💡Okay, so I have this problem here with two inequalities, p and q, and some logic conditions. Let me try to unpack it step by step.First, let's look at the given statements:- p: -2 ≤ 1 - (x - 1)/3 ≤ 2- q: x² - 2x + 1 - m² ≤ 0, where m > 0And the logical condition is that ¬p is a necessary but not sufficient condition for ¬q. Hmm, that sounds a bit tricky, but let's break it down.Starting with p: I need to solve the inequality -2 ≤ 1 - (x - 1)/3 ≤ 2. Maybe I can simplify this to find the range of x that satisfies p.So, let me rewrite the inequality:-2 ≤ 1 - (x - 1)/3 ≤ 2I can split this into two separate inequalities:1. -2 ≤ 1 - (x - 1)/32. 1 - (x - 1)/3 ≤ 2Let me solve the first one:-2 ≤ 1 - (x - 1)/3Subtract 1 from both sides:-3 ≤ - (x - 1)/3Multiply both sides by -3, remembering to flip the inequality sign:9 ≥ x - 1Which simplifies to:x - 1 ≤ 9So,x ≤ 10Wait, that doesn't seem right. Let me double-check my steps.Starting again:-2 ≤ 1 - (x - 1)/3Subtract 1:-3 ≤ - (x - 1)/3Multiply both sides by -3 (and flip inequality):9 ≥ x - 1So,x - 1 ≤ 9Thus,x ≤ 10Okay, that seems correct.Now, the second inequality:1 - (x - 1)/3 ≤ 2Subtract 1:- (x - 1)/3 ≤ 1Multiply both sides by -3 (and flip inequality):x - 1 ≥ -3So,x ≥ -2Therefore, combining both inequalities, x must satisfy:-2 ≤ x ≤ 10Wait, hold on, that doesn't align with my initial thought. Let me check my calculations again.Wait, maybe I made a mistake when solving the first inequality. Let's go through it again.First inequality:-2 ≤ 1 - (x - 1)/3Subtract 1:-3 ≤ - (x - 1)/3Multiply both sides by -3 (inequality flips):9 ≥ x - 1So,x - 1 ≤ 9Thus,x ≤ 10Second inequality:1 - (x - 1)/3 ≤ 2Subtract 1:- (x - 1)/3 ≤ 1Multiply both sides by -3 (inequality flips):x - 1 ≥ -3So,x ≥ -2Therefore, the solution for p is:-2 ≤ x ≤ 10Wait, that seems different from my initial thought. Maybe I confused the steps earlier. So, p is the interval [-2, 10].Therefore, ¬p would be x < -2 or x > 10.Okay, that makes sense.Now, moving on to q: x² - 2x + 1 - m² ≤ 0I can rewrite this as:(x - 1)² - m² ≤ 0Which is a difference of squares, so it factors as:[(x - 1) - m][(x - 1) + m] ≤ 0So,(x - 1 - m)(x - 1 + m) ≤ 0This inequality will hold when x is between the roots. The roots are:x = 1 + m and x = 1 - mSince m > 0, 1 - m < 1 + mTherefore, the solution for q is:1 - m ≤ x ≤ 1 + mSo, q is the interval [1 - m, 1 + m]Therefore, ¬q would be x < 1 - m or x > 1 + mSo, ¬q is the set of x such that x < 1 - m or x > 1 + mNow, the logical condition is that ¬p is a necessary but not sufficient condition for ¬q.In logical terms, this means:¬p is necessary for ¬q ⇒ ¬q ⇒ ¬pAnd ¬p is not sufficient for ¬q ⇒ ¬p does not imply ¬qIn set terms, this translates to:¬q is a subset of ¬p, but ¬p is not a subset of ¬qWhich means:A ⊂ B, where A is ¬q and B is ¬pSo, A is a proper subset of B.Therefore, the set of x where ¬q holds must be entirely contained within the set where ¬p holds, but not the other way around.So, in terms of intervals:¬q: x < 1 - m or x > 1 + m¬p: x < -2 or x > 10So, for ¬q to be a subset of ¬p, the intervals where ¬q holds must lie entirely within the intervals where ¬p holds.Therefore:1. The left interval of ¬q: x < 1 - m must be contained within x < -2Which implies:1 - m ≤ -2Similarly, the right interval of ¬q: x > 1 + m must be contained within x > 10Which implies:1 + m ≥ 10So, let's solve these inequalities.First inequality:1 - m ≤ -2Subtract 1:-m ≤ -3Multiply both sides by -1 (inequality flips):m ≥ 3Second inequality:1 + m ≥ 10Subtract 1:m ≥ 9So, combining both inequalities, m must satisfy both m ≥ 3 and m ≥ 9, which simplifies to m ≥ 9But wait, hold on. The problem states that ¬p is necessary but not sufficient for ¬q.Which means that ¬q implies ¬p, but ¬p does not imply ¬q.In terms of sets, that means ¬q is a subset of ¬p, but ¬p is not a subset of ¬q.So, in our case, ¬q is x < 1 - m or x > 1 + m, and ¬p is x < -2 or x > 10.Therefore, for ¬q to be a subset of ¬p, we must have:1 - m ≤ -2 and 1 + m ≥ 10Which gives m ≥ 3 and m ≥ 9, so m ≥ 9But also, since ¬p is not sufficient for ¬q, it means that ¬p does not imply ¬q, so ¬p is not a subset of ¬q.Which would mean that there exists some x in ¬p that is not in ¬q.Given that ¬p is x < -2 or x > 10, and ¬q is x < 1 - m or x > 1 + m.So, if m = 9, then ¬q is x < -8 or x > 10But ¬p is x < -2 or x > 10So, in this case, ¬q is x < -8 or x > 10, which is a subset of ¬p (x < -2 or x > 10). So, that's good.But is ¬p not a subset of ¬q? Let's see.¬p is x < -2 or x > 10¬q is x < -8 or x > 10So, x < -2 includes x between -8 and -2, which are not in ¬q. Therefore, ¬p is not a subset of ¬q, which satisfies the condition that ¬p is not sufficient for ¬q.Therefore, m must be at least 9.Wait, but earlier when solving, I got m ≥ 3 and m ≥ 9, so m ≥ 9.But let me double-check.If m = 9, then ¬q is x < -8 or x > 10, which is indeed a subset of ¬p (x < -2 or x > 10). So, that works.If m were less than 9, say m = 8, then ¬q would be x < -7 or x > 9.But ¬p is x < -2 or x > 10.So, in this case, ¬q is x < -7 or x > 9, which is not entirely contained within ¬p, because x > 9 includes x between 9 and 10, which are not in ¬p (since ¬p is x > 10). Therefore, ¬q is not a subset of ¬p when m = 8.Similarly, if m = 6, ¬q is x < -5 or x > 7, which is not a subset of ¬p (x < -2 or x > 10). So, only when m is at least 9, does ¬q become a subset of ¬p.Therefore, the range of m is m ≥ 9.Wait, but in the initial problem, the user wrote:"Given p: -2 ≤ 1 - (x - 1)/3 ≤ 2 and q: x² - 2x + 1 - m² ≤ 0 (m > 0), and ¬p is a necessary but not sufficient condition for ¬q, find the range of the real number m."But in my calculation, I got m ≥ 9, but in the initial assistant's answer, it was m ≥ 6.Hmm, that's a discrepancy. Let me check where I might have gone wrong.Wait, when solving p, I got x between -2 and 10, but in the initial assistant's solution, they got x between -5 and 7.Wait, that's a big difference. Let me re-examine the solution of p.Original p: -2 ≤ 1 - (x - 1)/3 ≤ 2Let me solve this correctly.First, let's subtract 1 from all parts:-2 - 1 ≤ - (x - 1)/3 ≤ 2 - 1Which is:-3 ≤ - (x - 1)/3 ≤ 1Now, multiply all parts by -3, remembering to flip the inequalities:When multiplying by a negative, inequalities reverse.So,-3 * (-3) ≥ x - 1 ≥ 1 * (-3)Which is:9 ≥ x - 1 ≥ -3Then, add 1 to all parts:10 ≥ x ≥ -2So, x is between -2 and 10.Wait, so the initial assistant's solution was incorrect. They got x between -5 and 7, but the correct solution is x between -2 and 10.Therefore, ¬p is x < -2 or x > 10.Similarly, for q: x² - 2x + 1 - m² ≤ 0Which is (x - 1)^2 ≤ m²So, |x - 1| ≤ mThus, x is between 1 - m and 1 + m.Therefore, ¬q is x < 1 - m or x > 1 + m.Now, ¬p is necessary but not sufficient for ¬q.Which means:¬q ⇒ ¬p, but ¬p does not imply ¬q.In set terms, ¬q is a subset of ¬p, but ¬p is not a subset of ¬q.Therefore, the intervals for ¬q must be entirely within ¬p.So, ¬q: x < 1 - m or x > 1 + m¬p: x < -2 or x > 10Therefore, for ¬q to be a subset of ¬p:1 - m ≤ -2 and 1 + m ≥ 10Solving:1 - m ≤ -2 ⇒ -m ≤ -3 ⇒ m ≥ 31 + m ≥ 10 ⇒ m ≥ 9Therefore, m must satisfy both, so m ≥ 9.Additionally, to ensure that ¬p is not sufficient for ¬q, we need that ¬p is not a subset of ¬q.Which means there exists some x in ¬p that is not in ¬q.Given that ¬p is x < -2 or x > 10, and ¬q is x < 1 - m or x > 1 + m.So, if m = 9, then ¬q is x < -8 or x > 10.Therefore, ¬p is x < -2 or x > 10.So, x between -8 and -2 are in ¬p but not in ¬q, hence ¬p is not a subset of ¬q.Therefore, m must be at least 9.But wait, in the initial assistant's answer, they got m ≥ 6, but that was based on an incorrect solution of p.So, the correct range is m ≥ 9.Wait, but let me double-check my calculations again.Given p: -2 ≤ 1 - (x - 1)/3 ≤ 2Let me solve it step by step.First, let's isolate the fraction:-2 ≤ 1 - (x - 1)/3 ≤ 2Subtract 1 from all parts:-3 ≤ - (x - 1)/3 ≤ 1Multiply all parts by -3, flipping inequalities:9 ≥ x - 1 ≥ -3Add 1 to all parts:10 ≥ x ≥ -2So, x ∈ [-2, 10]Therefore, ¬p is x < -2 or x > 10For q: x² - 2x + 1 - m² ≤ 0Which is (x - 1)^2 ≤ m²So, |x - 1| ≤ mThus, x ∈ [1 - m, 1 + m]Therefore, ¬q is x < 1 - m or x > 1 + mNow, ¬p is necessary for ¬q: ¬q ⇒ ¬pWhich means that whenever ¬q is true, ¬p is also true.In other words, the set of x where ¬q holds must be a subset of the set where ¬p holds.Therefore:x < 1 - m ⇒ x < -2andx > 1 + m ⇒ x > 10Which gives:1 - m ≤ -2 ⇒ m ≥ 3and1 + m ≥ 10 ⇒ m ≥ 9So, m must be at least 9.Additionally, since ¬p is not sufficient for ¬q, it means that ¬p does not imply ¬q.Which means there exists some x in ¬p that is not in ¬q.Given that ¬p is x < -2 or x > 10, and ¬q is x < 1 - m or x > 1 + m.So, if m = 9, then ¬q is x < -8 or x > 10.Therefore, x between -8 and -2 are in ¬p but not in ¬q, so ¬p does not imply ¬q.Hence, m must be at least 9.Therefore, the correct range is m ≥ 9.I think the initial assistant made a mistake in solving p, leading to an incorrect range of m ≥ 6. The correct range should be m ≥ 9.

💡First, determine Cheenu's running speed as a boy. He ran 18 miles in 4 hours. To find his speed in miles per hour, divide the distance by the time:[text{Speed as a boy} = frac{18 text{ miles}}{4 text{ hours}} = 4.5 text{ miles per hour}]Next, determine Cheenu's walking speed as an old man. He walks 12 miles in 5 hours. To find his speed in miles per hour, divide the distance by the time:[text{Speed as an old man} = frac{12 text{ miles}}{5 text{ hours}} = 2.4 text{ miles per hour}]Now, calculate the time it takes for Cheenu to run one mile as a boy. Since speed is distance divided by time, time is distance divided by speed. Convert hours to minutes by multiplying by 60:[text{Time per mile as a boy} = frac{1 text{ mile}}{4.5 text{ miles per hour}} times 60 text{ minutes} = frac{60}{4.5} approx 13.33 text{ minutes per mile}]Similarly, calculate the time it takes for Cheenu to walk one mile as an old man:[text{Time per mile as an old man} = frac{1 text{ mile}}{2.4 text{ miles per hour}} times 60 text{ minutes} = frac{60}{2.4} = 25 text{ minutes per mile}]Finally, find the difference in time per mile between walking and running:[text{Difference} = 25 text{ minutes per mile} - 13.33 text{ minutes per mile} approx 11.67 text{ minutes per mile}]Since the answer choices are in whole numbers, round 11.67 to the nearest whole number, which is 12 minutes.

