giving \(x = 10 + 2\sqrt{55} \approx 24.83\,\text{ft}\). 4. For the second ball, the vertex occurs at \(x = -\frac{0.8}{2 \cdot (-0.04)} = 10\), and \(k(10) = 10\,\text{ft}\). Its positive ground-level intercept solves \(-0.04x^2 + 0.8x + 6 = 0\), giving \(x = 10 + 5\sqrt{10} \approx 25.81\,\text{ft}\). 5. The second ball does not travel higher, but it travels farther.

a) Function (1), \(f(x) = -0.05x^2 + x + 6\) b) \(10 + 2\sqrt{5}\,\text{ft} \approx 14.47\,\text{ft}\) c) The second ball travels lower but farther. Its maximum height is \(10\,\text{ft}\), compared with \(11\,\text{ft}\) for the first; its range is about \(25.81\,\text{ft}\), compared with about \(24.83\,\text{ft}\).

- Use two points on the revenue line to find its slope. - Break-even occurs where revenue equals cost. - Factor the resulting quadratic equation. - Profit equals revenue minus cost.

1. The revenue graph is a line through \((0, 0)\) and \((10, 400)\). Its slope is \(\frac{400-0}{10-0}=40\), so \(E(x)=40x\). 2. At a break-even point, revenue equals cost: \(40x=0.5x^2+15x+200\). 3. Rearrange and multiply by \(2\): \(x^2-50x+400=0\). 4. Factor: \((x-10)(x-40)=0\), so the break-even production levels are \(x=10\) and \(x=40\). 5. At \(x=30\), the revenue is \(E(30)=40\cdot30=1200\), or \(\$1200\). 6. The cost is \(K(30)=0.5\cdot30^2+15\cdot30+200=1100\), or \(\$1100\). 7. The profit is \(\$1200-\$1100=\$100\).

a) \(E(x)=40x\) b) \(10\) bottles and \(40\) bottles c) \(\$100\)

- Find where the graph crosses the horizontal level \(h = 4\). - What is the height at \(t = 0\)? - Identify when the descending graph returns to that height and when it reaches the ground.

1. The release height is \(h(0) = 1\,\text{m}\). 2. The graph meets the height \(4\,\text{m}\) at \(t = 1\) and \(t = 3\). The parabola is on or above that height between those times, so \(1 \leq t \leq 3\). 3. The ball returns to its release height when \(-t^2 + 4t + 1 = 1\). This gives \(-t^2 + 4t = 0\), so \(t = 0\) or \(t = 4\). 4. The ball reaches the ground when \(-t^2 + 4t + 1 = 0\). The positive solution is \(t = 2 + \sqrt{5} \approx 4.24\). 5. Therefore, after release and before ground contact, the ball is below its release height for \(4 < t < 2 + \sqrt{5}\), approximately \(4 < t < 4.24\).

a) \(1 \leq t \leq 3\) b) \(4 < t < 2 + \sqrt{5}\), approximately \(4 < t < 4.24\)

- Substitute a clearly readable point from each curve into the quadratic model. - Keep the scale factor \(\frac{v}{10}\) together before squaring. - For the percent comparison, identify which curve supplies the reference value.

1. Curve \(f\) contains the point \((20,8)\). Substituting gives \(8=k\left(\frac{20}{10}\right)^2=4k\), so \(k_f=2\). 2. Curve \(g\) contains the point \((20,4)\). Substituting gives \(4=4k\), so \(k_g=1\). 3. At \(v=25\), \(F_f=2(2.5)^2=12.5\,\text{kN}\), and \(F_g=(2.5)^2=6.25\,\text{kN}\). 4. The percent increase from \(g\) to \(f\) is \(\frac{12.5-6.25}{6.25}\cdot100\%=100\%\).

a) Curve \(f\): \(k=2\); curve \(g\): \(k=1\) b) The force on curve \(f\) is \(100\%\) greater.

- Use two exact points on the linear graph to find its slope. - At an intersection, the two plans have equal balances. - Convert \(3\) years to months before evaluating \(s_1\).

1. The linear graph for \(s_2\) passes through \((0, 0)\) and \((30, 450)\). Its slope is \(\frac{450}{30} = 15\), so \(s_2(x) = 15x\). 2. Set the balances equal: \(0.5x^2 = 15x\). Factoring gives \(x(0.5x - 15) = 0\), so \(x = 0\) or \(x = 30\). After the start, the plans next have the same balance at \(30\) months. 3. Three years is \(36\) months. Plan \(s_1\) gives \(s_1(36) = 0.5 \cdot 36^2 = 648\). Since \(\$648 > \$600\), the purchase is affordable, with \(\$48\) left over.

a) \(s_2(x) = 15x\) b) \(30\) months c) Yes. The balance is \(\$648\), so \(\$48\) remains.

- Which graph has an increasing rate of change? - Use an exact point on the line to find its slope. - Add the reaction distance and braking distance at the given speed.

1. Braking distance grows quadratically, so it is represented by the curved graph \(p\). 2. The linear graph \(k\) passes through \((0, 0)\) and \((50, 60)\). Its slope is \(\frac{60}{50} = 1.2\), so \(k(v) = 1.2v\). 3. At \(50\,\text{mph}\), the reaction distance is \(k(50) = 1.2 \cdot 50 = 60\,\text{ft}\). 4. The braking distance is \(p(50) = 0.04 \cdot 50^2 = 100\,\text{ft}\). 5. The total stopping distance is \(60 + 100 = 160\,\text{ft}\).

a) Graph \(p\), because it is quadratic b) \(k(v) = 1.2v\) c) \(160\,\text{ft}\)

- Use the origin and another clear point on line \(a\) to find its slope. - Equal costs occur where the two graphs intersect. - Verify the positive intersection by setting the functions equal. - Substitute \(50\) into both functions and compare.

1. The graph for machine A is a line through the origin and \((40, 480)\). Its slope is \(\frac{480}{40}=12\), so \(A(x)=12x\). 2. Set the costs equal: \(0.3x^2=12x\). 3. Since \(x>0\), divide by \(x\): \(0.3x=12\), so \(x=40\). 4. At \(x=40\), both machines cost \(\$480\). 5. At \(x=50\), machine A costs \(A(50)=12\cdot50=600\), or \(\$600\). 6. Machine B costs \(B(50)=0.3\cdot50^2=750\), or \(\$750\). 7. Machine A is less expensive by \(\$150\) per day.

a) \(A(x)=12x\) b) The costs are equal at \(x=40\) units, when each costs \(\$480\). c) The company would save \(\$150\) per day by choosing machine A.

- Revenue is zero when no units are sold. - Fixed cost is the cost graph's y-intercept. - Profit is positive where the profit graph lies above the x-axis. - The maximum profit occurs at the vertex of the profit parabola.

1. Graph \(f\) is revenue because it is a line through the origin: with no sales, revenue is zero. Graph \(g\) is total cost because it has a positive y-intercept representing fixed cost. Graph \(h\) is profit because it equals revenue minus cost and is negative when cost exceeds revenue. 2. The cost graph has y-intercept \(10\). Since the y-axis is in hundreds of dollars, the fixed cost is \(\$1000\). 3. Profit is positive where graph \(h\) lies above the x-axis. Its zeros satisfy \(-0.1x^2+3x-10=0\), giving \(x=15\pm5\sqrt5\), or \(x\approx3.8\) and \(x\approx26.2\). 4. Therefore, the profit range is approximately \(3.8<x<26.2\), corresponding to about \(380\) to \(2620\) gallons. 5. The profit graph has its vertex at \((15, 12.5)\). Thus, the maximum profit is \(\$1250\) at \(1500\) gallons.

a) \(f\): revenue, \(g\): total cost, \(h\): profit b) \(\$1000\) c) Approximately \(3.8<x<26.2\), or about \(380\) to \(2620\) gallons d) A maximum profit of \(\$1250\) at \(1500\) gallons

- Use the vertex formula to find the horizontal coordinate of the maximum. - Ground contact means the height is \(0\). - Keep only the root that represents a positive horizontal distance.

