answer:First, we find the derivative of the given curve, y=e^{-5x}+2. The derivative is y'=-5e^{-5x}. Next, we find the slope of the tangent line at the point x=0. Substituting x=0 into the derivative, we get y'=-5e^{-5*0}=-5. Now, we know that the tangent line has a slope of -5 and passes through the point (0,3). Using the point-slope form of a linear equation, y-y_1=m(x-x_1), we can find the equation of the tangent line. Substituting the slope m=-5 and the point (x_1,y_1)=(0,3) into the point-slope form, we get y-3=-5(x-0). Simplifying this equation gives us y=-5x+3. Therefore, the equation of the tangent line to the curve y=e^{-5x}+2 at the point (0,3) is boxed{y=-5x+3}. This problem tests the understanding of the geometric meaning of a derivative: the slope of the tangent line to the curve at a given point. It also tests computational skills and is considered a basic problem.

answer:To determine the maximum number of candies that Carlson can eat by performing the described operations, we need to examine the pattern and constraints of these operations. 1. **Initialization**: Carlson starts with three boxes, each containing 10 candies. The numbers written on the boxes are: 4, 7, and 10. 2. **Understanding the operations**: - In each operation, Carlson will take a certain number of candies (denoted by the number on the box) from any one box. - He then eats 3 candies from the taken bunch. - The remaining candies (after eating 3) are placed into any of the other two boxes. 3. **Divisibility by 3**: Since Carlson eats 3 candies in each operation, the total number of candies eaten will always be a multiple of 3. This is a crucial observation because it sets the stage for proving the upper limits based on the total number of candies available. 4. **Total candies**: - Initially, there are (3 times 10 = 30) candies in total. - Given the process described, the object is to show that the maximum Carlson can eat is less than or equal to 27. 5. **Proof by contradiction**: To show that Carlson cannot eat 30 candies, assume that he could: - Before the last operation, there would be exactly 3 candies left. - Each of the numbers on the boxes (4, 7, 10) is greater than 3. Therefore, Carlson cannot perform the last operation to eat the remaining candies. This is a contradiction since he cannot make a move with these numbers if each box has at most 3 candies left. 6. **Construct an example showing Carlson can eat 27 candies**: - Here we provide a sequence of steps Carlson can follow to consume exactly 27 candies: [ begin{align*} text{Initial state}: & (10, 10, 10) text{After 1st operation}: & (6, 13, 10) quad text{(take 4 from the first box, eat 3, place 1 in the second)} text{After 2nd operation}: & (6, 6, 17) quad text{(take 7 from the second box, eat 3, place 4 in the third)} text{After 3rd operation}: & (6, 6, 7) quad text{(take 10 from the third box, eat 3, place 7 back in the same box)} text{After 4th operation}: & (3, 9, 7) quad text{(take 3 from the first box, eat 3, no leftover)} text{After 5th operation}: & (3, 2, 14) quad text{(take 7 from the second box, eat 3, place 4 in the third)} text{After 6th operation}: & (3, 2, 11) quad text{(take 10 from the third box, eat 3, place 7 back in the same box)} text{After 7th operation}: & (0, 2, 14) quad text{(take 3 from the first box, eat 3, no leftover)} text{After 8th operation}: & (0, 0, 16) quad text{(take 2 from the second box, eat 3, place 13 back in the third)} text{After 9th operation}: & (0, 0, 6) quad text{(take 10 from the third box, eat 3, place 7 in another box)} text{After 10th operation}: & (0, 0, 3) quad text{(take 4 from the last box, eat 3, place 1 back)} text{After 11th operation}: & (0, 0, 0) quad text{(take 3 from the last box, eat 3)} end{align*} ] This sequence shows that Carlson consumes exactly (3 times 9 = 27) candies. # Conclusion: [ boxed{27} ]

answer:(1) Since the right focus of the ellipse is F(m,0), the focus is on the x-axis. Let the equation of the ellipse be: frac{x^2}{a^2} + frac{y^2}{b^2} = 1 quad (a > b > 0) From this, we have c=m. The directrix equations are x=frac{a^2}{c}=m+1. Thus, we obtain a^2=m(m+1) and b^2=m. Given the eccentricity e=frac{c}{a} = sqrt{1-frac{b^2}{a^2}} = frac{sqrt{2}}{2}, we can derive b=c, which leads to m=1. Therefore, a=sqrt{2} and b=1. Hence, the equation of the ellipse is: boxed{frac{x^2}{2} + y^2 = 1} (2) From the problem, we can determine that A(-m-1,-m-1) and B(m+1,m+1). Thus, overrightarrow{AF} = (2m+1, m+1) and overrightarrow{FB} = (1, m+1). Calculate the dot product: overrightarrow{AF} cdot overrightarrow{FB} = 2m+1+(m+1)^2 = m^2 + 4m + 2 < 7. Solving for m, we get: boxed{0 < m < 1}. The eccentricity e = frac{c}{a} = frac{m}{sqrt{m(m+1)}} = frac{1}{sqrt{1+frac{1}{m}}}. Therefore, the range of the eccentricity is: boxed{left(0, frac{sqrt{2}}{2}right)}

answer:Let the two numbers be 8x and 3x, where x is a common factor. According to the problem, the sum of these two numbers is 143. So, we can write the equation as: 8x + 3x = 143 Combining like terms, we get: 11x = 143 Now, we solve for x: x = 143 / 11 x = 13 Now that we have the value of x, we can find the bigger number (which is 8x): 8x = 8 * 13 8x = 104 Therefore, the bigger number is boxed{104} .