answer:First, let's determine the equation of the line AB. The equation in intercept form is given by frac {x}{-8} + frac {y}{6} = 1. Converting this to the standard form of a linear equation, we get frac {x}{-8} + frac {y}{6} - 1 = 0, which simplifies to 3x - 4y + 24 = 0. The distance d from a point (x_0, y_0) to a line Ax + By + C = 0 can be found using the formula d = frac{|Ax_0 + By_0 + C|}{sqrt{A^2 + B^2}}. Applying this formula, the distance from the origin (0,0) to the line AB is d = frac{|3(0) - 4(0) + 24|}{sqrt{3^2 + 4^2}} = frac{24}{sqrt{9 + 16}} = frac{24}{sqrt{25}} = frac{24}{5}. Therefore, the distance from the origin to the line AB is boxed{frac{24}{5}}. This problem tests the student's understanding of the intercept form of the equation of a line as well as the formula for the distance from a point to a line. It is considered a basic level question.

answer:Given the 4x4 grid, we need to determine the number in the lower right-hand square with the constraint that each number 1 through 4 appears exactly once in each row and each column. 1. **First Row Analysis**: The first row contains 1 and 2. Missing numbers are 3 and 4. The fourth column already has a 2, so 3 and 4 cannot be placed there. Thus, 3 and 4 are placed in the second and third columns respectively. [ begin{array}{|c|c|c|c|} hline 1 & 3 & 4 & 2 hline & 2 & & hline & & 3 & hline & & & hline end{array} ] 2. **Second Row Analysis**: The second row already has a 2 in the second column. The remaining numbers are 1, 3, and 4. Since 1 is in the first column and 3 is in the third column in other rows, place 4 in the third column and 1 in the fourth column. [ begin{array}{|c|c|c|c|} hline 1 & 3 & 4 & 2 hline 4 & 2 & 1 & hline & & 3 & hline & & & hline end{array} ] 3. **Third Row Analysis**: The third row already has 3. Missing numbers are 1, 2, and 4. Since 1 and 4 are placed in other rows in the first and third columns, and 2 is in the second column of the second row, place 2 in the first column and 4 in the fourth column. [ begin{array}{|c|c|c|c|} hline 1 & 3 & 4 & 2 hline 4 & 2 & 1 & hline 2 & & 3 & 4 hline & & & hline end{array} ] 4. **Fourth Row Completion**: The fourth row is missing all numbers. Using the exclusion method based on existing numbers in each column, place 3 in the second column, and 1 in the third column. [ begin{array}{|c|c|c|c|} hline 1 & 3 & 4 & 2 hline 4 & 2 & 1 & 3 hline 2 & 1 & 3 & 4 hline 3 & 4 & 2 & 1 hline end{array} ] Thus, the number in the lower right-hand square is 1. The final answer is boxed{textbf{(A)} 1}.

answer:Let's denote the speed of the boat in still water as "b" km/hr and the speed of the stream as "s" km/hr. When the man is rowing downstream, the speed of the boat is the sum of the speed of the boat in still water and the speed of the stream. So, the downstream speed is (b + s) km/hr. When the man is rowing upstream, the speed of the boat is the difference between the speed of the boat in still water and the speed of the stream. So, the upstream speed is (b - s) km/hr. We are given that the man rows 60 km downstream and 30 km upstream, taking 3 hours each time. Using the formula for distance (Distance = Speed × Time), we can write two equations: For downstream: 60 km = (b + s) km/hr × 3 hrs 60 km = 3b + 3s 20 km/hr = b + s (1) For upstream: 30 km = (b - s) km/hr × 3 hrs 30 km = 3b - 3s 10 km/hr = b - s (2) Now we have a system of two equations with two variables. We can solve for "b" and "s" by adding the two equations: (1) + (2): 20 km/hr + 10 km/hr = b + s + b - s 30 km/hr = 2b b = 15 km/hr Now we can substitute the value of "b" into either equation (1) or (2) to find "s". Let's use equation (1): 20 km/hr = 15 km/hr + s s = 20 km/hr - 15 km/hr s = 5 km/hr So, the speed of the stream is boxed{5} km/hr.

answer:1. **Let the mass of the first alloy be ( x ) kg, and the mass of the second alloy be ( 8 - x ) kg.** 2. **Establish the constraints for ( x ):** - Since the first alloy is available in 3 kg, we need ( 0 leq x leq 3 ). - Since the second alloy is available in 7 kg, we need ( 0 leq 8 - x leq 7 ). Combining these inequalities: [ left{ begin{array}{l} 0 leq x leq 3 1 leq x leq 8 end{array} right. ] This simplifies to: [ 1 leq x leq 3 ] 3. **Calculate the amount of copper in each alloy:** - The first alloy contains ( 40% ) copper: [ 0.4x , text{kg of copper in ( x ) kg of the first alloy} ] - The second alloy contains ( 30% ) copper: [ 0.3(8 - x) , text{kg of copper in ( 8 - x ) kg of the second alloy} ] 4. **Find the total amount of copper in the resulting mixture:** [ 0.4x + 0.3(8 - x) ] 5. **Express the new alloy's copper concentration, which is ( p% ) of 8 kg:** [ frac{p}{100} cdot 8 ] 6. **Set up the equation for the copper content and solve for ( x ):** [ 0.4x + 0.3(8 - x) = frac{p}{100} cdot 8 ] [ 0.4x + 2.4 - 0.3x = frac{8p}{100} ] [ 0.1x + 2.4 = 0.08p ] [ 0.1x = 0.08p - 2.4 ] [ x = 0.8p - 24 ] 7. **Substitute ( x = 0.8p - 24 ) into the constraints ( 1 leq x leq 3 ) to find ( p ):** - For the lower bound: [ 1 leq 0.8p - 24 ] [ 25 leq 0.8p ] [ p geq 31.25 ] - For the upper bound: [ 0.8p - 24 leq 3 ] [ 0.8p leq 27 ] [ p leq 33.75 ] 8. **Combine the results to get the range for ( p ):** [ 31.25 leq p leq 33.75 ] # Conclusion: The possible values of ( p ) for which the problem has a solution lie within the range ( 31.25 leq p leq 33.75 ). [ boxed{31.25 leq p leq 33.75} ]