💡Alright, so I have this problem here: I need to prove that there exists an integer ( n ) such that both ( 2n + 1 ) and ( 3n + 1 ) are perfect squares, and that this ( n ) is a multiple of 40. Hmm, okay, let's break this down step by step.First, let's understand what it means for ( 2n + 1 ) and ( 3n + 1 ) to be perfect squares. A perfect square is an integer that is the square of another integer. So, if ( 2n + 1 ) is a perfect square, there must be some integer ( a ) such that ( 2n + 1 = a^2 ). Similarly, there must be some integer ( b ) such that ( 3n + 1 = b^2 ).So, we can write:[2n + 1 = a^2 quad text{and} quad 3n + 1 = b^2]for some integers ( a ) and ( b ).Now, let's try to express ( n ) in terms of ( a ) and ( b ). From the first equation:[2n + 1 = a^2 implies 2n = a^2 - 1 implies n = frac{a^2 - 1}{2}]Similarly, from the second equation:[3n + 1 = b^2 implies 3n = b^2 - 1 implies n = frac{b^2 - 1}{3}]So, we have two expressions for ( n ):[n = frac{a^2 - 1}{2} quad text{and} quad n = frac{b^2 - 1}{3}]Since both expressions equal ( n ), we can set them equal to each other:[frac{a^2 - 1}{2} = frac{b^2 - 1}{3}]To eliminate the denominators, let's multiply both sides by 6 (the least common multiple of 2 and 3):[3(a^2 - 1) = 2(b^2 - 1)]Expanding both sides:[3a^2 - 3 = 2b^2 - 2]Let's bring all terms to one side:[3a^2 - 2b^2 - 1 = 0]So, we have the equation:[3a^2 - 2b^2 = 1]This is a Diophantine equation, which is an equation that seeks integer solutions. Let's see if we can find integers ( a ) and ( b ) that satisfy this equation.Let me try plugging in some small integer values for ( a ) and see if ( b ) comes out as an integer.Start with ( a = 1 ):[3(1)^2 - 2b^2 = 1 implies 3 - 2b^2 = 1 implies 2b^2 = 2 implies b^2 = 1 implies b = pm1]So, ( a = 1 ) and ( b = pm1 ) are solutions. Let's check what ( n ) would be in this case:[n = frac{a^2 - 1}{2} = frac{1 - 1}{2} = 0]But ( n = 0 ) is a multiple of 40, technically, since 0 is a multiple of every integer. However, the problem might be looking for a positive integer ( n ), so let's check the next possible ( a ).Next, ( a = 3 ):[3(3)^2 - 2b^2 = 1 implies 27 - 2b^2 = 1 implies 2b^2 = 26 implies b^2 = 13]But 13 is not a perfect square, so ( b ) is not an integer here.Next, ( a = 5 ):[3(5)^2 - 2b^2 = 1 implies 75 - 2b^2 = 1 implies 2b^2 = 74 implies b^2 = 37]Again, 37 is not a perfect square.Next, ( a = 7 ):[3(7)^2 - 2b^2 = 1 implies 147 - 2b^2 = 1 implies 2b^2 = 146 implies b^2 = 73]73 is not a perfect square.Hmm, this isn't working. Maybe I need a different approach. Let's consider that ( a ) and ( b ) must satisfy ( 3a^2 - 2b^2 = 1 ). This resembles a Pell-type equation, which usually has infinitely many solutions. Pell equations are of the form ( x^2 - Dy^2 = N ), where ( D ) is a non-square positive integer and ( N ) is an integer.Our equation is ( 3a^2 - 2b^2 = 1 ). Let's rearrange it to resemble a Pell equation:[3a^2 - 2b^2 = 1 implies 3a^2 = 2b^2 + 1]Alternatively, we can write:[( sqrt{3}a )^2 - 2b^2 = 1]But this isn't quite a standard Pell equation because of the coefficient 3 on ( a^2 ). Maybe we can manipulate it further.Let me try to express it as:[( sqrt{3}a )^2 - 2b^2 = 1]But this still isn't helpful because of the square root. Maybe another substitution.Let me consider ( a ) and ( b ) in terms of other variables. Suppose ( a = k ) and ( b = m ). Then, the equation is:[3k^2 - 2m^2 = 1]This is a hyperbola, and we're looking for integer points on it.I know that Pell equations have solutions that can be generated from the minimal solution. Let's see if we can find the minimal solution for this equation.We already saw that ( a = 1 ), ( b = 1 ) is a solution. Let's see if there are larger solutions.Let me try ( a = 5 ):[3(5)^2 - 2b^2 = 1 implies 75 - 2b^2 = 1 implies 2b^2 = 74 implies b^2 = 37]Not a square.( a = 7 ):[3(7)^2 - 2b^2 = 1 implies 147 - 2b^2 = 1 implies 2b^2 = 146 implies b^2 = 73]Not a square.( a = 9 ):[3(9)^2 - 2b^2 = 1 implies 243 - 2b^2 = 1 implies 2b^2 = 242 implies b^2 = 121 implies b = pm11]Ah, here we go! So, ( a = 9 ) and ( b = 11 ) is a solution.Let's verify:[3(9)^2 - 2(11)^2 = 3(81) - 2(121) = 243 - 242 = 1]Yes, that works.So, now that we have a solution ( a = 9 ), ( b = 11 ), we can find ( n ):[n = frac{a^2 - 1}{2} = frac{81 - 1}{2} = frac{80}{2} = 40]Alternatively, using the other expression:[n = frac{b^2 - 1}{3} = frac{121 - 1}{3} = frac{120}{3} = 40]Perfect, both give ( n = 40 ).So, ( n = 40 ) is a solution. Now, the problem states that such an ( n ) is a multiple of 40. Well, 40 is obviously a multiple of itself, so that checks out.But wait, is 40 the only solution? Or are there other multiples of 40 that also satisfy the conditions? Let's see.Given that Pell-type equations have infinitely many solutions, it's likely that there are infinitely many ( n ) that satisfy the conditions, each being a multiple of 40. However, the problem only asks to prove that such an ( n ) exists and that it's a multiple of 40. So, finding ( n = 40 ) suffices for the proof.Just to be thorough, let's check ( n = 40 ):[2n + 1 = 2(40) + 1 = 81 = 9^2][3n + 1 = 3(40) + 1 = 121 = 11^2]Both are perfect squares, as required.Therefore, we've found that ( n = 40 ) satisfies the conditions, and since 40 is a multiple of 40, the proof is complete.**Final Answer**boxed{40}

💡First, I need to determine the number of sixes the batsman hit. I'll start by calculating the runs scored by running between the wickets, which is 60% of the total score.Next, I'll calculate the runs contributed by the boundaries. Since the batsman hit 5 boundaries, each worth 4 runs, the total runs from boundaries are 20.Then, I'll find out the total runs scored from both boundaries and sixes by subtracting the runs scored by running from the total score.Finally, I'll determine the number of sixes by dividing the runs from sixes by the value of each six.

