💡Okay, so I've got this problem here about finding the vertex of a parabola. The equation is given as y equals negative nine times (x minus seven) squared. Hmm, I remember that the vertex form of a parabola is y equals a times (x minus h) squared plus k, where (h, k) is the vertex. So, in this case, I need to identify h and k from the given equation.Let me write it down again to make sure I have it right: y = -9(x - 7)^2. Comparing this to the vertex form, y = a(x - h)^2 + k, I can see that h is 7 because it's inside the parentheses with x. But wait, there's no constant term added or subtracted outside the squared term, which means k must be zero. So, the vertex should be at (7, 0).Let me double-check to make sure I didn't miss anything. The equation is y = -9(x - 7)^2. There's no "+ k" at the end, so k is indeed zero. The h is 7 because it's x minus h, and here it's x minus 7. So, h is 7 and k is 0. That makes the vertex (7, 0).I think that's it. So, looking at the options given:A: (9,0)B: (-9,7)C: (9,-7)D: (7,0)The correct one should be D: (7,0). Yeah, that makes sense because the vertex form directly gives us h and k, and there's no need for any further calculations here.

💡First, I need to determine the total number of pencils Ashton initially had. He has three boxes with fourteen pencils each.Next, I'll calculate the total number of pencils he gave away. He gave six pencils to his brother and twelve pencils to his friends.Finally, I'll subtract the total number of pencils given away from the initial total to find out how many pencils Ashton has left.

💡Okay, so I've got this problem here where Jack has some money in different currencies: pounds, euros, and yen. I need to figure out how much Jack has altogether in yen. Hmm, let's see. The problem says Jack has 42 pounds, 11 euros, and 3000 yen. The exchange rates given are 2 pounds per euro and 100 yen per pound. Alright, first things first, I need to make sure I understand the exchange rates correctly. It says there are 2 pounds per euro. So, that means for every euro, you get 2 pounds. Similarly, it says there are 100 yen per pound, meaning for every pound, you get 100 yen. So, to find out how much Jack has in yen, I need to convert all his money into yen. That means converting the euros he has into pounds first, then converting those pounds into yen, and then adding that to the yen he already has. Let me break it down step by step. First, Jack has 11 euros. Since the exchange rate is 2 pounds per euro, I can calculate how many pounds that is. So, 11 euros times 2 pounds per euro. Let me write that out: 11 * 2 = 22 pounds. Okay, so 11 euros is equal to 22 pounds. Now, Jack already has 42 pounds. So, I need to add the pounds he got from converting euros to his existing pounds. That would be 42 pounds + 22 pounds = 64 pounds total. Next, I need to convert these 64 pounds into yen. The exchange rate is 100 yen per pound, so I multiply the number of pounds by 100 to get yen. So, 64 * 100 = 6400 yen. Finally, Jack already has 3000 yen. So, I need to add that to the 6400 yen I just calculated. That would be 6400 + 3000 = 9400 yen. Wait a minute, let me double-check my calculations to make sure I didn't make any mistakes. Starting with the euros: 11 euros * 2 pounds/euro = 22 pounds. That seems right. Adding that to the existing pounds: 42 + 22 = 64 pounds. Yep, that checks out. Converting pounds to yen: 64 pounds * 100 yen/pound = 6400 yen. That makes sense. Adding the existing yen: 6400 + 3000 = 9400 yen. Okay, that seems correct. So, after converting all of Jack's money into yen, he has a total of 9400 yen. I think I've covered all the steps and double-checked my calculations. It seems straightforward, but it's always good to verify each step to avoid any errors, especially when dealing with currency conversions which can sometimes be tricky.

