💡First, I need to determine the speed of the man in still water. I'll denote the man's speed in still water as ( V_m ) and the speed of the stream as ( V_s ).When the man swims downstream, the effective speed is the sum of his speed and the stream's speed, which is ( V_m + V_s ). Conversely, when he swims upstream, the effective speed is the difference between his speed and the stream's speed, which is ( V_m - V_s ).Given that the man swims 40 km downstream in 8 hours, I can set up the equation:[40 = (V_m + V_s) times 8]Simplifying this, I find:[V_m + V_s = 5 text{ km/h}]Similarly, for the upstream journey of 56 km in 8 hours, the equation becomes:[56 = (V_m - V_s) times 8]Simplifying this, I get:[V_m - V_s = 7 text{ km/h}]Now, I have a system of two equations:[begin{cases}V_m + V_s = 5 V_m - V_s = 7end{cases}]To solve for ( V_m ), I'll add the two equations together:[(V_m + V_s) + (V_m - V_s) = 5 + 7][2V_m = 12][V_m = 6 text{ km/h}]Therefore, the speed of the man in still water is 6 km/h.

💡Okay, so I have this inequality to solve: |2x - 3| < 5. Hmm, absolute value inequalities can be a bit tricky, but I think I remember that they involve two separate inequalities. Let me try to recall how to approach this.First, I know that the absolute value of something being less than a number means that the expression inside the absolute value is between the negative and positive versions of that number. So, for |A| < B, it translates to -B < A < B. Applying that to my problem, |2x - 3| < 5 should mean that -5 < 2x - 3 < 5. That seems right.Now, I need to solve this compound inequality for x. Let me write it down step by step. Starting with:-5 < 2x - 3 < 5I think the next step is to isolate the term with x in the middle. To do that, I'll add 3 to all three parts of the inequality. Adding 3 to -5 gives me -2, and adding 3 to 5 gives me 8. So, the inequality becomes:-5 + 3 < 2x - 3 + 3 < 5 + 3Which simplifies to:-2 < 2x < 8Okay, now I have -2 < 2x < 8. The next step is to solve for x by dividing all parts of the inequality by 2. Let's do that:-2 / 2 < 2x / 2 < 8 / 2Which simplifies to:-1 < x < 4So, the solution is all real numbers x such that x is greater than -1 and less than 4. In interval notation, that would be (-1, 4).Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the absolute value inequality, I correctly converted it into a compound inequality. Then, I added 3 to all parts, which seems correct. Dividing by 2 also seems straightforward. Yeah, I think that's right.Just to be thorough, maybe I should test a value within the interval to see if it satisfies the original inequality. Let's pick x = 0, which is between -1 and 4. Plugging into |2x - 3|, we get |2*0 - 3| = |-3| = 3, which is indeed less than 5. Good.What about a value just outside the interval, say x = -2. Plugging in, we get |2*(-2) - 3| = |-4 - 3| = |-7| = 7, which is greater than 5. That's correct because -2 is not in the solution set.Similarly, if I pick x = 5, which is greater than 4, |2*5 - 3| = |10 - 3| = |7| = 7, which is also greater than 5. So, that checks out too.I think I'm confident that the solution is correct. The steps make sense, and the test values confirm that the solution interval works.Another way to think about it is to graph the absolute value function |2x - 3| and see where it is below 5. The graph of |2x - 3| is a V-shape with its vertex at x = 3/2. The value of the function at the vertex is 0, and it increases as we move away from x = 3/2. So, the points where |2x - 3| = 5 would be the boundaries of the solution. Solving 2x - 3 = 5 gives x = 4, and solving 2x - 3 = -5 gives x = -1. Therefore, between x = -1 and x = 4, the function is below 5, which matches my earlier solution.I guess another thing to consider is whether the inequality is strict or not. Since it's |2x - 3| < 5, not ≤, the endpoints x = -1 and x = 4 are not included in the solution. That's why we use parentheses in the interval notation.Just to recap, the steps are:1. Start with |2x - 3| < 5.2. Rewrite as -5 < 2x - 3 < 5.3. Add 3 to all parts: -2 < 2x < 8.4. Divide by 2: -1 < x < 4.5. Express in interval notation: (-1, 4).Everything seems consistent. I don't think I missed anything here. The logic flows correctly, and the test cases support the solution. I feel good about this answer.

💡First, I recognize that the set consists of all positive integers that leave a remainder of 5 when divided by 8. This can be expressed as 8k + 5, where k is a non-negative integer.Given that the nth number in this set is 597, I set up the equation 8k + 5 = 597 to find the value of k.Subtracting 5 from both sides gives 8k = 592. Dividing both sides by 8, I find that k = 74.Since k represents the position of the number in the set, the nth number corresponds to k = 74. Therefore, n is 74.

