💡First, I need to determine the new prices for the jewelry and paintings after Mr. Rocky increases them. The jewelry was originally priced at 30 each, and he increases it by 10, making the new price 40 per piece. For the paintings, the original price is 100 each, and a 20% increase means each painting now costs 120.Next, I'll calculate the sales tax for each item. The jewelry has a 6% sales tax, so the tax per piece is 6% of 40, which is 2.40. Adding this tax to the new price, each piece of jewelry costs 42.40 after tax. For the paintings, the 8% sales tax amounts to 9.60 per painting, making the total price per painting 129.60 after tax.Now, I'll calculate the total cost for two pieces of jewelry and five paintings before any discounts. Two pieces of jewelry cost 2 times 42.40, totaling 84.80. Five paintings cost 5 times 129.60, which is 648.00. Adding these together, the total price before any discounts is 732.80.Finally, I need to check if the buyer qualifies for the 10% discount. The discount is available if the buyer purchases more than one piece of jewelry, at least three paintings, and spends a minimum of 800. In this case, the buyer meets the first two conditions but does not meet the minimum spending requirement of 800. Therefore, the buyer does not qualify for the discount, and the total price remains 732.80.

💡First, I recognize that the area of a triangle can be calculated using the formula:[text{Area} = frac{1}{2} times text{base} times text{height}]Given the area of the triangle is 24 square units and the base length is 6 units, I can substitute these values into the formula to find the height.[24 = frac{1}{2} times 6 times text{height}]Next, I simplify the equation by multiplying (frac{1}{2}) by 6:[24 = 3 times text{height}]Finally, to solve for the height, I divide both sides of the equation by 3:[text{height} = frac{24}{3} = 8]Therefore, the height of the triangle is 8 units.

💡Okay, so I have this problem about two circles, X and Y. Circle X has a radius of 120, and Circle Y has an integer radius r, which is less than 120 and a prime number. Circle Y remains internally tangent to Circle X as it rolls around the circumference of Circle X. They start and end at the same point of tangency. I need to find how many possible values r can have.Hmm, let me try to visualize this. Circle Y is rolling inside Circle X without slipping. Since they remain tangent throughout the journey, the distance traveled by Circle Y as it rolls around Circle X must be related to their circumferences.First, I remember that when a circle rolls around another without slipping, the distance it travels is equal to the circumference of the path it takes. In this case, since Circle Y is rolling inside Circle X, the path it takes is called a hypocycloid. The length of this path should be equal to the circumference of Circle X minus the circumference of Circle Y, but I'm not entirely sure. Wait, actually, when rolling around another circle internally, the center of Circle Y moves along a circle of radius (120 - r). So, the distance the center travels is the circumference of this path, which is 2π(120 - r).But Circle Y is also rotating around its own center as it rolls. So, the total rotation of Circle Y is a combination of its revolution around Circle X and its own spinning. For the circle to return to the same point of tangency after completing the journey, the total rotation must be an integer multiple of 360 degrees, or 2π radians.Let me think about the relationship between the rotation and the distance traveled. The number of rotations Circle Y makes as it rolls around Circle X is equal to the total angle rotated divided by 2π. This should be equal to the distance traveled divided by the circumference of Circle Y.So, the total distance traveled by the center of Circle Y is 2π(120 - r). The number of rotations Circle Y makes is then (2π(120 - r)) / (2πr) = (120 - r)/r. For the circle to return to the same point of tangency, this number of rotations must be an integer. Therefore, (120 - r)/r must be an integer.Let me denote this integer as k. So, (120 - r)/r = k. Rearranging this, we get 120 - r = kr, which leads to 120 = r(k + 1). Therefore, r must be a divisor of 120.But r is given to be a prime number less than 120. So, I need to find all prime divisors of 120 that are less than 120.Let me factorize 120. 120 can be factored as 2^3 * 3 * 5. So, the prime factors are 2, 3, and 5. These are the prime numbers that divide 120.Wait, are there any other prime numbers less than 120 that divide 120? Let me check. 2, 3, 5 are the prime factors. 7 doesn't divide 120 because 120 divided by 7 is approximately 17.14, which is not an integer. Similarly, 11 doesn't divide 120, as 120 divided by 11 is about 10.9, not an integer. So, the only prime divisors of 120 are 2, 3, and 5.Therefore, the possible values of r are 2, 3, and 5. That gives us three possible values.Wait, but let me double-check. If r is 2, then (120 - 2)/2 = 59, which is an integer. Similarly, for r=3, (120 - 3)/3 = 39, which is an integer. For r=5, (120 - 5)/5 = 23, which is also an integer. So, all three satisfy the condition.Are there any other prime numbers less than 120 that could work? Let's see. For example, r=7. Then, (120 - 7)/7 = 120/7 - 1 ≈ 17.14 - 1 = 16.14, which is not an integer. So, 7 doesn't work. Similarly, r=11: (120 - 11)/11 = 109/11 ≈ 9.909, not an integer. So, 11 doesn't work either.Therefore, only 2, 3, and 5 are the prime numbers less than 120 that divide 120, making them the possible values for r.So, the number of possible values r can have is 3.**Final Answer**boxed{B}

