💡First, I need to determine the distance traveled by the object at ( t = 4 ) seconds using the given equation ( s = 5t^2 + 2t ).I'll start by substituting ( t = 4 ) into the equation:[s = 5(4)^2 + 2(4)]Next, I'll calculate ( (4)^2 ):[(4)^2 = 16]Then, I'll multiply this result by 5:[5 times 16 = 80]After that, I'll calculate ( 2 times 4 ):[2 times 4 = 8]Finally, I'll add the two results together to find the total distance:[80 + 8 = 88]So, the distance traveled by the object when ( t = 4 ) seconds is 88 meters.

💡First, I need to simplify the expression ((x - 1)^2 + (x + 3)(x - 3) - (x - 3)(x - 1)). I'll start by expanding each term individually. Expanding ((x - 1)^2) gives (x^2 - 2x + 1).Next, expanding ((x + 3)(x - 3)) results in (x^2 - 9).Then, expanding ((x - 3)(x - 1)) yields (x^2 - 4x + 3).Now, I'll substitute these expanded forms back into the original expression:[(x^2 - 2x + 1) + (x^2 - 9) - (x^2 - 4x + 3)]Combining like terms, I simplify the expression to:[x^2 + 2x - 11]Given the equation (x^2 + 2x = 2), I substitute this into the simplified expression:[2 - 11 = -9]Therefore, the value of the expression is (-9).

💡To find the smallest positive integer ( a ) such that ( a^{-1} ) is undefined modulo 72 and modulo 90, we need to ensure that ( a ) shares a common factor greater than 1 with both 72 and 90.First, we factorize both moduli:- ( 72 = 2^3 times 3^2 )- ( 90 = 2 times 3^2 times 5 )The greatest common divisor (GCD) of 72 and 90 is:- ( gcd(72, 90) = 2 times 3^2 = 18 )The smallest positive integer ( a ) that shares a common factor greater than 1 with both 72 and 90 is the smallest prime factor of their GCD, which is 2. However, we need to ensure that ( a ) does not have an inverse in both moduli. Since 2 is a common factor, ( a = 2 ) satisfies the condition.Therefore, the smallest positive integer ( a ) is 2.