💡Okay, so I have this problem where I need to find the degree of the polynomial ( q(x) ) given some information about another polynomial ( i(x) ) and ( p(x) ). Let me try to break this down step by step.First, the problem states that ( i(x) = p(q(x))^2 - q(x)^3 ). I know that ( i(x) ) has a degree of 12, and ( p(x) ) has a degree of 4. I need to find the degree of ( q(x) ). Hmm, okay.Let me recall that when you compose two polynomials, the degree of the resulting polynomial is the product of their degrees. So, if ( p(x) ) is a degree 4 polynomial and ( q(x) ) is a degree ( m ) polynomial, then ( p(q(x)) ) would have a degree of ( 4 times m ). That makes sense because each term in ( p(x) ) is multiplied by ( q(x) ) raised to some power, so the highest degree term would be ( q(x)^4 ), which would have degree ( 4m ).Now, the expression ( p(q(x))^2 ) would then be squaring that composed polynomial. So, the degree of ( p(q(x))^2 ) would be twice the degree of ( p(q(x)) ). That would be ( 2 times 4m = 8m ). Okay, so ( p(q(x))^2 ) has a degree of ( 8m ).Next, looking at the other term in ( i(x) ), which is ( q(x)^3 ). The degree of this term would be ( 3m ) because it's ( q(x) ) raised to the third power.So, now we have ( i(x) = p(q(x))^2 - q(x)^3 ), and the degrees of these two terms are ( 8m ) and ( 3m ) respectively. The degree of ( i(x) ) is the maximum of these two degrees because when you subtract two polynomials, the degree of the result is the higher degree of the two, assuming the leading coefficients don't cancel each other out. The problem tells us that the degree of ( i(x) ) is 12. So, we have:[max(8m, 3m) = 12]Now, I need to figure out what ( m ) is. Let's think about this. Since ( 8m ) is larger than ( 3m ) for any positive ( m ), the maximum of the two will always be ( 8m ). Therefore, we can set up the equation:[8m = 12]Solving for ( m ):[m = frac{12}{8} = frac{3}{2}]Wait a minute, that gives me ( m = 1.5 ), which is not an integer. But the degree of a polynomial has to be an integer. Hmm, that's a problem. Did I make a mistake somewhere?Let me double-check my reasoning. The degree of ( p(q(x)) ) is ( 4m ), so squaring that gives ( 8m ). The degree of ( q(x)^3 ) is ( 3m ). The degree of ( i(x) ) is the maximum of these two, which is ( 8m ). So, ( 8m = 12 ) leads to ( m = 1.5 ). But degrees can't be fractions. Maybe I need to reconsider.Perhaps the leading terms could cancel out when subtracting ( q(x)^3 ) from ( p(q(x))^2 ). If that happens, the degree of ( i(x) ) would be less than ( 8m ). But the problem states that the degree is 12, so maybe ( 8m ) is not the degree after all because the leading terms canceled. Then, the degree would be ( 3m ). Let's explore that.If the leading terms canceled, then the degree of ( i(x) ) would be ( 3m ). So, setting ( 3m = 12 ):[3m = 12 m = 4]Okay, that gives ( m = 4 ), which is an integer. So, maybe the leading terms did cancel out. But how?Let me think about the leading coefficients. Suppose ( p(x) ) is a degree 4 polynomial, so it can be written as ( p(x) = a x^4 + text{lower degree terms} ). Similarly, ( q(x) ) is a degree ( m ) polynomial, so ( q(x) = b x^m + text{lower degree terms} ).Then, ( p(q(x)) ) would be ( a (q(x))^4 + text{lower degree terms} ). So, the leading term of ( p(q(x)) ) is ( a b^4 x^{4m} ).Squaring this, ( p(q(x))^2 ) would have a leading term of ( a^2 b^8 x^{8m} ).On the other hand, ( q(x)^3 ) has a leading term of ( b^3 x^{3m} ).So, when we subtract ( q(x)^3 ) from ( p(q(x))^2 ), the leading term is ( a^2 b^8 x^{8m} - b^3 x^{3m} ). For these leading terms to cancel, their coefficients must be equal and opposite, right? So, we would need:[a^2 b^8 = -b^3]But ( a ) and ( b ) are coefficients of polynomials, which are typically considered to be non-zero (since they are leading coefficients). So, ( a ) and ( b ) are non-zero. Therefore, ( a^2 b^8 = -b^3 ) implies:[a^2 b^5 = -1]Hmm, this is possible if ( b ) is negative, but ( a^2 ) is always positive because it's squared. So, ( a^2 b^5 = -1 ) would require ( b ) to be negative. Let me see:Let me denote ( b = -c ), where ( c ) is a positive real number. Then, substituting back:[a^2 (-c)^5 = -1 a^2 (-c^5) = -1 - a^2 c^5 = -1 a^2 c^5 = 1]So, ( a^2 c^5 = 1 ). This is possible for some positive real numbers ( a ) and ( c ). For example, if ( a = 1 ) and ( c = 1 ), then ( 1^2 times 1^5 = 1 ). So, yes, it's possible for the leading terms to cancel if the coefficients are chosen appropriately.Therefore, in this case, the degree of ( i(x) ) would be ( 3m ) because the ( 8m ) degree terms canceled out. So, setting ( 3m = 12 ):[m = 4]That makes sense because ( m = 4 ) is an integer, and it satisfies the condition that the degree of ( i(x) ) is 12. So, the degree of ( q(x) ) is 4.Wait, but earlier I thought that ( 8m ) would always be larger than ( 3m ), so the degree should be ( 8m ). But in this case, because the leading terms canceled, the degree became ( 3m ). So, it's possible for the degree to drop if the leading coefficients cancel each other out. That's an important point.Let me verify this with an example. Suppose ( p(x) = x^4 ) and ( q(x) = x^4 ). Then, ( p(q(x)) = (x^4)^4 = x^{16} ). Squaring that, ( p(q(x))^2 = x^{32} ). On the other hand, ( q(x)^3 = (x^4)^3 = x^{12} ). So, ( i(x) = x^{32} - x^{12} ), which has a degree of 32, not 12. So, in this case, the degree is ( 8m = 32 ).But in our problem, the degree is 12, which is ( 3m ). So, that means that in our problem, the leading terms must have canceled out. Therefore, the degree of ( i(x) ) is determined by the next highest degree term, which is ( 3m ).So, to summarize, the degree of ( i(x) ) is 12, which comes from ( 3m ), so ( m = 4 ). Therefore, the degree of ( q(x) ) is 4.I think that makes sense. The key here was realizing that even though ( p(q(x))^2 ) would normally have a higher degree, the leading terms could cancel out, making the degree of ( i(x) ) determined by the next highest term, which is ( q(x)^3 ). So, the degree of ( q(x) ) must be 4.