💡First, I need to determine the total space available for placing the posters and the spaces between them. The wall is 20 meters wide, and the posters are placed starting 1 meter away from each end. This means I need to subtract the 1-meter gaps at both ends from the total width.20 meters (total width) - 1 meter (left end gap) - 1 meter (right end gap) = 18 metersNext, I need to figure out how many spaces there will be between the 7 posters. Since there are 7 posters, there will be 6 spaces between them.Finally, to find the space between each poster, I divide the available space by the number of spaces.18 meters / 6 spaces = 3 metersTherefore, the space between each poster is 3 meters.

💡Okay, so I need to figure out how many pairs of two-digit positive integers have a difference of 50. Hmm, let's break this down step by step.First, let's understand what a two-digit number is. A two-digit number ranges from 10 to 99, right? So both numbers in the pair must be within this range.Now, the problem says the difference between the two numbers is 50. That means if I have two numbers, say, a and b, then either a - b = 50 or b - a = 50. But since difference is a positive value, I can just consider one case, say a - b = 50, and then account for the other case if needed. Wait, actually, since the difference is 50, it doesn't matter which one is bigger; the absolute difference is 50. So, I can just focus on one scenario where a is larger than b, and then I can double the count if necessary. Hmm, but maybe I don't need to because each pair is unique regardless of order. Let me think about that later.So, let's assume a is the larger number and b is the smaller one. Therefore, a - b = 50. Since both a and b are two-digit numbers, a must be at least 50 more than b. But wait, b has to be a two-digit number, so b must be at least 10. Therefore, the smallest possible value for a would be when b is 10, so a would be 60. Similarly, the largest possible value for a is 99, so b would be 99 - 50 = 49.So, b can range from 10 to 49, and for each value of b, a is uniquely determined as b + 50. Let's see how many values that is. If b starts at 10 and goes up to 49, that's 40 numbers because 49 - 10 = 39, and adding 1 gives 40.Wait, let me double-check that. If b is 10, then a is 60. If b is 11, a is 61, and so on, up to b = 49, which gives a = 99. So, yes, that's 40 pairs.But hold on, does the order matter? For example, is (60, 10) considered a different pair than (10, 60)? The problem says "pairs," which usually means unordered pairs, so (60, 10) and (10, 60) would be the same pair. But in my initial assumption, I considered a as the larger number, so I only counted each pair once. Therefore, the total number of unique pairs is 40.Alternatively, if the problem had considered ordered pairs, meaning (a, b) and (b, a) as different, then the number would be double, but I don't think that's the case here.Let me also think about whether there are any other constraints or possibilities I might have missed. For instance, could a be less than b? Well, if a is less than b, then b - a = 50, but since a and b are both two-digit numbers, the maximum b can be is 99, so a would have to be 99 - 50 = 49. But wait, that's the same as when a is larger than b. So, actually, whether a is larger or smaller, the pairs are the same, just reversed. Therefore, counting them once is sufficient.Another thing to consider: are there any two-digit numbers where the difference is 50 but one of them is not a two-digit number? For example, if b were 9, then a would be 59, but b = 9 is not a two-digit number, so that's invalid. Similarly, if b were 50, then a would be 100, which is a three-digit number, so that's also invalid. Therefore, b must be between 10 and 49, as I initially thought, to ensure both a and b are two-digit numbers.So, to recap, b can be any two-digit number from 10 to 49, inclusive. For each of these 40 values of b, there is exactly one corresponding a = b + 50, which is also a two-digit number. Therefore, there are 40 such pairs.I think that's it. I don't see any other possibilities or constraints that I've missed. So, the answer should be 40 pairs.

