answer:Let's denote the speed of the stream as "v" km/h. When the man swims downstream, his effective speed is the sum of his own speed and the speed of the stream. So, the downstream speed is (7 + v) km/h. When he swims upstream, his effective speed is his own speed minus the speed of the stream. So, the upstream speed is (7 - v) km/h. We know that the man swims 32 km downstream in 4 hours. Therefore, we can write the equation for downstream as: Distance = Speed * Time 32 km = (7 + v) km/h * 4 h From this, we can solve for v: 32 km = 28 km + 4v km 4v km = 32 km - 28 km 4v km = 4 km v = 1 km/h Now that we know the speed of the stream is 1 km/h, we can find out how far he swam upstream. We know that he took 4 hours to swim upstream. So, using the upstream speed (7 - v) km/h, which is (7 - 1) km/h = 6 km/h, we can write the equation for upstream as: Distance = Speed * Time Distance = 6 km/h * 4 h Distance = 24 km Therefore, the man swam boxed{24} km upstream.

answer:Let's denote the price of each basketball as ( P ). Coach A bought 10 basketballs, so the total cost for Coach A is ( 10P ). Coach B bought 14 baseballs at 2.50 each and a baseball bat for 18. So the total cost for Coach B is ( 14 times 2.50 + 18 ). According to the information given, Coach A spent 237 more than Coach B. We can set up the following equation: [ 10P = (14 times 2.50 + 18) + 237 ] Now, let's calculate the total cost for Coach B's purchases: [ 14 times 2.50 = 35 ] [ 35 + 18 = 53 ] So the equation becomes: [ 10P = 53 + 237 ] [ 10P = 290 ] Now, we divide both sides by 10 to find the price of each basketball: [ P = frac{290}{10} ] [ P = 29 ] Therefore, the price of each basketball is boxed{29} .

answer:First, we determine the distance (d) from the circle's center (0,0) to the line 3x+4y-25=0. This can be calculated using the formula for the distance between a point and a line: d = frac{|Ax_1 + By_1 + C|}{sqrt{A^2 + B^2}} where (x_1, y_1) is the point, and Ax + By + C = 0 is the line equation. In our case, (x_1, y_1) = (0,0), A = 3, B = 4, and C = -25. Plugging these values into the formula, we get: d = frac{|3(0) + 4(0) - 25|}{sqrt{3^2 + 4^2}} = frac{25}{sqrt{9+16}} = 5 Now, let r be the radius of the circle. Since the circle's equation is x^{2}+y^{2}=4, we have r=sqrt{4}=2. The minimum distance from a point on the circle to the line is equal to the difference between the distance from the center to the line and the radius of the circle, which is AC = d - r = 5 - 2 = 3. Therefore, the minimum distance from a point on the circle x^{2}+y^{2}=4 to the line 3x+4y-25=0 is boxed{3}.

answer:Let's analyze the problem step by step to prove that a^{2} = 2(a + 2b) - 1. 1. Recognizing symmetry: Observe that the given system is: [ begin{cases} y = x^2 + ax + b x = y^2 + ay + b end{cases} ] If (x, y) is a solution to this system, then (y, x) is also a solution due to the symmetry of the equations. 2. Unique solution implication by symmetry: Given that the system has exactly one solution, this implies that this solution must satisfy x = y. Otherwise, if x neq y, (x, y) and (y, x) would form two different solutions due to the symmetry, contradicting the uniqueness. 3. Setting x = y: Substituting y = x into the first equation: [ x = x^2 + ax + b ] This can be rewritten as: [ x^2 + (a-1)x + b = 0 ] 4. Condition for a quadratic equation to have a unique solution: For a quadratic equation (Ax^2 + Bx + C = 0) to have a unique solution, its discriminant must be zero. The discriminant of the equation (x^2 + (a-1)x + b = 0) is given by: [ D = (a-1)^2 - 4b ] Since there is a unique solution, we set the discriminant to zero: [ (a-1)^2 - 4b = 0 ] 5. Solving for (a^2): Expand and simplify the equation: [ a^2 - 2a + 1 - 4b = 0 implies a^2 - 2a + 1 = 4b ] Rearrange to match the required form: [ a^2 = 4b + 2a - 1 ] 6. Conclusion: Hence, we have proven that: [ a^2 = 2(a + 2b) - 1 ] (boxed{a^2 = 2(a + 2b) - 1})