💡First, I need to determine the number of days between the 15th day and the 365th day of the year 2003. This can be calculated by subtracting 15 from 365, which gives 350 days.Next, to find out how these 350 days translate into weeks and extra days, I'll divide 350 by 7 (since there are 7 days in a week). 350 divided by 7 equals exactly 50 weeks, with no remainder. This means that 350 days is exactly 50 weeks, and there are no extra days beyond complete weeks.Since there are no extra days, the day of the week on the 365th day will be the same as the day of the week on the 15th day. Given that the 15th day is a Monday, the 365th day will also be a Monday.Therefore, the final day of the year 2003 falls on a Monday.

💡Okay, so I have this geometry problem here about an acute triangle ABC. The problem says that the altitudes AX and BY intersect at H. I know that in a triangle, the point where all three altitudes intersect is called the orthocenter, so H must be the orthocenter of triangle ABC. The angles given are angle BAC, which is 61 degrees, and angle ABC, which is 73 degrees. I need to find the measure of angle CHX. Hmm, I'm not entirely sure where point X is located. Since AX is an altitude, X must be the foot of the altitude from A to BC. Similarly, Y is the foot of the altitude from B to AC. So, X is on BC, and Y is on AC.Let me try to visualize triangle ABC. Since it's an acute triangle, all altitudes are inside the triangle, and the orthocenter H is also inside the triangle. So, AX is perpendicular to BC, and BY is perpendicular to AC. Their intersection is H.I think I should start by finding the measure of the third angle in triangle ABC, which is angle ACB. Since the sum of the angles in a triangle is 180 degrees, angle ACB = 180 - 61 - 73. Let me calculate that: 180 - 61 is 119, and 119 - 73 is 46. So, angle ACB is 46 degrees.Now, I need to find angle CHX. Let me think about point X. Since AX is the altitude from A to BC, X is on BC, and AX is perpendicular to BC. So, triangle AXH is a right triangle with right angle at X. Similarly, triangle BXH is also a right triangle with right angle at X.Wait, maybe I should consider triangle CHX. To find angle CHX, I need to look at triangle CHX. I know that H is the orthocenter, so CH is another altitude. But wait, CH is the altitude from C to AB, right? So, CH is perpendicular to AB.But in triangle CHX, we have point X on BC, and H somewhere inside the triangle. So, angle CHX is the angle at point H between points C and X. Hmm, I'm not sure how to approach this directly.Maybe I can use some properties of the orthocenter or cyclic quadrilaterals. I remember that in a triangle, the orthocenter creates several cyclic quadrilaterals because of the right angles. For example, since AX and BY are altitudes, quadrilateral AHBX is cyclic because both angles at X and Y are right angles.Wait, actually, quadrilateral AHBX has two right angles at X and Y, but I'm not sure if that's directly helpful here. Maybe I should consider the angles around point H.Since H is the orthocenter, the angles at H are related to the angles of the original triangle. For example, angle BHC is equal to 180 degrees minus angle BAC. Let me verify that: in triangle ABC, angle BAC is 61 degrees, so angle BHC should be 180 - 61 = 119 degrees. Is that correct? I think so because the orthocenter reflects the original angles in a way.Similarly, angle AHB would be 180 - angle ACB, which is 180 - 46 = 134 degrees. And angle AHC would be 180 - angle ABC, which is 180 - 73 = 107 degrees. Hmm, I'm not sure if that's directly helpful, but it's good to note.Let me try to find some relationships in triangle CHX. I know that CH is perpendicular to AB, so angle CHX is related to the angles in triangle ABC. Maybe I can find some similar triangles or use trigonometric ratios.Alternatively, perhaps I can use coordinate geometry. Let me assign coordinates to the triangle to