💡First, calculate the initial number of red and blue balls. With 40% of 150 balls being red, there are 60 red balls and 90 blue balls. To make red balls 80% of the urn, let ( x ) be the number of blue balls to remove. The equation becomes ( frac{60}{150 - x} = 0.80 ). Solving for ( x ) gives ( x = 75 ). Therefore, 75 blue balls must be removed.The final answer is (boxed{textbf{(C)} 75}).

💡Okay, so I have this problem about a zookeeper feeding animals, and I need to figure out how many ways he can do it under certain conditions. Let me try to break it down step by step.First, the problem says there are five pairs of different animals, each pair consisting of one male and one female. So, that means there are 5 males and 5 females, making a total of 10 animals. The zookeeper starts feeding the male lion, and he must finish with the female bear. Also, he has to alternate between genders each time he feeds an animal. So, male, female, male, female, and so on.Alright, let me visualize this. The feeding sequence will have 10 animals, starting with a male (the male lion) and ending with a female (the female bear). Since he alternates genders, the sequence will go M, F, M, F, ..., ending with F. So, positions 1, 3, 5, 7, 9 are males, and positions 2, 4, 6, 8, 10 are females.But wait, the first male is fixed as the male lion, and the last female is fixed as the female bear. So, positions 1 and 10 are already determined. That leaves positions 3,5,7,9 for the remaining males and positions 2,4,6,8 for the remaining females.So, how many males and females are left? Initially, there are 5 males and 5 females. Since the male lion is already fed first, we have 4 males left. Similarly, since the female bear is fed last, we have 4 females left for the other female positions.Now, let's think about the sequence. Starting with the male lion (position 1), the next position (position 2) must be a female. But it can't be the female bear because she has to be last. So, for position 2, we have 4 choices (since one female is reserved for the end).After choosing a female for position 2, we move to position 3, which must be a male. We have 4 males left (excluding the male lion already fed). So, 4 choices here.Then, position 4 must be a female. Now, we've already used one female for position 2, so we have 3 females left. So, 3 choices for position 4.Position 5 is another male. We've used one male for position 3, so 3 males left. So, 3 choices here.Position 6 is a female. We've used two females so far (positions 2 and 4), so 2 females left. So, 2 choices.Position 7 is a male. We've used two males so far (positions 3 and 5), so 2 males left. So, 2 choices.Position 8 is a female. We've used three females so far (positions 2, 4, 6), so 1 female left. So, 1 choice here.Position 9 is a male. We've used three males so far (positions 3, 5, 7), so 1 male left. So, 1 choice.Finally, position 10 is the female bear, which is already determined.So, putting it all together, the number of ways is the product of the number of choices at each step:Position 2: 4 choicesPosition 3: 4 choicesPosition 4: 3 choicesPosition 5: 3 choicesPosition 6: 2 choicesPosition 7: 2 choicesPosition 8: 1 choicePosition 9: 1 choiceSo, total ways = 4 * 4 * 3 * 3 * 2 * 2 * 1 * 1Let me compute that:4 * 4 = 1616 * 3 = 4848 * 3 = 144144 * 2 = 288288 * 2 = 576576 * 1 = 576576 * 1 = 576So, the total number of ways is 576.Wait, let me double-check if I considered all constraints correctly. The male lion is fixed at position 1, and the female bear is fixed at position 10. For the other positions, we alternate between males and females, and each time, we have one less choice because we can't repeat an animal.Yes, that seems correct. So, the calculation should be accurate.

