To cover the walls twice, double the total area \(w\) to get \(2w\). Since each gallon covers 180 square feet, divide \(2w\) by 180 to get \(P = \frac{w}{90}\).

To have no real solution, the discriminant of \(x(q x – 64) + 20 = 0\) must be negative. Solving gives \(\)q^2 < 16[/latex], so the smallest integer for [latex]q[/latex] is 4.

Using the quadratic formula, \(x = \frac{12 \pm \sqrt{144 – 44}}{2} = 6 \pm \sqrt{25}\). So, the solution can be written as \(6 + \sqrt{25}\) and the correct value of \(k\) is 25.

To find \( p \): \( 50 = \frac{p}{100} \times 80 \implies p = \frac{50 \times 100}{80} = 62.5 \). Calculate \( 62.5 \% \) of \( 50 \): \( \frac{62.5}{100} \times 50 = 31.25 \).

Data set B has higher frequencies around the center value (20) compared to Data set A, resulting in equal means for both data sets.

To find the surface area, first find the side length using \( V = s^3 \). Given \( V = 27000 \), find \( s = \sqrt[3]{27000} = 30 \). The surface area is given by \( 6s^2 \). Therefore, the surface area is \( 6(30)^2 = 5400 \) square units.

The half-life is 3 years, so the remaining quantity after \( x \) minutes is \( Q = 5 \cdot \left(\frac{1}{2}\right)^{\frac{x}{3 \cdot 365 \cdot 24 \cdot 60}} \).

Calculate percentages: Candidate X received \( \frac{350}{800} = 0.4375 \), or 43.75%. Candidate Y received \( \frac{450}{800} = 0.5625 \), or 56.25%. If 5,600 people vote, Candidate X would receive \( 0.4375 \times 5600 = 2450 \) votes, and Candidate Y \( 0.5625 \times 5600 = 3150 \) votes. The difference is \( 3150 – 2450 = 700 \) votes.

The leakage rate is 5 gallons per hour. Over one day (24 hours), the volume leaked is 5 * 24 = 120 gallons. Converting to pints (1 gallon = 8 pints), the answer is 960 pints.

First, find the measure of angle B in radians: \( \frac{\pi}{4} + \frac{3\pi}{8} = \frac{5\pi}{8} \). Then, convert to degrees by multiplying by \( \frac{180}{\pi} \): \( \frac{5\pi}{8} \cdot \frac{180}{\pi} = 112.5 \) degrees.

Since \( f(3) = 1024 \), substitute into \( f(x) = 8 \cdot d^x \): \( 8 \cdot d^3 = 1024 \Rightarrow d^3 = 128 \Rightarrow d = 4 \). Now find \( f(-2) \): \( f(-2) = 8 \cdot 4^{-2} = 8 \cdot \frac{1}{16} = 0.5 \).

Expanding the equation, we obtain \( 2x^2 + 3x – 9 = 4x^2 – 24x \). Rearranging terms, we get \( -2x^2 + 27x – 9 = 0 \). Using the quadratic formula, the sum of the solutions is given by \( -b/a \), where \( a = -2 \) and \( b = 27 \), which yields \( 27/2 \).

The percentage can be calculated using the formula: \( \frac{350}{700} \times 100 = 50\% \).

Both sides simplify to \( 8z + 4 \), making the equation true for all values of z. Therefore, there are infinitely many solutions.

Since the side lengths of square C are 8 times those of square D, the area of square C is \( 8^2 = 64 \) times the area of square D. Therefore, \( k = 64 \).

Using the quadratic formula, \(x = \frac{6 \pm \sqrt{36 – 12}}{2} = 3 \pm \sqrt{6}\). So, the solution can be written as \(3 + \sqrt{6}\) and the correct value of \(k\) is 6.

Using the point-slope form \( y – y_1 = m(x – x_1) \) with the slope \( m = \frac{3}{5} \) and point \( (2, -4) \), the resulting equation is \( y = \frac{3}{5}x – \frac{22}{5} \).

Expanding the right side results in \( 10a – 3 \), which is identical to the left side. Therefore, the equation holds for all values of a, implying infinitely many solutions.

The function increases by 20%, which means it becomes 1.20 times its previous value at each step. The multiplier is 1.20. Therefore, the correct equation is \( g(x) = 30(1.20)^x \).

Since the function decreases by 60%, it retains 40% of its value at each step. Thus, the multiplier for the decay is 0.40. Therefore, the correct equation is \( p(x) = 50(0.40)^x \).

In the function f(x) = -5x + 30, the y-intercept (where x = 0) represents the initial amount of juice Caleb had before he made any popsicles. When x = 0, f(0) = 30, indicating that Caleb started with 30 fluid ounces of juice.