make it easier. Let's place point A at (0, 0), point B at (c, 0), and point C somewhere in the plane. Since angle BAC is 61 degrees and angle ABC is 73 degrees, I can use the Law of Sines to find the lengths of the sides.But this might get complicated. Maybe there's a simpler way. Let me think about the properties of the orthocenter again. Since H is the orthocenter, the reflections of H over the sides lie on the circumcircle of triangle ABC. But I'm not sure if that helps here.Wait, maybe I can consider triangle CHX. Since CH is perpendicular to AB, and AX is perpendicular to BC, perhaps there's a relationship between these two altitudes. Maybe triangle CHX is similar to some other triangle in the figure.Alternatively, I can consider the fact that in triangle ABC, the orthocenter H creates several smaller triangles, each of which is similar to the original triangle. Maybe triangle CHX is similar to triangle ABC or some other triangle.Let me try to find the measure of angle HXC. Since AX is perpendicular to BC, angle AXH is 90 degrees. Similarly, since CH is perpendicular to AB, angle CHX is part of that right angle. Maybe I can find some relationship between these angles.Wait, perhaps I can use the fact that the sum of the angles in triangle CHX is 180 degrees. If I can find two angles, I can find the third one. I know that angle CHX is what I'm looking for, and if I can find angles at C and X, I can find angle CHX.But I'm not sure about the measures of those angles. Maybe I need to find some other angles first. Let me think about the angles at point H. Since H is the orthocenter, the angles at H are related to the original angles of the triangle.I remember that in the orthocenter configuration, the angles at H are equal to 180 degrees minus the original angles of the triangle. So, angle BHC is 180 - angle BAC = 119 degrees, as I calculated earlier. Similarly, angle AHB is 180 - angle ACB = 134 degrees, and angle AHC is 180 - angle ABC = 107 degrees.Now, focusing on angle CHX, which is part of triangle CHX. If I can find the measure of angle HCX, then I can find angle CHX because I know that angle CHX is 90 degrees minus angle HCX, since CH is perpendicular to AB.Wait, is that correct? Let me think. Since CH is perpendicular to AB, and X is on BC, then angle CHX is part of the right angle at H. So, angle CHX is equal to 90 degrees minus angle HCX.But angle HCX is the same as angle ACB, which is 46 degrees. Wait, is that true? Let me see. Since CH is perpendicular to AB, and AX is perpendicular to BC, maybe there's a relationship between these angles.Alternatively, maybe angle HCX is equal to angle ABC, which is 73 degrees. Hmm, I'm getting confused here. Let me try to draw a diagram to visualize this better.[Imagining a diagram: Triangle ABC with A at the top, B at the bottom left, and C at the bottom right. Altitudes AX and BY intersect at H inside the triangle. X is on BC, and Y is on AC.]Looking at point X on BC, and H inside the triangle. CH is perpendicular to AB, so from C, the altitude goes to AB at some point, say Z. So, CZ is perpendicular to AB, and H lies somewhere along CZ.Similarly, AX is perpendicular to BC, so from A, the altitude goes to BC at X. So, AX is perpendicular to BC, and H lies somewhere along AX.Now, in triangle CHX, we have point H on AX and on CZ. So, angle CHX is the angle at H between points C and X.Since CH is perpendicular to AB, and AX is perpendicular to BC, maybe there's a way to relate these angles.Wait, perhaps I can consider the quadrilateral formed by points A, H, X, and Z. Since both AX and CZ are altitudes, and they intersect at H, maybe this quadrilateral has some special properties.Alternatively, maybe I can use the fact that in triangle ABC, the orthocenter H creates several right angles, and these can be used to find angle CHX.Let me try to find angle HCX. Since CH is perpendicular to AB, and AX is perpendicular to BC, maybe