💡First, I need to find the horizontal and vertical distances between the two points ( (2,2) ) and ( (5,9) ).The horizontal distance is calculated by subtracting the x-coordinates: ( 5 - 2 = 3 ) units.The vertical distance is calculated by subtracting the y-coordinates: ( 9 - 2 = 7 ) units.These two distances form the legs of a right triangle, with the segment connecting the points as the hypotenuse.Using the Pythagorean theorem, the distance between the points can be found by taking the square root of the sum of the squares of the horizontal and vertical distances.So, the distance is ( sqrt{3^2 + 7^2} = sqrt{9 + 49} = sqrt{58} ).Therefore, the distance between the points ( (2,2) ) and ( (5,9) ) in simplest radical form is ( sqrt{58} ).

💡Okay, so I have this problem about kangaroos in a zoo. There are 2016 kangaroos, and each one is either grey or pink. I know that there's at least one grey and at least one pink kangaroo, so it's not all one color. The task is to calculate a fraction for each kangaroo, where the fraction is the number of kangaroos of the other color divided by the number of kangaroos of the same color as that kangaroo, including itself. Then, I need to find the sum of all these 2016 fractions.Hmm, let me break this down. Let's say there are G grey kangaroos and P pink kangaroos. So, G + P = 2016. Since there's at least one of each, G and P are both at least 1.Now, for each grey kangaroo, the fraction would be the number of pink kangaroos divided by the number of grey kangaroos. So, for each grey kangaroo, the fraction is P/G. Similarly, for each pink kangaroo, the fraction is G/P.Since there are G grey kangaroos, each contributing P/G, the total sum from the grey kangaroos would be G*(P/G) = P. Similarly, for the pink kangaroos, each contributing G/P, the total sum from the pink kangaroos would be P*(G/P) = G.So, adding both sums together, we get P + G, which is equal to 2016.Wait, that seems too straightforward. Let me double-check. If I have G grey kangaroos, each with a fraction of P/G, then summing over all G grey kangaroos gives me G*(P/G) = P. Similarly, for the P pink kangaroos, each with a fraction of G/P, summing over all P pink kangaroos gives me P*(G/P) = G. So, total sum is P + G = 2016.Yeah, that makes sense. So, the sum of all the fractions is 2016.

💡First, I need to determine the type of triangle with sides 5, 12, and 13 units. Recognizing that (5^2 + 12^2 = 25 + 144 = 169 = 13^2), I confirm it's a right triangle.Next, I calculate the area of the triangle using the formula for the area of a right triangle: (frac{1}{2} times text{base} times text{height}). Substituting the values, the area is (frac{1}{2} times 5 times 12 = 30) square units.Given that the rectangle has the same area as the triangle, which is 30 square units, and the width of the rectangle is 5 units, I can find the length of the rectangle by dividing the area by the width: (30 div 5 = 6) units.Finally, to find the perimeter of the rectangle, I use the formula (2 times (text{length} + text{width})). Substituting the values, the perimeter is (2 times (6 + 5) = 2 times 11 = 22) units.

💡Alright, let me try to figure out this problem step by step. So, we have the function ( f(x) = sin(omega x + varphi) ) where ( omega > 0 ) and ( |varphi| < frac{pi}{2} ). The graph of this function passes through the point ( (0, frac{1}{2}) ), which means when ( x = 0 ), ( f(x) = frac{1}{2} ). First, let's plug in ( x = 0 ) into the function:[ f(0) = sin(omega cdot 0 + varphi) = sin(varphi) = frac{1}{2} ]So, ( sin(varphi) = frac{1}{2} ). Now, since ( |varphi| < frac{pi}{2} ), the possible values for ( varphi ) are within the first and fourth quadrants. The sine of an angle is ( frac{1}{2} ) at ( frac{pi}{6} ) and ( frac{5pi}{6} ), but since ( |varphi| < frac{pi}{2} ), ( varphi ) must be ( frac{pi}{6} ) because ( frac{5pi}{6} ) is greater than ( frac{pi}{2} ).So, we have ( varphi = frac{pi}{6} ).Next, the problem states that ( f(x) leqslant fleft( frac{pi}{12} right) ) for all ( x in mathbb{R} ). This means that the maximum value of the function ( f(x) ) occurs at ( x = frac{pi}{12} ). Since the sine function has a maximum value of 1, we can set up the equation:[ sinleft( omega cdot frac{pi}{12} + varphi right) = 1 ]We already know ( varphi = frac{pi}{6} ), so plugging that in:[ sinleft( frac{omega pi}{12} + frac{pi}{6} right) = 1 ]The sine function equals 1 at ( frac{pi}{2} + 2pi k ) where ( k ) is an integer. Therefore:[ frac{omega pi}{12} + frac{pi}{6} = frac{pi}{2} + 2pi k ]Let's solve for ( omega ):1. Subtract ( frac{pi}{6} ) from both sides:[ frac{omega pi}{12} = frac{pi}{2} - frac{pi}{6} + 2pi k ]2. Simplify the right side:[ frac{pi}{2} - frac{pi}{6} = frac{3pi}{6} - frac{pi}{6} = frac{2pi}{6} = frac{pi}{3} ]So,[ frac{omega pi}{12} = frac{pi}{3} + 2pi k ]3. Divide both sides by ( pi ):[ frac{omega}{12} = frac{1}{3} + 2k ]4. Multiply both sides by 12:[ omega = 4 + 24k ]Since ( omega > 0 ) and we're looking for the minimum value, we set ( k = 0 ):[ omega = 4 ]So, the minimum value of ( omega ) is 4.**Final Answer**boxed{B}

