💡First, I need to understand the pricing structures of both plans. Under Plan A, there is a fixed charge of 0.60 for the first 4 minutes, and then an additional charge of 0.06 per minute for any time beyond that. Under Plan B, the charge is a straightforward 0.08 per minute for the entire duration of the call.To find the duration at which both plans cost the same, I'll set up an equation where the total cost under Plan A equals the total cost under Plan B. For Plan A, the cost can be expressed as 0.60 plus 0.06 multiplied by the number of minutes beyond the initial 4 minutes. For Plan B, the cost is simply 0.08 multiplied by the total number of minutes.By setting these two expressions equal to each other, I can solve for the number of minutes where the costs are the same. This involves simplifying the equation by distributing and combining like terms. After solving, I find that the duration at which both plans cost the same is 18 minutes.

💡Okay, so I have this problem: "Is it possible to cut a non-convex pentagon into two equal pentagons?" Hmm, interesting. Let me try to wrap my head around it. First off, what exactly is a non-convex pentagon? Well, a pentagon is a five-sided polygon, right? And convex means that all the interior angles are less than 180 degrees, and none of the sides bend inwards. So, a non-convex pentagon must have at least one interior angle greater than 180 degrees, making it concave. Got it.Now, the question is about cutting this non-convex pentagon into two equal pentagons. Equal, I assume, means congruent—so the two resulting pentagons should be identical in shape and size. That sounds tricky because cutting a shape into two congruent parts usually requires some symmetry, doesn't it? But non-convex shapes can be pretty irregular, so I'm not sure if they have the necessary symmetry.Let me think about how I would approach this. Maybe I should start by visualizing a non-convex pentagon. Imagine a five-sided shape where one of the sides bends inward, creating a sort of "dent." Now, if I want to cut this into two equal pentagons, I need to find a line or a way to divide it such that both parts are also pentagons and are congruent.I wonder if there's a specific way to make the cut. Maybe along a line of symmetry? But non-convex pentagons aren't necessarily symmetric. Wait, some non-convex pentagons might still have lines of symmetry, though. For example, if the indentation is placed symmetrically, maybe there's a line that goes through the middle, splitting it into two mirror images.But then again, even if there is a line of symmetry, cutting along that line might not necessarily result in two pentagons. It could result in two different shapes, maybe quadrilaterals or something else. So, I need to be careful about how the cut affects the number of sides.Let me think about the properties of the resulting shapes. If I have a pentagon and I make a straight cut, the resulting shapes will each have one additional side compared to the original, right? Because the cut introduces a new edge. So, if I start with a pentagon and make a straight cut, the two resulting shapes would each have six sides—hexagons. But the question asks for pentagons, so that doesn't work.Hmm, maybe the cut isn't straight? Or perhaps it's not a single cut? The problem doesn't specify the type of cut, so maybe it's allowed to be a more complex cut, not just a straight line. But even then, ensuring that both resulting shapes are pentagons and congruent seems challenging.Let me try to think of an example. Suppose I have a non-convex pentagon that's almost like a house shape—like a square with a triangular roof, but with one side indented. If I make a cut from one corner to another, maybe I can split it into two congruent pentagons. But I'm not sure if that would work because the indentation might mess up the symmetry.Wait, maybe if the non-convex pentagon has a certain kind of symmetry, like rotational symmetry, then it might be possible to cut it into two congruent pentagons. For example, if you have a shape that looks the same after a 180-degree rotation, then cutting it along the axis of rotation could result in two congruent parts. But would those parts still be pentagons?I'm not entirely sure. Let me try to sketch this out in my mind. Imagine a non-convex pentagon that's symmetric about a vertical axis. If I cut it along that vertical axis, each half would have half of the original sides plus the new edge from the cut. But since the original shape is a pentagon, cutting it in half would give each piece three sides from the original and two new sides from the cut, making a total of five sides. Wait, that actually works! So, each resulting shape would be a pentagon.But hold on, does this always hold true? What if the non-convex pentagon doesn't have such symmetry? Then, cutting it might not result in congruent pentagons. So, maybe it's only possible for certain non-convex pentagons that have the right kind of symmetry.But the original question is asking if it's possible to cut a non-convex pentagon into two equal pentagons, not necessarily any non-convex pentagon. So, if there exists at least one non-convex pentagon that can be cut into two congruent pentagons, then the answer is yes.From my earlier reasoning, if a non-convex pentagon has a line of symmetry, then cutting along that line would result in two congruent pentagons. Therefore, it is possible for some non-convex pentagons.But wait, I should verify this. Let me think of a specific example. Suppose I have a non-convex pentagon that's shaped like a rectangle with an isosceles triangle attached to one side, creating an indentation. If this shape is symmetric along the vertical axis, then cutting it along that axis would indeed split it into two congruent pentagons.Each half would have three sides from the original rectangle and two sides from the triangle, plus the new edge from the cut, making five sides in total. And since the original shape was symmetric, both halves would be congruent.Okay, that makes sense. So, at least for non-convex pentagons with a line of symmetry, it's possible to cut them into two equal pentagons. Therefore, the answer should be yes, it is possible.But just to be thorough, are there non-convex pentagons without such symmetry that can still be cut into two congruent pentagons? Maybe, but the question doesn't specify "any" non-convex pentagon, just "a" non-convex pentagon. So, as long as there exists at least one example, the answer is yes.In conclusion, yes, it is possible to cut a non-convex pentagon into two equal pentagons, provided the original pentagon has the necessary symmetry.