angle HCX is equal to angle ABC, which is 73 degrees.Wait, that might make sense because angle ABC is at point B, and angle HCX is at point C, but they might be related through the orthocenter.If angle HCX is 73 degrees, then angle CHX would be 90 - 73 = 17 degrees. But that seems too small. Maybe I'm making a mistake here.Alternatively, if angle HCX is equal to angle ACB, which is 46 degrees, then angle CHX would be 90 - 46 = 44 degrees. That seems more reasonable, but I'm not sure.Wait, maybe I should consider the triangle CHX more carefully. Since CH is perpendicular to AB, and AX is perpendicular to BC, perhaps triangle CHX is similar to triangle ABC or some other triangle.Alternatively, maybe I can use the fact that the angles in triangle CHX add up to 180 degrees. If I can find two angles, I can find the third one.But I'm still stuck on finding the measure of angle HCX. Maybe I need to find some other angles first.Let me think about the angles at point H. Since H is the orthocenter, angle BHC is 119 degrees, as I calculated earlier. Similarly, angle AHB is 134 degrees, and angle AHC is 107 degrees.Now, focusing on angle CHX, which is part of triangle CHX. If I can find the measure of angle HCX, then I can find angle CHX because I know that angle CHX is 90 degrees minus angle HCX.Wait, is that correct? Let me think again. Since CH is perpendicular to AB, angle CHX is part of the right angle at H. So, angle CHX is equal to 90 degrees minus angle HCX.But angle HCX is the same as angle ACB, which is 46 degrees. So, angle CHX would be 90 - 46 = 44 degrees.But earlier, I thought angle HCX might be equal to angle ABC, which is 73 degrees, leading to angle CHX being 17 degrees. Now I'm confused because I have two different possibilities.Wait, maybe I need to clarify the relationship between angle HCX and the other angles. Let me consider triangle HCX. Since CH is perpendicular to AB, and AX is perpendicular to BC, perhaps there's a way to relate these angles through the orthocenter.Alternatively, maybe I can use the fact that in triangle ABC, the orthocenter H creates several cyclic quadrilaterals. For example, quadrilateral AHBX is cyclic because both angles at X and Y are right angles.Wait, actually, quadrilateral AHBX has two right angles at X and Y, so it is cyclic. Therefore, angle AHB is equal to angle AXB. But I'm not sure if that helps directly.Alternatively, maybe I can consider the angles around point H. Since angle BHC is 119 degrees, and angle AHB is 134 degrees, and angle AHC is 107 degrees, perhaps I can use these to find angle CHX.Wait, angle CHX is part of angle AHC, which is 107 degrees. So, if I can find the other angle in triangle CHX, maybe I can find angle CHX.But I'm still not making progress. Maybe I need to approach this differently. Let me try to use coordinate geometry.Let me assign coordinates to the triangle. Let me place point A at (0, 0), point B at (c, 0), and point C somewhere in the plane. Since angle BAC is 61 degrees and angle ABC is 73 degrees, I can use the Law of Sines to find the lengths of the sides.Let me denote the sides as follows: side opposite angle A is BC = a, side opposite angle B is AC = b, and side opposite angle C is AB = c.Using the Law of Sines: a / sin A = b / sin B = c / sin C.We know angle A = 61 degrees, angle B = 73 degrees, and angle C = 46 degrees.So, a / sin 61 = b / sin 73 = c / sin 46.Let me assign a specific length to one side to make calculations easier. Let's assume that side AB = c = 1 unit. Then, using the Law of Sines:a = (sin 61 / sin 46) * c = (sin 61 / sin 46) * 1 ≈ (0.8746 / 0.7193) ≈ 1.216.Similarly, b = (sin 73 / sin 46) * c ≈ (0.9563 / 0.7193) ≈ 1.329.So, sides are approximately: AB = 1, BC ≈ 1.216, AC ≈ 1.329.Now, let me assign coordinates. Let me place point A at (0, 0), point B at (1, 0). Now, I need to find coordinates of point C.Since AC ≈ 1.329 and angle BAC = 61 degrees, I can find the