1. The vertex occurs at \(x = -\frac{0.6}{2 \cdot (-0.05)} = 6\). 2. Evaluate the height: \(h(6) = -0.05 \cdot 6^2 + 0.6 \cdot 6 + 2 = 3.8\). The vertex is \((6, 3.8)\). This means the shot reaches its maximum height of \(3.8\,\text{m}\) after traveling \(6\,\text{m}\) horizontally. 3. Set the height equal to \(0\): \(-0.05x^2 + 0.6x + 2 = 0\). Dividing by \(-0.05\) gives \(x^2 - 12x - 40 = 0\). 4. Using the quadratic formula, \(x = 6 \pm \sqrt{76}\). The context-valid solution is \(x = 6 + \sqrt{76} \approx 14.72\,\text{m}\).

a) \((6, 3.8)\). The shot reaches its maximum height of \(3.8\,\text{m}\) after \(6\,\text{m}\) of horizontal travel. b) \(6 + \sqrt{76}\,\text{m} \approx 14.72\,\text{m}\)

- Use the formula for the number of pairs in a group. - Solve the equation in part a) before doubling the group size. - Recalculate the model for the doubled number of computers. - Compare the new result with twice the original value.

1. If there are \(n\) computers, the number of cables is \(\frac{n(n-1)}{2}\). For part a), solve \(\frac{n(n-1)}{2}=66\). 2. Rearranging gives \(n^2-n-132=0\). The quadratic formula gives \(n=\frac{1\pm23}{2}\), so \(n=12\) or \(n=-11\). Thus, the network has \(12\) computers. 3. Doubling the number of computers gives \(24\) computers. The required number of cables is \(\frac{24\cdot23}{2}=276\). 4. Twice the original number of cables is \(2\cdot66=132\). Since \(276\ne132\), the claim is false. The number of connections follows a quadratic, not linear, model.

a) \(12\) computers b) The claim is false. With \(24\) computers, the network needs \(276\) cables, not \(132\).

- Use the three data points to create a system for \(a\), \(b\), and \(c\). - How can subtracting equations reduce the system? - What height represents landing? - Which quadratic-equation solution is meaningful as time?

1. Let \(h(t) = at^2 + bt + c\). Substituting the three data points gives \(a + b + c = 15\), \(4a + 2b + c = 18\), and \(9a + 3b + c = 17\). 2. Subtract consecutive equations: \(3a + b = 3\) and \(5a + b = -1\). Their difference gives \(2a = -4\), so \(a = -2\). Then \(b = 9\), and \(c = 8\). 3. The model is \(h(t) = -2t^2 + 9t + 8\). 4. The initial height is \(h(0) = 8\,\text{m}\). 5. Solve \(-2t^2 + 9t + 8 = 0\): \(t = \frac{9 \pm \sqrt{145}}{4}\). The positive solution is \(t \approx 5.26\), so the drone reaches the ground after about \(5.26\,\text{s}\).

a) \(h(t) = -2t^2 + 9t + 8\) b) \(8\,\text{m}\) c) About \(5.26\,\text{s}\)

- Convert the span and maximum height into points on the graph. - Where are the road-level intercepts when the center is \(x = 0\)? - How far are the vehicle's outer edges from the center? - Which arch height determines whether the full width clears?

1. The highest point is \((0, 10)\), so \(c = 10\). A width of \(40\,\text{m}\) gives road-level points at \(x = -20\) and \(x = 20\). 2. Substitute \((20, 0)\): \(0 = 400a + 10\), so \(a = -0.025\). Thus, \(f(x) = -0.025x^2 + 10\). 3. Evaluate \(f(15) = -0.025 \cdot 15^2 + 10 = 4.375\). The arch is \(4.375\,\text{m}\) high there. 4. A centered \(6\,\text{m}\)-wide vehicle extends from \(x = -3\) to \(x = 3\). The lowest clearance above the vehicle occurs at its outer edges: \(f(3) = -0.025 \cdot 3^2 + 10 = 9.775\,\text{m}\). 5. Since \(9.775\,\text{m} > 9.5\,\text{m}\), the vehicle fits under the arch.

a) \(f(x) = -0.025x^2 + 10\) b) \(4.375\,\text{m}\) c) Yes. The clearance at the vehicle's outer edges is about \(9.78\,\text{m}\), which exceeds \(9.5\,\text{m}\).

- How do you find the maximum of each downward-opening quadratic? - What does a positive zero represent in this context? - Keep only the positive solution because distance is measured forward from \(x = 0\). - Compare the two landing distances.

1. Athlete A's vertex occurs at \(x = -\frac{0.6}{2 \cdot (-0.05)} = 6\). The maximum height is \(h_A(6) = 3.8\,\text{m}\). 2. Athlete B's vertex occurs at \(x = -\frac{0.5}{2 \cdot (-0.04)} = 6.25\). The maximum height is \(h_B(6.25) = 3.6625\,\text{m}\). Therefore, Athlete A's shot reaches the greater maximum height. 3. For Athlete A, solve \(-0.05x^2 + 0.6x + 2 = 0\). The positive zero is \(x = 6 + \sqrt{76} \approx 14.72\,\text{m}\). 4. For Athlete B, solve \(-0.04x^2 + 0.5x + 2.1 = 0\). The positive zero is \(x = \frac{25 + \sqrt{1465}}{4} \approx 15.82\,\text{m}\). 5. Since \(15.82\,\text{m} > 14.72\,\text{m}\), Athlete B's shot travels farther.

a) Athlete A, with a maximum height of \(3.8\,\text{m}\), compared with Athlete B's \(3.6625\,\text{m}\) b) Athlete B; its shot travels about \(15.82\,\text{m}\), compared with Athlete A's \(14.72\,\text{m}\).

- Draw the enclosure and identify which three sides require fencing. - Use the fencing constraint to express one dimension in terms of the other. - Write the area as a quadratic function and find its vertex. - For the square comparison, all three fenced sides have the same length.

1. Let \(x\) be the length of each side perpendicular to the barn and \(y\) the fenced side parallel to the barn. The fencing constraint is \(2x + y = 120\), so \(y = 120 - 2x\). 2. The area is \(A(x) = x(120 - 2x) = -2x^2 + 120x\). 3. The maximum occurs at the vertex: \(x = -\frac{120}{2 \cdot (-2)} = 30\). Then \(y = 120 - 2 \cdot 30 = 60\). 4. The maximum area is \(30 \cdot 60 = 1800\,\text{ft}^2\). 5. For a square with three fenced sides, \(3s = 120\), so \(s = 40\,\text{ft}\). Its area is \(40^2 = 1600\,\text{ft}^2\), which is smaller than the maximum rectangular area.

The maximizing dimensions are \(30\,\text{ft}\) perpendicular to the barn and \(60\,\text{ft}\) parallel to it, giving \(1800\,\text{ft}^2\). A square would have side length \(40\,\text{ft}\) and area \(1600\,\text{ft}^2\), so it has less area.

- Profit equals revenue minus fixed cost. - Use the vertex of the profit function for part a. - For part b, find the zeros and determine where the downward-opening parabola is on or above the x-axis.

1. Revenue is \(xN(x)\), so profit is \(P(x) = x(200 - 10x) - 750 = -10x^2 + 200x - 750\). 2. The maximum occurs at the vertex: \(x = -\frac{200}{2 \cdot (-10)} = 10\). The profit-maximizing ticket price is \(\$10\). 3. To find the no-loss interval, solve \(-10x^2 + 200x - 750 = 0\). Dividing by \(-10\) gives \(x^2 - 20x + 75 = 0\). 4. Factor: \((x - 5)(x - 15) = 0\). The break-even prices are \(\$5\) and \(\$15\). 5. Since the profit parabola opens downward, profit is nonnegative between the zeros, including the endpoints. Thus, \(5 \leq x \leq 15\).

a) \(\$10\) b) Ticket prices from \(\$5\) through \(\$15\), including both break-even prices

- Use the value at \(t = 0\) as the constant term in both models. - Substitute the remaining data points to solve for the unknown coefficients. - Compare each model's prediction at \(t = 15\) with the measured value.

