💡First, I need to find the value of ( x ) that satisfies the equation ( frac{144}{x} = frac{14.4}{0.0144} ).To simplify the equation, I'll start by calculating the right side. I'll multiply both the numerator and the denominator by 100 to eliminate the decimals:[frac{14.4 times 100}{0.0144 times 100} = frac{1440}{1.44}]Next, I'll divide 1440 by 1.44:[frac{1440}{1.44} = 1000]Now, the equation becomes:[frac{144}{x} = 1000]To solve for ( x ), I'll multiply both sides of the equation by ( x ):[144 = 1000x]Finally, I'll divide both sides by 1000 to find the value of ( x ):[x = frac{144}{1000} = 0.144]

💡First, I need to simplify the expression (2^{8} cdot 2^{32}) using the properties of exponents. By the product of powers rule, (2^{8} cdot 2^{32} = 2^{8+32} = 2^{40}).Next, I want to express (2^{40}) as an integer raised to the fifth power. To do this, I can rewrite the exponent 40 as a multiple of 5. Since (40 = 8 times 5), I can express (2^{40}) as ((2^{8})^5).Calculating (2^8), I find that (2^8 = 256).Therefore, (2^{40}) can be written as (256^5), meaning the integer is 256.

💡Alright, so I have this fraction here: (frac{k^2 - 5k + 8}{k^2 + 6k + 19}), and I need to show that if one of the terms in the fraction is divisible by 11, then the fraction can be reduced by 11, given that (k) is a natural number. Hmm, okay, let me try to break this down step by step.First, let's understand what it means for a term in the fraction to be divisible by 11. The fraction has two terms: the numerator (k^2 - 5k + 8) and the denominator (k^2 + 6k + 19). So, if either the numerator or the denominator is divisible by 11, then the fraction can be reduced by 11. That means both the numerator and the denominator should be divisible by 11, right? Because if only one of them is divisible by 11, you can't really reduce the fraction by 11 unless both are.Wait, actually, the problem says "if one of the terms in the fraction is divisible by 11, then the fraction can be reduced by 11." So, maybe it's implying that if either the numerator or the denominator is divisible by 11, then both are, which would allow the fraction to be simplified by 11. That makes sense because if only one term is divisible by 11, you can't reduce the fraction by 11 unless the other term is also divisible by 11.So, let's assume that the numerator (k^2 - 5k + 8) is divisible by 11. That means (k^2 - 5k + 8 equiv 0 mod 11). Similarly, if the denominator is divisible by 11, then (k^2 + 6k + 19 equiv 0 mod 11). I need to show that if one of these congruences holds, then the other must also hold, which would imply that both the numerator and the denominator are divisible by 11, allowing the fraction to be reduced by 11.Let me try subtracting the numerator from the denominator to see if there's a relationship. So, ((k^2 + 6k + 19) - (k^2 - 5k + 8)). Let's compute that:[(k^2 + 6k + 19) - (k^2 - 5k + 8) = k^2 + 6k + 19 - k^2 + 5k - 8 = 11k + 11]Oh, interesting! The difference between the denominator and the numerator is (11k + 11), which is clearly divisible by 11. So, if the numerator is divisible by 11, then the denominator is equal to the numerator plus (11k + 11), which is also divisible by 11. Therefore, if the numerator is divisible by 11, the denominator must also be divisible by 11.Similarly, if the denominator is divisible by 11, then the numerator is equal to the denominator minus (11k + 11), which is also divisible by 11. So, in either case, if one term is divisible by 11, the other term must also be divisible by 11. Therefore, both the numerator and the denominator are divisible by 11, which means the fraction can be reduced by 11.Wait, let me make sure I didn't skip any steps here. So, if (k^2 - 5k + 8) is divisible by 11, then (k^2 - 5k + 8 equiv 0 mod 11). Then, the denominator (k^2 + 6k + 19) is equal to the numerator plus (11k + 11). Since (11k + 11) is divisible by 11, adding it to the numerator, which is already divisible by 11, will result in the denominator also being divisible by 11. That seems correct.Similarly, if the denominator is divisible by 11, then subtracting (11k + 11) from it will give the numerator, which is also divisible by 11. So, in both cases, both terms are divisible by 11, meaning the fraction can be reduced by 11.Let me test this with an example to make sure. Suppose (k = 