💡Okay, so I need to find the repetend in the decimal representation of 5/17. Hmm, I remember that the repetend is the repeating part of a decimal that goes on forever. Since 17 is a prime number, I think the decimal might have a long repeating cycle. Let me try to figure this out step by step.First, I know that to convert a fraction to its decimal form, I can perform long division. So, I'll set up the division of 5 by 17. Since 5 is less than 17, I'll add a decimal point and some zeros to make it 5.000000... and start dividing.Alright, 17 goes into 5 zero times, so I write a 0 before the decimal point. Then, I bring down a 0, making it 50. Now, how many times does 17 go into 50? Let me calculate: 17 times 2 is 34, and 17 times 3 is 51, which is too big. So, it goes in 2 times. I write down 2, subtract 34 from 50, and get a remainder of 16.Next, I bring down another 0, making it 160. How many times does 17 go into 160? Let's see: 17 times 9 is 153, and 17 times 10 is 170, which is too big. So, it goes in 9 times. I write down 9, subtract 153 from 160, and get a remainder of 7.I bring down another 0, making it 70. 17 goes into 70 how many times? 17 times 4 is 68, and 17 times 5 is 85, which is too big. So, it goes in 4 times. I write down 4, subtract 68 from 70, and get a remainder of 2.Bringing down another 0, I have 20. 17 goes into 20 once, because 17 times 1 is 17, and 17 times 2 is 34, which is too big. I write down 1, subtract 17 from 20, and get a remainder of 3.Next, bring down another 0, making it 30. 17 goes into 30 once again, since 17 times 1 is 17, and 17 times 2 is 34, which is too big. I write down another 1, subtract 17 from 30, and get a remainder of 13.Bring down another 0, making it 130. 17 goes into 130 how many times? Let's see: 17 times 7 is 119, and 17 times 8 is 136, which is too big. So, it goes in 7 times. I write down 7, subtract 119 from 130, and get a remainder of 11.Wait, I think I made a mistake here. Let me check: 17 times 7 is actually 119, and 130 minus 119 is 11. So, the remainder is 11, not 9. Hmm, okay, so I need to correct that.Now, I bring down another 0, making it 110. 17 goes into 110 how many times? 17 times 6 is 102, and 17 times 7 is 119, which is too big. So, it goes in 6 times. I write down 6, subtract 102 from 110, and get a remainder of 8.Bring down another 0, making it 80. 17 goes into 80 how many times? 17 times 4 is 68, and 17 times 5 is 85, which is too big. So, it goes in 4 times. I write down 4, subtract 68 from 80, and get a remainder of 12.Bring down another 0, making it 120. 17 goes into 120 how many times? 17 times 7 is 119, and 17 times 8 is 136, which is too big. So, it goes in 7 times. I write down 7, subtract 119 from 120, and get a remainder of 1.Now, I bring down another 0, making it 10. 17 goes into 10 zero times, so I write down 0, and the remainder is still 10.Wait, this seems like it's starting to repeat because I had a remainder of 10 before. Let me check: earlier, after dividing 5 by 17, I had a remainder of 16, then 7, then 2, then 3, then 13, then 11, then 8, then 12, then 1, and now 10. Hmm, I don't see a repetition yet.Let me continue. Bring down another 0, making it 100. 17 goes into 100 how many times? 17 times 5 is 85, and 17 times 6 is 102, which is too big. So, it goes in 5 times. I write down 5, subtract 85 from 100, and get a remainder of 15.Bring down another 0, making it 150. 17 goes into 150 how many times? 