1. For the linear model, the y-intercept is \(400\). Using \((20, 480)\), \(480 = 20m + 400\), so \(m = 4\). Thus, \(k_{\text{lin}}(t) = 4t + 400\). 2. Let \(k_{\text{quad}}(t) = at^2 + bt + 400\). Using \((10, 430)\) gives \(100a + 10b = 30\), or \(10a + b = 3\). Using \((20, 480)\) gives \(400a + 20b = 80\), or \(20a + b = 4\). 3. Subtracting gives \(10a = 1\), so \(a = 0.1\). Then \(10 \cdot 0.1 + b = 3\), so \(b = 2\). Therefore, \(k_{\text{quad}}(t) = 0.1t^2 + 2t + 400\). 4. At \(t = 15\), the linear model gives \(k_{\text{lin}}(15) = 460\,\text{ppm}\), an error of \(7.5\,\text{ppm}\). The quadratic model gives \(k_{\text{quad}}(15) = 452.5\,\text{ppm}\), an error of \(0\,\text{ppm}\). The quadratic model is more accurate at that time. 5. Using the quadratic model, \(k_{\text{quad}}(30) = 0.1 \cdot 30^2 + 2 \cdot 30 + 400 = 550\,\text{ppm}\).

a) \(k_{\text{lin}}(t) = 4t + 400\) b) \(k_{\text{quad}}(t) = 0.1t^2 + 2t + 400\) c) The quadratic model; it predicts exactly \(452.5\,\text{ppm}\). d) \(550\,\text{ppm}\)

- Evaluate both functions at \(x = 0\). - Find the vertex of each downward-opening parabola. - To find horizontal distance, solve each equation when the height is \(0\) and keep the positive solution.

1. At release, \(x = 0\). Since \(h_1(0) = h_2(0) = 6\), both balls are released from \(6\,\text{ft}\). 2. For the first ball, the vertex occurs at \(x = -\frac{1.2}{2 \cdot (-0.1)} = 6\). Its maximum height is \(h_1(6) = -0.1 \cdot 6^2 + 1.2 \cdot 6 + 6 = 9.6\,\text{ft}\). 3. For the second ball, the vertex occurs at \(x = -\frac{0.8}{2 \cdot (-0.05)} = 8\). Its maximum height is \(h_2(8) = -0.05 \cdot 8^2 + 0.8 \cdot 8 + 6 = 9.2\,\text{ft}\). Therefore, the first ball reaches the greater maximum height. 4. For the first ball, solve \(-0.1x^2 + 1.2x + 6 = 0\). This is equivalent to \(x^2 - 12x - 60 = 0\), so the positive zero is \(x = 6 + 4\sqrt{6} \approx 15.80\,\text{ft}\). 5. For the second ball, solve \(-0.05x^2 + 0.8x + 6 = 0\). This is equivalent to \(x^2 - 16x - 120 = 0\), so the positive zero is \(x = 8 + 2\sqrt{46} \approx 21.56\,\text{ft}\). Therefore, the second ball travels farther.

a) \(6\,\text{ft}\) b) The first ball; its maximum height is \(9.6\,\text{ft}\), compared with \(9.2\,\text{ft}\) for the second ball. c) The second ball travels farther. The first travels about \(15.80\,\text{ft}\), and the second travels about \(21.56\,\text{ft}\).

- Split the fraction in the numerator into two terms. - Relate \(\frac{b}{a}\) to \(\delta\). - A geometric length ratio must be positive.

1. Split the fraction on the left: \(\frac{2a+b}{a}=2+\frac{b}{a}\). 2. Since \(\delta=\frac{a}{b}\), its reciprocal is \(\frac{b}{a}=\frac1\delta\). Thus, \(2+\frac1\delta=\delta\). 3. Multiply by \(\delta\): \(2\delta+1=\delta^2\), so \(\delta^2-2\delta-1=0\). 4. Apply the quadratic formula: \(\delta=\frac{2\pm\sqrt{(-2)^2-4\cdot1\cdot(-1)}}{2}=1\pm\sqrt2\). 5. A ratio of positive lengths must be positive. Since \(1-\sqrt2<0\), the silver ratio is \(\delta=1+\sqrt2\).

a) \(\delta^2-2\delta-1=0\) b) \(\delta=1+\sqrt2\). The solution \(1-\sqrt2\) is rejected because a ratio of positive lengths cannot be negative.

- Use the maximum point to write vertex form. - Substitute the starting point to find \(a\). - If the new range is doubled, where is the second zero? - The vertex lies midway between the zeros.

1. Use vertex form with vertex \((2, 4)\): \(f(x) = a(x - 2)^2 + 4\). 2. Since the stream passes through \((0, 0)\), \(0 = a(0 - 2)^2 + 4\), so \(4a = -4\) and \(a = -1\). 3. Expand: \(f(x) = -(x - 2)^2 + 4 = -x^2 + 4x\). 4. Factor \(f(x) = -x(x - 4)\). The zeros are \(x = 0\) and \(x = 4\), so the water returns to the surface \(4\,\text{m}\) from the outlet. 5. Doubling the range places the second zero at \(x = 8\). With the same leading coefficient, the new function is \(g(x) = -x(x - 8) = -x^2 + 8x\). 6. The new vertex lies midway between the zeros at \(x = 4\). Its height is \(g(4) = -4^2 + 8 \cdot 4 = 16\). The maximum height increases from \(4\,\text{m}\) to \(16\,\text{m}\), so it is quadrupled.

a) \(f(x) = -x^2 + 4x\) b) \(4\,\text{m}\) c) The maximum height quadruples, increasing from \(4\,\text{m}\) to \(16\,\text{m}\).

- Factor out \(x\) to find each pair of zeros. - The vertex lies midway between the zeros. - Write the nonzero zero and the vertex height in terms of \(a\) and \(b\).

1. For \(f\), the vertex occurs at \(x = -\frac{2}{2 \cdot (-0.05)} = 20\). Then \(f(20) = -0.05 \cdot 20^2 + 2 \cdot 20 = 20\), so the maximum height is \(20\,\text{m}\). 2. Factor \(f(x) = x(-0.05x + 2)\). The zeros are \(x = 0\) and \(x = 40\), so the range is \(40\,\text{m}\). 3. For \(g\), the vertex occurs at \(x = -\frac{4}{2 \cdot (-0.05)} = 40\). Then \(g(40) = -0.05 \cdot 40^2 + 4 \cdot 40 = 80\), so the maximum height is \(80\,\text{m}\). 4. Factor \(g(x) = x(-0.05x + 4)\). The zeros are \(x = 0\) and \(x = 80\), so the range is \(80\,\text{m}\). 5. In general, the nonzero zero of \(ax^2 + bx = x(ax + b)\) is \(x = -\frac{b}{a}\). With \(a\) fixed, doubling \(b\) doubles the range. 6. The vertex height is \(-\frac{b^2}{4a}\). With \(a < 0\) fixed, doubling \(b\) multiplies the maximum height by \(2^2 = 4\).

a) For \(f\): maximum height \(20\,\text{m}\), range \(40\,\text{m}\) For \(g\): maximum height \(80\,\text{m}\), range \(80\,\text{m}\) b) Doubling \(b\) doubles the range and quadruples the maximum height when \(a\) remains fixed and negative.

- Use the starting and landing positions as zeros. - The maximum lies midway between the two zeros. - For part c, set the function equal to the specified height and solve the resulting quadratic equation.

1. The zeros are \(x = 0\) and \(x = 8\), so write \(f(x) = ax(x - 8)\). 2. Use the point \((2, 3)\): \(3 = a \cdot 2 \cdot (2 - 8) = -12a\), so \(a = -0.25\). Thus, \(f(x) = -0.25x^2 + 2x\). 3. The vertex lies midway between the zeros at \(x = 4\). Evaluate \(f(4) = -0.25 \cdot 4^2 + 2 \cdot 4 = 4\). The maximum height is \(4\,\text{m}\). 4. Set the height equal to \(2\): \(-0.25x^2 + 2x = 2\). This is equivalent to \(x^2 - 8x + 8 = 0\). 5. Using the quadratic formula, \(x = \frac{8 \pm \sqrt{64 - 32}}{2} = 4 \pm 2\sqrt{2}\). Therefore, \(x \approx 1.17\,\text{m}\) and \(x \approx 6.83\,\text{m}\).

a) \(f(x) = -0.25x^2 + 2x\) b) \(4\,\text{m}\) c) Approximately \(1.17\,\text{m}\) and \(6.83\,\text{m}\)

- Place the center of the tunnel on the y-axis and use the roadway width to locate the zeros. - How far is each side of the truck from the centerline? - Compare the tunnel height at an outer edge with the truck's height.