💡Okay, so I have this geometry problem here about triangle ABC. It says that BP and BQ trisect angle ABC. Hmm, trisect means they divide the angle into three equal parts, right? So if angle ABC is being trisected by BP and BQ, that means angle ABP, angle PBQ, and angle QBC are all equal. Let me write that down.Let’s denote angle ABC as θ. Since BP and BQ trisect it, each of the three angles ABP, PBQ, and QBC must be θ/3 each. So, angle ABP = θ/3, angle PBQ = θ/3, and angle QBC = θ/3.Now, the problem mentions two more lines: BM and CN. BM bisects angle PBQ, and CN bisects angle QBC. I need to find the ratio of the measure of angle MBQ to the measure of angle NCQ.First, let me visualize this. Triangle ABC with point B. From B, there are two trisectors BP and BQ, dividing angle ABC into three equal parts. Then, from B, another line BM bisects angle PBQ, which is already one of the trisected angles. Similarly, from C, a line CN bisects angle QBC.Wait, angle PBQ is θ/3, right? So if BM bisects angle PBQ, it splits θ/3 into two equal parts. That means angle MBP and angle MBQ are each (θ/3)/2 = θ/6. So angle MBQ is θ/6.Similarly, angle QBC is θ/3, and CN bisects it. So angle QCN and angle NCB are each (θ/3)/2 = θ/6. Therefore, angle NCQ is θ/6.So, the measure of angle MBQ is θ/6, and the measure of angle NCQ is also θ/6. Therefore, the ratio of angle MBQ to angle NCQ is (θ/6) : (θ/6), which simplifies to 1:1.Wait, that seems straightforward. Did I miss anything? Let me double-check.1. Angle ABC is trisected by BP and BQ, so each trisected angle is θ/3.2. BM bisects angle PBQ, so each of the resulting angles is θ/6.3. Similarly, CN bisects angle QBC, so each of those angles is θ/6.4. Therefore, both angle MBQ and angle NCQ are θ/6, so their ratio is 1.I think that's correct. Maybe I should draw a diagram to visualize it better.[Imagining a diagram: Triangle ABC with point B. BP and BQ are trisectors, dividing angle B into three equal parts. BM is the bisector of angle PBQ, so it splits that angle into two equal parts. Similarly, CN is the bisector of angle QBC, splitting it into two equal parts.]Yes, in the diagram, angle MBQ would be half of angle PBQ, which is θ/6, and angle NCQ would be half of angle QBC, which is also θ/6. So both are equal, making their ratio 1.I don't see any complications here. It seems like a straightforward problem once you break it down into the trisected angles and then the bisected angles. So, the ratio is 1:1.**Final Answer**The ratio of the measure of angle MBQ to the measure of angle NCQ is boxed{1}.

💡First, I need to evaluate ( f(-1) ). Since (-1 < 0), according to the function definition, ( f(-1) = pi^2 + 1 ).Next, I need to evaluate ( f(f(-1)) = f(pi^2 + 1) ). Since (pi^2 + 1 > 0), the function definition tells me that ( f(pi^2 + 1) = 0 ).Finally, I need to evaluate ( f(f(f(-1))) = f(0) ). According to the function definition, when ( x = 0 ), ( f(0) = pi ).Therefore, the value of ( f(f(f(-1))) ) is (pi).