17 times 8 is 136, and 17 times 9 is 153, which is too big. So, it goes in 8 times. I write down 8, subtract 136 from 150, and get a remainder of 14.Bring down another 0, making it 140. 17 goes into 140 how many times? 17 times 8 is 136, and 17 times 9 is 153, which is too big. So, it goes in 8 times again. I write down 8, subtract 136 from 140, and get a remainder of 4.Bring down another 0, making it 40. 17 goes into 40 how many times? 17 times 2 is 34, and 17 times 3 is 51, which is too big. So, it goes in 2 times. I write down 2, subtract 34 from 40, and get a remainder of 6.Bring down another 0, making it 60. 17 goes into 60 how many times? 17 times 3 is 51, and 17 times 4 is 68, which is too big. So, it goes in 3 times. I write down 3, subtract 51 from 60, and get a remainder of 9.Bring down another 0, making it 90. 17 goes into 90 how many times? 17 times 5 is 85, and 17 times 6 is 102, which is too big. So, it goes in 5 times. I write down 5, subtract 85 from 90, and get a remainder of 5.Wait a minute, the remainder is now 5, which was our original numerator. That means the cycle is about to repeat. So, let me see: starting from the beginning, the decimal expansion was 0.2941176470588235... and now we're back to a remainder of 5, which was where we started. So, the repeating part, the repetend, is the sequence of digits we've written down since the decimal point.Let me list out the digits I've written down so far: 2, 9, 4, 1, 1, 7, 6, 4, 7, 0, 5, 8, 8, 2, 3, 5. Hmm, that's 16 digits. Wait, but I thought the repetend was shorter. Maybe I made a mistake in counting.Let me recount the digits after the decimal point:1. 22. 93. 44. 15. 16. 77. 68. 49. 710. 011. 512. 813. 814. 215. 316. 5Yes, that's 16 digits. But I thought the repetend for 1/17 is 16 digits long, so maybe 5/17 also has a 16-digit repetend. Let me check if the cycle repeats correctly.Starting from the beginning, after the decimal, the digits are 2941176470588235, and then it repeats. So, the repetend is indeed 16 digits long.Wait, but in my earlier steps, I had a remainder of 5 at the end, which brings us back to the start. So, the cycle length is 16, which is the maximum possible for a prime number like 17. That makes sense because 17 is a full reptend prime in base 10, meaning its reciprocal has the maximum possible period of 16 digits.Therefore, the repetend in the decimal representation of 5/17 is the 16-digit sequence that repeats indefinitely. To confirm, I can write out the decimal expansion:5 ÷ 17 = 0.29411764705882352941176470588235...Yes, the sequence 2941176470588235 repeats over and over. So, the repetend is 2941176470588235.But wait, earlier in my long division, I got a remainder of 5 after 16 steps, which matches the start. So, the repetend is indeed 16 digits long, and it's 2941176470588235.I think that's it. The repetend for 5/17 is 2941176470588235.

💡First, I need to determine Bethany's original speed and the time she took for her first trip. She traveled 120 miles at a constant speed ( s ) miles per hour. The time taken for the first trip can be calculated using the formula:[t_1 = frac{120}{s} text{ hours}]For the second trip, Bethany increased her speed to four times her original speed, making her new speed ( 4s ) miles per hour. She traveled 480 miles at this increased speed. The time taken for the second trip is:[t_2 = frac{480}{4s} = frac{120}{s} text{ hours}]Comparing the two times:[t_1 = frac{120}{s} text{ hours}][t_2 = frac{120}{s} text{ hours}]Since both times are equal, the new time is the same as the old time.