💡Okay, so I have two logarithmic expressions to calculate. Let me take them one by one. Starting with the first one: (2log_{5}10 + log_{5}0.25). Hmm, logarithms can sometimes be tricky, but I remember some properties that might help here. First, I recall that (a cdot log_b c = log_b c^a). So, applying that to the first term, (2log_{5}10) becomes (log_{5}10^2), which is (log_{5}100). That seems manageable.Now, the expression is (log_{5}100 + log_{5}0.25). Another logarithm property is that (log_b a + log_b c = log_b (a cdot c)). So, combining these two logs, I get (log_{5}(100 times 0.25)). Let me calculate that product: 100 times 0.25 is 25. So now, the expression simplifies to (log_{5}25).Alright, (log_{5}25) is asking, "5 raised to what power gives 25?" Well, 5 squared is 25, so that's 2. So, the first part is 2. That wasn't too bad.Moving on to the second expression: (0.027^{-frac{1}{3}} - (-frac{1}{7})^{-1} + (2frac{7}{9})^{frac{1}{2}} - (sqrt{2} - 1)^0). Wow, that's a mouthful. Let me break it down term by term.Starting with (0.027^{-frac{1}{3}}). Negative exponents mean reciprocals, so this is the same as (frac{1}{0.027^{frac{1}{3}}}). The exponent (frac{1}{3}) is the cube root. So, I need to find the cube root of 0.027. Hmm, 0.027 is a decimal. Let me think: 0.027 is equal to (27 times 10^{-3}), right? So, the cube root of 27 is 3, and the cube root of (10^{-3}) is (10^{-1}) because ((10^{-1})^3 = 10^{-3}). So, putting that together, the cube root of 0.027 is 3 times (10^{-1}), which is 0.3. Therefore, (0.027^{-frac{1}{3}}) is (frac{1}{0.3}). Calculating that, 1 divided by 0.3 is approximately 3.333... or (frac{10}{3}). Let me keep it as a fraction for accuracy: (frac{10}{3}).Next term: (-(-frac{1}{7})^{-1}). Let me parse this carefully. The negative exponent here is -1, so it's the reciprocal. But there's also a negative sign in front. So, first, let's compute ((- frac{1}{7})^{-1}). The reciprocal of (-frac{1}{7}) is (-7). So, the term becomes (-(-7)), which simplifies to +7. Okay, so that term is 7.Third term: ((2frac{7}{9})^{frac{1}{2}}). This is a mixed number, so I should convert it to an improper fraction first. 2 and 7/9 is the same as (frac{25}{9}). So, now it's ((frac{25}{9})^{frac{1}{2}}), which is the square root of (frac{25}{9}). The square root of 25 is 5, and the square root of 9 is 3, so this simplifies to (frac{5}{3}).Last term: ((sqrt{2} - 1)^0). Any number raised to the power of 0 is 1, right? So, this term is just 1.Putting it all together, the expression becomes: (frac{10}{3} + 7 + frac{5}{3} - 1). Let me combine these step by step.First, combine the fractions: (frac{10}{3} + frac{5}{3} = frac{15}{3} = 5). Then, add the whole numbers: 7 - 1 = 6. So, adding those results together: 5 + 6 = 11.Wait, let me double-check that. So, (frac{10}{3} + 7 + frac{5}{3} - 1). Combining (frac{10}{3} + frac{5}{3}) gives (frac{15}{3} = 5). Then, 7 - 1 is 6. So, 5 + 6 is indeed 11. Okay, that seems correct.So, summarizing:1. (2log_{5}10 + log_{5}0.25 = 2)2. (0.027^{-frac{1}{3}} - (-frac{1}{7})^{-1} + (2frac{7}{9})^{frac{1}{2}} - (sqrt{2} - 1)^0 = 11)I think that's all. I tried to break it down step by step, making sure each operation was clear and checking my work as I went along. It's easy to make a mistake with negative exponents and reciprocals, so I was careful with the signs. Also, remembering that any number to the power of 0 is 1 was key for the last term. Overall, it makes sense now.

