💡First, I need to understand the problem clearly. Rachel has two identical basil plants and one aloe plant, making a total of three plants. She also has two identical white lamps and two identical red lamps. The goal is to determine the number of ways she can place each plant under a lamp, with the possibility of having multiple plants under a single lamp.Since the basil plants are identical, swapping them under the same lamp doesn't create a new arrangement. However, the aloe plant is distinct, so its placement will affect the total count differently.I'll start by considering the different ways the plants can be grouped under the lamps. There are three plants, so they can either all be under the same lamp, or they can be split between two lamps.**Case 1: All three plants under the same lamp**- **Subcase 1a:** All three plants under a single white lamp.- **Subcase 1b:** All three plants under a single red lamp.Since the lamps of the same color are identical, these are the only two possibilities in this case.**Case 2: Two plants under one lamp and one plant under another lamp**- **Subcase 2a:** Both basil plants under one lamp and the aloe plant under another lamp. - **Subcase 2a1:** Basil plants under a white lamp and the aloe plant under a red lamp. - **Subcase 2a2:** Basil plants under a red lamp and the aloe plant under a white lamp. - **Subcase 2b:** One basil plant and the aloe plant under one lamp, and the other basil plant under another lamp. - **Subcase 2b1:** One basil plant and the aloe plant under a white lamp, and the other basil plant under a red lamp. - **Subcase 2b2:** One basil plant and the aloe plant under a red lamp, and the other basil plant under a white lamp.**Case 3: Each plant under a different lamp**Since there are only two lamps of each color, and we have three plants, it's not possible to have each plant under a different lamp without overlapping colors. Therefore, this case is not applicable.Now, I'll count the number of distinct arrangements in each applicable subcase:- **Subcase 1a:** 1 way- **Subcase 1b:** 1 way- **Subcase 2a1:** 1 way- **Subcase 2a2:** 1 way- **Subcase 2b1:** 1 way- **Subcase 2b2:** 1 wayAdding these up: 1 + 1 + 1 + 1 + 1 + 1 = 6 ways.However, I realize that in **Subcase 2b**, swapping the basil plants doesn't create a new arrangement because they are identical. Therefore, **Subcase 2b1** and **Subcase 2b2** are actually the same, reducing the total count by 1.So, the corrected total number of ways is 5.But wait, I think I made a mistake in considering the identical nature of the basil plants. Let me re-examine the cases.Upon re-evaluation, I realize that in **Subcase 2b**, even though the basil plants are identical, the aloe plant is distinct, so placing it with one basil plant under a white lamp and the other basil plant under a red lamp is different from placing it with one basil plant under a red lamp and the other basil plant under a white lamp. Therefore, these are indeed two distinct arrangements.Thus, the initial count of 6 ways stands.But I'm still unsure because the lamps of the same color are identical. Does this affect the count further?Considering that the white lamps are identical and the red lamps are identical, but the colors themselves are distinct, the arrangements under different colors are unique. Therefore, the total number of distinct ways remains 6.However, I recall that the problem allows for multiple plants under a single lamp, but each plant must be under exactly one lamp. I need to ensure that I haven't missed any arrangements where multiple plants are under the same lamp.Reviewing the cases again:- All three under one lamp: 2 ways (white or red).- Two under one lamp and one under another: 4 ways (as calculated in Subcases 2a and 2b).Adding these gives a total of 6 ways.But I'm still uncertain because sometimes in combinatorial problems, identical items can lead to overcounting or undercounting. Let me try a different approach using combinations.Since the basil plants are identical, the number of ways to distribute them is limited. The aloe plant is distinct, so its placement affects the count.Possible distributions:1. All three plants under a white lamp.2. All three plants under a red lamp.3. Both basil plants under a white lamp and the aloe plant under a red lamp.4. Both basil plants under a red lamp and the aloe plant under a white lamp.5. One basil plant and the aloe plant under a white lamp, and the other basil plant under a red lamp.6. One basil plant and the aloe plant under a red lamp, and the other basil plant under a white lamp.This confirms the 6 