answer:To find out how much shorter Chad's trip is if he canoes across the lake, we need to compare the distance he would travel by walking around the lake with the distance he would travel by canoeing across it. If Chad walks around the lake, he would be walking along the circumference of the circular lake. The formula to calculate the circumference (C) of a circle is: C = 2 * pi * r where r is the radius of the circle. Since the diameter of the lake is 2 miles, the radius (r) is half of that, which is 1 mile. So, the circumference of the lake is: C = 2 * 3.14 * 1 C = 6.28 miles If Chad canoes across the lake, he would be traveling along the diameter of the lake, which is given as 2 miles. To find out how much shorter the canoe trip is compared to walking around the lake, we subtract the canoeing distance from the walking distance: Difference = Walking distance - Canoeing distance Difference = Circumference - Diameter Difference = 6.28 miles - 2 miles Difference = 4.28 miles So, Chad's trip is boxed{4.28} miles shorter if he canoes across the lake rather than walks around it.

answer:To solve the system of equations: [ begin{cases} x^2 - y^2 = 0 (x - a)^2 + y^2 = 1 end{cases} ] we will analyze the geometric interpretations of the equations. 1. The first equation ( x^2 - y^2 = 0 ) represents two lines: [ x = y quad text{and} quad x = -y. ] 2. The second equation ( (x - a)^2 + y^2 = 1 ) represents a circle with radius 1 and center at ( (a, 0) ). Next, we determine the number of intersection points between the lines and the circle: # Case Analysis: Case a: ( a > sqrt{2} ) - For values ( a > sqrt{2} ), the center of the circle is sufficiently far from the origin. Each line can intersect the circle at two points (one on each side of the axis). - Thus, the total number of intersections is: [ 2 , text{lines} times 2 , text{points per line} = 4 , text{points}. ] Case b: ( a = sqrt{2} ) - For ( a = sqrt{2} ), the center of the circle is at ( (sqrt{2}, 0) ). The circle touches the lines exactly at points ( (sqrt{2}, pm frac{1}{sqrt{2}}) ) or ( (pm sqrt{2}, frac{1}{sqrt{2}}) ). - Each line will only touch the circle at one distinct point. - Thus, the number of distinct intersection points is: [ 2 , text{lines} times 1 , text{point per line} = 2 , text{points}. ] Case c: ( 1 < |a| < sqrt{2} ) - When ( 1 < |a| < sqrt{2} ), the circle intersects each of the two lines at two different points. - Thus, the total number of intersections is: [ 2 , text{lines} times 2 , text{points per line} = 4 , text{points}. ] Case d: ( a = 1 ) - When ( a = 1 ), the circle is tangent to each of the lines at one point. Hence each line gives 1 intersection point. - Thus, the total number of intersections is: [ 2 , text{lines} times 1 , text{point per line} = 2 , text{points}. ] Case e: ( |a| < 1 ) - For values ( |a| < 1 ), the circle intersects each line at exactly two points. - Thus, the total number of intersections is: [ 2 , text{lines} times 2 , text{points per line} = 4 , text{points}. ] # Conclusion: Summarizing the results: a) For ( a > sqrt{2} ): ( quad boxed{4} text{ points} ). b) For ( a = sqrt{2} ): ( quad boxed{2} text{ points} ). c) For ( 1 < |a| < sqrt{2} ): ( quad boxed{4} text{ points} ). d) For ( a = 1 ): ( quad boxed{2} text{ points} ). e) For ( |a| < 1 ): ( quad boxed{4} text{ points} ).

answer:**Analysis** This problem primarily tests the ability to determine the range of a parameter using the monotonicity of a function. It is a moderate difficulty problem. By using the trigonometric identity and substitution method, we can transform the analytical expression of f(x), and then solve it using the knowledge of quadratic functions. **Solution** Given f(x)=cos 2x+acos left( dfrac{pi }{2}+x right)=1-2sin ^{2}x-asin x=-2left( sin x+ dfrac {a}{4} right)^{2}+1+ dfrac {a^{2}}{8}, Let t=sin x, then f(x)=g(t)=-2left( t+ dfrac {a}{4} right)^{2}+1+ dfrac {a^{2}}{8}. Since t=sin x is an increasing function on the interval left( dfrac{pi }{6},dfrac{pi }{2} right), we have tin left( dfrac{1}{2},1 right). In order for f(x) to be an increasing function, we need -2left( t+ dfrac {a}{4} right)^{2} geq 0, which leads to -dfrac {a}{4} geq 1, Thus, aleq -4. Therefore, the correct answer is boxed{D: left( -infty ,-4 right]}.

answer:Since the function f(x)=-x^2+2ax+4a+1 has one root less than -1 and another root greater than 3, we have the following conditions: 1. f(-1) > 0, which ensures that there is a root less than -1. 2. f(3) > 0, which ensures that there is a root greater than 3. 3. The value of a should be such that the parabola is open downwards (since the coefficient of x^2 is negative) and the vertex of the parabola lies between -1 and 3. By satisfying the first two conditions, we calculate f(-1) and f(3) to ensure they are greater than zero: For f(-1) > 0: [f(-1) = -(-1)^2 + 2a(-1) + 4a + 1 > 0] [1 - 2a + 4a + 1 > 0] [2a + 2 > 0] [a > -1] For f(3) > 0: [f(3) = -(3)^2 + 2a(3) + 4a + 1 > 0] [-9 + 6a + 4a + 1 > 0] [10a - 8 > 0] [a > frac{4}{5}] The intersection of the two conditions a > -1 and a > frac{4}{5} is a > frac{4}{5}. However, we also have to consider the third condition, which limits the range of a further. The vertex of the parabola given by the formula -frac{b}{2a} (where b is the coefficient of x) should satisfy: [-1 < -frac{2a}{-2} < 3] This simplifies to: [-1 < a < 3] Combining all three conditions, the range for a becomes frac{4}{5} < a < 1. Thus, the range of real numbers for a is boxed{left(frac{4}{5}, 1right)}.