💡Okay, so I have this problem about a parabola and a line intersecting it. Let me try to figure it out step by step. First, the parabola is given by the equation ( y^2 = 4x ). I remember that for a parabola in the form ( y^2 = 4ax ), the focus is at ( (a, 0) ). In this case, comparing to the standard form, ( 4a = 4 ), so ( a = 1 ). That means the focus ( F ) is at ( (1, 0) ). Got that down.Next, the problem says there's a line passing through the focus ( F ) with an inclination angle of ( frac{pi}{4} ). I think an inclination angle is the angle the line makes with the positive x-axis. So, ( frac{pi}{4} ) radians is 45 degrees. The slope of a line is equal to the tangent of its inclination angle. So, the slope ( m ) is ( tan(frac{pi}{4}) ). Since ( tan(frac{pi}{4}) = 1 ), the slope is 1. Now, I need the equation of the line that passes through ( F(1, 0) ) with slope 1. Using the point-slope form of a line equation, which is ( y - y_1 = m(x - x_1) ), where ( (x_1, y_1) ) is a point the line passes through. Plugging in the values, we get ( y - 0 = 1 cdot (x - 1) ), which simplifies to ( y = x - 1 ). So, the equation of the line is ( y = x - 1 ).Now, this line intersects the parabola ( y^2 = 4x ) at points ( A ) and ( B ). To find these points of intersection, I need to solve the system of equations:1. ( y^2 = 4x )2. ( y = x - 1 )I can substitute equation 2 into equation 1 to eliminate ( y ). So, replacing ( y ) with ( x - 1 ) in equation 1 gives:( (x - 1)^2 = 4x )Let me expand the left side:( x^2 - 2x + 1 = 4x )Now, bring all terms to one side to form a quadratic equation:( x^2 - 2x + 1 - 4x = 0 )( x^2 - 6x + 1 = 0 )So, the quadratic equation is ( x^2 - 6x + 1 = 0 ). Let me denote the roots of this equation as ( x_1 ) and ( x_2 ). These correspond to the x-coordinates of points ( A ) and ( B ).From the quadratic equation ( ax^2 + bx + c = 0 ), the sum of the roots is ( -b/a ) and the product is ( c/a ). Here, ( a = 1 ), ( b = -6 ), and ( c = 1 ). So,Sum of roots: ( x_1 + x_2 = -(-6)/1 = 6 )Product of roots: ( x_1 x_2 = 1/1 = 1 )Now, I need to find the distance between points ( A ) and ( B ). To do this, I can use the distance formula, which is ( |AB| = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ).But to use this, I need the coordinates of ( A ) and ( B ). Alternatively, since I know the x-coordinates, I can express the y-coordinates using the line equation ( y = x - 1 ). So, for each ( x ), ( y = x - 1 ). Therefore, the coordinates are ( A(x_1, x_1 - 1) ) and ( B(x_2, x_2 - 1) ).So, the distance squared between ( A ) and ( B ) is:( (x_2 - x_1)^2 + [(x_2 - 1) - (x_1 - 1)]^2 )Simplify the y-component:( (x_2 - x_1)^2 + (x_2 - x_1)^2 )Which is:( 2(x_2 - x_1)^2 )Therefore, the distance ( |AB| = sqrt{2(x_2 - x_1)^2} = sqrt{2} |x_2 - x_1| )So, I need to find ( |x_2 - x_1| ). I know that for a quadratic equation ( ax^2 + bx + c = 0 ), the difference of the roots is ( sqrt{(x_1 + x_2)^2 - 4x_1x_2} ). Let me verify that:( (x_2 - x_1)^2 = (x_1 + x_2)^2 - 4x_1x_2 )Yes, that's correct because:( (x_2 - x_1)^2 = x_2^2 - 2x_1x_2 + x_1^2 = (x_1 + x_2)^2 - 4x_1x_2 )So, substituting the known values:( (x_2 - x_1)^2 = (6)^2 - 4(1) = 36 - 4 = 32 )Therefore, ( |x_2 - x_1| = sqrt{32} = 4sqrt{2} )Now, plug this back into the distance formula:( |AB| = sqrt{2} times 4sqrt{2} = sqrt{2} times 4sqrt{2} )Multiplying these together:( sqrt{2} times sqrt{2} = 2 ), so ( 4 times 2 = 8 )Therefore, the distance between points ( A ) and ( B ) is 8 units.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Found the focus correctly at (1,0).2. Calculated the slope as 1, which seems right since tan(45°) is 1.3. Equation of the line: y = x - 1, that looks correct.4. Substituted into the parabola equation: (x - 1)^2 = 4x, expanded correctly to x² - 6x + 1 = 0.5. Sum and product of roots: 6 and 1, correct.6. Distance formula: I considered the difference in x and y, which are both (x2 - x1), so squared and added, giving 2(x2 - x1)^2. That seems right.7. Calculated (x2 - x1)^2 as 32, so |x2 - x1| is 4√2. Then multiplied by √2 to get 8. That all adds up.I think that's solid. So, the length of segment AB is 8.**Final Answer**The length of segment ( AB ) is boxed{8}.