answer:Since the triangle ABC has a right angle at point B, we can assume that point C lies on a line that is perpendicular to the line passing through points A and B. Since A and B have the same x-coordinate (0), the line AB is vertical. Therefore, the line that is perpendicular to AB will be horizontal, and point C will have the same y-coordinate as point B. So, the y-coordinate of point C will be the same as the y-coordinate of point B, which is 4a - 2. Now, let's denote the x-coordinate of point C as x. Since the area of the triangle is given as 63, we can use the formula for the area of a right-angled triangle, which is: Area = (1/2) * base * height In this case, the base of the triangle is the distance between points A and C, which is x (since A is at the origin), and the height is the distance between points A and B, which is 4a - 2. So, we have: 63 = (1/2) * x * (4a - 2) Multiplying both sides by 2 to get rid of the fraction, we have: 126 = x * (4a - 2) Now, we need to find the value of x. To do this, we need to know the value of a. However, since a is not given, we cannot find a unique solution for x. If you have a value for a, we can proceed to find the x-coordinate of point C. Otherwise, we can express x in terms of a: x = 126 / (4a - 2) So, the coordinates of point C are (x, 4a - 2), where x = boxed{126} / (4a - 2).

answer:**Step 1**: Set ( x = y = 0 ): [ f(0)^2 = (f(0) + f(0))^2 = 4f(0)^2 Rightarrow f(0)^2 = 0 Rightarrow f(0) = 0 ] **Step 2**: Set ( x = y ): [ f(2x)f(0) = (f(x) + f(x))^2 - 2x^2 f(x) Rightarrow 0 = 4f(x)^2 - 2x^2 f(x) ] [ f(x)(2f(x) - x^2) = 0 Rightarrow f(x) = 0 text{ or } f(x) = frac{x^2}{2} ] **Step 3**: Verification of possible functions ( f(x) = 0 ) and ( f(x) = frac{x^2}{2} ). - **For ( f(x) = 0 )**: [ f(x + y)f(x - y) = 0 = (0 + 0)^2 - 2xy cdot 0 ] - **For ( f(x) = frac{x^2}{2} )**: [ f(x+y)f(x-y) = frac{(x+y)^2}{2} cdot frac{(x-y)^2}{2} = frac{(x^2 + 2xy + y^2)(x^2 - 2xy + y^2)}{4} = frac{x^4 - y^4}{4} ] [ (f(x) + f(y))^2 - 2xy f(y) = left(frac{x^2}{2} + frac{y^2}{2}right)^2 - 2xy cdot frac{y^2}{2} = frac{x^4 + 2x^2y^2 + y^4}{4} - xyy^2 = frac{x^4 - y^4}{4} ] Both sides are equal. Therefore, the two functions that satisfy the equation are ( f(x) = 0 ) and ( f(x) = frac{x^2}{2} ). Conclusion: [ boxed{2} ]

answer:1. **Understanding the Problem**: - We have a square divided into five rectangles of equal area. - We know the width of one of these rectangles is 5. 2. **Analyzing the Equal Areas**: - Let the central rectangle and the rectangle below it share a common horizontal side. - Let the vertical sides of these two rectangles be x. This means the height of the central rectangle is x and the height of the rectangle below it is also x. 3. **Setting Up Variables**: - Let the width of the rectangle to the left of the central rectangle be y. - Since the area of all rectangles is equal, let’s express the central rectangle's horizontal side. It is divided such that the height of the central rectangle becomes 2y (since 2xy must match the area of the lower and upper rectangles). 4. **Calculating Area**: - The left lower rectangle has a vertical side 2x. - The horizontal side of this rectangle is y. - The area of this rectangle is 2xy. - This area equals twice the area of the central rectangle, meaning the horizontal side of the central rectangle must be 2y. 5. **Analyzing the Left Upper Rectangle**: - The horizontal side of the left upper rectangle can be broken into 3y (going up vertically) since this rectangle extends to cover the vertical slices of 3 rectangles. - The product 3y cdot 5 = 15y must equals the area of the lower rectangle, 2xy. 6. **Solving for x**: [ 2xy = 15y implies x = frac{15y}{2y} = 7.5 ] 7. **Calculating the Side of the Square**: - The total side of the square equals the sum of the height of the larger upper rectangle and the side of the horizontal stripe below it. [ text{Total Side} = 5 + 2x = 5 + 2 cdot 7.5 = 5 + 15 = 20 ] 8. **Finding the Area of the Square**: [ text{Area} = 20^2 = 400 ] # Conclusion: [boxed{400}]

answer:We start by noting that x^2 + 4 geq 4 for all x, because x^2 geq 0 and adding 4 makes it strictly positive and at least 4. Thus, g(x) = frac{1}{x^2 + 4} is always positive and becomes smaller as x^2 increases because the denominator grows larger. The smallest value the denominator can take is when x^2 = 0, which gives us x^2 + 4 = 4. So, the maximum value of g(x) occurs at x = 0 and g(0) = frac{1}{4}. As x^2 to infty, x^2 + 4 to infty and thus g(x) to 0. Therefore, g(x) can get arbitrarily close to 0 but never actually reaches 0. Putting this together, the range of g(x) is boxed{(0, frac{1}{4}]}.