answer:Let's denote the positions of P, Q, and R as mathbf{p} = begin{pmatrix} 0 3 0 end{pmatrix}, mathbf{q} = begin{pmatrix} 2 0 0 end{pmatrix}, mathbf{r} = begin{pmatrix} 2 6 6 end{pmatrix}. First, find the normal vector to the plane using the cross product of (mathbf{p} - mathbf{q}) and (mathbf{p} - mathbf{r}): [ (mathbf{p} - mathbf{q}) = begin{pmatrix} -2 3 0 end{pmatrix}, quad (mathbf{p} - mathbf{r}) = begin{pmatrix} -2 -3 -6 end{pmatrix} ] [ (mathbf{p} - mathbf{q}) times (mathbf{p} - mathbf{r}) = begin{pmatrix} -2 3 0 end{pmatrix} times begin{pmatrix} -2 -3 -6 end{pmatrix} = begin{pmatrix} -18 12 -6 end{pmatrix} ] We can simplify this normal vector to begin{pmatrix} 3 -2 1 end{pmatrix}. Hence, the equation of the plane is 3x - 2y + z = d. Substituting point mathbf{p} into this equation gives d = 6, so [ 3x - 2y + z = 6 ] Now, check the intersections with cube edges. Consider the edge connecting (0,6,0) and (0,6,6), and the edge connecting (0,0,6) and (6,0,6). For (0,6,z): [ -12 + z = 6 Rightarrow z = 18 ] This value is outside of the cube's edge, so check another edge. For (x,0,6): [ 3x + 6 = 6 Rightarrow x = 0 ] One intersection is at (0,0,6). For (x,6,0): [ 3x - 12 = 6 Rightarrow x = 6 ] The other intersection is at (6,6,0). Calculate the distance between these two points: [ sqrt{(6-0)^2 + (6-0)^2 + (0-6)^2} = sqrt{108} = boxed{6sqrt{3}} ]

answer:1. **Identify the Coin Values**: The coins available are pennies (1 cent), nickels (5 cents), dimes (10 cents), and quarters (25 cents). 2. **Analyze the Options**: Determine which of the total values (22, 30, 40, 48, 55 cents) could not be the sum of five coins. 3. **Minimum Value with Five Coins (Using Only Nickels as Divisibility by 5)**: - The smallest coin value divisible by 5 is a nickel (5 cents). - The minimum sum using five nickels is (5 times 5 = 25) cents. 4. **Maximum Value with Five Coins (Using All Quarters)**: - The maximum sum using five quarters is (5 times 25 = 125) cents. 5. **Checking each option**: - **22 cents**: This value cannot be achieved since it breaks the divisibility rule of 5 when only considering nickels, dimes, and quarters. - **30 cents**: Achievable with three dimes and two nickels ((10+10+10 + 5 + 5)). - **40 cents**: Achievable with four dimes and one nickel ((10+10+10+10+5)). - **48 cents**: It cannot be made divisibly by 5 using combinations that include only nickels, dimes, and quarters. - **55 cents**: Achievable with two quarters, two nickels, and one dime ((25+25+5+5+10)). 6. **Final Conclusion**: - The values 22 and 48 cents cannot be achieved with five coins under the divisibility by 5 constraint (excluding pennies), so the correct choices that cannot be achieved are textbf{(A) 22 text{ and } textbf{(D)} 48}. The final answer is boxed{textbf{(A)} 22 text{ and } textbf{(D)} 48}

answer:Given the equation |z^2 + 4| = |z(z + 2i)|, we start by expressing z^2 + 4 in a different form: [z^2 + 4 = (z + 2i)(z - 2i).] This allows us to rewrite the given equation as: [|z + 2i||z - 2i| = |z||z + 2i|.] Now, we consider two cases based on the value of |z + 2i|. **Case 1:** If |z + 2i| = 0, then z = -2i. In this scenario, we calculate |z + i| as follows: [|z + i| = |-2i + i| = |-i| = 1.] Therefore, in this case, the smallest possible value of |z + i| is 1. **Case 2:** If |z + 2i| neq 0, we can divide both sides of the equation by |z + 2i| to obtain: [|z - 2i| = |z|.] This implies that the distance of z from the origin is equal to its distance from 2i. Geometrically, this means z lies on the perpendicular bisector of the segment connecting the origin and 2i, which is the line where the imaginary part is 1. Representing z as x + i (where x is a real number), we find: [|z + i| = |x + 2i| = sqrt{x^2 + 4}.] To minimize |z + i|, we look at the expression under the square root: [|z + i| = sqrt{x^2 + 4} geq sqrt{0 + 4} = 2.] Thus, the smallest possible value of |z + i| considering both cases is boxed{1}, which occurs when z = -2i.

answer:Since text{work} = text{rate} times text{time}, let r be the rate at which one worker can complete the project. It follows that 1 project requires: [1text{ project} = (60r) times (3 text{days})] Thus, r = frac{1}{3 cdot 60}. If only 30 workers were available, then: [1text{ project} = (30r) times (t text{days})] Solving for t: [t = frac{1}{30 cdot frac{1}{3 cdot 60}} = frac{180}{30} = boxed{6} text{days}.]