answer：First, I recognize that the denominator is √125, which can be simplified by factoring out perfect squares.I know that 125 is 5 cubed, so √125 is equal to √(25 × 5), which simplifies to 5√5.Substituting this back into the original expression, the fraction becomes 5 divided by 5√5.Next, I can simplify the fraction by canceling out the common factor of 5 in the numerator and the denominator, resulting in 1 over √5.To rationalize the denominator, I multiply both the numerator and the denominator by √5. This gives me √5 in the numerator and 5 in the denominator.Therefore, the simplified expression with a rationalized denominator is √5 divided by 5.

answer：First, determine the total number of slices in the cheesecake by dividing the total calories by the calories per slice: 2800 ÷ 350 = 8 slices.Next, calculate 25% of the total slices to find out how many slices Kiley ate: 0.25 × 8 = 2 slices.Therefore, Kiley ate 2 slices of cheesecake.

answer：Okay, so I have this problem about a caravan with 100 camels, some one-humped and some two-humped. There's at least one of each kind, so that means the number of two-humped camels, which is N, can be from 1 to 99. The condition given is that if you take any 62 camels, they will have at least half of the total number of humps in the caravan. I need to figure out how many possible values N can take.First, let me parse the problem step by step.We have 100 camels in total. Let me denote:- ( N ) as the number of two-humped camels.- Then, the number of one-humped camels would be ( 100 - N ).Total number of humps in the caravan would be ( 1 times (100 - N) + 2 times N = 100 - N + 2N = 100 + N ).So, total humps = ( 100 + N ).The condition is that any 62 camels will have at least half of the total humps. Half of the total humps is ( frac{100 + N}{2} ).So, any subset of 62 camels must have at least ( frac{100 + N}{2} ) humps.Wait, that seems a bit counterintuitive because 62 camels is more than half of 100, which is 50. So, if we take 62 camels, which is 62% of the caravan, they should have at least half the humps. Hmm, okay.I need to translate this into an inequality. Let me think about the worst-case scenario. To minimize the number of humps in 62 camels, we would take as many one-humped camels as possible. So, the minimum number of humps in 62 camels would be when we take all the one-humped camels and then some two-humped camels.Wait, so the minimum number of humps in 62 camels is ( (100 - N) times 1 + (62 - (100 - N)) times 2 ). Wait, let me make sure.If we have ( 100 - N ) one-humped camels, then if we take all of them, that's ( 100 - N ) camels, each contributing 1 hump. Then, the remaining camels we take would be two-humped. The number of two-humped camels we take would be ( 62 - (100 - N) ), but only if ( 62 - (100 - N) ) is positive. Otherwise, if ( 100 - N geq 62 ), then all 62 camels could be one-humped, which would give only 62 humps.Wait, so let me formalize this.The minimum number of humps in 62 camels is:If ( 100 - N geq 62 ), then the minimum humps is 62.Otherwise, if ( 100 - N < 62 ), then the minimum humps is ( (100 - N) times 1 + (62 - (100 - N)) times 2 ).So, let me compute that.Case 1: ( 100 - N geq 62 ) which implies ( N leq 38 ).In this case, the minimum humps in 62 camels is 62.Case 2: ( 100 - N < 62 ) which implies ( N > 38 ).In this case, the minimum humps is ( (100 - N) + 2 times (62 - (100 - N)) ).Simplify that:( (100 - N) + 2 times (62 - 100 + N) )= ( 100 - N + 2 times (N - 38) )= ( 100 - N + 2N - 76 )= ( (100 - 76) + ( -N + 2N ) )= ( 24 + N )So, in Case 2, the minimum humps is ( 24 + N ).Now, the condition is that this minimum humps must be at least half of the total humps.Total humps is ( 100 + N ), so half is ( frac{100 + N}{2} ).So, in Case 1: ( 62 geq frac{100 + N}{2} )In Case 2: ( 24 + N geq frac{100 + N}{2} )Let me solve both inequalities.Starting with Case 1:( 62 geq frac{100 + N}{2} )Multiply both sides by 2:( 124 geq 100 + N )Subtract 100:( 24 geq N )But in Case 1, we already have ( N leq 38 ). So, this inequality tells us that ( N leq 24 ).So, in Case 1, ( N leq 24 ).Now, for Case 2:( 24 + N geq frac{100 + N}{2} )Multiply both