ways calculated earlier.But I think I need to consider that the lamps are identical within their colors. So, for example, placing both basil plants under one white lamp and the aloe under another white lamp is the same as placing them under the other white lamp. Therefore, some arrangements might be duplicates.Wait, no, because the lamps are identical within their colors, but the colors themselves are distinct. So, placing plants under white lamps versus red lamps are distinct arrangements.Thus, the initial count of 6 ways seems correct.However, I'm still not entirely confident. Let me try listing all possible arrangements explicitly.1. All three under a white lamp.2. All three under a red lamp.3. Both basil plants under a white lamp, aloe under a red lamp.4. Both basil plants under a red lamp, aloe under a white lamp.5. One basil and aloe under a white lamp, other basil under a red lamp.6. One basil and aloe under a red lamp, other basil under a white lamp.Yes, that's 6 distinct arrangements.But I think I might have missed some arrangements where the basil plants are split differently. Let me think again.If I consider that the basil plants are identical, splitting them between lamps doesn't create new arrangements beyond what's already listed. The aloe plant being distinct allows for the differentiation in cases 3, 4, 5, and 6.Therefore, the total number of ways is indeed 6.But wait, I'm still unsure because sometimes in combinatorial problems, identical items can lead to overcounting. Let me try using the multiplication principle.For each plant, there are two choices: white or red. However, since the basil plants are identical, the number of distinct assignments is limited.The aloe plant has 2 choices: white or red.For the basil plants:- Both under white.- Both under red.- One under white and one under red.But since the basil plants are identical, the last option is just one way.So, combining with the aloe plant's choices:1. Aloe under white: - Both basil under white. - Both basil under red. - One basil under white, one under red.2. Aloe under red: - Both basil under white. - Both basil under red. - One basil under white, one under red.This gives a total of 3 (for aloe under white) + 3 (for aloe under red) = 6 ways.Yes, this confirms the earlier count of 6 ways.But I'm still not entirely confident. Let me think of it as assigning each plant to a lamp color, considering the basil plants are identical.The possible assignments are:- All three plants under white.- All three plants under red.- Two basil under white, aloe under red.- Two basil under red, aloe under white.- One basil and aloe under white, one basil under red.- One basil and aloe under red, one basil under white.That's 6 distinct assignments.Therefore, the total number of ways is 6.

💡Alright, so I need to figure out how to show that for any number ( n > 5 ), it's possible to divide a square into ( n ) smaller squares. Hmm, okay. Let me start by understanding what the problem is asking. It wants me to take a square and split it into smaller squares, and the number of these smaller squares should be any number greater than 5. That means for ( n = 6, 7, 8, ) and so on, I should be able to find a way to divide the original square accordingly.First, I should think about simpler cases. Maybe if I can figure out how to divide a square into 6, 7, or 8 smaller squares, I can find a pattern or a method that can be generalized for any ( n > 5 ). Let's start with ( n = 6 ).For ( n = 6 ), how can I divide a square into 6 smaller squares? Well, one approach is to divide the original square into smaller squares of different sizes. Maybe I can have one large square and then divide the remaining space into smaller squares. But I need to ensure that all the smaller squares fit perfectly without overlapping or leaving gaps.Wait, maybe I can use a known method for dividing squares into smaller squares. I remember something about using a grid system or recursively dividing squares. Let me think. If I divide the square into four equal smaller squares, that gives me ( n = 4 ). But I need ( n = 6 ). Maybe I can further divide some of these smaller squares into even smaller squares.Okay, so if I start with four equal smaller squares, and then take one of them and divide it into four even smaller squares, that would give me a total of ( 4 - 1 + 4 = 7 ) squares. Hmm, that's ( n = 7 ). But I wanted ( n = 6 ). Maybe I can adjust this method.Instead of dividing one of the four equal squares into four smaller squares, what if I divide one of them into nine smaller squares? That would give me ( 4 - 1 + 9 = 12 ) squares, which is way more than 6. That's not helpful. Maybe I need a different approach.Perhaps I can divide the original square into unequal smaller squares. Let's say I divide the original square into one large square on one side and then divide the remaining space into smaller squares. For example, if I place a smaller square in one corner, the remaining area would be a rectangle. Then, I can divide that rectangle into smaller squares.But dividing a rectangle into squares is not straightforward because the rectangle has different side lengths. I might need to use a method called "squaring the rectangle," which involves tiling a rectangle with squares of different sizes. This could be complicated, but maybe it's necessary.Alternatively, I recall that there's a theorem related to squaring the square, which states that a square can be divided into smaller squares, all of different sizes, and this is called a perfect squared square. However, I'm not sure if that's applicable here since the problem doesn't specify that the smaller squares need to be different sizes. They can be the same size or different sizes.Wait, if the smaller squares can be the same size, then it's simpler. For example, if I divide the original square into a grid of smaller squares, like a 2x2 grid, that gives me 4 squares. A 3x3 grid gives me 9 squares. But I need numbers between 4 and 9, like 6, 7, or 8.Maybe I can combine different grid sizes. For instance, if I have a 2x2 grid on one side and a 3x3 grid on the other, but that might not fit perfectly. I need to ensure that the entire original square is covered without overlaps or gaps.Another idea is to use a recursive approach. If I can divide a square into ( n ) smaller squares, then I can further divide one of those smaller squares into even more squares to increase the total number. This could help me reach higher numbers.Let me try to formalize this. Suppose I have a square divided into ( n ) smaller squares. If I take one of those smaller squares and divide it into ( k ) even smaller squares, then the total number of squares becomes ( n - 1 + k ). So, if I choose ( k = 4 ), which is dividing one square into four equal smaller squares, then the total number increases by 3, making it ( n + 3 ).This seems useful. If I can show that for some initial values of ( n ), I can divide the square into ( n ) smaller squares, and then use this method to increase ( n ) by 3 each time, I can cover all numbers greater than 5.So, what are the initial values I need to check? If I can show that ( n = 6, 7, 8 ) are possible, then using the recursive method of adding 3 each time, I can cover all ( n > 5 ).Let's start with ( n = 6 ). How can I divide a square into 6 smaller squares? One way is to divide the original square into one large square and five smaller squares arranged around it. But I need to ensure that all squares fit perfectly.Alternatively, I can divide the original square into a 2x2 grid, giving me 4 squares, and then divide one of those into four smaller squares, resulting in a total of 7 squares. But that's ( n = 7 ). To get ( n = 6 ), maybe I can divide one of the 2x2 squares into three smaller squares instead of four. But dividing a square into three smaller squares equally is not straightforward because 3 doesn't divide evenly into a square's area in terms of equal smaller squares.Wait, maybe I can have unequal smaller squares. If I divide one of the 2x2 squares into three smaller squares of different sizes, that could work. But this might be more complex.Alternatively, I can divide the original square into a 3x3 grid, giving me 9 squares, and then combine some of them to reduce the total number to 6. But combining squares would mean they are no longer smaller squares, so that doesn't work.Hmm, perhaps I need a different approach. Maybe I can divide the original square into a combination of different grid sizes. For example, divide it into a 2x2 grid on one side and a 1x1 grid on the other, but that might not cover the entire area.Wait, maybe I can use a method where I divide the square into smaller squares of varying sizes, not necessarily in a grid pattern. For instance, place a smaller square in one corner, and then divide the remaining area into smaller squares.Let me try to visualize this. If I place a smaller square in the top-left corner, the remaining area would be a rectangle. Then, I can divide that rectangle into smaller squares. But dividing a rectangle into squares requires careful planning to ensure all squares fit without gaps or overlaps.I think this is getting too complicated. Maybe