Detailed Solution:
Add 5 to both sides: \( 3x = 16 + 5 \implies 3x = 21 \).
Divide by 3: \( x = 21 / 3 = 7 \).

Detailed Solution:
Factor the quadratic: find two numbers that multiply to 6 and add to -5. These are -2 and -3.
\( (x – 2)(x – 3) = 0 \). Setting each factor to zero gives \( x = 2 \) and \( x = 3 \).

Detailed Solution:
Set up a proportion: \( \frac{3 \text{ cups}}{12 \text{ cookies}} = \frac{x \text{ cups}}{30 \text{ cookies}} \).
Cross-multiply: \( 12x = 90 \).
Divide by 12: \( x = 90 / 12 = 7.5 \) cups.

Detailed Solution:
The formula for circumference is \( C = 2\pi r \).
\( C = 2 \times \frac{22}{7} \times 7 = 2 \times 22 = 44 \).

Detailed Solution:
Average = \( \frac{\text{Sum of scores}}{3} \). Let the third score be \( x \).
\( \frac{72 + 88 + x}{3} = 85 \).
\( 160 + x = 255 \implies x = 255 – 160 = 95 \).

Detailed Solution:
Since \( 2^4 = 16 \), we know that \( x = 4 \).
The question asks for \( x^2 \), so \( 4^2 = 16 \).

Detailed Solution:
Slope \( m = \frac{y_2 – y_1}{x_2 – x_1} \).
\( m = \frac{9 – 3}{5 – 2} = \frac{6}{3} = 2 \).

Detailed Solution:
Surface Area = \( 2(LW + WH + LH) \).
\( SA = 2(10 \times 4 + 4 \times 5 + 10 \times 5) \).
\( SA = 2(40 + 20 + 50) = 2(110) = 220 \).

Detailed Solution:
Substitute \( x = 4 \) into the function:
\( f(4) = 2(4)^2 – 3(4) + 1 \).
\( f(4) = 2(16) – 12 + 1 = 32 – 12 + 1 = 21 \).

Detailed Solution:
Let the numbers be \( x \) and \( y \).
\( x + y = 25 \)
\( x – y = 7 \)
Add the equations: \( 2x = 32 \implies x = 16 \).
Substitute back: \( 16 + y = 25 \implies y = 9 \). The larger number is 16.

Detailed Solution:
The formula for arc length is \( S = r\theta \), where \( \theta \) is in radians.
\( 4\pi = 12 \cdot \theta \implies \theta = \frac{4\pi}{12} = \frac{\pi}{3} \text{ radians} \)
To convert radians to degrees, multiply by \( \frac{180}{\pi} \):
\( \frac{\pi}{3} \cdot \frac{180}{\pi} = 60^\circ \)

Detailed Solution:
Rewrite \( \frac{1}{27} \) as a power of 3: \( \frac{1}{27} = 3^{-3} \).
The equation becomes:
\( 3^{x-2} = 3^{-3} \)
Since the bases are the same, set the exponents equal:
\( x – 2 = -3 \implies x = -3 + 2 = -1 \)

Detailed Solution:
Volume \( V = L \times W \times H = 72 \).
The new volume \( V’ = (2L) \times (3W) \times (0.5H) \).
\( V’ = (2 \times 3 \times 0.5) \times (L \times W \times H) = 3 \times 72 = 216 \)

Detailed Solution:
For infinitely many solutions, the equations must be proportional.
Comparing the \( y \)-coefficients: \( -3 \) was multiplied by \( -2 \) to get \( 6 \).
To maintain the ratio, multiply the first equation by \( -2 \):
\( 2 \times (-2) = k \implies k = -4 \).
Also check the constants: \( 8 \times (-2) = -16 \), which matches the second equation.

Detailed Solution:
Case 1: \( 2x – 5 = 11 \implies 2x = 16 \implies x = 8 \).
Case 2: \( 2x – 5 = -11 \implies 2x = -6 \implies x = -3 \).
Sum of solutions: \( 8 + (-3) = 5 \).

