💡First, I need to determine the order of the sine values for the angles 1, 2, 3, and 4 radians from greatest to least.I know that the sine function increases from 0 to π/2 radians (approximately 1.5708) and then decreases from π/2 to π radians (approximately 3.1416). Since all the given angles (1, 2, 3, and 4 radians) are within the range of 0 to π radians, I can use this property to compare their sine values.Starting with the smallest angle, 1 radian, which is less than π/2, the sine value will be relatively low but positive.Next, 2 radians is still less than π/2, so the sine value will be higher than sin(1) because the sine function is increasing in this interval.Moving on to 3 radians, this angle is greater than π/2 but less than π. In this interval, the sine function starts to decrease. Therefore, sin(3) will be less than sin(2) but still positive.Finally, 4 radians is greater than π, which means it's in the decreasing part of the sine function's cycle. However, since 4 radians is less than 2π, the sine value will still be positive but significantly smaller compared to the previous values.By comparing these values, I can conclude the order from greatest to least.

💡First, I need to determine the number of ways to choose pears. Since there are 8 identical pears, I can choose anywhere from 0 to 8 pears, which gives me 9 options.Next, I calculate the number of ways to choose bananas. With 12 identical bananas, I can choose from 0 to 12 bananas, resulting in 13 options.To find the total number of possible fruit baskets, I multiply the number of pear options by the number of banana options:[ 9 times 13 = 117 ]However, this total includes the scenario where no fruits are chosen. Since the basket must contain at least one piece of fruit, I subtract this empty basket scenario:[ 117 - 1 = 116 ]Therefore, the final answer is 116 possible fruit baskets.

💡Alright, so I've got this problem here about a new operation called "⊕" defined on real numbers. It's kind of interesting because it changes based on the relationship between two numbers, a and b. If a is greater than or equal to b, then a ⊕ b is just a. But if a is less than b, then a ⊕ b becomes b squared. Okay, that's a bit unusual, but I think I get it.The function we're dealing with is f(x) = (1 ⊕ x) · x, where "·" is just regular multiplication. We need to find the range of this function on the interval [0, 2]. So, basically, we need to figure out all the possible values f(x) can take when x is between 0 and 2, inclusive.First, let's break down the operation "⊕" in the context of our function. Since the function is (1 ⊕ x) multiplied by x, we need to see how 1 and x relate to each other. There are two cases here:1. When 1 is greater than or equal to x (i.e., x ≤ 1)2. When 1 is less than x (i.e., x > 1)Let's tackle each case one by one.**Case 1: x ≤ 1**In this scenario, since 1 is greater than or equal to x, according to the definition of "⊕", 1 ⊕ x should just be 1. So, the function simplifies to:f(x) = 1 · x = xSo, for all x in [0, 1], f(x) is just x. That means when x is 0, f(x) is 0; when x is 1, f(x) is 1. Since f(x) = x is a straight line in this interval, the function increases from 0 to 1 as x goes from 0 to 1.**Case 2: x > 1**Here, since 1 is less than x, the operation "⊕" changes. According to the definition, 1 ⊕ x becomes x squared. So, the function becomes:f(x) = x² · x = x³So, for all x in (1, 2], f(x) is x cubed. Let's see what happens here. When x is just above 1, say 1.1, f(x) is approximately 1.331. When x is 2, f(x) is 8. So, in this interval, the function increases from just above 1 to 8.Wait a second, that seems like a big jump. At x = 1, f(x) is 1, and just a tiny bit above 1, it's already more than 1. So, there's a discontinuity at x = 1? Or is it continuous?Let me check. At x = 1, from the left side (x approaching 1 from below), f(x) approaches 1. From the right side (x approaching 1 from above), f(x) approaches 1³ = 1. So, actually, it's continuous at x = 1. There's no jump; it smoothly transitions from f(x) = x to f(x) = x³ at x = 1.But wait, when x is exactly 1, both cases give f(x) = 1, so that's consistent. Okay, good.Now, let's summarize:- For x in [0, 1], f(x) = x, which ranges from 0 to 1.- For x in (1, 2], f(x) = x³, which ranges from just above 1 to 8.So, combining both intervals, the function f(x) starts at 0 when x = 0, increases linearly to 1 when x = 1, and then increases more rapidly (cubically) to 8 when x = 2.Therefore, the range of f(x) on [0, 2] should be all the values from 0 up to 8. So, the range is [0, 8].But hold on, let me double-check. Is there any value between 0 and 8 that f(x) doesn't reach? For x in [0, 1], f(x) covers [0, 1], and for x in (1, 2], f(x) covers (1, 8]. Since 1 is included in both intervals, there's no gap. So, yes, the range is indeed [0, 8].I think that makes sense. The function starts at 0, goes up to 1 linearly, and then shoots up to 8 as x approaches 2. So, all values in between are covered without any missing points.**Final Answer**The range of the function ( f(x) ) on the interval ([0, 2]) is (boxed{[0, 8]}).