I should look for a known result or theorem that addresses this problem. I recall that any square can be divided into smaller squares, and there's a method involving induction or some form of recursive division.Wait, I think the key is to use induction. If I can show that for some base cases ( n = 6, 7, 8 ), it's possible to divide the square into ( n ) smaller squares, and then show that if it's possible for ( n ), it's also possible for ( n + 3 ), then by induction, it would be possible for all ( n > 5 ).So, let's try to establish the base cases. For ( n = 6 ), how can I divide the square? Maybe divide it into one large square and five smaller squares arranged around it. But I need to ensure that all squares fit perfectly.Alternatively, I can divide the square into a 2x2 grid, giving me 4 squares, and then divide one of those into four smaller squares, resulting in 7 squares. But that's ( n = 7 ). To get ( n = 6 ), I need to find a way to have one square divided into three smaller squares instead of four. But as I thought earlier, dividing a square into three equal smaller squares isn't straightforward.Wait, maybe I can have unequal smaller squares. If I divide one of the 2x2 squares into three smaller squares of different sizes, that could work. For example, one square could be larger, and the other two could be smaller, fitting into the remaining space.But this might require precise calculations to ensure that all squares fit perfectly. It could be done, but it's more complex.Alternatively, I can consider that for ( n = 6 ), it's possible to divide the square into six smaller squares by having one large square and five smaller squares arranged in a specific pattern. I think this is possible, but I need to visualize it.Similarly, for ( n = 7 ), I can use the method of dividing one of the 2x2 squares into four smaller squares, resulting in 7 squares. And for ( n = 8 ), I can divide two of the 2x2 squares into four smaller squares each, resulting in 8 squares.Wait, no. If I start with a 2x2 grid (4 squares), and then divide one square into four, I get 7 squares. If I divide two squares into four each, I get 4 - 2 + 4 + 4 = 10 squares, which is more than 8. So, that's not helpful.Maybe for ( n = 8 ), I can divide the original square into a 3x3 grid, giving me 9 squares, and then combine one of them with another to make a larger square, reducing the total to 8. But combining squares would mean they are no longer smaller squares, so that doesn't work.Hmm, perhaps I need to find a different way to divide the square into 8 smaller squares. Maybe by having a combination of different grid sizes or arranging smaller squares in a specific pattern.I think I'm overcomplicating this. Maybe I should look for a simpler method. I remember that if you can divide a square into ( n ) smaller squares, you can always divide it into ( n + 3 ) squares by further dividing one of the smaller squares into four. So, if I can establish the base cases for ( n = 6, 7, 8 ), then I can use this method to cover all higher numbers.Let me try to accept that for ( n = 6, 7, 8 ), it's possible to divide the square into that many smaller squares, even if I can't visualize it perfectly right now. Then, using the method of adding 3 each time, I can cover all ( n > 5 ).So, assuming that ( n = 6, 7, 8 ) are possible, then for any ( n > 5 ), I can keep adding 3 to reach that number. For example, if I need ( n = 9 ), I can take the ( n = 6 ) division and add 3 more squares by dividing one of the smaller squares into four. Similarly, for ( n = 10 ), I can take ( n = 7 ) and add 3, and so on.This seems like a viable approach. Therefore, by establishing the base cases and using the recursive method of adding 3 squares each time, I can show that for any ( n > 5 ), it's possible to divide a square into ( n ) smaller squares.I think I've got a rough idea. Now, I need to formalize this into a proper proof, making sure that each step is logically sound and that I cover all necessary cases.

💡First, I need to determine the amount of money given at the second interest rate. Since the total amount is ₹80,000 and ₹70,000 is given at 10%, the remaining amount is ₹10,000.Next, I'll calculate the interest earned from the first part. Using the formula for simple interest, I can find the interest from ₹70,000 at 10%.After finding the interest from the first part, I'll subtract this amount from the total profit to determine the interest earned from the second part.Finally, I'll use the interest earned from the second part and the amount given at that rate to calculate the second interest rate.