1. Place the vertex at \((0, 4)\). Since the roadway is \(8\,\text{m}\) wide, the tunnel meets the ground at \(x = -4\) and \(x = 4\). 2. Write the model in intercept form: \(f(x) = a(x + 4)(x - 4)\). Use the vertex: \(4 = a \cdot 4 \cdot (-4) = -16a\), so \(a = -0.25\). Thus, \(f(x) = -0.25x^2 + 4\). 3. A centered \(3\,\text{m}\)-wide truck extends \(1.5\,\text{m}\) to either side of the center. Evaluate the tunnel height at an outer edge: \(f(1.5) = -0.25 \cdot 1.5^2 + 4 = 3.4375\,\text{m}\). 4. Since \(3.4375\,\text{m} < 3.50\,\text{m}\), the tunnel is too low at the truck's outer top corners. The truck cannot pass through while centered.

No. At the truck's outer top corners, the tunnel height is only \(3.4375\,\text{m}\), which is less than the truck's height of \(3.50\,\text{m}\).

- Draw the two plots and count the two outer side fences plus the shared divider. - Write the fencing constraint using the depth and total width. - Express the total area as a quadratic in one variable. - Compare the two maximizing dimensions to decide whether the layout is square.

1. Let \(x\) be the depth of the layout. Three fence segments have length \(x\): the two outer sides and the divider. Let \(y\) be the total width, so two outer fence segments have length \(y\). 2. The fencing constraint is \(3x + 2y = 120\), so \(y = 60 - 1.5x\). 3. The total area is \(A(x) = xy = x(60 - 1.5x) = -1.5x^2 + 60x\). 4. The maximum occurs at the vertex: \(x = -\frac{60}{2 \cdot (-1.5)} = 20\). Then \(y = 60 - 1.5 \cdot 20 = 30\). 5. The maximum total area is \(20 \cdot 30 = 600\,\text{ft}^2\). Since \(20 \neq 30\), the entire layout is not a square.

The maximizing layout is \(20\,\text{ft}\) by \(30\,\text{ft}\), with total area \(600\,\text{ft}^2\). It is not a square.

- Set the height equal to \(0\) to find the ground-level intercepts. - A centered vehicle extends half its width to each side of the y-axis. - For part c, find the two horizontal positions where the tunnel height equals the vehicle height.

1. At ground level, set \(h(x) = 0\): \(0 = -0.15x^2 + 5.4\). Then \(x^2 = 36\), so \(x = \pm 6\). The total ground-level width is \(12\,\text{m}\). 2. A centered \(3.20\,\text{m}\)-wide truck extends \(1.60\,\text{m}\) on either side of the center. At an outer edge, \(h(1.60) = -0.15 \cdot 1.60^2 + 5.4 = 5.016\,\text{m}\). 3. Since \(5.016\,\text{m} > 3.80\,\text{m}\), the truck can pass through the center. 4. For a vehicle height of \(4.50\,\text{m}\), solve \(4.50 = -0.15x^2 + 5.4\). Then \(0.15x^2 = 0.90\), so \(x^2 = 6\) and \(x = \pm\sqrt{6}\). 5. The greatest possible width is \(2\sqrt{6}\,\text{m} \approx 4.90\,\text{m}\).

a) \(12\,\text{m}\) b) Yes. The tunnel is \(5.016\,\text{m}\) high at the truck's outer edges, which exceeds \(3.80\,\text{m}\). c) \(2\sqrt{6}\,\text{m} \approx 4.90\,\text{m}\)

- Use the two ground-level supports as the zeros of the quadratic. - Substitute the known clearance point to determine the leading coefficient. - Find the horizontal positions of the truck's outer edges and evaluate the arch height there.

1. Place the left support at \((0, 0)\) and the right support at \((12, 0)\). Write the arch in intercept form: \(f(x) = ax(x - 12)\). 2. Use the point \((2, 4)\): \(4 = a \cdot 2 \cdot (2 - 12) = -20a\), so \(a = -0.2\). Therefore, \(f(x) = -0.2x(x - 12) = -0.2x^2 + 2.4x\). 3. The bridge center is at \(x = 6\). A centered \(3.2\,\text{m}\)-wide truck extends from \(x = 6 - 1.6 = 4.4\) to \(x = 6 + 1.6 = 7.6\). 4. The lowest clearance above the truck occurs at its outer edges. By symmetry, evaluate one edge: \(f(4.4) = -0.2 \cdot 4.4^2 + 2.4 \cdot 4.4 = 6.688\,\text{m}\). 5. Since \(6.688\,\text{m} > 4.5\,\text{m}\), the truck can pass beneath the bridge while centered.

Yes. The clearance at the truck's outer top corners is \(6.688\,\text{m}\), which is greater than the truck's height of \(4.5\,\text{m}\).

- Relate daily production, number of days, and total production. - Express the planned number of days in terms of the daily quota. - Determine how many days were worked at each production rate. - Add the production from the two phases.

1. Let \(x\) be the planned number of components per day and \(t\) the planned number of days. Then \(xt=400\), so \(t=\frac{400}{x}\). 2. The team works \(t-1\) days in total. After the first two days, it works \(t-3\) days at \(x+10\) components per day. 3. Model the actual production: \(2x+(t-3)(x+10)=430\). 4. Substitute \(t=\frac{400}{x}\): \(2x+\left(\frac{400}{x}-3\right)(x+10)=430\). 5. Simplifying gives \(x^2+60x-4000=0\). 6. Apply the quadratic formula: \(x=\frac{-60\pm\sqrt{60^2-4\cdot1\cdot(-4000)}}{2}=\frac{-60\pm140}{2}\). 7. The solutions are \(x=40\) and \(x=-100\). A production rate must be positive, so the planned quota was \(40\) components per day.

\(40\) components per day

- Use the area formula for a rectangle before and after the change. - Express one dimension in terms of the other. - Substitute to obtain a quadratic equation. - Reject any solution that cannot represent a length.

1. Let \(l\) be the original length and \(w\) the original width, in feet. Then \(lw=1500\) and \((l+10)(w-5)=1500\). 2. Since \(w=\frac{1500}{l}\), substitute into the second equation: \((l+10)\left(\frac{1500}{l}-5\right)=1500\). 3. Multiply by \(l\) and simplify: \((l+10)(1500-5l)=1500l\), so \(l^2+10l-3000=0\). 4. Factor: \((l-50)(l+60)=0\). Thus, \(l=50\) or \(l=-60\). 5. A length must be positive, so \(l=50\,\text{ft}\). Then \(w=\frac{1500}{50}=30\,\text{ft}\).

The original lot was \(50\,\text{ft}\) long and \(30\,\text{ft}\) wide.

- Represent the total number of spaces as rows times spaces per row. - Express the original number of spaces per row in terms of the original number of rows. - Use the renovated configuration to form a quadratic equation. - Remember that the question asks for the number of rows after the renovation.

1. Let \(r\) be the original number of rows and \(s\) the original number of spaces per row. Then \(rs=300\) and \((r+5)(s+2)=440\). 2. Since \(s=\frac{300}{r}\), substitute into the second equation: \((r+5)\left(\frac{300}{r}+2\right)=440\). 3. Multiply by \(r\) and simplify: \((r+5)(300+2r)=440r\), which gives \(r^2-65r+750=0\). 4. Factor: \((r-15)(r-50)=0\). Thus, the original number of rows could be \(15\) or \(50\). 5. After adding \(5\) rows, the renovated lot could have \(20\) rows or \(55\) rows. 6. Both configurations work: \(20\cdot22=440\) and \(55\cdot8=440\).

The renovated parking lot could have \(20\) rows or \(55\) rows.

- a) Write the per-person contribution before and after the additional students join. - b) Test the proposed doubled group directly; you do not need to solve a second quadratic equation. - Compare the two per-person contributions for the \(\$144\) gift.

1. For part a), let \(x\) be the original number of students. The original contribution is \(\frac{72}{x}\), and the new contribution is \(\frac{72}{x+3}\). 2. The contribution decreases by \(\$2\), so \(\frac{72}{x}-\frac{72}{x+3}=2\). 3. Multiply by \(x(x+3)\): \(72(x+3)-72x=2x(x+3)\). 4. Simplify: \(x^2+3x-108=0\). 5. Factor: \((x-9)(x+12)=0\). The candidates are \(x=9\) and \(x=-12\), so \(9\) students originally planned to participate. 6. For part b), if \(18\) students divide \(\$144\), each pays \(\$8\). If \(3\) more students join, each pays \(\frac{144}{21}=\frac{48}{7}\) dollars. 7. The decrease is \(8-\frac{48}{7}=\frac{8}{7}\approx1.14\) dollars, not \(\$2\). Therefore, doubling the original group does not preserve the stated decrease.

a) \(9\) students b) No. With \(18\) students, adding \(3\) more would reduce each contribution by only \(\frac{8}{7}\) dollars, or about \(\$1.14\), not \(\$2\).