coordinates of C using trigonometry.The coordinates of C can be found as follows: from point A, moving along AC at an angle of 61 degrees. So, the x-coordinate is AC * cos(angle BAC) ≈ 1.329 * cos(61) ≈ 1.329 * 0.4848 ≈ 0.643.The y-coordinate is AC * sin(angle BAC) ≈ 1.329 * sin(61) ≈ 1.329 * 0.8746 ≈ 1.163.So, point C is approximately at (0.643, 1.163).Now, I need to find the coordinates of H, the orthocenter. To find H, I need to find the intersection of the altitudes AX and BY.First, let's find the equation of altitude AX. Since AX is perpendicular to BC, I need the slope of BC to find the slope of AX.Coordinates of B: (1, 0), coordinates of C: (0.643, 1.163).Slope of BC: (1.163 - 0) / (0.643 - 1) ≈ 1.163 / (-0.357) ≈ -3.258.Therefore, the slope of BC is approximately -3.258. Therefore, the slope of AX, which is perpendicular to BC, is the negative reciprocal: 1 / 3.258 ≈ 0.307.Since AX passes through point A (0, 0), its equation is y = 0.307x.Next, let's find the equation of altitude BY. Since BY is perpendicular to AC, I need the slope of AC to find the slope of BY.Coordinates of A: (0, 0), coordinates of C: (0.643, 1.163).Slope of AC: (1.163 - 0) / (0.643 - 0) ≈ 1.163 / 0.643 ≈ 1.809.Therefore, the slope of AC is approximately 1.809. Therefore, the slope of BY, which is perpendicular to AC, is the negative reciprocal: -1 / 1.809 ≈ -0.552.Now, I need a point on BY. Since BY is the altitude from B to AC, it passes through point B (1, 0). So, the equation of BY is y - 0 = -0.552(x - 1), which simplifies to y = -0.552x + 0.552.Now, to find the orthocenter H, I need to find the intersection of AX and BY. So, solve the system:y = 0.307xy = -0.552x + 0.552Set them equal:0.307x = -0.552x + 0.5520.307x + 0.552x = 0.5520.859x = 0.552x ≈ 0.552 / 0.859 ≈ 0.643Then, y ≈ 0.307 * 0.643 ≈ 0.197So, the orthocenter H is approximately at (0.643, 0.197).Now, point X is the foot of the altitude from A to BC. We already have the equation of AX: y = 0.307x. We need to find the coordinates of X, which is the intersection of AX and BC.We have the equation of BC. Let's find it. Coordinates of B: (1, 0), coordinates of C: (0.643, 1.163). The slope of BC is approximately -3.258, as calculated earlier. So, the equation of BC is y - 0 = -3.258(x - 1), which simplifies to y = -3.258x + 3.258.Now, set the equations of AX and BC equal to find X:0.307x = -3.258x + 3.2580.307x + 3.258x = 3.2583.565x = 3.258x ≈ 3.258 / 3.565 ≈ 0.914Then, y ≈ 0.307 * 0.914 ≈ 0.281So, point X is approximately at (0.914, 0.281).Now, we have coordinates for points C (0.643, 1.163), H (0.643, 0.197), and X (0.914, 0.281). We need to find angle CHX, which is the angle at point H between points C and X.To find angle CHX, we can use vector analysis or calculate the angle using coordinates.First, let's find the vectors HC and HX.Coordinates of H: (0.643, 0.197)Coordinates of C: (0.643, 1.163)Coordinates of X: (0.914, 0.281)Vector HC = C - H = (0.643 - 0.643, 1.163 - 0.197) = (0, 0.966)Vector HX = X - H = (0.914 - 0.643, 0.281 - 0.197) = (0.271, 0.084)Now, to find angle between vectors HC and HX at point H, we can use the dot product formula:cos(theta) = (HC . HX) / (|HC| |HX|)First, calculate the dot product:HC . HX = (0)(0.271) + (0.966)(0.084) ≈ 0 + 0.0814 ≈ 0.0814Now, calculate |HC|:|HC| = sqrt(0^2 + 0.966^2) ≈ 0.966Calculate |HX|:|HX| = sqrt(0.271^2 + 0.084^2) ≈ sqrt(0.0734 + 0.0071) ≈ sqrt(0.0805) ≈ 0.284Now, cos(theta) ≈ 0.0814 / (0.966 * 0.284) ≈ 0.0814 / 0.274 ≈ 0.297Therefore, theta ≈ arccos(0.297) ≈ 72.8 degrees.Hmm, that's approximately 73 degrees, which is very close to angle ABC, which is 73 degrees. That seems significant.So, angle CHX is approximately 73 degrees. Given that the problem likely expects an exact value, and considering the approximate calculation led us to 73 degrees, which is one of the given angles, I think the exact measure of angle CHX is 73 degrees.Therefore, the measure of angle CHX is 73 degrees.