💡Okay, so I have this geometry problem here. It says that segments AB and CD intersect at a right angle, and AC equals AD. I need to prove that BC equals BD and that angle ACB equals angle ADB. Hmm, let me try to visualize this first.Alright, let me draw a rough sketch in my mind. There's segment AB, and segment CD intersects it at a right angle. Let's say they intersect at point A. So, point A is where AB and CD cross each other, forming a right angle. Now, AC equals AD. That means point A is equidistant from points C and D. So, A is like the center between C and D.Wait, if AC equals AD, then point A must be the midpoint of CD? Or is it just that A is equidistant from C and D, but not necessarily the midpoint? Hmm, no, if AC equals AD, then A is equidistant from C and D, which makes it the midpoint if CD is a straight line. But CD is a segment, so yeah, A should be the midpoint of CD.But hold on, AB and CD intersect at A, which is the midpoint of CD. So, AB is intersecting CD at its midpoint, and they intersect at a right angle. That makes AB the perpendicular bisector of CD. Okay, that's a key point.So, if AB is the perpendicular bisector of CD, then any point on AB is equidistant from C and D. That means point B, which is on AB, is equidistant from C and D. Therefore, BC equals BD. That seems straightforward. So, that's the first part done.Now, for the second part, angle ACB equals angle ADB. Hmm, okay. Let me think about triangles ACB and ADB. So, in triangle ACB, we have points A, C, B, and in triangle ADB, we have points A, D, B. We already know that AC equals AD, and we just proved that BC equals BD. Also, AB is common to both triangles. So, triangles ACB and ADB have two sides equal: AC equals AD, BC equals BD, and they share side AB.Wait, so by the Side-Side-Side (SSS) congruence theorem, if all three sides of one triangle are equal to all three sides of another triangle, then the triangles are congruent. So, triangles ACB and ADB are congruent because AC equals AD, BC equals BD, and AB is common. Therefore, all corresponding angles are equal. So, angle ACB equals angle ADB.But let me double-check if I can use another congruence theorem. Maybe Side-Angle-Side (SAS)? We know AC equals AD, angle at A is 90 degrees for both triangles because AB and CD intersect at a right angle. So, angle BAC equals angle BAD, both are 90 degrees. And AB is common. So, by SAS, triangles ACB and ADB are congruent. Therefore, angle ACB equals angle ADB.Yeah, that seems solid. So, both parts are proved. BC equals BD because B is on the perpendicular bisector of CD, and angle ACB equals angle ADB because triangles ACB and ADB are congruent by SAS or SSS.Wait, just to make sure I didn't miss anything. Let me recap:1. AB and CD intersect at A at a right angle.2. AC equals AD, so A is the midpoint of CD.3. Therefore, AB is the perpendicular bisector of CD.4. Any point on the perpendicular bisector is equidistant from C and D, so BC equals BD.5. Triangles ACB and ADB have sides AC=AD, BC=BD, and AB common, so they are congruent by SSS.6. Therefore, corresponding angles are equal, so angle ACB equals angle ADB.Alternatively, using SAS:- AC=AD, angle BAC=angle BAD=90 degrees, AB common.- Therefore, triangles ACB and ADB are congruent by SAS.- Hence, angle ACB equals angle ADB.Yeah, both approaches work. I think I've covered all the necessary steps and checked my reasoning. I don't see any gaps or mistakes here. It all ties together neatly.

💡To find the largest four-digit number whose digits add up to 23, I'll start by maximizing the leftmost digit, which has the highest place value. Setting the thousands digit to 9 gives me the largest possible starting point.Next, I'll subtract this 9 from the total sum of 23, leaving 14 to be distributed among the remaining three digits. To continue maximizing the number, I'll set the hundreds digit to 9 as well. This reduces the remaining sum to 5.Finally, I'll allocate the remaining 5 to the tens digit, making it 5, and set the units digit to 0 to keep the number as large as possible. This results in the number 9950, which is the largest four-digit number with digits that add up to 23.