1). Then, the numerator is (1 - 5 + 8 = 4), and the denominator is (1 + 6 + 19 = 26). Neither 4 nor 26 is divisible by 11, so the fraction cannot be reduced by 11. Okay, that makes sense.What about (k = 2)? Numerator: (4 - 10 + 8 = 2). Denominator: (4 + 12 + 19 = 35). Again, neither is divisible by 11.Let me try (k = 3). Numerator: (9 - 15 + 8 = 2). Denominator: (9 + 18 + 19 = 46). Still not divisible by 11.Hmm, maybe I need to find a (k) where one of the terms is divisible by 11. Let's try (k = 5). Numerator: (25 - 25 + 8 = 8). Denominator: (25 + 30 + 19 = 74). Not divisible by 11.Wait, maybe (k = 10). Numerator: (100 - 50 + 8 = 58). Denominator: (100 + 60 + 19 = 179). Neither 58 nor 179 is divisible by 11.Hmm, maybe (k = 11). Numerator: (121 - 55 + 8 = 74). Denominator: (121 + 66 + 19 = 206). 74 divided by 11 is approximately 6.727, and 206 divided by 11 is approximately 18.727. So, still not divisible by 11.Wait, maybe I need to find a (k) such that (k^2 - 5k + 8) is divisible by 11. Let's solve for (k) in the congruence (k^2 - 5k + 8 equiv 0 mod 11).So, (k^2 - 5k + 8 equiv 0 mod 11). Let's rewrite this as (k^2 - 5k + 8 equiv 0 mod 11). To solve this quadratic congruence, I can try completing the square or using the quadratic formula modulo 11.Let me try completing the square. The equation is (k^2 - 5k + 8 equiv 0 mod 11). Let's complete the square for (k):(k^2 - 5k = (k - frac{5}{2})^2 - frac{25}{4}).But since we're working modulo 11, I need to find the equivalent of (frac{5}{2}) and (frac{25}{4}) modulo 11.First, (frac{5}{2}) modulo 11. Since 2*6 = 12 ≡ 1 mod 11, so the inverse of 2 modulo 11 is 6. Therefore, (frac{5}{2} equiv 5*6 = 30 ≡ 8 mod 11).Similarly, (frac{25}{4}) modulo 11. 25 mod 11 is 3, and 4^{-1} mod 11 is 3 because 4*3 = 12 ≡ 1 mod 11. So, (frac{25}{4} ≡ 3*3 = 9 mod 11).So, rewriting the equation:((k - 8)^2 - 9 + 8 ≡ 0 mod 11)Simplify:((k - 8)^2 - 1 ≡ 0 mod 11)So,((k - 8)^2 ≡ 1 mod 11)Taking square roots modulo 11, we have:(k - 8 ≡ pm1 mod 11)Therefore,(k ≡ 8 + 1 ≡ 9 mod 11) or (k ≡ 8 - 1 ≡ 7 mod 11)So, the solutions are (k ≡ 7 mod 11) or (k ≡ 9 mod 11).Therefore, for (k = 7, 9, 18, 20, ...), the numerator (k^2 - 5k + 8) is divisible by 11.Let's test (k = 7). Numerator: (49 - 35 + 8 = 22), which is divisible by 11. Denominator: (49 + 42 + 19 = 110), which is also divisible by 11. So, the fraction becomes (frac{22}{110} = frac{1}{5}), which is reduced by 11.Similarly, (k = 9). Numerator: (81 - 45 + 8 = 44), which is divisible by 11. Denominator: (81 + 54 + 19 = 154), which is also divisible by 11. So, the fraction becomes (frac{44}{154} = frac{4}{14} = frac{2}{7}), reduced by 11.Okay, that seems to work. So, when (k ≡ 7 or 9 mod 11), both the numerator and the denominator are divisible by 11, allowing the fraction to be reduced by 11.Similarly, if I assume that the denominator is divisible by 11, then (k^2 + 6k + 19 ≡ 0 mod 11). Let's solve this congruence as well.(k^2 + 6k + 19 ≡ 0 mod 11)Simplify 19 mod 11: 19 - 11 = 8, so 19 ≡ 8 mod 11.So, the equation becomes:(k^2 + 6k + 8 ≡ 0 mod 11)Again, let's complete the square.(k^2 + 6k = (k + 3)^2 - 9)So, the equation becomes:((k + 3)^2 - 9 + 8 ≡ 0 mod 11)Simplify:((k + 3)^2 - 1 ≡ 0 mod 11)Thus,((k + 3)^2 ≡ 1 mod 11)Taking square roots:(k + 3 ≡ pm1 mod 11)Therefore,(k ≡ -3 + 1 ≡ -2 ≡ 9 mod 11) or (k ≡ -3 - 1 ≡ -4 ≡ 7 mod 11)So, again, (k ≡ 7 or 9 mod 11).This confirms that if the denominator is divisible by 11, then (k ≡ 7 or 9 mod 11), which are the same solutions as when the numerator is divisible by 11. Therefore, in both cases, both the numerator and the denominator are divisible by 11, allowing the fraction to be reduced by 11.So, to summarize, if either the numerator or the denominator of the fraction (frac{k^2 - 5k + 8}{k^2 + 6k + 19}) is divisible by 11, then both must be divisible by 11. This is because the difference between the denominator and the numerator is (11k + 11), which is always divisible by 11. Therefore, if one term is divisible by 11, the other term must also be divisible by 11, allowing the fraction to be reduced by 11.I think that covers it. I've checked with specific examples and solved the congruences to confirm that the only solutions where one term is divisible by 11 are when both terms are divisible by 11, which means the fraction can indeed be reduced by 11.