💡Okay, so I have this problem: I need to find the relationship between arcsin(cos(arcsin x)) and arccos(sin(arccos x)). Hmm, that sounds a bit complicated, but maybe I can break it down step by step.First, let me recall what arcsin and arccos functions do. Arcsin x gives me an angle whose sine is x, and arccos x gives me an angle whose cosine is x. Both of these functions have specific ranges to make sure they're functions: arcsin x is between -π/2 and π/2, and arccos x is between 0 and π.Alright, so let's tackle the first expression: arcsin(cos(arcsin x)). Let me break it down. Let’s say θ = arcsin x. That means sin θ = x, and θ is between -π/2 and π/2. Now, I need to find cos(arcsin x), which is cos θ. Since sin²θ + cos²θ = 1, cos θ = sqrt(1 - sin²θ) = sqrt(1 - x²). But wait, cos θ is always positive in the range of arcsin x, right? Because θ is between -π/2 and π/2, so cos θ is non-negative. So cos(arcsin x) is sqrt(1 - x²).Now, I need to find arcsin of that. So arcsin(sqrt(1 - x²)). Let me call this value α. So α = arcsin(sqrt(1 - x²)). That means sin α = sqrt(1 - x²), and α is between -π/2 and π/2.Okay, now let's look at the second expression: arccos(sin(arccos x)). Let me do the same thing. Let’s set φ = arccos x. That means cos φ = x, and φ is between 0 and π. Now, I need to find sin(arccos x), which is sin φ. Using the identity sin²φ + cos²φ = 1, sin φ = sqrt(1 - cos²φ) = sqrt(1 - x²). But wait, sin φ is non-negative in the range of arccos x, which is between 0 and π, so sin(arccos x) is sqrt(1 - x²).Now, I need to find arccos of that. So arccos(sqrt(1 - x²)). Let me call this value β. So β = arccos(sqrt(1 - x²)). That means cos β = sqrt(1 - x²), and β is between 0 and π.Now, I have α = arcsin(sqrt(1 - x²)) and β = arccos(sqrt(1 - x²)). I need to find the relationship between α and β.I remember that for any y between 0 and 1, arcsin(y) + arccos(y) = π/2. Is that true? Let me check. If I have y = sin α, then α = arcsin y, and arccos y would be the angle whose cosine is y. Since sin α = y, cos(π/2 - α) = y, so arccos y = π/2 - α. Therefore, arcsin y + arccos y = π/2. Yes, that seems correct.In our case, y is sqrt(1 - x²). So arcsin(sqrt(1 - x²)) + arccos(sqrt(1 - x²)) = π/2. That means α + β = π/2.So, putting it all together, arcsin(cos(arcsin x)) + arccos(sin(arccos x)) = π/2.Wait, let me make sure I didn't make any mistakes. Let me go through it again.For arcsin(cos(arcsin x)):- Let θ = arcsin x, so sin θ = x.- cos θ = sqrt(1 - x²).- arcsin(cos θ) = arcsin(sqrt(1 - x²)) = α.For arccos(sin(arccos x)):- Let φ = arccos x, so cos φ = x.- sin φ = sqrt(1 - x²).- arccos(sin φ) = arccos(sqrt(1 - x²)) = β.Then, using the identity arcsin y + arccos y = π/2, with y = sqrt(1 - x²), we get α + β = π/2.Yes, that seems consistent. I think that's the correct relationship.

💡Okay, so Dolly wants to ride three different attractions at the amusement park: the Ferris wheel, the roller coaster, and the log ride. She plans to go on the Ferris wheel twice, the roller coaster three times, and the log ride seven times. Each of these rides costs a certain number of tickets: the Ferris wheel costs 2 tickets per ride, the roller coaster costs 5 tickets per ride, and the log ride costs 1 ticket per ride. Dolly currently has 20 tickets, and she needs to figure out how many more tickets she should buy to have enough for all her planned rides.First, I need to calculate the total number of tickets Dolly will need for each ride. For the Ferris wheel, since it costs 2 tickets per ride and she wants to ride it twice, that would be 2 tickets multiplied by 2 rides. So, 2 times 2 equals 4 tickets needed for the Ferris wheel.Next, for the roller coaster, which costs 5 tickets per ride, and she wants to ride it three times. So, 5 tickets multiplied by 3 rides equals 15 tickets needed for the roller coaster.Then, for the log ride, which is cheaper at 1 ticket per ride, and she wants to ride it seven times. So, 1 ticket multiplied by 7 rides equals 7 tickets needed for the log ride.Now, to find out the total number of tickets Dolly needs for all the rides, I should add up the tickets needed for each ride. That would be 4 tickets for the Ferris wheel plus 15 tickets for the roller coaster plus 7 tickets for the log ride. Adding those together: 4 plus 15 is 19, and 19 plus 7 is 26. So, Dolly needs a total of 26 tickets.But Dolly currently has 20 tickets. To find out how many more tickets she needs to buy, I need to subtract the number of tickets she already has from the total number of tickets required. So, 26 tickets needed minus 20 tickets she has equals 6 tickets. Therefore, Dolly needs to buy 6 more tickets to have enough for all her rides.Wait, let me double-check that to make sure I didn't make a mistake. So, Ferris wheel: 2 rides times 2 tickets each is 4 tickets. Roller coaster: 3 rides times 5 tickets each is 15 tickets. Log ride: 7 rides times 1 ticket each is 7 tickets. Adding those together: 4 plus 15 is 19, and 19 plus 7 is indeed 26. She has 20 tickets, so 26 minus 20 is 6. Yep, that seems correct.I think that's all there is to it. Dolly needs to buy 6 more tickets to enjoy all the rides she wants.