💡Okay, so I have this problem here that says if there are triangles with sides (a, b, c) and another triangle with sides (A, B, C), then there must also be a triangle with sides (sqrt{a^2 + A^2}), (sqrt{b^2 + B^2}), and (sqrt{c^2 + C^2}). Hmm, interesting. I need to show that these new sides can form a triangle. First, I remember that for three lengths to form a triangle, they must satisfy the triangle inequality. That is, the sum of any two sides must be greater than the third side. So, I need to check if (sqrt{a^2 + A^2} + sqrt{b^2 + B^2} > sqrt{c^2 + C^2}), and similarly for the other combinations.But before jumping into that, maybe I should think about what these sides represent. The original triangles have sides (a, b, c) and (A, B, C). So, each new side is the hypotenuse of a right triangle with legs (a) and (A), (b) and (B), and (c) and (C) respectively. That makes me think of vectors or maybe coordinate systems, but I'm not sure yet.Let me try to visualize this. If I have two triangles, one with sides (a, b, c) and another with sides (A, B, C), maybe I can think of them as being in perpendicular planes or something? Like, if I have one triangle in the x-y plane and another in the y-z plane, then combining them somehow might give me a new triangle in 3D space. But I'm not sure if that's the right approach.Wait, maybe it's simpler. Since each new side is (sqrt{a^2 + A^2}), which is like combining the lengths from both triangles, perhaps there's a way to relate the triangle inequalities of the original triangles to the new one.Let me write down the triangle inequalities for the original triangles. For the first triangle with sides (a, b, c), we have:1. (a + b > c)2. (a + c > b)3. (b + c > a)Similarly, for the second triangle with sides (A, B, C), we have:1. (A + B > C)2. (A + C > B)3. (B + C > A)Now, I need to show that the new sides (sqrt{a^2 + A^2}), (sqrt{b^2 + B^2}), (sqrt{c^2 + C^2}) satisfy the triangle inequalities. Let's focus on one inequality first: (sqrt{a^2 + A^2} + sqrt{b^2 + B^2} > sqrt{c^2 + C^2}).To prove this, maybe I can square both sides to eliminate the square roots. Let's try that.Starting with:[sqrt{a^2 + A^2} + sqrt{b^2 + B^2} > sqrt{c^2 + C^2}]Squaring both sides:[(sqrt{a^2 + A^2} + sqrt{b^2 + B^2})^2 > (sqrt{c^2 + C^2})^2]Expanding the left side:[a^2 + A^2 + b^2 + B^2 + 2sqrt{(a^2 + A^2)(b^2 + B^2)} > c^2 + C^2]Simplify:[(a^2 + b^2) + (A^2 + B^2) + 2sqrt{(a^2 + A^2)(b^2 + B^2)} > c^2 + C^2]Now, from the triangle inequalities of the original triangles, we know that (a + b > c) and (A + B > C). If I square these inequalities, maybe I can relate them to the terms here.Squaring (a + b > c):[a^2 + 2ab + b^2 > c^2]Similarly, squaring (A + B > C):[A^2 + 2AB + B^2 > C^2]Adding these two inequalities:[(a^2 + b^2 + A^2 + B^2) + 2ab + 2AB > c^2 + C^2]Comparing this with our earlier expression:[(a^2 + b^2) + (A^2 + B^2) + 2sqrt{(a^2 + A^2)(b^2 + B^2)} > c^2 + C^2]It looks similar, but instead of (2ab + 2AB), we have (2sqrt{(a^2 + A^2)(b^2 + B^2)}). So, if I can show that (2sqrt{(a^2 + A^2)(b^2 + B^2)} geq 2ab + 2AB), then the inequality would hold.Let me check that. Is (sqrt{(a^2 + A^2)(b^2 + B^2)} geq ab + AB)?Hmm, let's square both sides to see:[(a^2 + A^2)(b^2 + B^2) geq (ab + AB)^2]Expanding both sides:Left side: (a^2b^2 + a^2B^2 + A^2b^2 + A^2B^2)Right side: (a^2b^2 + 2abAB + A^2B^2)Subtracting right side from left side:[a^2B^2 + A^2b^2 - 2abAB = (aB - Ab)^2 geq 0]Yes, that's always true because it's a square. So, (sqrt{(a^2 + A^2)(b^2 + B^2)} geq ab + AB). Therefore, the earlier inequality holds:[(a^2 + b^2) + (A^2 + B^2) + 2sqrt{(a^2 + A^2)(b^2 + B^2)} > c^2 + C^2]Which means:[sqrt{a^2 + A^2} + sqrt{b^2 + B^2} > sqrt{c^2 + C^2}]Great, so that's one inequality. I need to check the other two as well, but I think the process will be similar. Let's try the next one: (sqrt{a^2 + A^2} + sqrt{c^2 + C^2} > sqrt{b^2 + B^2}).Following the same steps:1. Start with the inequality:[sqrt{a^2 + A^2} + sqrt{c^2 + C^2} > sqrt{b^2 + B^2}]2. Square both sides:[(a^2 + A^2) + (c^2 + C^2) + 2sqrt{(a^2 + A^2)(c^2 + C^2)} > b^2 + B^2]3. From the triangle inequalities of the original triangles, we have (a + c > b) and (A + C > B). Squaring these:[a^2 + 2ac + c^2 > b^2][A^2 + 2AC + C^2 > B^2]4. Adding these two:[(a^2 + c^2 + A^2 + C^2) + 2ac + 2AC > b^2 + B^2]5. Comparing with our earlier expression:[(a^2 + c^2) + (A^2 + C^2) + 2sqrt{(a^2 + A^2)(c^2 + C^2)} > b^2 + B^2]Again, we have (2sqrt{(a^2 + A^2)(c^2 + C^2)} geq 2ac + 2AC) because:[sqrt{(a^2 + A^2)(c^2 + C^2)} geq ac + AC]Which, when squared, gives:[(a^2 + A^2)(c^2 + C^2) geq (ac + AC)^2]Expanding both sides:Left: (a^2c^2 + a^2C^2 + A^2c^2 + A^2C^2)Right: (a^2c^2 + 2acAC + A^2C^2)Subtracting right from left:[a^2C^2 + A^2c^2 - 2acAC = (aC - Ac)^2 geq 0]So, the inequality holds.Therefore, the second triangle inequality is satisfied:[sqrt{a^2 + A^2} + sqrt{c^2 + C^2} > sqrt{b^2 + B^2}]Now, the last inequality: (sqrt{b^2 + B^2} + sqrt{c^2 + C^2} > sqrt{a^2 + A^2}).Following the same method:1. Start with:[sqrt{b^2 + B^2} + sqrt{c^2 + C^2} > sqrt{a^2 + A^2}]2. Square both sides:[(b^2 + B^2) + (c^2 + C^2) + 2sqrt{(b^2 + B^2)(c^2 + C^2)} > a^2 + A^2]3. From the original triangle inequalities, (b + c > a) and (B + C > A). Squaring these:[b^2 + 2bc + c^2 > a^2][B^2 + 2BC + C^2 > A^2]4. Adding these two:[(b^2 + c^2 + B^2 + C^2) + 2bc + 2BC > a^2 + A^2]5. Comparing with our earlier expression:[(b^2 + c^2) + (B^2 + C^2) + 2sqrt{(b^2 + B^2)(c^2 + C^2)} > a^2 + A^2]Again, we need to show that (2sqrt{(b^2 + B^2)(c^2 + C^2)} geq 2bc + 2BC). As before, this is true because:[sqrt{(b^2 + B^2)(c^2 + C^2)} geq bc + BC]Which, when squared, gives:[(b^2 + B^2)(c^2 + C^2) geq (bc + BC)^2]Expanding both sides:Left: (b^2c^2 + b^2C^2 + B^2c^2 + B^2C^2)Right: (b^2c^2 + 2bcBC + B^2C^2)Subtracting right from left:[b^2C^2 + B^2c^2 - 2bcBC = (bC - Bc)^2 geq 0]So, the inequality holds.Therefore, the third triangle inequality is satisfied:[sqrt{b^2 + B^2} + sqrt{c^2 + C^2} > sqrt{a^2 + A^2}]Since all three triangle inequalities are satisfied, the lengths (sqrt{a^2 + A^2}), (sqrt{b^2 + B^2}), and (sqrt{c^2 + C^2}) can indeed form a triangle. I think that covers all the necessary steps. I started by recalling the triangle inequality, then applied it to the new sides by squaring and using the Cauchy-Schwarz inequality in a way. It took a bit of manipulation, but breaking it down step by step helped me see how the original triangle inequalities lead to the new ones. It's interesting how combining the sides in this way preserves the triangle property.**Final Answer**boxed{sqrt{a^2 + A^2}}, boxed{sqrt{b^2 + B^2}}, boxed{sqrt{c^2 + C^2}} form a triangle.**Final Answer**boxed{sqrt{a^2 + A^2}}, boxed{sqrt{b^2 + B^2}}, boxed{sqrt{c^2 + C^2}} form a triangle.

💡First, I need to calculate the multiplication of 245.738 and 104.251. This will give me the numerator of the expression.Next, I will add 15.934 and 137.086 to find the denominator of the expression.After obtaining both the numerator and the denominator, I will divide the numerator by the denominator to get the final result.Finally, I will present the result in a clear and concise manner.