💡Alright, so I've got this problem here about the function ( f(x) = |2x - a| + |2x - 1| ) where ( a ) is a real number. There are two parts to this problem. Let me try to tackle them one by one.Starting with part (1): When ( a = -1 ), find the solution set for ( f(x) leq 2 ).Okay, so if ( a = -1 ), then the function becomes ( f(x) = |2x - (-1)| + |2x - 1| ), which simplifies to ( f(x) = |2x + 1| + |2x - 1| ). I need to solve the inequality ( |2x + 1| + |2x - 1| leq 2 ).Hmm, absolute value functions can sometimes be tricky because they change their behavior based on the value inside. Maybe I can consider different cases based on the critical points where the expressions inside the absolute values change sign. For ( |2x + 1| ), the critical point is when ( 2x + 1 = 0 ), which is ( x = -frac{1}{2} ). For ( |2x - 1| ), the critical point is when ( 2x - 1 = 0 ), which is ( x = frac{1}{2} ).So, these critical points divide the real number line into three intervals:1. ( x < -frac{1}{2} )2. ( -frac{1}{2} leq x leq frac{1}{2} )3. ( x > frac{1}{2} )I'll need to analyze each interval separately.**Case 1: ( x < -frac{1}{2} )**In this interval, ( 2x + 1 ) is negative because ( 2x < -1 ), so ( |2x + 1| = -(2x + 1) = -2x - 1 ).Similarly, ( 2x - 1 ) is also negative because ( 2x < -1 < 1 ), so ( |2x - 1| = -(2x - 1) = -2x + 1 ).Therefore, ( f(x) = (-2x - 1) + (-2x + 1) = -4x ).So, the inequality becomes ( -4x leq 2 ).Solving for ( x ), we get ( -4x leq 2 ) ⇒ ( x geq -frac{1}{2} ).But wait, in this case, we're considering ( x < -frac{1}{2} ). So, the solution here is the intersection of ( x < -frac{1}{2} ) and ( x geq -frac{1}{2} ), which is just ( x = -frac{1}{2} ).But since ( x = -frac{1}{2} ) is the boundary point between Case 1 and Case 2, maybe I should check it in the next case.**Case 2: ( -frac{1}{2} leq x leq frac{1}{2} )**In this interval, ( 2x + 1 ) is non-negative because ( 2x geq -1 ), so ( |2x + 1| = 2x + 1 ).However, ( 2x - 1 ) is still negative because ( 2x leq 1 ), so ( |2x - 1| = -(2x - 1) = -2x + 1 ).Therefore, ( f(x) = (2x + 1) + (-2x + 1) = 2 ).So, the inequality ( f(x) leq 2 ) becomes ( 2 leq 2 ), which is always true.Therefore, all ( x ) in this interval satisfy the inequality.**Case 3: ( x > frac{1}{2} )**In this interval, both ( 2x + 1 ) and ( 2x - 1 ) are positive because ( 2x > 1 ).So, ( |2x + 1| = 2x + 1 ) and ( |2x - 1| = 2x - 1 ).Therefore, ( f(x) = (2x + 1) + (2x - 1) = 4x ).So, the inequality becomes ( 4x leq 2 ).Solving for ( x ), we get ( x leq frac{1}{2} ).But in this case, we're considering ( x > frac{1}{2} ). So, the solution here is the intersection of ( x > frac{1}{2} ) and ( x leq frac{1}{2} ), which is empty. So, no solutions in this interval.Putting it all together:- From Case 1: Only ( x = -frac{1}{2} ) is a potential solution, but since it's the boundary, it's included in Case 2.- From Case 2: All ( x ) between ( -frac{1}{2} ) and ( frac{1}{2} ) satisfy the inequality.