answer：Okay, so I have this problem about traffic on a highway. Let me try to understand it step by step. First, the problem says that traffic on an east-west highway is moving at a constant speed. The eastbound lane is going at 60 miles per hour, and the westbound lane is going at 50 miles per hour. There's an eastbound driver who passes 40 westbound vehicles in ten minutes. The vehicles in the westbound lane are equally spaced. I need to find out how many westbound vehicles are present in a 150-mile section of the highway. The options are 100, 150, 300, 450, or 600.Hmm, okay. So, let's break this down. The eastbound driver is moving at 60 mph, and the westbound vehicles are moving at 50 mph. Since they're moving towards each other, their relative speed is the sum of their speeds. That makes sense because when two objects move towards each other, their relative speed is the sum of their individual speeds.So, the relative speed is 60 mph + 50 mph = 110 mph. That means that from the perspective of the eastbound driver, the westbound vehicles are approaching at 110 mph. Now, the driver passes 40 westbound vehicles in ten minutes. I need to figure out how many vehicles are in a 150-mile section. So, first, maybe I should find out how many vehicles pass a certain point in an hour, and then scale that up to 150 miles.But wait, the driver is moving, so in ten minutes, the driver covers some distance. Let me calculate how far the driver goes in ten minutes. Since the driver is moving at 60 mph, in one hour, they cover 60 miles. In ten minutes, which is 1/6 of an hour, they cover 60 * (10/60) = 10 miles. So, the driver moves 10 miles in ten minutes.But the relative speed is 110 mph, so in ten minutes, the distance covered relative to the westbound lane is 110 * (10/60) = 110 * (1/6) ≈ 18.33 miles. So, in ten minutes, the eastbound driver effectively covers 18.33 miles of the westbound lane. During this time, they pass 40 vehicles. So, the density of vehicles in the westbound lane is 40 vehicles per 18.33 miles. Let me calculate that density: 40 / 18.33 ≈ 2.18 vehicles per mile. Now, if the density is about 2.18 vehicles per mile, then in a 150-mile section, the number of vehicles would be 2.18 * 150 ≈ 327 vehicles. Looking at the options, 300 is the closest to 327. So, the answer should be 300.Wait, let me double-check my calculations to make sure I didn't make a mistake. First, relative speed: 60 + 50 = 110 mph. Correct.Distance covered in ten minutes at relative speed: 110 * (10/60) = 110 * (1/6) ≈ 18.33 miles. Correct.Number of vehicles passed: 40. So, density is 40 / 18.33 ≈ 2.18 vehicles per mile. Correct.Number of vehicles in 150 miles: 2.18 * 150 ≈ 327. Correct.Yes, 300 is the closest option. So, I think 300 is the right answer.

answer：First, I recognize that increasing each side of a rectangle by 100% means doubling the length and the width.Let’s denote the original length as ( L ) and the original width as ( W ). The original area of the rectangle is ( A_{text{original}} = L times W ).After the increase, the new length becomes ( 2L ) and the new width becomes ( 2W ). Therefore, the new area is ( A_{text{new}} = 2L times 2W = 4LW ).To find the percentage increase in area, I use the formula:[text{Percentage Increase} = left( frac{A_{text{new}} - A_{text{original}}}{A_{text{original}}} right) times 100%]Substituting the values:[text{Percentage Increase} = left( frac{4LW - LW}{LW} right) times 100% = 3 times 100% = 300%]

answer：Alright, so I've got this geometry problem here about triangle ABC. It says that AB equals AC equals BC, and then it's asking for the degree of angle A. The options are 40°, 50°, 60°, and 70°. Hmm, okay, let me think about this step by step.First off, the problem mentions that AB equals AC equals BC. That means all three sides of the triangle are equal in length. Wait, if all sides are equal, doesn't that make it an equilateral triangle? I remember that in an equilateral triangle, all sides are equal, and all angles are equal too. So, if it's equilateral, then each angle should be the same.Now, I recall that the sum of the angles in any triangle is always 180 degrees. So, if all three angles are equal, I can find each angle by dividing 180 by 3. Let me do that: 180 divided by 3 is 60. So, each angle in the triangle should be 60 degrees. That means angle A is 60 degrees.Looking back at the options, 60° is option C. So, I think the answer should be C: 60°. But wait, let me make sure I didn't skip any steps or make any mistakes. Sometimes, problems can be tricky, and maybe there's something I'm missing here.Let me visualize the triangle. If AB, AC, and BC are all equal, then it's definitely an equilateral triangle. In such a triangle, not only are the sides equal, but all the angles are equal as well. So, there's no way angle A could be anything other than 60°. The other options, like 40°, 50°, or 70°, don't make sense because they don't add up to 180° when multiplied by 3. For example, 40 times 3 is 120, which is less than 180. Similarly, 50 times 3 is 150, and 70 times 3 is 210, which is way over 180. So, none of those options work except for 60°.I think I'm confident now that the correct answer is 60°, which is option C. It all lines up with what I know about equilateral triangles and the sum of angles in a triangle. There's no conflicting information here, so I don't see any reason to doubt this conclusion.Just to recap: equal sides mean equilateral triangle, equilateral triangle means all angles are 60°, so angle A is 60°. Yep, that seems solid. I don't think I made any mistakes in my reasoning. It's straightforward once you recognize that all sides being equal leads to all angles being equal.

