💡To determine the smallest three-digit multiple of 17, I start by identifying the smallest three-digit number, which is 100.Next, I divide 100 by 17 to see how many times 17 fits into 100. The division gives a quotient of 5 and a remainder of 15, as shown in the equation:100 = 17 cdot 5 + 15Since the remainder is not zero, 100 is not a multiple of 17. To find the next multiple, I increase the quotient by 1, making it 6.Finally, I multiply 17 by 6 to find the smallest three-digit multiple:17 cdot 6 = 102Therefore, the smallest three-digit multiple of 17 is 102.

💡First, I need to determine the total number of candies in the bag. There are 8 green, 5 blue, 9 red, 10 yellow, and 6 pink candies, which adds up to 38 candies in total.Next, I'll calculate the probability of picking a green candy first. Since there are 8 green candies out of 38 total, the probability is 8/38.After picking a green candy, there are now 37 candies left. The number of yellow candies remains 10, so the probability of picking a yellow candy next is 10/37.Finally, after picking a yellow candy, there are 36 candies left, and there are still 6 pink candies. The probability of picking a pink candy last is 6/36, which simplifies to 1/6.To find the overall probability of this specific sequence of events, I'll multiply the probabilities of each individual event:(8/38) * (10/37) * (1/6) = 80/8436Simplifying this fraction by dividing both the numerator and the denominator by 4 gives:20/2109So, the probability that Violet picks a green candy first, then a yellow candy, and finally a pink candy is 20/2109.

💡Okay, so I have this problem about a rectangle with two given vertices, (-3, 6) and (1, 1), which share a common side. There's another vertex adjacent to (1, 1) that's not on the side connecting (-3, 6) and (1, 1), and it's located at (1, -6). I need to find the area of this rectangle.First, I should probably visualize the rectangle to understand the positions of these points. Let me sketch a rough coordinate system in my mind. The point (-3, 6) is somewhere in the second quadrant, and (1, 1) is in the first quadrant. The side connecting these two points is one side of the rectangle.Since (-3, 6) and (1, 1) are adjacent vertices, the distance between them will give me the length of one side of the rectangle. To find this distance, I can use the distance formula, which is sqrt[(x2 - x1)^2 + (y2 - y1)^2]. Plugging in the coordinates, that would be sqrt[(1 - (-3))^2 + (1 - 6)^2]. Let me compute that:1 - (-3) is 4, and 1 - 6 is -5. Squaring these gives 16 and 25, respectively. Adding them together gives 41, so the distance is sqrt(41). Okay, so one side of the rectangle is sqrt(41) units long.Now, the problem mentions another vertex adjacent to (1, 1) that's not on the side connecting (-3, 6) and (1, 1). This vertex is at (1, -6). Since it's adjacent to (1, 1), it must be connected by another side of the rectangle. Let me find the length of this side as well.Looking at the coordinates, both points have the same x-coordinate, which is 1. That means this side is vertical. The distance between (1, 1) and (1, -6) is just the difference in the y-coordinates. So, 1 - (-6) is 7. Therefore, the length of this side is 7 units.Now that I have both side lengths, sqrt(41) and 7, I can find the area of the rectangle by multiplying these two lengths together. So, the area would be sqrt(41) multiplied by 7, which is 7*sqrt(41).Wait, let me double-check to make sure I didn't make any mistakes. The distance between (-3, 6) and (1, 1) is indeed sqrt[(1 - (-3))^2 + (1 - 6)^2] = sqrt[16 + 25] = sqrt[41]. The vertical distance from (1, 1) to (1, -6) is 7 units. Multiplying these gives the area as 7*sqrt(41). That seems correct.I think I've covered all the necessary steps. I found the lengths of two adjacent sides of the rectangle and multiplied them to get the area. There doesn't seem to be any missing information or steps I overlooked. So, I'm confident that the area of the rectangle is 7*sqrt(41) square units.

💡First, I need to determine the total weight of the dumbbells Parker initially found at the gym. There are 4 dumbbells, each weighing 20 pounds.Next, I'll calculate the total weight Parker is using for his exercises, which is 120 pounds.To find out how much additional weight Parker added, I'll subtract the initial weight from the total weight he is using: 120 pounds minus 80 pounds equals 40 pounds.Since each dumbbell weighs 20 pounds, I'll divide the additional weight by the weight of one dumbbell to find out how many dumbbells Parker added: 40 pounds divided by 20 pounds per dumbbell equals 2 dumbbells.Therefore, Parker added 2 dumbbells to the setup.