💡Okay, so I have this triangle ABC, and the sides opposite angles A, B, and C are a, b, and c respectively. The problem says that cos A equals b over c, and I need to figure out what kind of triangle ABC is—whether it's right-angled, acute-angled, or obtuse-angled.Hmm, let me recall some trigonometry and triangle properties. I remember that in any triangle, the Law of Cosines relates the sides of a triangle to the cosine of one of its angles. The formula is:cos A = (b² + c² - a²) / (2bc)So, the problem gives me that cos A = b/c. That means I can set these two expressions equal to each other:b/c = (b² + c² - a²) / (2bc)Alright, let me write that down:b/c = (b² + c² - a²) / (2bc)Now, I need to solve for the sides a, b, and c. Maybe I can cross-multiply to get rid of the denominators. Let's try that.Multiplying both sides by 2bc to eliminate the denominator on the right side:2bc * (b/c) = b² + c² - a²Simplify the left side:2b² = b² + c² - a²Okay, so now I have 2b² = b² + c² - a². Let me subtract b² from both sides to simplify:2b² - b² = c² - a²Which simplifies to:b² = c² - a²Wait, that seems a bit odd. Let me rearrange it:c² = a² + b²Oh! That looks familiar. That's the Pythagorean theorem. So, if c² equals a² plus b², then the triangle must be a right-angled triangle, with the right angle opposite side c. In this case, since side c is opposite angle C, does that mean angle C is the right angle?Wait, hold on. Let me think again. The Law of Cosines was applied to angle A, so the sides related are a, b, and c. If c² = a² + b², then according to the Pythagorean theorem, the triangle is right-angled at the angle opposite side c, which is angle C.But in the problem, we were given information about angle A. So, does that mean angle A is related to the right angle? Hmm, maybe not directly. Let me double-check my steps.Starting again:Given cos A = b/cLaw of Cosines: cos A = (b² + c² - a²) / (2bc)Set them equal:b/c = (b² + c² - a²) / (2bc)Cross-multiplying:2b² = b² + c² - a²Subtract b²:b² = c² - a²So, c² = a² + b²Yes, that's correct. So, regardless of which angle we're considering, the relationship c² = a² + b² tells us that the triangle is right-angled. Since c is the longest side (as it's equal to the square root of a² + b²), the right angle must be opposite side c, which is angle C.Therefore, triangle ABC is a right-angled triangle.Wait a second, but the problem didn't specify which angle is the right angle. It just asks for the shape of the triangle. So, as long as one angle is 90 degrees, it's a right-angled triangle, regardless of which angle it is.So, even though we started with angle A, the conclusion is that the triangle is right-angled. Therefore, the answer should be right-angled triangle.Let me just visualize this to make sure. If I have a triangle where c² = a² + b², then it's a right-angled triangle with sides a, b, c, where c is the hypotenuse. So, angle C is 90 degrees. Therefore, regardless of what angle A is, the triangle is right-angled.I think that makes sense. So, even though angle A was given with a specific cosine value, the relationship ends up showing that the triangle satisfies the Pythagorean theorem, making it a right-angled triangle.Yeah, I'm pretty confident now. The key was applying the Law of Cosines and then manipulating the equation to see if it fits the Pythagorean theorem, which it does. So, the triangle must be right-angled.**Final Answer**boxed{text{right-angled triangle}}