sides by 2:( 48 + 2N geq 100 + N )Subtract N from both sides:( 48 + N geq 100 )Subtract 48:( N geq 52 )But in Case 2, we have ( N > 38 ). So, combining these, ( N geq 52 ).So, putting it all together:From Case 1: ( N leq 24 )From Case 2: ( N geq 52 )But wait, the problem states that there is at least one of each kind, so ( N ) must be at least 1 and at most 99.But according to these results, the possible values of N are either ( N leq 24 ) or ( N geq 52 ).But wait, is that all? Let me think.Wait, actually, in Case 1, when ( N leq 38 ), the minimum humps is 62, which must be at least ( frac{100 + N}{2} ). So, 62 >= (100 + N)/2, which simplifies to N <= 24.Similarly, in Case 2, when ( N > 38 ), the minimum humps is 24 + N, which must be >= (100 + N)/2, leading to N >= 52.So, the possible N are N <=24 or N >=52.But wait, the problem says "at least one of each kind", so N can be from 1 to 99.So, N can be 1 to 24, inclusive, or 52 to 99, inclusive.But wait, let me check if N=25 is allowed.If N=25, then 100 - N =75, so 75 one-humped camels. So, if we take 62 camels, the minimum humps would be 62, since 75 >=62.So, 62 humps must be >= (100 +25)/2=62.5.But 62 is less than 62.5, which violates the condition. So, N=25 is not allowed.Similarly, N=24:If N=24, then 100 -24=76 one-humped camels. So, taking 62 camels, all one-humped, gives 62 humps.Half of total humps is (100 +24)/2=62.So, 62 >=62, which is okay.So, N=24 is allowed.Similarly, N=52:100 -52=48 one-humped camels. So, taking 62 camels, we can take all 48 one-humped and 14 two-humped.Total humps would be 48*1 +14*2=48+28=76.Half of total humps is (100 +52)/2=76.So, 76 >=76, which is okay.So, N=52 is allowed.Similarly, N=53:100 -53=47 one-humped. Taking 62 camels, 47 one-humped and 15 two-humped.Total humps:47 +30=77.Half of total humps is (100 +53)/2=76.5.77 >=76.5, which is okay.Similarly, N=51:Wait, N=51 is in the middle. Let's check N=51.100 -51=49 one-humped.Taking 62 camels, 49 one-humped and 13 two-humped.Total humps:49 +26=75.Half of total humps is (100 +51)/2=75.5.75 <75.5, which doesn't satisfy the condition. So, N=51 is not allowed.So, N must be either <=24 or >=52.Thus, the possible N are 1 to24 and52 to99.So, how many values is that?From1 to24:24 numbers.From52 to99:99 -52 +1=48 numbers.Total:24 +48=72.Wait, but let me confirm.Wait, 1 to24 is 24 numbers.52 to99: Let's compute 99 -52=47, so 47 +1=48 numbers.So, total is24 +48=72.But wait, the problem says "within the range from1 to99". So, that's correct.But let me think again.Wait, is N=24 allowed? Yes, as we saw earlier.N=52 is allowed.N=25 is not allowed, as 62 <62.5.Similarly, N=51 is not allowed.So, the possible N are 1-24 and52-99.So, total possible values:24 +48=72.Wait, but let me check N=0 and N=100. But the problem says at least one of each, so N cannot be0 or100. So, N is from1 to99.Thus, the answer is72.But wait, let me think again.Wait, is there a possibility that in the case when N is between25 to51, the condition is not satisfied, so those N are invalid.Hence, the number of possible N is72.Wait, but let me check another value, say N=30.100 -30=70 one-humped.Taking 62 camels, all one-humped, gives62 humps.Total humps:100 +30=130, half is65.62 <65, so condition not satisfied. So, N=30 is invalid.Similarly, N=50:100 -50=50 one-humped.Taking 62 camels, 50 one-humped and12 two-humped.Total humps:50 +24=74.Half of total humps: (100 +50)/2=75.74 <75, so condition not satisfied.So, N=50 is invalid.Similarly, N=51:As before, 49 one-humped,13 two-humped.Total humps:49 +26=75.Half of total humps:75.5.75 <75.5, invalid.N=52:48 one-humped,14 two-humped.Total humps:48 +28=76.Half of total humps:76, so 76 >=76, valid.So, yes, N=52 is valid.Similarly, N=24:76 one-humped,62 camels can be all one-humped, giving62 humps.Total humps:124, half is62, so 62 >=62, valid.N=25:75 one-humped,62 camels can be all one-humped, giving62 humps.Total humps:125, half is62.5, so 62 <62.5, invalid.So, yes, N=25 is invalid.Therefore, the possible N are1-24 and52-99, which is24 +48=72.Hence, the number of possible values N can take is72.**Final Answer**The number of possible values ( N ) can take is boxed{72}.