- Account for both the change in motor speed and the river current. - Write one time expression for the upstream trip and one for the downstream trip. - Set the sum of the times equal to \(6\) hours. - Check the solutions against the requirement that the boat move upstream.

1. Let \(v\) be the boat's original speed in still water, in miles per hour. The domain requires \(v>3\). 2. The upstream ground speed is \(v-3\). 3. On the return trip, the boat's still-water speed is \(v+3\), so its downstream ground speed is \(v+6\). 4. The total-time equation is \(\frac{36}{v-3}+\frac{36}{v+6}=6\). 5. Divide by \(6\): \(\frac{6}{v-3}+\frac{6}{v+6}=1\). 6. Multiply by \((v-3)(v+6)\): \(6(v+6)+6(v-3)=(v-3)(v+6)\). 7. Simplify: \(v^2-9v-36=0\). 8. Factor: \((v-12)(v+3)=0\), so \(v=12\) or \(v=-3\). 9. Only \(v=12\) satisfies \(v>3\), so the original still-water speed was \(12\,\text{mph}\).

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

- First subtract the break from the total elapsed time. - Add the wind speed on one trip and subtract it on the other. - Write a time expression for each direction. - Check the solution against the requirement that the cyclist can ride into the wind.

1. Subtract the break from the total time. The actual riding time is \(5\) hours. 2. Let \(v\) be the cyclist's speed in still air, in miles per hour. The domain requires \(v>3\). 3. The outbound speed is \(v+3\), and the return speed is \(v-3\). Therefore, \(\frac{36}{v+3}+\frac{36}{v-3}=5\). 4. Multiply by \((v+3)(v-3)\): \(36(v-3)+36(v+3)=5(v^2-9)\). 5. Simplify: \(5v^2-72v-45=0\). 6. Factor: \((5v+3)(v-15)=0\), so \(v=-\frac35\) or \(v=15\). 7. Only \(v=15\) satisfies the domain, so the cyclist's speed in still air is \(15\,\text{mph}\).

\(15\,\text{mph}\)

- Express each pump's hourly rate as the reciprocal of its individual filling time. - Determine how many hours each pump runs. - Add the fractions of the tank filled by the two pumps. - Check each solution against the relationship between the two individual times.

1. Let \(x\) be the number of hours the first pump would need alone. The second pump would need \(x-4\) hours, so the domain requires \(x>4\). 2. The first pump runs for \(5\) hours in all, and the second pump runs for \(3\) hours. 3. Their contributions fill one tank: \(\frac{5}{x}+\frac{3}{x-4}=1\). 4. Multiply by \(x(x-4)\): \(5(x-4)+3x=x(x-4)\). 5. Simplify: \(x^2-12x+20=0\). 6. Factor: \((x-10)(x-2)=0\), so \(x=10\) or \(x=2\). 7. Only \(x=10\) satisfies \(x>4\). Thus, the first pump takes \(10\) hours alone, and the second takes \(10-4=6\) hours alone.

The first pump takes \(10\) hours, and the second pump takes \(6\) hours.

- a) Add the machines' hourly work rates. - Express machine B's time in terms of machine A's time. - b) Multiply each hourly operating cost by the number of hours that machine runs. - When both machines run, include both hourly costs.

1. For part a), let \(x\) be the number of hours machine A would need alone. Machine B would need \(x+5\) hours. 2. Add their work rates: \(\frac{1}{x}+\frac{1}{x+5}=\frac16\). 3. Multiply by \(6x(x+5)\): \(6(x+5)+6x=x(x+5)\). 4. Simplify: \(x^2-7x-30=0\). 5. Factor: \((x-10)(x+3)=0\), so \(x=10\) or \(x=-3\). Time must be positive, so machine A takes \(10\) hours and machine B takes \(15\) hours. 6. For part b), operating both machines for \(6\) hours gives \((110+70)\cdot6=1080\), so the cost is \(\$1080\). 7. Operating machine B alone for \(15\) hours gives \(70\cdot15=1050\), so the cost is \(\$1050\). 8. Because \(1050<1080\), using machine B alone costs \(\$30\) less.

a) Machine A takes \(10\) hours, and machine B takes \(15\) hours. b) Using machine B alone costs less: \(\$1050\), compared with \(\$1080\) for using both machines.

- Determine how cutting squares from both ends changes each base dimension. - Use the volume formula for a rectangular prism. - Express both original dimensions using one variable. - Check that the dimensions allow the corner squares to be removed.

1. Let \(x\) be the original width in centimeters. The original length is \(x+10\). 2. After the corner squares are removed, the box is \(2\,\text{cm}\) high. Its base dimensions are \(x-4\) and \(x+6\). The domain requires \(x>4\). 3. Use the volume: \(2(x-4)(x+6)=192\). 4. Divide by \(2\) and expand: \((x-4)(x+6)=96\), so \(x^2+2x-120=0\). 5. Factor: \((x-10)(x+12)=0\), so \(x=10\) or \(x=-12\). 6. Only \(x=10\) satisfies the domain. The original width was \(10\,\text{cm}\), and the original length was \(20\,\text{cm}\).

The cardboard was \(20\,\text{cm}\) long and \(10\,\text{cm}\) wide.

- a) Substitute \(90\) for \(d\) and solve the resulting quadratic equation. - b) Replace \(n\) with \(n+1\) in the diagonal formula. - Subtract the original formula from the new formula. - Consider why adding a vertex also changes one former side into a diagonal.

1. For part a), set the diagonal formula equal to \(90\): \(\frac{n(n-3)}{2}=90\). 2. Rearrange: \(n^2-3n-180=0\). 3. Factor: \((n-15)(n+12)=0\), so \(n=15\) or \(n=-12\). 4. A polygon cannot have a negative number of vertices, so \(n=15\). 5. For part b), a polygon with \(n+1\) vertices has \(d_{n+1}=\frac{(n+1)(n-2)}{2}\) diagonals. 6. Find the increase: \(d_{n+1}-d_n=\frac{(n+1)(n-2)-n(n-3)}{2}=n-1\). 7. For \(n=15\), the increase is \(15-1=14\) diagonals.

a) \(15\) vertices b) The increase is \(n-1\) diagonals. For the polygon in part a), the increase is \(14\) diagonals.

- Express the first-year profit as a fraction of the starting capital. - Determine the capital available at the beginning of the second year. - Apply the same percentage return to that new amount. - Clear the denominator before solving the quadratic equation.

1. The first-year rate of return, written as a decimal, is \(\frac{500}{x}\). 2. At the beginning of the second year, the capital is \(x+500\). 3. Apply the same rate of return during the second year: \((x+500)\left(1+\frac{500}{x}\right)=3600\). 4. Rewrite and clear the denominator: \(\frac{(x+500)^2}{x}=3600\), so \((x+500)^2=3600x\). 5. Expand and rearrange: \(x^2-2600x+250{,}000=0\). 6. Factor: \((x-100)(x-2500)=0\). 7. Therefore, the mathematically possible starting amounts are \(x=100\) and \(x=2500\). Both are positive and satisfy the model.

The original starting capital could have been \(\$100\) or \(\$2500\).

- a) Subtract the border area from the total poster area. - The inner dimensions each decrease by \(2x\). - b) Substitute half of the border width into the area model. - The border area is not a linear function of its width.

1. The poster's total area is \(24\cdot16=384\,\text{in}^2\), so the printed design has area \(384-144=240\,\text{in}^2\). 2. The design dimensions are \(24-2x\) and \(16-2x\). Therefore, \((24-2x)(16-2x)=240\). 3. Expand and simplify: \(4x^2-80x+144=0\), so \(x^2-20x+36=0\). 4. Factor: \((x-2)(x-18)=0\), so \(x=2\) or \(x=18\). 5. The border must satisfy \(0<x<8\), so \(x=2\,\text{in}\). 6. If the width is halved to \(1\,\text{in}\), the design area is \((24-2)(16-2)=22\cdot14=308\,\text{in}^2\). 7. The new border area is \(384-308=76\,\text{in}^2\). Half of \(144\) is \(72\), so the claim is false.

a) \(2\,\text{in}\) b) The claim is false. A \(1\)-inch border has area \(76\,\text{in}^2\), not \(72\,\text{in}^2\).