Detailed Solution:
Using vertex form: \( y = a(x – h)^2 + k \).
Plug in vertex \( (3, -4) \): \( y = a(x – 3)^2 – 4 \).
Plug in point \( (5, 0) \): \( 0 = a(5 – 3)^2 – 4 \)
\( 0 = a(2^2) – 4 \implies 4a = 4 \implies a = 1 \).

Detailed Solution:
In a right triangle, \( \tan = \frac{\text{opp}}{\text{adj}} \). Using Pythagoras \( 3^2 + 4^2 = 5^2 \), the hypotenuse is 5.
\( \cos(\theta) = \frac{\text{adj}}{\text{hyp}} = \frac{4}{5} \).
In the 3rd quadrant, both sine and cosine are negative, while tangent is positive. Therefore, \( \cos(\theta) = -4/5 \).

Detailed Solution:
Sum of 5 numbers = \( 12 \times 5 = 60 \).
Ordered set: \( \{a, b, 10, d, e\} \). Since mode is 8, we must have at least two 8s: \( a=8 \) and \( b=8 \).
Set: \( \{8, 8, 10, d, e\} \).
Sum = \( 8+8+10+d+e = 26 + d + e = 60 \), so \( d + e = 34 \).
To maximize \( e \), we minimize \( d \). Since \( d \geq 10 \) (to keep 10 as the median), we let \( d = 10 \).
\( 10 + e = 34 \implies e = 24 \).

Detailed Solution:
Use log properties: \( \log_2(x(x-2)) = 3 \).
Convert to exponential form: \( x(x-2) = 2^3 \implies x^2 – 2x = 8 \).
\( x^2 – 2x – 8 = 0 \implies (x-4)(x+2) = 0 \).
Possible solutions are \( x=4 \) and \( x=-2 \). Since the domain of \( \log(x) \) is \( x > 0 \), we must discard \( -2 \). Thus, \( x=4 \).

Detailed Solution:
\( P(\text{not A}) = 1 – 0.4 = 0.6 \).
\( P(\text{not B}) = 1 – 0.5 = 0.5 \).
Since they are independent, the probability of neither occurring is the product of their individual probabilities:
\( P(\text{neither}) = 0.6 \times 0.5 = 0.3 \).

Detailed Solution:
First, find \( g(2) \):
\( g(2) = 2(2) + 1 = 5 \).
Next, substitute this result into \( f(x) \):
\( f(5) = \sqrt{5^2 – 9} = \sqrt{25 – 9} = \sqrt{16} = 4 \).

Detailed Solution:
Complete the square for both \( x \) and \( y \):
\( (x^2 – 6x + 9) + (y^2 + 10y + 25) = 2 + 9 + 25 \)
\( (x – 3)^2 + (y + 5)^2 = 36 \)
The standard form is \( (x-h)^2 + (y-k)^2 = r^2 \), so \( r^2 = 36 \).
Taking the square root, \( r = 6 \).

Detailed Solution:
Let the three consecutive even integers be \( n \), \( n+2 \), and \( n+4 \).
\( n + (n+2) + (n+4) = 72 \)
\( 3n + 6 = 72 \implies 3n = 66 \implies n = 22 \).
The integers are 22, 24, and 26. The largest is 26.

Detailed Solution:
Use the FOIL method:
\( (3)(4) + (3)(-i) + (2i)(4) + (2i)(-i) \)
\( 12 – 3i + 8i – 2i^2 \).
Since \( i^2 = -1 \), the last term becomes \( -2(-1) = +2 \):
\( 12 + 2 + 5i = 14 + 5i \).

Detailed Solution:
The slope of a perpendicular line is the negative reciprocal. The negative reciprocal of \( -1/3 \) is 3.
Use the slope formula: \( m = \frac{y_2 – y_1}{x_2 – x_1} \).
\( 3 = \frac{k – 5}{4 – 2} \)
\( 3 = \frac{k – 5}{2} \implies 6 = k – 5 \implies k = 11 \).