💡Okay, so I'm trying to figure out how many slots are available in the photo album that Cristina, John, Clarissa, and Sarah are putting together for their mother's birthday. Each of them has brought a certain number of photos, and I need to find out the total number of slots required to fit all these photos.First, let's list out how many photos each person has brought:- Cristina has brought 7 photos.- John has brought 10 photos.- Sarah has brought 9 photos.- Clarissa has brought 14 photos.Alright, so to find the total number of photos, I need to add up all these numbers. Let's do that step by step to make sure I don't make any mistakes.Starting with Cristina's 7 photos and John's 10 photos:7 (Cristina) + 10 (John) = 17 photos.Okay, so together, Cristina and John have brought 17 photos. Now, let's add Sarah's 9 photos to this total:17 (Cristina + John) + 9 (Sarah) = 26 photos.So, now we have 26 photos from Cristina, John, and Sarah. Finally, we need to add Clarissa's 14 photos to this total:26 (Cristina + John + Sarah) + 14 (Clarissa) = 40 photos.So, altogether, there are 40 photos that need to be placed in the photo album. Therefore, the photo album must have at least 40 slots to accommodate all the photos brought by Cristina, John, Sarah, and Clarissa.Wait a minute, let me double-check my addition to make sure I didn't make any errors. Starting over:7 (Cristina) + 10 (John) = 17.17 + 9 (Sarah) = 26.26 + 14 (Clarissa) = 40.Yes, that seems correct. Alternatively, I can add them in a different order to verify:10 (John) + 14 (Clarissa) = 24.7 (Cristina) + 9 (Sarah) = 16.24 + 16 = 40.Okay, that still gives me 40. So, it looks like the total number of slots needed is indeed 40.I wonder if there's any other factor I need to consider, like whether the album has any additional slots for titles or special messages. But the problem doesn't mention anything like that, so I think it's safe to assume that the total number of slots is just the sum of all the photos brought by the four siblings.So, to summarize:- Cristina: 7 photos- John: 10 photos- Sarah: 9 photos- Clarissa: 14 photosTotal photos = 7 + 10 + 9 + 14 = 40.Therefore, the photo album must have 40 slots available.

💡Okay, so I need to figure out which point the graph of the function ( y = a^{x-3} + 1 ) definitely goes through. The options are A: (0,1), B: (3,1), C: (3,2), and D: (0,2). Hmm, let's see how to approach this.First, I remember that for exponential functions like ( a^x ), there are some key points that they always pass through. For example, when ( x = 0 ), ( a^0 = 1 ) no matter what the base ( a ) is, as long as ( a > 0 ) and ( a neq 1 ). So that's a useful property.Looking at the given function ( y = a^{x-3} + 1 ), it seems like it's a transformation of the basic exponential function. Specifically, it's shifted horizontally and vertically. Let me break it down:1. The exponent is ( x - 3 ), which means the graph is shifted 3 units to the right compared to ( a^x ).2. The entire function is shifted up by 1 unit because of the "+1" at the end.So, if the original function ( a^x ) passes through (0,1), then after shifting it 3 units to the right, the point (0,1) would move to (3,1). But wait, there's also a vertical shift of 1 unit up. So, does that mean the point (3,1) would then move up to (3,2)?Let me test this by plugging in ( x = 3 ) into the function:[ y = a^{3 - 3} + 1 = a^0 + 1 = 1 + 1 = 2 ]So, when ( x = 3 ), ( y = 2 ). That means the point (3,2) is definitely on the graph, regardless of the value of ( a ).Just to be thorough, let me check the other options as well:- For option A: (0,1). Plugging ( x = 0 ):[ y = a^{0 - 3} + 1 = a^{-3} + 1 ]Since ( a^{-3} ) is ( frac{1}{a^3} ), which isn't necessarily 0, so ( y ) would be ( frac{1}{a^3} + 1 ). This isn't always 1 unless ( a ) is infinity, which isn't the case here. So, (0,1) isn't guaranteed.- For option B: (3,1). As I calculated earlier, when ( x = 3 ), ( y = 2 ), not 1. So, (3,1) isn't correct.- For option D: (0,2). Plugging ( x = 0 ):[ y = a^{-3} + 1 ]Again, this is ( frac{1}{a^3} + 1 ), which isn't necessarily 2. For example, if ( a = 2 ), then ( y = frac{1}{8} + 1 = frac{9}{8} ), which isn't 2. So, (0,2) isn't guaranteed either.Therefore, the only point that the graph certainly passes through is (3,2), which is option C.

💡First, I recognize that for two monomials to be like terms, their corresponding variables must have the same exponents.Looking at the given monomials, I compare the exponents of ( a ) and ( b ) in both.For the exponent of ( a ):In the first monomial, the exponent is ( 3 ).In the second monomial, the exponent is ( n + 1 ).Setting these equal gives ( n + 1 = 3 ), which simplifies to ( n = 2 ).For the exponent of ( b ):In the first monomial, the exponent is ( m ).In the second monomial, the exponent is ( 1 ).Setting these equal gives ( m = 1 ).Finally, to find ( m - n ), I substitute the values:( m - n = 1 - 2 = -1 ).