- a) Subtract the photo area from the outside rectangle's area. - b) Expand the general frame-area expression before substituting dimensions. - Compare the sums of the two photos' side lengths. - The two photos have equal area, but they do not have equal perimeters.

1. For part a), the frame area is \((4+2x)(6+2x)-4\cdot6\). Set it equal to \(24\): \((4+2x)(6+2x)-24=24\). 2. Expand and simplify: \(4x^2+20x-24=0\), so \(x^2+5x-6=0\). 3. Factor: \((x-1)(x+6)=0\). Since a width must be positive, \(x=1\,\text{in}\). 4. For a rectangle with side lengths \(a\) and \(b\), a frame of width \(x\) has area \((a+2x)(b+2x)-ab=2x(a+b)+4x^2\). 5. The first photo has \(a+b=4+6=10\), while the second has \(a+b=3+8=11\). 6. With the same positive width \(x\), the larger value of \(a+b\) gives a larger frame area. In fact, the second frame area is \((3+2)(8+2)-24=26\,\text{in}^2\).

a) \(1\,\text{in}\) b) The second frame area is greater. It is \(26\,\text{in}^2\), because the second photo has the larger sum of side lengths.

- Find the amount and fraction of vinegar after the first replacement. - The second removal contains the same vinegar fraction as the mixed tank. - Write an expression for the vinegar remaining after both removals. - Check the possible values against the tank capacity.

1. Let \(x\) be the number of gallons removed each time. 2. After the first removal and replacement, \(18-x\) gallons of vinegar remain, so the vinegar fraction is \(\frac{18-x}{18}\). 3. During the second removal, the amount of vinegar removed is \(x\left(\frac{18-x}{18}\right)\). 4. The vinegar remaining is \((18-x)-x\left(\frac{18-x}{18}\right)=\frac{(18-x)^2}{18}\). 5. Set this equal to \(8\): \(\frac{(18-x)^2}{18}=8\), so \((18-x)^2=144\). 6. Thus, \(18-x=12\) or \(18-x=-12\), giving \(x=6\) or \(x=30\). 7. The amount removed cannot exceed the tank capacity, so \(x=6\) gallons.

\(6\) gallons were removed each time.

- Think of the route as two segments separated by the meeting point. - Each hiker's pre-meeting segment is traveled by the other hiker after the meeting. - Use distance equals rate times time for those matching segments. - Equate two expressions for the speed ratio.

1. Let \(t\) be the number of hours before the hikers meet, and let their speeds be \(v_A\) and \(v_B\). 2. Avery's pre-meeting distance is the distance Blake covers in \(9\) hours after the meeting, so \(v_A t=9v_B\). 3. Blake's pre-meeting distance is the distance Avery covers in \(4\) hours after the meeting, so \(v_B t=4v_A\). 4. The first equation gives \(\frac{v_A}{v_B}=\frac{9}{t}\), while the second gives \(\frac{v_A}{v_B}=\frac{t}{4}\). 5. Set the ratios equal: \(\frac{9}{t}=\frac{t}{4}\), so \(t^2=36\). Since time is positive, \(t=6\) hours. 6. Avery's total time is \(6+4=10\) hours, and Blake's total time is \(6+9=15\) hours.

Avery's total travel time was \(10\) hours, and Blake's was \(15\) hours.

- Convert both remaining times to hours. - Express each side of the meeting point using the person who walks it after the meeting. - Use the equal pre-meeting times to find a speed ratio. - Then use the total route length.

1. Convert the remaining times to hours: Alex walks for \(\frac34\) hour after the meeting, and Brooke walks for \(\frac43\) hours. 2. Let their speeds be \(v_A\) and \(v_B\). The segment from \(A\) to the meeting point has length \(\frac43v_B\), and the segment from the meeting point to \(B\) has length \(\frac34v_A\). 3. Because they traveled for the same amount of time before meeting, \(\frac{\frac43v_B}{v_A}=\frac{\frac34v_A}{v_B}\). 4. Simplify: \(\frac{v_A^2}{v_B^2}=\frac{16}{9}\), so \(v_A=\frac43v_B\) because both speeds are positive. 5. Use the total distance: \(\frac43v_B+\frac34v_A=7\). 6. Substitute \(v_A=\frac43v_B\): \(\frac43v_B+v_B=7\), so \(\frac73v_B=7\) and \(v_B=3\). 7. Therefore, \(v_A=\frac43\cdot3=4\).

Alex walks at \(4\,\text{mph}\), and Brooke walks at \(3\,\text{mph}\).

- Match each train's pre-meeting segment with the segment traveled by the other train afterward. - Express those segments as rate times time. - First find the ratio of the squared speeds. - Equal travel times before the meeting make the distance ratio equal to the speed ratio.

1. Let \(t\) be the common travel time before the meeting. The distances traveled before the meeting are \(d_1=v_1t\) and \(d_2=v_2t\). 2. After the meeting, train 1 travels the segment \(d_2\) in time \(t_1\), so \(d_2=v_1t_1\). Train 2 travels the segment \(d_1\) in time \(t_2\), so \(d_1=v_2t_2\). 3. Substitute into the common-time relationship: \(\frac{v_2t_2}{v_1}=\frac{v_1t_1}{v_2}\). 4. Rearranging gives \(\frac{v_1^2}{v_2^2}=\frac{t_2}{t_1}\). Because speeds are positive, \(\frac{v_1}{v_2}=\sqrt{\frac{t_2}{t_1}}\). 5. For part a), \(\frac{v_1}{v_2}=\sqrt{\frac{4.5}{2}}=\sqrt{2.25}=1.5=\frac32\). Thus, \(v_1:v_2=3:2\). 6. For part b), both trains travel for the same time before meeting, so their distance ratio also is \(3:2\). The faster train travels \(\frac{3}{3+2}=\frac35=60\%\) of the total distance.

a) \(v_1:v_2=3:2\) b) \(60\%\) c) The matching route segments give \(v_1t=v_2t_2\) and \(v_2t=v_1t_1\). Eliminating \(t\) yields \(\frac{v_1^2}{v_2^2}=\frac{t_2}{t_1}\), and positive speeds give the stated square-root relationship.

- Convert the time difference to hours. - Use distance divided by speed for each travel time. - For part b), examine the time-difference expression when both speeds increase but their difference stays fixed. - A fixed speed difference has a smaller effect on travel time at higher speeds.

1. For part a), let \(v\) be the bus's speed in miles per hour. The car's speed is \(v+20\). 2. Convert \(30\) minutes to \(\frac12\) hour and write the time-difference equation: \(\frac{120}{v}-\frac{120}{v+20}=\frac12\). 3. Multiply by \(2v(v+20)\): \(240(v+20)-240v=v(v+20)\). 4. Simplify: \(v^2+20v-4800=0\). 5. Factor: \((v-60)(v+80)=0\), so \(v=60\) or \(v=-80\). 6. Speed must be positive, so the bus travels at \(60\,\text{mph}\), and the car travels at \(80\,\text{mph}\). 7. For part b), with a fixed speed difference of \(20\,\text{mph}\), the time advantage is \(\frac{120}{v}-\frac{120}{v+20}=\frac{2400}{v(v+20)}\). 8. Increasing both speeds increases the denominator while the numerator stays fixed, so the time advantage decreases.

a) The bus travels at \(60\,\text{mph}\), and the car travels at \(80\,\text{mph}\). b) The car's time advantage would decrease.

- Express each post-meeting route segment using speed times time. - The hikers traveled for equal amounts of time before meeting. - Use the \(1\,\text{mph}\) difference to relate their speeds. - Add the two route segments after finding the speeds.

1. Let \(v\) be the slower hiker's speed in miles per hour. The faster hiker's speed is \(v+1\). 2. The segment from \(P\) to the meeting point has length \(4(v+1)\), and the segment from the meeting point to \(Q\) has length \(9v\). 3. The hikers spent the same amount of time walking before they met, so \(\frac{4(v+1)}{v}=\frac{9v}{v+1}\). 4. Cross-multiply: \(4(v+1)^2=9v^2\). 5. Because speeds are positive, \(2(v+1)=3v\), so \(v=2\). The faster speed is \(3\,\text{mph}\). 6. The two route segments are \(4\cdot3=12\) miles and \(9\cdot2=18\) miles. 7. The total distance is \(12+18=30\) miles.