💡Okay, so I have this problem here about finding the distance between two parallel lines. The lines are given by the equations ( l_1: 3x + 4y + 6 = 0 ) and ( l_2: (a+1)x + 2ay + 1 = 0 ). Hmm, I need to find the distance between these two lines, but first, I think I need to make sure they're actually parallel. Alright, I remember that two lines are parallel if their slopes are equal. But these equations are in standard form, not slope-intercept form, so maybe I should convert them to slope-intercept form to check. Let me try that.Starting with ( l_1: 3x + 4y + 6 = 0 ). Let me solve for y:( 4y = -3x - 6 )Divide both sides by 4:( y = (-3/4)x - 6/4 )Simplify:( y = (-3/4)x - 3/2 )So the slope of ( l_1 ) is -3/4.Now, let's do the same for ( l_2: (a+1)x + 2ay + 1 = 0 ). Solving for y:( 2ay = -(a+1)x - 1 )Divide both sides by 2a:( y = [-(a+1)/(2a)]x - 1/(2a) )So the slope of ( l_2 ) is ( -(a+1)/(2a) ).Since the lines are parallel, their slopes must be equal. So set the slopes equal to each other:( -3/4 = -(a+1)/(2a) )Let me solve for a. First, I can multiply both sides by 2a to eliminate the denominator:( -3/4 * 2a = -(a+1) )Simplify the left side:( (-3/4)*2a = (-3/2)a )So:( (-3/2)a = -(a + 1) )Multiply both sides by -1 to make it positive:( (3/2)a = a + 1 )Now, subtract a from both sides:( (3/2)a - a = 1 )Simplify:( (1/2)a = 1 )Multiply both sides by 2:( a = 2 )Okay, so a is 2. Now, let me substitute a = 2 back into the equation of ( l_2 ):( l_2: (2+1)x + 2*2y + 1 = 0 )Simplify:( 3x + 4y + 1 = 0 )So now, both lines are:( l_1: 3x + 4y + 6 = 0 )( l_2: 3x + 4y + 1 = 0 )Great, now that they are in the same form, I can use the distance formula between two parallel lines. I remember the formula is:( d = frac{|c_2 - c_1|}{sqrt{A^2 + B^2}} )Where the lines are in the form ( Ax + By + c_1 = 0 ) and ( Ax + By + c_2 = 0 ).In this case, A = 3, B = 4, ( c_1 = 6 ), and ( c_2 = 1 ).Plugging into the formula:( d = frac{|1 - 6|}{sqrt{3^2 + 4^2}} )Simplify the numerator:( |1 - 6| = |-5| = 5 )Denominator:( sqrt{9 + 16} = sqrt{25} = 5 )So,( d = frac{5}{5} = 1 )Therefore, the distance between the two parallel lines is 1 unit.Wait, let me double-check everything to make sure I didn't make a mistake. First, I found that a = 2, which makes the coefficients of x and y in both lines the same, confirming they are parallel. Then, using the distance formula, I subtracted the constants, took the absolute value, and divided by the square root of the sum of the squares of the coefficients. The calculations seem correct: 5 divided by 5 is indeed 1. So, I think the answer is correct.**Final Answer**The distance between the two parallel lines is boxed{1}.