To have no real solution, the discriminant of \(x(m x – 40) + 25 = 0\) must be negative. Solving gives \(\)m^2 < 25[/latex], so the smallest integer for [latex]m[/latex] is 5.

To isolate \(2p + 9q\), multiply both sides by \((2p + 9q)\), resulting in \(g = w(2p + 9q)\). Dividing by \(w\) gives \(2p + 9q = \frac{g}{w}\).

Substitute \( (1, 5) \) into both inequalities. Both inequalities are satisfied.

The percentage can be calculated using the formula: \( \frac{\text{Part}}{\text{Whole}} \times 100 \). Here, \( \frac{50}{200} \times 100 = 25\% \).

The decay factor is 0.90 per three months, equating to a 10% drop per quarter. After one year, this approximates to a 34% annual decay rate, so \(m = 34\).

To find \( f(0) \), substitute \( 0 \) for \( x \) in the function: \( f(0) = 300(0.2)^0 \). Since any non-zero number raised to the power of \( 0 \) is \( 1 \), the expression becomes \( 300 imes 1 = 300 \).

In the function k(x) = -8x + 64, the y-intercept (where x = 0) represents the initial amount of concrete the worker had before building any foundations. When x = 0, k(0) = 64, indicating that the worker started with 64 cubic feet of concrete.

Rewrite \(f(x)\) in vertex form as \(5(x – 2)^2 + 75\). The minimum of \(f(x)\) is at \(x = 2\), so \(g(x) = f(x + 3)\) shifts this to \(x = -1\).

Calculate percentages: Candidate P received \( \frac{725}{1000} = 0.725 \), or 72.5%. Candidate Q received \( \frac{275}{1000} = 0.275 \), or 27.5%. If 8,000 people vote, Candidate P would receive \( 0.725 \times 8000 = 5800 \) votes, and Candidate Q \( 0.275 \times 8000 = 2200 \) votes. The difference is \( 5800 – 2200 = 3600 \) votes.

Start by setting up the equation: \( 32,000 = 4,000(2)^{3r} \). Divide both sides by 4,000 to obtain \( 8 = (2)^{3r} \). Taking the logarithm base 2 of both sides yields \( 3 = 3r \). Therefore, \( r = 1 \).

Isolate \(10a + 3b\) by multiplying both sides by \((10a + 3b)\), resulting in \(m = q(10a + 3b)\). Dividing by \(q\) gives \(10a + 3b = \frac{m}{q}\).

To have no real solution, the discriminant of \(x(r x – 120) + 64 = 0\) must be negative. Solving gives \(\)r^2 < 64[/latex], so the smallest integer for [latex]r[/latex] is 8.

Expanding the equation, we obtain \( x^2 + 2x – 10 = 3x^2 – 15x \). Rearranging terms, we get \( -2x^2 + 17x – 10 = 0 \). Using the quadratic formula, the sum of the solutions is given by \( -b/a \), where \( a = -2 \) and \( b = 17 \), which yields \( 17/2 \).

Since the rate of drainage is 3 liters per second, over one minute (60 seconds), the volume drained would be 3 * 60 = 180 liters. Converting to milliliters (1 liter = 1000 mL), the answer is 180,000 mL.

Solving for \( y \) using substitution or elimination, we find that \( y = 30 \).

Dividing the equation by 36, completing the square, and simplifying gives n = 12.

Start by setting up the equation with the given values: \( 64,000 = 8,000(2)^{6r} \). Divide both sides by 8,000 to get \( 8 = (2)^{6r} \). Taking the logarithm base 2 of both sides gives \( 3 = 6r \). Therefore, \( r = \frac{1}{2} \).

Find the slope using the points, then derive the line’s equation. Set \(x = n + 2\) to find \(b = 48\).

The function \(q(x) = 1.3x\) represents a proportional increase, showing a steady, linear growth with respect to \(x\). Therefore, ‘Increasing Linear’ is correct.

Factor out the greatest common factor, \(3x\), from the expression: \(3x(x^2 + 9x + 18)\). Then factor \(x^2 + 9x + 18\) as \((x + 3)(x + 6)\). The smallest possible value of \(b\) is 3.

To have no real solution, the discriminant of \(x(n x – 72) + 36 = 0\) must be negative. Solving gives \(\)n^2 < 36[/latex], so the smallest integer for [latex]n[/latex] is 6.

Since \(\frac{c}{d} = 2\), we have \(c = 2d\). Substitute into \(\frac{40c}{qd} = 2\) to get \(\frac{40(2d)}{qd} = 2\). Simplify to \(\frac{80d}{qd} = 2\), cancel \(d\) and solve to find \(q = 40\).

Only \( (3, 3) \) is a solution that satisfies both equations.