answer:1. **Analyze the given problem:** A circle intersects a line at two points. We need to prove that the circle forms equal angles with the line at these points of intersection. 2. **Define the points of intersection and draw tangents:** Let's denote the points of intersection between the circle and the line as (A) and (B). According to the definition of the angle between a line (tangent) and a circle, at each intersection point (A) and (B), draw tangents to the circle. Let these tangents be denoted as line segments (t_A) at (A) and (t_B) at (B). 3. **Utilize the fact about tangents from a common external point:** Each tangent from a common external point to a circle has equal lengths. This property implies that the lengths of the tangents (t_A) and (t_B) are equal if they are drawn from points (A) and (B), respectively. 4. **Identify the key geometric properties:** Since (t_A) and (t_B) are tangents to the circle at the points (A) and (B), they form right angles with the radii drawn to the points of tangency. That is, [ t_A perp OA quad text{and} quad t_B perp OB ] where (O) is the center of the circle. 5. **Form an isosceles triangle:** By considering the segment of the line that intersects the circle as the base, and the points of tangency forming the legs, you create an isosceles triangle. Therefore, since the tangents (t_A) and (t_B) are equal, and radii are equal, the triangles ( triangle OAC) and ( triangle OBC) are isosceles with (OA = OB). 6. **Prove angles at the base of isosceles triangle are equal:** In (triangle OAC) and (triangle OBC), since they are both isosceles triangles, it follows from the properties of isosceles triangles (base angles are equal) that: [ angle OAC = angle OBC ] 7. **Conclude that the angles between the line and the circle are equal:** Therefore, the angles formed between the line and the tangents at points (A) and (B), (angle t_AA) and (angle t_BB) are equal due to the equal base angles of the isosceles triangles. # Conclusion: [ boxed{text{Thus, a circle intersecting a line at two points forms equal angles with the line at these points of intersection.}} ]

answer:Solution: (1) From the given information, we know e= dfrac{c}{a}= dfrac{1}{2}, thus e^2= dfrac{c^2}{a^2}= dfrac{a^2-b^2}{a^2}= dfrac{1}{4}, which implies a^2= dfrac{4}{3}b^2, Since the circle with the ellipse's minor axis as its radius is tangent to the line x - y + sqrt{6} = 0, therefore b= dfrac{sqrt{6}}{sqrt{1+1}}= sqrt{3}, thus a^2=4, b^2=3, so the equation of the ellipse is dfrac{x^2}{4} + dfrac{y^2}{3} = 1; (2) Given that the slope of line AB exists, let the equation of line AB be y=k(x-4). Substituting the line equation y=k(x-4) into the ellipse equation yields: (3+4k^2)x^2-32k^2x+64k^2-12=0, From triangle > 0, we get: 1024k^4-4(3+4k^2)(64k^2-12) > 0, solving this gives k^2 < dfrac{1}{4}, Let A(x_1,y_1), B(x_2,y_2), then x_1+x_2= dfrac{32k^2}{3+4k^2}, x_1x_2= dfrac{64k^2-12}{3+4k^2}, therefore overrightarrow{OA} cdot overrightarrow{OB}=x_1x_2+y_1y_2=(1+k^2) cdot dfrac{64k^2-12}{4k^2+3}-4k^2 cdot dfrac{32k^2}{4k^2+3}+16k^2=25- dfrac{87}{4k^2+3}, Since 0 leqslant k^2 < dfrac{1}{4}, therefore overrightarrow{OA} cdot overrightarrow{OB} in [-4, dfrac{13}{4}), therefore The range of values for overrightarrow{OA} cdot overrightarrow{OB} is boxed{[-4, dfrac{13}{4})}.

answer:Simple interest can be calculated using the formula: Simple Interest (SI) = Principal (P) × Rate of Interest (R) × Time (T) Given: Principal (P) = 400 Rate of Interest (R) = 20% per annum Time (T) = 2 years First, convert the rate of interest from a percentage to a decimal by dividing by 100: R = 20% ÷ 100 = 0.20 Now, plug the values into the formula: SI = P × R × T SI = 400 × 0.20 × 2 SI = 80 × 2 SI = 160 The simple interest for 2 years at a rate of 20% on a principal amount of 400 is boxed{160} .

answer:Let's denote the initial number of Ace cards that Nell had as A. According to the information given, after giving some cards to Jeff, Nell has 55 Ace cards and 178 baseball cards left. We also know that she now has 123 more baseball cards than Ace cards. This can be represented by the equation: 178 (baseball cards left) = 55 (Ace cards left) + 123 Now, let's find out how many baseball cards Nell gave to Jeff: Nell had 438 baseball cards initially and now has 178 left, so she gave away: 438 (initial baseball cards) - 178 (baseball cards left) = 260 baseball cards Now, let's find out how many Ace cards she gave to Jeff: Since Nell now has 123 more baseball cards than Ace cards, we can say that initially, she had: A (initial Ace cards) - 55 (Ace cards left) = 260 (baseball cards given away) Now we can solve for the initial number of Ace cards (A): A - 55 = 260 A = 260 + 55 A = 315 Nell had initially boxed{315} Ace cards.