answer:To solve this problem, we first need to understand the given hyperbola equation and then use the information to find the equation of the ellipse. The given hyperbola equation is: [ dfrac{x^{2}}{4} - dfrac{y^{2}}{12} = -1 ] **Step 1: Identifying the Vertex of the Hyperbola** The standard form of a hyperbola is dfrac{x^{2}}{a^{2}} - dfrac{y^{2}}{b^{2}} = 1. Comparing this with the given equation, we can rewrite the given equation as: [ dfrac{x^{2}}{4} - dfrac{y^{2}}{12} = -1 implies dfrac{x^{2}}{4} - dfrac{y^{2}}{12} + 1 = 0 ] The vertices of the hyperbola are at points where x=0, so we find y by setting x=0: [ - dfrac{y^{2}}{12} + 1 = 0 implies y^{2} = 12 implies y = pm 2sqrt{3} ] Therefore, the vertices of the hyperbola are at left(0, -2sqrt{3}right) and left(0, 2sqrt{3}right). **Step 2: Identifying the Foci of the Hyperbola** The distance between the center and the foci of a hyperbola is c, where c^2 = a^2 + b^2. Here, a^2 = 4 and b^2 = 12, so: [ c^2 = 4 + 12 = 16 implies c = 4 ] Therefore, the foci of the hyperbola are at left(0, -4right) and left(0, 4right). **Step 3: Formulating the Equation of the Ellipse** Given that the ellipse has the focus of the hyperbola as its vertex and vice versa, the major axis length of the ellipse is 2c = 8, and the distance between the foci of the ellipse is 2a = 4sqrt{3}. Therefore, the semi-major axis a = 4 and the semi-minor axis b = 2sqrt{3}. The equation of an ellipse is given by: [ dfrac{x^{2}}{a^{2}} + dfrac{y^{2}}{b^{2}} = 1 ] Substituting a = 4 and b = 2sqrt{3}: [ dfrac{x^{2}}{4^{2}} + dfrac{y^{2}}{(2sqrt{3})^{2}} = 1 implies dfrac{x^{2}}{16} + dfrac{y^{2}}{12} = 1 ] However, considering the roles of a and b are reversed due to the orientation of the ellipse (with the major axis along the y-axis), the correct equation is: [ dfrac{x^{2}}{4} + dfrac{y^{2}}{16} = 1 ] Therefore, the correct option is: [ boxed{text{D: } dfrac{x^{2}}{4} + dfrac{y^{2}}{16} = 1} ]

answer:The problem states that the leading digit of the numbers (2^n) and (5^n) are the same for some natural numbers (n). We want to determine what this leading digit can be. 1. **Understanding the inherent properties**: - First, observe that for specific powers of 2 and 5: [ 2^5 = 32 quad text{and} quad 5^5 = 3125 ] Both numbers have the same leading digit, which is (3). 2. **Assume leading digit (a)**: - Assume that (a) is the common leading digit for both (2^n) and (5^n). - Suppose (2^n) has (s+1) digits and (5^n) has (t+1) digits. Then we can write: [ a cdot 10^{s} < 2^n < (a+1) cdot 10^{s} ] [ a cdot 10^{t} < 5^n < (a+1) cdot 10^{t} ] 3. **Multiply these inequalities**: [ (a cdot 10^{s}) times (a cdot 10^{t}) < 2^n times 5^n < left((a+1) cdot 10^{s}right) times left((a+1) cdot 10^{t}right) ] [ a^2 cdot 10^{s+t} < 10^n < (a+1)^2 cdot 10^{s+t} ] 4. **Simplify the inequality**: - Since (10^n = 10^{s+t} cdot 10^{n-s-t}), we use this property to write: [ a^2 cdot 10^{s+t} < 10^{s+t} cdot 10^{n-s-t} < (a+1)^2 cdot 10^{s+t} ] [ a^2 < 10^{n-s-t} < (a+1)^2 ] 5. **Constraint analysis**: - Given that (a geq 1) and (a+1 leq 10), we infer: [ 1 leq a^2 < 10^{n-s-t} < (a+1)^2 leq 100 ] - This implies that for (10^{n-s-t}) to be between (a^2) and ((a+1)^2), the term (10^{n-s-t} = 10). 6. **Conclusion**: - Thus, solving the equation (n-s-t = 1) gives us the constraints: [ a^2 < 10 quad text{and} quad (a+1)^2 > 10 ] - These inequalities hold if: [ a = 3 ] Therefore, the only possible common leading digit for (2^n) and (5^n) is: [ boxed{3} ]

answer:1. **Set up the relationships**: - Let the width be w and the length be l. - From the problem, l = w + 6. - Area A = w times l and Perimeter P = 2(w + l). - Given: A = 2P. 2. **Substitute the length and simplify**: - Substituting for l, we have: [ A = w times (w + 6) ] - Perimeter becomes: [ P = 2(w + w + 6) = 4w + 12 ] 3. **Apply the area condition**: - Substitute for A = 2P: [ w(w + 6) = 2(4w + 12) ] - Simplify and solve for w: [ w^2 + 6w = 8w + 24 ] [ w^2 - 2w - 24 = 0 ] - Factorizing the quadratic: [ (w - 6)(w + 4) = 0 ] - Since w is positive, w = 6. 4. **Calculate the area A for w = 3 units**: - Substitute w = 3 into the area formula: [ A = 3 times (3 + 6) = 3 times 9 = 27 ] Conclusion with boxed answer: [ 27 text{ square units} ] The final answer is boxed{textbf{(C)} 27text{ square units}}

answer:Since overrightarrow {a} and overrightarrow {b} are not collinear, overrightarrow {m}=2 overrightarrow {a}-3 overrightarrow {b} neq overrightarrow {0}. Given that overrightarrow {m} is parallel to overrightarrow {n}, there exists a real number lambda such that overrightarrow {n}=lambda overrightarrow {m}. This implies that 3 overrightarrow {a}+k overrightarrow {b}=2lambda overrightarrow {a}-3lambda overrightarrow {b}. By the fundamental theorem of plane vectors, we have the system of equations: begin{cases} 3=2lambda k=-3lambda end{cases} Solving for k, we find that k=- frac {9}{2}. Therefore, the answer is: boxed{k = -frac{9}{2}}.