\(30\) miles

- Use the post-meeting times to express the two route segments. - Account for the one-hour difference in starting times. - Introduce a variable for the ratio of the two speeds. - Use the difference between the pre-meeting distances after finding the ratio.

1. Let the walking speeds be \(v_A\) and \(v_B\), in miles per hour. 2. The segment from the meeting point to \(B\) is \(v_A\) miles because Alex walks it in \(1\) hour. The segment from \(A\) to the meeting point is \(6v_B\) miles because Blake walks it in \(6\) hours. 3. Alex's pre-meeting time is \(\frac{6v_B}{v_A}\), and Blake's is \(\frac{v_A}{v_B}\). Since Alex started one hour earlier, \(\frac{6v_B}{v_A}=\frac{v_A}{v_B}+1\). 4. Let \(r=\frac{v_A}{v_B}\). Then \(\frac{6}{r}=r+1\), so \(r^2+r-6=0\). 5. Factor: \((r-2)(r+3)=0\). A speed ratio is positive, so \(r=2\), meaning \(v_A=2v_B\). 6. Alex had walked \(8\) miles farther at the meeting: \(6v_B-v_A=8\). 7. Substitute \(v_A=2v_B\): \(4v_B=8\), so \(v_B=2\,\text{mph}\) and \(v_A=4\,\text{mph}\). 8. The total distance is \(6\cdot2+1\cdot4=16\) miles.

Alex walks at \(4\,\text{mph}\), Blake walks at \(2\,\text{mph}\), and the total distance is \(16\) miles.

- Convert the time difference to hours. - Write a time expression for each segment. - The second speed is \(2\,\text{mph}\) greater than the first. - Check both positive solutions in the original equation.

1. Let \(v\) be the speed on the first segment, in miles per hour. The speed on the second segment is \(v+2\). 2. The travel times are \(\frac{12}{v}\) and \(\frac{18}{v+2}\) hours. 3. Since the second segment takes \(15\) minutes, or \(\frac14\) hour, longer, \(\frac{18}{v+2}-\frac{12}{v}=\frac14\). 4. Multiply by \(4v(v+2)\): \(72v-48(v+2)=v(v+2)\). 5. Simplify: \(v^2-22v+96=0\). 6. Factor: \((v-6)(v-16)=0\), so \(v=6\) or \(v=16\). 7. Both values are positive and satisfy the original time relationship.

The first-segment speed could be \(6\,\text{mph}\) or \(16\,\text{mph}\).

- Express the planned total travel time using the unknown speed. - Find the distance remaining after the first \(2\) hours. - Convert the stop time to hours. - Set the planned time equal to the actual travel-and-stop time.

1. Let \(v\) be the planned speed in miles per hour. The planned total travel time is \(\frac{180}{v}\) hours. 2. In the first \(2\) hours, the train travels \(2v\) miles, leaving \(180-2v\) miles. 3. Convert \(12\) minutes to \(\frac15\) hour. The actual time is \(2+\frac15+\frac{180-2v}{v+15}\). 4. Since the train arrives on time, \(\frac{180}{v}=\frac{11}{5}+\frac{180-2v}{v+15}\). 5. Clear the denominators and simplify: \(v^2+165v-13{,}500=0\). 6. Factor: \((v-60)(v+225)=0\), so \(v=60\) or \(v=-225\). 7. Speed must be positive, so the planned speed was \(60\,\text{mph}\).

\(60\,\text{mph}\)

- Express the total distance using both speeds and the time before the meeting. - The distance hiker B covers before the meeting is the distance hiker A covers afterward. - Use the \(1\,\text{mph}\) speed difference to eliminate one variable. - Check both algebraic solutions against the stated speed restriction.

1. Let the speeds be \(v_A\) and \(v_B\), and let \(t\) be the time before they meet. Then \((v_A+v_B)t=22.5\) and \(v_A=v_B+1\). 2. After the meeting, hiker A covers in \(2\) hours the distance hiker B covered before the meeting. Thus, \(v_Bt=2v_A\), so \(t=\frac{2v_A}{v_B}\). 3. Substitute \(v_B=v_A-1\) into the distance equation: \((2v_A-1)\frac{2v_A}{v_A-1}=22.5\). 4. Clear the denominator and simplify: \(4v_A^2-24.5v_A+22.5=0\). 5. The solutions are \(v_A=5\) and \(v_A=1.125\). 6. Since \(v_B=v_A-1\), the corresponding pairs are \((v_A,v_B)=(5,4)\) and \((1.125,0.125)\). 7. The second pair violates the condition that both speeds exceed \(1\,\text{mph}\). Therefore, \(v_A=5\,\text{mph}\) and \(v_B=4\,\text{mph}\).

Hiker A walks at \(5\,\text{mph}\), and hiker B walks at \(4\,\text{mph}\).

- Express each individual time using the shared time \(x\). - Add the individual filling rates. - Identify the slowest pipe from the individual times. - After removing that pipe, add the two remaining rates and take the reciprocal.

1. The individual filling times are \(x+4\), \(x+10\), and \(2x\) hours. 2. Add the individual rates: \(\frac{1}{x}=\frac{1}{x+4}+\frac{1}{x+10}+\frac{1}{2x}\). 3. Subtract \(\frac{1}{2x}\): \(\frac{1}{2x}=\frac{1}{x+4}+\frac{1}{x+10}\). 4. Multiply by \(2x(x+4)(x+10)\): \((x+4)(x+10)=2x(x+10)+2x(x+4)\). 5. Simplify: \(3x^2+14x-40=0\). 6. Factor: \((3x+20)(x-2)=0\), so \(x=2\) or \(x=-\frac{20}{3}\). 7. Time must be positive, so all three pipes together take \(2\) hours. The individual times are \(6\) hours for A, \(12\) hours for B, and \(4\) hours for C. 8. Pipe B is the slowest. Without B, the combined rate is \(\frac16+\frac14=\frac{5}{12}\) tank per hour. 9. The new filling time is \(\frac{12}{5}\) hours, or \(2\) hours \(24\) minutes.

All three pipes together take \(2\) hours. Pipe A takes \(6\) hours, pipe B takes \(12\) hours, and pipe C takes \(4\) hours alone. Without pipe B, the tank takes \(\frac{12}{5}\) hours, or \(2\) hours \(24\) minutes, to fill.

- Add the current speed downstream and subtract it upstream. - Add the two travel times for part a). - For part b), compare the effect of equal speed increases and decreases on travel time. - The slower leg lasts longer than the faster leg.

1. For part a), let \(v\) be the boat's speed in still water, in miles per hour. The domain requires \(v>4\). 2. The downstream and upstream speeds are \(v+4\) and \(v-4\), so \(\frac{32}{v+4}+\frac{32}{v-4}=6\). 3. Multiply by \((v+4)(v-4)\): \(32(v-4)+32(v+4)=6(v^2-16)\). 4. Simplify: \(3v^2-32v-48=0\). 5. Factor: \((3v+4)(v-12)=0\), so \(v=-\frac43\) or \(v=12\). 6. Only \(v=12\) satisfies the domain, so the still-water speed is \(12\,\text{mph}\). 7. For part b), the boat on the calm lake returns sooner. The speed lost while traveling upstream causes more added time than the equal speed gained downstream removes, because the boat spends longer traveling at the lower speed.

a) \(12\,\text{mph}\) b) The boat on the calm lake returns sooner.

- Write one equation for the combined hourly rate. - Write another equation for the time spent filling one-third and two-thirds of the tank. - Solve one equation for one individual time and substitute. - Check both positive solution pairs.

1. Let \(t_1\) and \(t_2\) be the individual filling times, in hours. 2. Their combined-rate equation is \(\frac{1}{t_1}+\frac{1}{t_2}=\frac16\). 3. The sequential procedure gives \(\frac13t_1+\frac23t_2=15\), so \(t_1=45-2t_2\). 4. Substitute into the rate equation: \(\frac{1}{45-2t_2}+\frac{1}{t_2}=\frac16\). 5. Clear the denominators and simplify: \(2t_2^2-51t_2+270=0\). 6. Factor: \((2t_2-15)(t_2-18)=0\), so \(t_2=\frac{15}{2}\) or \(t_2=18\). 7. If \(t_2=18\), then \(t_1=9\). If \(t_2=\frac{15}{2}\), then \(t_1=30\). 8. Both pairs satisfy the two original conditions.