answer：To find the value of ( j ), we start by determining the slope of the given line ( 2x + 3y = -6 ). First, we rearrange the equation into slope-intercept form ( y = mx + c ):[3y = -2x - 6 implies y = -frac{2}{3}x - 2]This gives us a slope ( m = -frac{2}{3} ).Since the line through the points ( (4, -3) ) and ( (j, 19) ) is parallel, it must have the same slope. Using the slope formula:[frac{19 - (-3)}{j - 4} = -frac{2}{3}]Simplifying:[frac{22}{j - 4} = -frac{2}{3}]Cross-multiplying:[3 times 22 = -2 times (j - 4) implies 66 = -2j + 8]Solving for ( j ):[66 - 8 = -2j implies 58 = -2j implies j = -29]

answer：First, I need to understand the problem. I have a three-digit number that starts with 63, and the third digit is missing. The entire number must be a multiple of 3. I need to find the greatest possible difference between two possible values for the missing digit.I recall that a number is divisible by 3 if the sum of its digits is divisible by 3. So, I'll start by adding the known digits: 6 and 3, which gives me 9. Since 9 is already divisible by 3, the missing digit must also be a multiple of 3 to keep the entire number divisible by 3.The possible single-digit multiples of 3 are 0, 3, 6, and 9. These are the potential values for the missing digit. To find the greatest possible difference between any two of these digits, I'll subtract the smallest digit from the largest digit: 9 - 0 = 9.Therefore, the greatest possible difference between two possible values for the missing digit is 9.

answer：Alright, so I have this problem about an arithmetic sequence. Let me try to figure it out step by step. First, I know that in an arithmetic sequence, each term increases by a constant difference. The problem gives me the first term, a₁ = 1, and the seventh term, a₇ = 19. I need to find a₃ + a₅ and the sum of the first seven terms, S₇.Okay, let's start by recalling the formula for the nth term of an arithmetic sequence. It's aₙ = a₁ + (n - 1)d, where d is the common difference. So, if I can find d, I can find any term in the sequence.Given that a₁ = 1 and a₇ = 19, let's plug these into the formula to find d. For the seventh term, n = 7, so:a₇ = a₁ + (7 - 1)d 19 = 1 + 6dNow, I can solve for d:19 - 1 = 6d 18 = 6d d = 18 / 6 d = 3Alright, so the common difference d is 3. Now that I have d, I can find any term in the sequence. Let's find a₃ and a₅.For a₃:a₃ = a₁ + (3 - 1)d a₃ = 1 + 2*3 a₃ = 1 + 6 a₃ = 7For a₅:a₅ = a₁ + (5 - 1)d a₅ = 1 + 4*3 a₅ = 1 + 12 a₅ = 13Now, adding a₃ and a₅ together:a₃ + a₅ = 7 + 13 = 20So, the sum of the third and fifth terms is 20.Next, I need to find the sum of the first seven terms, S₇. The formula for the sum of the first n terms of an arithmetic sequence is Sₙ = n/2 * (a₁ + aₙ). In this case, n = 7, a₁ = 1, and a₇ = 19.Plugging these into the formula:S₇ = 7/2 * (1 + 19) S₇ = 7/2 * 20 S₇ = 7 * 10 S₇ = 70So, the sum of the first seven terms is 70.Wait, let me double-check my calculations to make sure I didn't make any mistakes. For a₃, starting at 1 and adding 3 each time: 1, 4, 7, 10, 13, 16, 19. Yep, that looks right. So a₃ is 7 and a₅ is 13, which adds up to 20.For the sum, S₇, using the formula, it's 7 times the average of the first and seventh terms. The average of 1 and 19 is 10, and 7 times 10 is 70. That seems correct.I think I've got it. The sum of the third and fifth terms is 20, and the sum of the first seven terms is 70.**Final Answer**a₃ + a₅ = boxed{20} and S₇ = boxed{70}