💡Okay, so I have this trigonometric expression that I need to simplify and show that it equals -1. The expression is:[frac{sin ^{2}left(135^{circ}-alpharight)-sin ^{2}left(210^{circ}-alpharight)-sin 195^{circ} cos left(165^{circ}-2 alpharight)}{cos ^{2}left(225^{circ}+alpharight)-cos ^{2}left(210^{circ}-alpharight)+sin 15^{circ} sin left(75^{circ}-2 alpharight)}=-1]Alright, let me break this down step by step. First, I notice that both the numerator and the denominator have squared sine and cosine terms, as well as products of sine and cosine with other angles. Maybe I can use some trigonometric identities to simplify these terms.Starting with the numerator:1. **Simplify (sin^2(135^circ - alpha)):** I know that (135^circ) is (90^circ + 45^circ), so maybe I can use the identity (sin(90^circ + x) = cos(x)). Let me try that: [ sin(135^circ - alpha) = sin(90^circ + 45^circ - alpha) = cos(45^circ - alpha) ] So, (sin^2(135^circ - alpha) = cos^2(45^circ - alpha)).2. **Simplify (sin^2(210^circ - alpha)):** (210^circ) is (180^circ + 30^circ), so using the identity (sin(180^circ + x) = -sin(x)): [ sin(210^circ - alpha) = sin(180^circ + 30^circ - alpha) = -sin(30^circ - alpha) ] Therefore, (sin^2(210^circ - alpha) = sin^2(30^circ - alpha)).3. **Simplify (sin 195^circ):** (195^circ) is (180^circ + 15^circ), so: [ sin(195^circ) = sin(180^circ + 15^circ) = -sin(15^circ) ]4. **Simplify (cos(165^circ - 2alpha)):** (165^circ) is (180^circ - 15^circ), so using (cos(180^circ - x) = -cos(x)): [ cos(165^circ - 2alpha) = cos(180^circ - 15^circ - 2alpha) = -cos(15^circ + 2alpha) ]Putting these together, the numerator becomes:[cos^2(45^circ - alpha) - sin^2(30^circ - alpha) - (-sin 15^circ)(-cos(15^circ + 2alpha))]Wait, hold on. Let me make sure I substitute correctly. The term is (-sin 195^circ cos(165^circ - 2alpha)), which becomes:[-(-sin 15^circ)(-cos(15^circ + 2alpha)) = -sin 15^circ cos(15^circ + 2alpha)]So the numerator is:[cos^2(45^circ - alpha) - sin^2(30^circ - alpha) - sin 15^circ cos(15^circ + 2alpha)]Now, moving on to the denominator:1. **Simplify (cos^2(225^circ + alpha)):** (225^circ) is (180^circ + 45^circ), so using (cos(180^circ + x) = -cos(x)): [ cos(225^circ + alpha) = cos(180^circ + 45^circ + alpha) = -cos(45^circ + alpha) ] Therefore, (cos^2(225^circ + alpha) = cos^2(45^circ + alpha)).2. **Simplify (cos^2(210^circ - alpha)):** Similar to earlier, (210^circ) is (180^circ + 30^circ), so: [ cos(210^circ - alpha) = cos(180^circ + 30^circ - alpha) = -cos(30^circ - alpha) ] Hence, (cos^2(210^circ - alpha) = cos^2(30^circ - alpha)).3. **Simplify (sin 15^circ sin(75^circ - 2alpha)):** Hmm, I don't see an immediate identity here, but maybe I can express (75^circ) as (90^circ - 15^circ). Let's see: [ sin(75^circ - 2alpha) = sin(90^circ - 15^circ - 2alpha) = cos(15^circ + 2alpha) ] So, (sin 15^circ sin(75^circ - 2alpha) = sin 15^circ cos(15^circ + 2alpha)).Putting these together, the denominator becomes:[cos^2(45^circ + alpha) - cos^2(30^circ - alpha) + sin 15^circ cos(15^circ + 2alpha)]Now, let me write the entire expression with these simplifications:Numerator:[cos^2(45^circ - alpha) - sin^2(30^circ - alpha) - sin 15^circ cos(15^circ + 2alpha)]Denominator:[cos^2(45^circ + alpha) - cos^2(30^circ - alpha) + sin 15^circ cos(15^circ + 2alpha)]Looking at both numerator and denominator, I notice that they have similar structures. Let me see if I can factor or rearrange terms to see a relationship.First, let's recall the identity (cos^2 x - sin^2 y). Hmm, not sure if that helps directly. Maybe I can express (cos^2) in terms of double angles?Wait, another approach: let's consider the numerator and denominator separately.Starting with the numerator:[cos^2(45^circ - alpha) - sin^2(30^circ - alpha) - sin 