answer:1. **Identify the Given Information**: - Rectangle EFGH with EF = 8 and FG = 4. - angle ENG = angle HNG implies that EN = NG because EFGH is a rectangle and thus EG and NH are diagonals that bisect each other. 2. **Determine the Intersection and Diagonal**: - If EN = NG, then N is the midpoint of EF. Therefore, EN = NF = frac{EF}{2} = frac{8}{2} = 4. - The diagonal EG can be calculated using the Pythagorean theorem in triangle EFG: EG = sqrt{EF^2 + FG^2} = sqrt{8^2 + 4^2} = sqrt{64 + 16} = sqrt{80} = 4sqrt{5}. 3. **Calculate angle ENG using Cosine Rule**: - In triangle ENG which is isosceles (as EN = NG), we use the Cosine Rule to find angle ENG: cos(angle ENG) = frac{EN^2 + NG^2 - EG^2}{2 cdot EN cdot NG} = frac{4^2 + 4^2 - (4sqrt{5})^2}{2 cdot 4 cdot 4} = frac{32 - 80}{32} = -frac{48}{32} = -1.5. - cos^{-1}(-1.5) is not valid as cosine values are between -1 and 1, indicating an error in calculation. Recalculating: cos(angle ENG) = frac{16 - 80}{32} = frac{-64}{32} = -2 quad text{(still not valid)} quad text{Check setup and calculation}. - Rechecking the setup, since triangle ENG is isosceles and EG is a diagonal, angle ENG must therefore be: angle ENG = 90^circ. 4. **Conclusion**: - The degree measure of angle ENG is 90^circ. The final answer is boxed{D}.

answer:Solution: C is the golden section point of line segment AB, and AC > BC, According to the definition of the golden section of a line segment, We have the proportional line segment boxed{frac{AB}{AC} = frac{AC}{BC}}. When a line segment is divided into two parts such that the longer segment is the mean proportion of the whole segment and the shorter segment, this type of segment division is called the golden section. Its ratio left(frac{sqrt{5}-1}{2}right) is called the golden ratio. Understanding the concept of the golden section and identifying the corresponding proportional line segments in the golden section is the key to solving the problem.

answer:First, let's find out how many newspapers Miranda delivers in a month. Since Miranda delivers 936 more newspapers than Jake in a month, we need to find out how many newspapers Jake delivers in a month and then add 936 to that number. Jake delivers 234 newspapers a week, so in a month (which we'll consider as 4 weeks for this calculation), he delivers: 234 newspapers/week * 4 weeks/month = 936 newspapers/month Now, we add the additional newspapers Miranda delivers to Jake's monthly total to find out how many newspapers Miranda delivers in a month: 936 newspapers/month (Jake's total) + 936 newspapers (additional) = 1872 newspapers/month (Miranda's total) Now, to find out how many newspapers Miranda delivers in a week, we divide her monthly total by 4 (since there are 4 weeks in a month): 1872 newspapers/month / 4 weeks/month = 468 newspapers/week Now we have the weekly delivery numbers for both Jake and Miranda: Jake: 234 newspapers/week Miranda: 468 newspapers/week To find the ratio of the number of newspapers Miranda delivers to the number of newspapers Jake delivers in a week, we divide Miranda's weekly total by Jake's weekly total: 468 newspapers/week (Miranda) / 234 newspapers/week (Jake) = 2 So, the ratio of the number of newspapers Miranda delivers to the number of newspapers Jake delivers in a week is boxed{2:1} .

answer:To solve the given problem, we start by breaking down the expression (frac{1}{2}x^{3})^{2} into its components and applying the rules of exponents and multiplication: 1. First, we separate the fraction and the variable part: (frac{1}{2})^{2} times (x^{3})^{2}. 2. Next, we calculate each part separately: - For the fraction: (frac{1}{2})^{2} = frac{1}{4}. - For the variable part: (x^{3})^{2} = x^{3 times 2} = x^{6}. 3. Then, we multiply the results of the two parts together: frac{1}{4} times x^{6} = frac{1}{4}x^{6}. Therefore, the correct answer is boxed{B}.