- From Case 3: No solutions.Therefore, the solution set is ( x in left[ -frac{1}{2}, frac{1}{2} right] ).Wait, let me double-check. When I considered Case 1, I found that ( x geq -frac{1}{2} ), but since ( x < -frac{1}{2} ) in that case, the only solution is ( x = -frac{1}{2} ). But in Case 2, the entire interval ( left[ -frac{1}{2}, frac{1}{2} right] ) satisfies the inequality. So, combining these, the solution set is indeed ( left[ -frac{1}{2}, frac{1}{2} right] ).Alright, that seems solid.Moving on to part (2): If the solution set for ( f(x) leq |2x + 1| ) contains the set ( left[ frac{1}{2}, 1 right] ), find the range of values for the real number ( a ).Hmm, okay. So, the inequality ( f(x) leq |2x + 1| ) must hold for all ( x ) in ( left[ frac{1}{2}, 1 right] ). Therefore, for every ( x ) in that interval, ( |2x - a| + |2x - 1| leq |2x + 1| ).Let me write that down:For all ( x in left[ frac{1}{2}, 1 right] ), ( |2x - a| + |2x - 1| leq |2x + 1| ).I need to find the values of ( a ) such that this inequality holds for all ( x ) in ( left[ frac{1}{2}, 1 right] ).First, let's analyze ( |2x - 1| ) and ( |2x + 1| ) in the interval ( left[ frac{1}{2}, 1 right] ).For ( x in left[ frac{1}{2}, 1 right] ):- ( 2x ) ranges from 1 to 2.- Therefore, ( 2x - 1 ) ranges from 0 to 1, so ( |2x - 1| = 2x - 1 ).- Similarly, ( 2x + 1 ) ranges from 2 to 3, so ( |2x + 1| = 2x + 1 ).So, substituting these into the inequality:( |2x - a| + (2x - 1) leq (2x + 1) ).Simplify this:Subtract ( 2x ) from both sides:( |2x - a| - 1 leq 1 ).Then, add 1 to both sides:( |2x - a| leq 2 ).So, the inequality reduces to ( |2x - a| leq 2 ) for all ( x in left[ frac{1}{2}, 1 right] ).This means that for every ( x ) in ( left[ frac{1}{2}, 1 right] ), ( 2x - a ) must lie within the interval ( [-2, 2] ).In other words, ( -2 leq 2x - a leq 2 ) for all ( x in left[ frac{1}{2}, 1 right] ).Let me rewrite this as two separate inequalities:1. ( 2x - a geq -2 ) ⇒ ( a leq 2x + 2 )2. ( 2x - a leq 2 ) ⇒ ( a geq 2x - 2 )So, for each ( x ) in ( left[ frac{1}{2}, 1 right] ), ( a ) must satisfy both ( a leq 2x + 2 ) and ( a geq 2x - 2 ).But since this must hold for all ( x ) in ( left[ frac{1}{2}, 1 right] ), ( a ) must be less than or equal to the minimum of ( 2x + 2 ) over ( x in left[ frac{1}{2}, 1 right] ), and greater than or equal to the maximum of ( 2x - 2 ) over the same interval.Let me compute these:First, find the minimum of ( 2x + 2 ) on ( x in left[ frac{1}{2}, 1 right] ).Since ( 2x + 2 ) is an increasing function (as the coefficient of ( x ) is positive), its minimum occurs at the left endpoint ( x = frac{1}{2} ):( 2*(1/2) + 2 = 1 + 2 = 3 ).Similarly, the maximum of ( 2x - 2 ) on ( x in left[ frac{1}{2}, 1 right] ).Again, ( 2x - 2 ) is an increasing function, so its maximum occurs at the right endpoint ( x = 1 ):( 2*1 - 2 = 0 ).Therefore, ( a ) must satisfy:( 0 leq a leq 3 ).Wait, let me make sure I got that right. So, for all ( x ) in ( left[ frac{1}{2}, 1 right] ), ( a ) must be less than or equal to ( 2x + 2 ). The smallest value that ( 2x + 2 ) takes in this interval is 3 (at ( x = frac{1}{2} )), so ( a leq 3 ).Similarly, ( a ) must be greater than or equal to ( 2x - 2 ). The largest value that ( 2x - 2 ) takes in this interval is 0 (at ( x = 1 )), so ( a geq 0 ).Therefore, combining these, ( a ) must be between 0 and 3, inclusive.Let me test this with some values to make sure.If ( a = 0 ), then ( f(x) = |2x| + |2x - 1| ). Let's check at ( x = frac{1}{2} ):( f(frac{1}{2}) = |1| + |0| = 1 leq |2*(1/2) + 1| = |2| = 2 ). Okay.At ( x = 1 ):( f(1) = |2| + |1| = 3 leq |3| = 3 ). That's equal, so it's okay.If ( a = 3 ), then ( f(x) = |2x - 3| + |2x - 1| ).At ( x = frac{1}{2} ):( f(frac{1}{2}) = |1 - 3| + |0| = 2 leq |2| = 2 ). Good.At ( x = 1 ):( f(1) = |2 - 3| + |1| = 1 + 1 = 2 leq |3| = 3 ). Also good.What about a value inside the interval, say ( x = frac{3}{4} ):For ( a = 0 ):( f(frac{3}{4}) = |1.5| + |1.5 - 1| = 1.5 + 0.5 = 2 leq |1.5 + 1| = 2.5 ). Okay.For ( a = 3 ):( f(frac{3}{4}) = |1.5 - 3| + |1.5 - 1| = 1.5 + 0.5 = 2 leq |1.5 + 1| = 2.5 ). Still holds.What if ( a ) is outside this range? Let's try ( a = 4 ):At ( x = 1 ):( f(1) = |2 - 4| + |1| = 2 + 1 = 3 leq |3| = 3 ). Hmm, that's still equal.Wait, but maybe at another point. Let's try ( x = frac{1}{2} ):( f(frac{1}{2}) = |1 - 4| + |0| = 3 leq |2| = 2 ). That's not true. So, ( a = 4 ) doesn't work.Similarly, if ( a = -1 ):At ( x = 1 ):( f(1) = |2 - (-1)| + |1| = 3 + 1 = 4 leq |3| = 3 ). Not true.So, indeed, ( a ) must be between 0 and 3.Therefore, the range of ( a ) is ( [0, 3] ).**Final Answer**(1) The solution set is (boxed{left[-frac{1}{2}, frac{1}{2}right]}).(2) The range of values for ( a ) is (boxed{[0, 3]}).

💡First, I need to determine how long it takes for pipe R to fill the tank alone. I know the rates at which pipes P and Q can fill the tank, and I also know the combined rate when all three pipes are open.I'll start by assigning variables to the rates of each pipe. Let’s denote the rate of pipe R as ( frac{1}{x} ) tanks per hour, where ( x ) is the time it takes for pipe R to fill the tank alone.Next, I'll write down the rates for pipes P and Q based on the given information. Pipe P can fill the tank in 3 hours, so its rate is ( frac{1}{3} ) tanks per hour. Similarly, pipe Q can fill the tank in 9 hours, giving it a rate of ( frac{1}{9} ) tanks per hour.When all three pipes are open, their combined rate fills the tank in 2 hours, which means the combined rate is ( frac{1}{2} ) tanks per hour. I can set up an equation that adds up the individual rates of pipes P, Q, and R to equal the combined rate:[frac{1}{3} + frac{1}{9} + frac{1}{x} = frac{1}{2}]To solve for ( x ), I'll find a common denominator for the fractions on the left side of the equation, which is ( 9x ):[frac{3x}{9x} + frac{x}{9x} + frac{9}{9x} = frac{1}{2}]Simplifying the fractions:[frac{3x + x + 9}{9x} = frac{1}{2}]Combining like terms:[frac{4x + 9}{9x} = frac{1}{2}]Cross-multiplying to solve for ( x ):[2(4x + 9) = 9x]Expanding the left side:[8x + 18 = 9x]Subtracting ( 8x ) from both sides:[18 = x]Therefore, pipe R can fill the tank alone in 18 hours.

💡First, I need to determine the cost price of the cricket bat for A, given the final selling price to D.I'll start by letting ( CP_A ) represent A's cost price.A sells the bat to B at a 20% profit. This means B's cost price is 120% of ( CP_A ):[SP_B = 1.20 times CP_A]Next, B sells the bat to C at a 25% profit. Therefore, C's cost price is 125% of ( SP_B ):[SP_C = 1.25 times SP_B = 1.25 times 1.20 times CP_A = 1.50 times CP_A]Then, C sells the bat to D at a 30% profit. This means D's cost price is 130% of ( SP_C ):[SP_D = 1.30 times SP_C = 1.30 times 1.50 times CP_A = 1.95 times CP_A]Given that D pays 420.90, I can set up the equation:[1.95 times CP_A = 420.90]Solving for ( CP_A ):[CP_A = frac{420.90}{1.95} = 216.00]Therefore, the cost price of the cricket bat for A is 216.00.