💡Okay, so I have this problem where I need to find the equation of a line that passes through the point (-3, 3) and is tangent to the circle given by the equation (x + 2)^2 + y^2 = 1. Hmm, okay, let me try to figure this out step by step.First, I remember that a tangent line to a circle touches the circle at exactly one point. So, the line we're looking for should intersect the circle at only one point. Also, the line passes through (-3, 3), which is a specific point, so that gives us another condition.I think the general approach here is to use the point-slope form of a line. Since the line passes through (-3, 3), if I can find the slope, I can write the equation of the line. Let me denote the slope as 'k'. So, using point-slope form, the equation of the line would be:y - 3 = k(x + 3)Simplifying this, it becomes:y = kx + 3k + 3Or, rearranged:kx - y + 3k + 3 = 0Now, since this line is tangent to the circle (x + 2)^2 + y^2 = 1, the distance from the center of the circle to the line should be equal to the radius of the circle. The center of the circle is at (-2, 0), and the radius is 1.The formula for the distance from a point (x0, y0) to the line Ax + By + C = 0 is |Ax0 + By0 + C| / sqrt(A^2 + B^2). Applying this to our line and the center of the circle:Distance = |k*(-2) + (-1)*0 + 3k + 3| / sqrt(k^2 + (-1)^2)Simplifying the numerator:|-2k + 0 + 3k + 3| = |k + 3|So, the distance becomes:|k + 3| / sqrt(k^2 + 1)Since this distance must equal the radius, which is 1, we set up the equation:|k + 3| / sqrt(k^2 + 1) = 1To solve this, I can square both sides to eliminate the absolute value and the square root:(k + 3)^2 / (k^2 + 1) = 1Multiplying both sides by (k^2 + 1):(k + 3)^2 = k^2 + 1Expanding the left side:k^2 + 6k + 9 = k^2 + 1Subtracting k^2 from both sides:6k + 9 = 1Subtracting 9 from both sides:6k = -8Dividing both sides by 6:k = -8/6 = -4/3Okay, so the slope k is -4/3. Plugging this back into the equation of the line:y - 3 = (-4/3)(x + 3)Multiplying both sides by 3 to eliminate the fraction:3(y - 3) = -4(x + 3)Expanding both sides:3y - 9 = -4x - 12Bringing all terms to one side:4x + 3y + 3 = 0So, that's one tangent line. But wait, I remember that when dealing with tangent lines from a point outside the circle, there are usually two tangent lines. So, did I miss something?Let me think. When I assumed the slope was 'k', I considered the case where the slope exists, meaning the line is not vertical. But there might be another tangent line that's vertical, which would have an undefined slope. Let's check if such a line exists.A vertical line has the equation x = constant. Since the line passes through (-3, 3), the vertical line would be x = -3. Let me see if this line is tangent to the circle.The circle is centered at (-2, 0) with radius 1. The distance from the center (-2, 0) to the line x = -3 is |(-3) - (-2)| = |-1| = 1. Since the radius is also 1, this means the line x = -3 is indeed tangent to the circle.So, there are two tangent lines: one with slope -4/3 and the vertical line x = -3.Let me just double-check my calculations to make sure I didn't make a mistake. Starting from the distance formula:|k + 3| / sqrt(k^2 + 1) = 1Squaring both sides:(k + 3)^2 = k^2 + 1Expanding:k^2 + 6k + 9 = k^2 + 1Subtracting k^2:6k + 9 = 16k = -8k = -4/3That seems correct. And for the vertical line, the distance from the center to the line x = -3 is 1, which is equal to the radius, so it's tangent.Therefore, the two equations of the tangent lines are x = -3 and 4x + 3y + 3 = 0.I think that's it. I don't see any mistakes in my reasoning, so I'm confident with this answer.**Final Answer**The equations of the tangent lines are boxed{x = -3} and boxed{4x + 3y + 3 = 0}.

💡To solve (5^{-1} + 5^{-2} pmod{13}), we first need to find the modular inverses of 5 and (5^2) modulo 13.1. **Finding (5^{-1} pmod{13}):** We look for an integer (x) such that (5x equiv 1 pmod{13}). By testing values, we find that (5 times 8 = 40 equiv 1 pmod{13}). Thus, (5^{-1} equiv 8 pmod{13}).2. **Finding (5^{-2} pmod{13}):** Since (5^{-2} = (5^{-1})^2), we square the result from step 1: [ 8^2 = 64 equiv 12 pmod{13} ] Therefore, (5^{-2} equiv 12 pmod{13}).3. **Adding the inverses modulo 13:** [ 5^{-1} + 5^{-2} equiv 8 + 12 = 20 equiv 7 pmod{13} ]Thus, the final answer is (boxed{7}).