💡Okay, so I have this expression to simplify: ( dfrac {1}{cos 80 ^{circ} }- dfrac {sqrt {3}}{sin 80 ^{circ} } ). Hmm, I need to figure out how to make this simpler. Let me think about trigonometric identities that might help here.First, I notice that both terms have denominators involving cosine and sine of the same angle, 80 degrees. Maybe I can combine these two fractions into one by finding a common denominator. The common denominator would be ( cos 80^{circ} sin 80^{circ} ). So, let me rewrite the expression:( dfrac {1}{cos 80 ^{circ} } - dfrac {sqrt {3}}{sin 80 ^{circ} } = dfrac{sin 80^{circ} - sqrt{3} cos 80^{circ}}{cos 80^{circ} sin 80^{circ}} ).Alright, now I have a single fraction. The numerator is ( sin 80^{circ} - sqrt{3} cos 80^{circ} ). This looks familiar. I think it's similar to the sine of a difference formula, which is ( sin(A - B) = sin A cos B - cos A sin B ). Let me see if I can factor something out to make it fit that formula.If I factor out a 2 from the numerator, I get:( 2 left( dfrac{1}{2} sin 80^{circ} - dfrac{sqrt{3}}{2} cos 80^{circ} right) ).Wait, ( dfrac{1}{2} ) is ( cos 60^{circ} ) and ( dfrac{sqrt{3}}{2} ) is ( sin 60^{circ} ). So, substituting those in, the numerator becomes:( 2 left( cos 60^{circ} sin 80^{circ} - sin 60^{circ} cos 80^{circ} right) ).Now, using the sine difference identity ( sin(A - B) = sin A cos B - cos A sin B ), this simplifies to:( 2 sin(80^{circ} - 60^{circ}) = 2 sin 20^{circ} ).So, the numerator is ( 2 sin 20^{circ} ). Now, let's look at the denominator: ( cos 80^{circ} sin 80^{circ} ). I remember there's a double-angle identity for sine: ( sin 2A = 2 sin A cos A ). So, if I set ( A = 80^{circ} ), then:( sin 160^{circ} = 2 sin 80^{circ} cos 80^{circ} ).Therefore, ( sin 80^{circ} cos 80^{circ} = dfrac{1}{2} sin 160^{circ} ). But ( sin 160^{circ} ) is the same as ( sin(180^{circ} - 20^{circ}) ), and since ( sin(180^{circ} - x) = sin x ), this becomes ( sin 20^{circ} ). So, the denominator simplifies to:( dfrac{1}{2} sin 20^{circ} ).Putting it all together, the expression becomes:( dfrac{2 sin 20^{circ}}{dfrac{1}{2} sin 20^{circ}} ).Simplifying this, the ( sin 20^{circ} ) terms cancel out, and we have:( 2 div dfrac{1}{2} = 4 ).So, the simplified value of the original expression is 4.**Final Answer**The simplified expression is boxed{4}.

💡Okay, so I have this function f(x) = log_a(x + 1) - log_a(1 - x), and I need to find its domain and determine if it's even or odd. Alright, let's start with the domain.First, I remember that for logarithmic functions, the argument inside the log must be positive. So, for log_a(x + 1), the expression inside the log is (x + 1), which must be greater than 0. That gives me x + 1 > 0, which simplifies to x > -1. Similarly, for the second logarithm, log_a(1 - x), the argument is (1 - x), which must also be greater than 0. So, 1 - x > 0, which simplifies to x < 1. Putting these two inequalities together, x must be greater than -1 and less than 1. So, the domain of f(x) is all real numbers x such that -1 < x < 1. That seems straightforward.Now, moving on to determining if the function is even or odd. I recall that a function is even if f(-x) = f(x) for all x in its domain, and it's odd if f(-x) = -f(x) for all x in its domain. Let me compute f(-x) to see which one it is. So, f(-x) would be log_a(-x + 1) - log_a(1 - (-x)). Simplifying that, it becomes log_a(1 - x) - log_a(1 + x). Wait a minute, that looks familiar. The original function f(x) is log_a(x + 1) - log_a(1 - x). If I factor out a negative sign from f(-x), I get -[log_a(1 + x) - log_a(1 - x)], which is exactly -f(x). So, f(-x) = -f(x), which means the function is odd. Hmm, let me double-check that. If I plug in a specific value, say x = 0.5, then f(0.5) = log_a(1.5) - log_a(0.5). And f(-0.5) should be log_a(0.5) - log_a(1.5), which is indeed the negative of f(0.5). Yep, that seems to confirm it.I think that's solid. The function is odd because f(-x) equals negative f(x), and the domain is symmetric about the origin, which is a requirement for a function to be classified as even or odd.**Final Answer**(1) The domain of ( f(x) ) is (boxed{{x mid -1 < x < 1}}).(2) The function ( f(x) ) is odd, so the final answer is (boxed{f(-x) = -f(x)}).