15^circ cos(15^circ + 2alpha)]I can write (cos^2(45^circ - alpha)) as (frac{1 + cos(90^circ - 2alpha)}{2}) using the identity (cos^2 x = frac{1 + cos 2x}{2}). Similarly, (sin^2(30^circ - alpha)) can be written as (frac{1 - cos(60^circ - 2alpha)}{2}).Let me try that:Numerator becomes:[frac{1 + cos(90^circ - 2alpha)}{2} - frac{1 - cos(60^circ - 2alpha)}{2} - sin 15^circ cos(15^circ + 2alpha)]Simplify:[frac{1 + cos(90^circ - 2alpha) - 1 + cos(60^circ - 2alpha)}{2} - sin 15^circ cos(15^circ + 2alpha)][= frac{cos(90^circ - 2alpha) + cos(60^circ - 2alpha)}{2} - sin 15^circ cos(15^circ + 2alpha)]Now, (cos(90^circ - 2alpha) = sin(2alpha)), so:[= frac{sin(2alpha) + cos(60^circ - 2alpha)}{2} - sin 15^circ cos(15^circ + 2alpha)]Similarly, for the denominator:[cos^2(45^circ + alpha) - cos^2(30^circ - alpha) + sin 15^circ cos(15^circ + 2alpha)]Express (cos^2(45^circ + alpha)) as (frac{1 + cos(90^circ + 2alpha)}{2}) and (cos^2(30^circ - alpha)) as (frac{1 + cos(60^circ - 2alpha)}{2}).Denominator becomes:[frac{1 + cos(90^circ + 2alpha)}{2} - frac{1 + cos(60^circ - 2alpha)}{2} + sin 15^circ cos(15^circ + 2alpha)][= frac{cos(90^circ + 2alpha) - cos(60^circ - 2alpha)}{2} + sin 15^circ cos(15^circ + 2alpha)]Now, (cos(90^circ + 2alpha) = -sin(2alpha)), so:[= frac{-sin(2alpha) - cos(60^circ - 2alpha)}{2} + sin 15^circ cos(15^circ + 2alpha)]Hmm, so now the numerator is:[frac{sin(2alpha) + cos(60^circ - 2alpha)}{2} - sin 15^circ cos(15^circ + 2alpha)]And the denominator is:[frac{-sin(2alpha) - cos(60^circ - 2alpha)}{2} + sin 15^circ cos(15^circ + 2alpha)]Looking at these, I notice that the numerator is almost the negative of the denominator, except for the last term. Let me check:If I factor out a negative sign from the denominator:[frac{-sin(2alpha) - cos(60^circ - 2alpha)}{2} + sin 15^circ cos(15^circ + 2alpha) = -left( frac{sin(2alpha) + cos(60^circ - 2alpha)}{2} right) + sin 15^circ cos(15^circ + 2alpha)]So, the denominator is:[- left( frac{sin(2alpha) + cos(60^circ - 2alpha)}{2} right) + sin 15^circ cos(15^circ + 2alpha)]Comparing this to the numerator:[frac{sin(2alpha) + cos(60^circ - 2alpha)}{2} - sin 15^circ cos(15^circ + 2alpha)]It seems like the numerator is the negative of the denominator. Let me write that:Numerator = (-) DenominatorTherefore, the entire fraction becomes:[frac{text{Numerator}}{text{Denominator}} = frac{-text{Denominator}}{text{Denominator}} = -1]So, the expression simplifies to -1, which is what we needed to show.**Final Answer**boxed{-1}

💡First, I need to represent the four consecutive even integers. Let's denote the first integer as ( n ). Since they are consecutive even integers, the next three integers will be ( n + 2 ), ( n + 4 ), and ( n + 6 ).The problem states that the sum of the first and fourth integers is 160. So, I can set up the equation:[n + (n + 6) = 160]Simplifying this equation:[2n + 6 = 160]Subtract 6 from both sides:[2n = 154]Divide both sides by 2:[n = 77]Now, the four consecutive even integers are 77, 79, 81, and 83. To find the sum of the second and third integers:[79 + 81 = 160]Therefore, the sum of the second and third integers is 160.

💡To determine the number of vertices in a cube, I start by visualizing the cube. A cube is a three-dimensional shape with all sides equal and all angles right angles.I know that a cube has 6 faces, each of which is a square. Each square face has 4 vertices. However, each vertex is shared by 3 different faces.To find the total number of vertices, I can use the formula for polyhedrons, which relates the number of vertices (V), edges (E), and faces (F) through Euler's formula: V - E + F = 2.For a cube:- Faces (F) = 6- Edges (E) = 12Plugging these into Euler's formula:V - 12 + 6 = 2V - 6 = 2V = 8Therefore, a cube has 8 vertices.