There are two possibilities: - Pump 1 takes \(9\) hours and pump 2 takes \(18\) hours. - Pump 1 takes \(30\) hours and pump 2 takes \(7.5\) hours.

- Convert the sequential time to a fraction of an hour. - Use the sequential half-job condition to express one individual time in terms of the other. - Substitute that expression into the combined-rate equation. - Factor the resulting quadratic equation.

1. Convert \(4\) hours \(10\) minutes to \(\frac{25}{6}\) hours. 2. Let \(x\) and \(y\) be the individual times, in hours. The combined-rate equation is \(\frac1x+\frac1y=\frac12\). 3. Printing half the job on each copier sequentially gives \(\frac{x}{2}+\frac{y}{2}=\frac{25}{6}\), so \(x+y=\frac{25}{3}\). Thus, \(y=\frac{25}{3}-x\). 4. Substitute into the combined-rate equation: \(\frac1x+\frac{1}{\frac{25}{3}-x}=\frac12\). 5. Clear the denominators and simplify: \(3x^2-25x+50=0\). 6. Factor: \((3x-10)(x-5)=0\), so \(x=\frac{10}{3}\) or \(x=5\). 7. The corresponding value of \(y\) is \(5\) or \(\frac{10}{3}\). Therefore, one copier takes \(5\) hours, and the other takes \(\frac{10}{3}\) hours, or \(3\) hours \(20\) minutes.

One copier takes \(5\) hours, and the other takes \(3\) hours \(20\) minutes.

- Use the midpoint condition to compare the two travel times for \(40\) miles. - Use the remaining distance after \(2\) hours to find the sum of the speeds. - Substitute one speed into the other equation. - Reject a negative speed.

1. Let the speeds be \(v_A\) and \(v_B\), in miles per hour. 2. To meet at the midpoint, each cyclist travels \(40\) miles. Since cyclist B starts \(2\) hours earlier, \(\frac{40}{v_B}=\frac{40}{v_A}+2\). 3. If they start together, they cover \(80-20=60\) miles in \(2\) hours, so \(2(v_A+v_B)=60\). Thus, \(v_B=30-v_A\). 4. Substitute into the midpoint equation: \(\frac{40}{30-v_A}=\frac{40}{v_A}+2\). 5. Clear the denominators and simplify: \(v_A^2+10v_A-600=0\). 6. Factor: \((v_A-20)(v_A+30)=0\), so \(v_A=20\) or \(v_A=-30\). 7. Speed must be positive, so \(v_A=20\,\text{mph}\). Then \(v_B=30-20=10\,\text{mph}\).

The cyclist from city A rides at \(20\,\text{mph}\), and the cyclist from city B rides at \(10\,\text{mph}\).

- Express each pace as total minutes divided by miles traveled. - Use the one-minute-per-mile difference to form an equation. - After finding both distances to \(Z\), consider the possible orders of the three points on the line. - The starting points may be on the same side or opposite sides of the destination.

1. Let \(x\) be the distance from \(A\) to \(Z\), in miles. Then the distance from \(B\) to \(Z\) is \(x+5\). 2. The cyclists' average paces are \(\frac{30}{x}\) and \(\frac{30}{x+5}\) minutes per mile. 3. The pace difference is \(1\) minute per mile, so \(\frac{30}{x}-\frac{30}{x+5}=1\). 4. Clear the denominators: \(30(x+5)-30x=x(x+5)\). 5. Simplify: \(x^2+5x-150=0\). 6. Factor: \((x-10)(x+15)=0\), so \(x=10\) or \(x=-15\). 7. Distance must be positive, so \(AZ=10\) miles and \(BZ=15\) miles. 8. If \(A\) and \(B\) are on the same side of \(Z\), then \(AB=15-10=5\) miles. If they are on opposite sides of \(Z\), then \(AB=15+10=25\) miles.

The distance between \(A\) and \(B\) could be \(5\) miles or \(25\) miles.

- Their speed sum equals the total distance because they meet after one hour. - Each cyclist's remaining distance equals the distance the other cyclist traveled before the meeting. - Use a variable for the ratio of their speeds. - Convert \(50\) minutes to a fraction of an hour.

1. Because they meet after \(1\) hour, their speeds satisfy \(v_J+v_M=25\). 2. At the meeting point, Julia has traveled \(v_J\) miles and Marcus has traveled \(v_M\) miles. Julia's remaining distance is \(v_M\), while Marcus's remaining distance is \(v_J\). 3. Their remaining travel times are \(\frac{v_M}{v_J}\) and \(\frac{v_J}{v_M}\) hours. 4. Julia arrives \(50\) minutes, or \(\frac56\) hour, later: \(\frac{v_M}{v_J}-\frac{v_J}{v_M}=\frac56\). 5. Let \(r=\frac{v_J}{v_M}\). Then \(\frac1r-r=\frac56\), which gives \(6r^2+5r-6=0\). 6. Factor: \((3r-2)(2r+3)=0\). A speed ratio is positive, so \(r=\frac23\). 7. Thus, \(v_J=\frac23v_M\). Combining this with \(v_J+v_M=25\) gives \(v_M=15\,\text{mph}\) and \(v_J=10\,\text{mph}\).

Julia cycles at \(10\,\text{mph}\), and Marcus cycles at \(15\,\text{mph}\).

- Use the meeting time to find the sum of the speeds. - Each boat's remaining distance is the distance the other boat traveled before the meeting. - Use a speed ratio to simplify the remaining-time difference. - Multiply the slower speed by \(2\) hours for part b).

1. Because the boats cover \(60\) miles together in \(2\) hours, their speeds satisfy \(v_s+v_f=30\), where \(v_s\) is the slower speed and \(v_f\) is the faster speed. 2. Before the meeting, the slower boat travels \(2v_s\) miles and the faster boat travels \(2v_f\) miles. 3. After the meeting, the faster boat must travel \(2v_s\) miles, while the slower boat must travel \(2v_f\) miles. 4. The slower boat's remaining time exceeds the faster boat's by \(3\) hours: \(\frac{2v_f}{v_s}-\frac{2v_s}{v_f}=3\). 5. Let \(r=\frac{v_s}{v_f}\). Then \(\frac{2}{r}-2r=3\), so \(2r^2+3r-2=0\). 6. Factor: \((2r-1)(r+2)=0\). Since the ratio is positive, \(r=\frac12\), so \(v_f=2v_s\). 7. From \(v_s+v_f=30\), \(3v_s=30\), giving \(v_s=10\,\text{mph}\) and \(v_f=20\,\text{mph}\). 8. The slower boat travels \(2\cdot10=20\) miles before the meeting.

a) The boats travel at \(10\,\text{mph}\) and \(20\,\text{mph}\). b) The slower boat has traveled \(20\) miles when they meet.

- Complete the square to identify the vertex. - Eliminate \(k\) between the vertex coordinates. - Minimize the squared distance instead of the distance. - Express the squared distance as a quadratic in the vertex x-coordinate.

1. Complete the square: \(f_k(x)=k\left(x-\frac{1}{2k}\right)^2-\frac{1}{4k}\). Therefore, the vertex is \(\left(\frac{1}{2k}, -\frac{1}{4k}\right)\). 2. Since the vertex x-coordinate is \(x=\frac{1}{2k}\), we have \(k=\frac{1}{2x}\). Substituting into the y-coordinate gives \(y=-\frac{x}{2}\). 3. A point on this line has coordinates \(\left(x,-\frac{x}{2}\right)\). Its squared distance from \(P\) is \(D(x)=(x-5)^2+\left(-\frac{x}{2}\right)^2=\frac{5}{4}x^2-10x+25\). 4. Completing the square gives \(D(x)=\frac{5}{4}(x-4)^2+5\), so the minimum occurs at \(x=4\). Then \(y=-2\), and \(k=\frac{1}{2x}=\frac{1}{8}\).

a) \(\left(\frac{1}{2k}, -\frac{1}{4k}\right)\) b) \(y=-\frac{1}{2}x\) c) The closest vertex is \((4, -2)\), corresponding to \(k=\frac{1}{8}\).