answer：First, I need to convert the percentage to a decimal. 30% is equal to 0.3.Next, I'll convert the mixed number (1 frac{2}{5}) to an improper fraction. This becomes (frac{7}{5}).Then, I'll calculate the sum inside the parentheses: (frac{1}{3} + frac{1}{7}). Finding a common denominator, which is 21, the sum is (frac{10}{21}).Now, I'll perform the division: (0.3 div frac{7}{5}). Dividing by a fraction is the same as multiplying by its reciprocal, so this becomes (0.3 times frac{5}{7} = frac{3}{14}).Finally, I'll multiply (frac{3}{14}) by (frac{10}{21}) to get the final result: (frac{30}{294}), which simplifies to (frac{5}{49}).

answer：Alright, so I've got this equation here: √(a + 3) + √(2 - b) = 0. Hmm, okay, I need to find the value of a^b. Let me think about how to approach this.First off, I remember that square roots are always non-negative, right? So both √(a + 3) and √(2 - b) have to be greater than or equal to zero. That means each of these terms can't be negative. Now, if I have two non-negative numbers adding up to zero, the only way that can happen is if both of them are zero individually. Because if one was positive and the other was negative, they could cancel each other out, but since they can't be negative, the only possibility is that both are zero. So, that gives me two separate equations:1. √(a + 3) = 02. √(2 - b) = 0Let me solve the first equation: √(a + 3) = 0. To get rid of the square root, I'll square both sides. That gives me a + 3 = 0. Then, solving for a, I subtract 3 from both sides: a = -3.Okay, now the second equation: √(2 - b) = 0. Again, I'll square both sides to eliminate the square root. That gives me 2 - b = 0. Solving for b, I add b to both sides and subtract 2: b = 2.So, now I have a = -3 and b = 2. The question asks for the value of a^b, which is (-3)^2. Hmm, let me compute that. (-3) squared is 9 because when you multiply two negative numbers, the result is positive. So, (-3) * (-3) = 9.Wait, just to make sure I didn't make any mistakes. Let me double-check my steps.1. I started with the equation √(a + 3) + √(2 - b) = 0.2. Since both square roots are non-negative, each must be zero individually.3. Solved √(a + 3) = 0 to get a = -3.4. Solved √(2 - b) = 0 to get b = 2.5. Calculated a^b as (-3)^2 = 9.Everything seems to check out. I don't think I made any errors in my reasoning. The key here was recognizing that the sum of two non-negative numbers can only be zero if both are zero. That led me to solve for a and b separately, and then compute the exponent.I wonder if there's another way to approach this problem. Maybe by substituting or something else? Let me think. If I tried to isolate one of the square roots, say √(a + 3) = -√(2 - b), but wait, square roots can't be negative, so the right side would have to be non-positive. But the left side is non-negative, so the only way they can be equal is if both sides are zero. So, that brings me back to the same conclusion: both square roots must be zero. So, yeah, that reinforces my initial approach.Another thought: could there be any other solutions? For example, if a + 3 or 2 - b were negative, would that affect the equation? But no, because square roots of negative numbers aren't real, and I assume we're dealing with real numbers here. So, a + 3 and 2 - b must both be non-negative. Therefore, a must be greater than or equal to -3, and b must be less than or equal to 2. But in this case, they have to be exactly -3 and 2, respectively, because the sum is zero.So, I think I'm confident that a = -3 and b = 2, so a^b is 9. I don't see any other possibilities or mistakes in my logic.**Final Answer**The value of ( a^{b} ) is boxed{9}.

answer：First, I'll calculate the total number of tickets Dolly needs for each ride based on her desired number of times she wants to ride each attraction.For the Ferris wheel, she wants to ride it twice, and each ride costs 2 tickets. So, the total tickets needed for the Ferris wheel are 2 rides multiplied by 2 tickets per ride, which equals 4 tickets.Next, for the roller coaster, she plans to ride it three times, and each ride costs 5 tickets. Therefore, the total tickets needed for the roller coaster are 3 rides multiplied by 5 tickets per ride, totaling 15 tickets.Then, for the log ride, she wants to ride it seven times, and each ride costs 1 ticket. This means the total tickets needed for the log ride are 7 rides multiplied by 1 ticket per ride, which equals 7 tickets.Adding up all the tickets needed for each ride, the total number of tickets Dolly requires is 4 (Ferris wheel) + 15 (roller coaster) + 7 (log ride) = 26 tickets.Dolly currently has 20 tickets. To find out how many more tickets she needs to buy, I'll subtract the tickets she has from the total tickets required: 26 total tickets minus 20 tickets she has equals 6 tickets.Therefore, Dolly needs to buy 6 more tickets to enjoy all the rides as planned.