💡Okay, so I have this trigonometry problem here. It says that cos(α - π/4) equals √2 over 4, and I need to find tan²(α + π/4). The options are A: -7, B: 7, C: 1/7, D: -1/7. Hmm, let me try to figure this out step by step.First, I remember that cosine of a difference can be expanded using the cosine difference identity. The identity is cos(A - B) = cosA cosB + sinA sinB. So, maybe I can apply that here. Let me write that down:cos(α - π/4) = cosα cos(π/4) + sinα sin(π/4).I know that cos(π/4) and sin(π/4) are both √2/2. So substituting those in:cos(α - π/4) = cosα*(√2/2) + sinα*(√2/2).Factor out √2/2:cos(α - π/4) = (√2/2)(cosα + sinα).We are given that this equals √2/4. So:(√2/2)(cosα + sinα) = √2/4.Hmm, let me solve for (cosα + sinα). If I divide both sides by √2/2, that's the same as multiplying by 2/√2, which simplifies to √2. So:cosα + sinα = (√2/4) * (2/√2) = (√2 * 2)/(4 * √2) = 2/4 = 1/2.So, cosα + sinα = 1/2. Okay, that's a useful equation.Now, I need to find tan²(α + π/4). Let me recall that tan(A + B) can be expanded using the tangent addition formula:tan(A + B) = (tanA + tanB)/(1 - tanA tanB).So, tan(α + π/4) = (tanα + tan(π/4))/(1 - tanα tan(π/4)).I know that tan(π/4) is 1, so this simplifies to:tan(α + π/4) = (tanα + 1)/(1 - tanα).Therefore, tan²(α + π/4) = [(tanα + 1)/(1 - tanα)]².So, if I can find tanα, I can compute this expression.But how do I find tanα? I have the equation cosα + sinα = 1/2. Maybe I can square both sides to get an equation involving sin²α and cos²α, which can be related to tanα.Let's square both sides:(cosα + sinα)² = (1/2)²Expanding the left side:cos²α + 2 sinα cosα + sin²α = 1/4.I know that cos²α + sin²α = 1, so substituting that in:1 + 2 sinα cosα = 1/4.Subtract 1 from both sides:2 sinα cosα = 1/4 - 1 = -3/4.So, sinα cosα = -3/8.Hmm, that's interesting. Now, I need to find tanα. Let me recall that tanα = sinα / cosα. Let me denote t = tanα. Then, sinα = t cosα.But I also have sinα cosα = -3/8. Substituting sinα = t cosα:(t cosα) * cosα = t cos²α = -3/8.So, t cos²α = -3/8.But I also know that cos²α = 1 / (1 + t²) because 1 + tan²α = 1 / cos²α.So, substituting cos²α:t * [1 / (1 + t²)] = -3/8.Therefore:t / (1 + t²) = -3/8.Multiply both sides by (1 + t²):t = (-3/8)(1 + t²).Bring all terms to one side:t + (3/8)(1 + t²) = 0.Multiply through by 8 to eliminate denominators:8t + 3(1 + t²) = 0.Expanding:8t + 3 + 3t² = 0.Rearranged:3t² + 8t + 3 = 0.So, we have a quadratic equation in t: 3t² + 8t + 3 = 0.Let me solve for t using the quadratic formula. The quadratic is at² + bt + c = 0, so a = 3, b = 8, c = 3.t = [-b ± √(b² - 4ac)] / (2a) = [-8 ± √(64 - 36)] / 6 = [-8 ± √28] / 6.Simplify √28: √28 = 2√7.So, t = [-8 ± 2√7]/6 = [-4 ± √7]/3.Therefore, tanα = (-4 + √7)/3 or tanα = (-4 - √7)/3.Hmm, so tanα has two possible values. Let me note that down.Now, going back to tan(α + π/4) = (tanα + 1)/(1 - tanα). So, let's compute this for both values of tanα.First, let's take tanα = (-4 + √7)/3.Compute numerator: tanα + 1 = [(-4 + √7)/3] + 1 = [(-4 + √7) + 3]/3 = (-1 + √7)/3.Compute denominator: 1 - tanα = 1 - [(-4 + √7)/3] = [3 + 4 - √7]/3 = (7 - √7)/3.So, tan(α + π/4) = [(-1 + √7)/3] / [(7 - √7)/3] = (-1 + √7)/(7 - √7).Similarly, for tanα = (-4 - √7)/3.Compute numerator: tanα + 1 = [(-4 - √7)/3] + 1 = [(-4 - √7) + 3]/3 = (-1 - √7)/3.Compute denominator: 1 - tanα = 1 - [(-4 - √7)/3] = [3 + 4 + √7]/3 = (7 + √7)/3.So, tan(α + π/4) = [(-1 - √7)/3] / [(7 + √7)/3] = (-1 - √7)/(7 + √7).Now, I need to compute tan²(α + π/4) for both cases.First case: tan(α + π/4) = (-1 + √7)/(7 - √7).Let me rationalize the denominator:Multiply numerator and denominator by (7 + √7):[(-1 + √7)(7 + √7)] / [(7 - √7)(7 + √7)].Denominator: 7² - (√7)² = 49 - 7 = 42.Numerator: (-1)(7) + (-1)(√7) + √7*7 + √7*√7 = -7 - √7 + 7√7 + 7.Simplify numerator:-7 + 7 = 0.-√7 + 7√7 = 6√7.So, numerator is 6√7.Therefore, tan(α + π/4) = (6√7)/42 = √7/7.Thus, tan²(α + π/4) = (√7/7)² = 7/49 = 1/7.Second case: tan(α + π/4) = (-1 - √7)/(7 + √7).Again, rationalize the denominator by multiplying numerator and denominator by (7 - √7):[(-1 - √7)(7 - √7)] / [(7 + √7)(7 - √7)].Denominator is again 42.Numerator: (-1)(7) + (-1)(-√7) + (-√7)(7) + (-√7)(-√7) = -7 + √7 - 7√7 + 7.Simplify numerator:-7 + 7 = 0.√7 - 7√7 = -6√7.So, numerator is -6√7.Therefore, tan(α + π/4) = (-6√7)/42 = -√7/7.Thus, tan²(α + π/4) = (-√7/7)² = 7/49 = 1/7.So, in both cases, tan²(α + π/4) equals 1/7. Therefore, the answer is C: 1/7.Wait, but let me double-check my steps to make sure I didn't make any mistakes.Starting from cos(α - π/4) = √2/4, I used the cosine difference identity correctly, expanded it, and found that cosα + sinα = 1/2. That seems right.Then, I squared both sides to get 1 + 2 sinα cosα = 1/4, leading to sinα cosα = -3/8. That also seems correct.Then, I set t = tanα, expressed sinα in terms of t and cosα, substituted into sinα cosα = -3/8, and ended up with t / (1 + t²) = -3/8. That seems okay.Then, I solved the quadratic equation 3t² + 8t + 3 = 0, which gave me t = [-4 ± √7]/3. That seems correct.Then, I computed tan(α + π/4) for both t values, rationalized the denominators, and found that tan²(α + π/4) was 1/7 in both cases. That seems consistent.So, I think my answer is correct. The answer is C: 1/7.**Final Answer**boxed{C}

💡To determine the length of the train, we'll first calculate the speed of the train using the time it takes to cross the lamp post. When crossing the lamp post, the train covers its own length in 2.5 seconds. Let ( L ) be the length of the train. The speed of the train is:[text{Speed} = frac{L}{2.5}]Next, we'll use the time it takes to cross the bridge to set up an equation. The train covers its own length plus the length of the bridge (150 meters) in 7.5 seconds.Using the same speed:[frac{L + 150}{7.5} = frac{L}{2.5}]Solving for ( L ):[2.5(L + 150) = 7.5L][2.5L + 375 = 7.5L][375 = 5L][L = 75]Therefore, the length of the train is 75 meters.

💡Okay, so Vika has 60 cards numbered from 1 to 60, and she wants to pair all of them such that each pair has the same modulus of the difference between the numbers. I need to figure out how many ways she can do this.First, I should understand what it means for each pair to have the same modulus difference. The modulus difference is just the absolute difference between two numbers. So, if Vika pairs the cards, each pair should have the same difference between the two numbers. For example, if one pair is (1, 3), the difference is 2, so all other pairs should also have a difference of 2.Let me denote this common difference as ( d ). So, each pair will be of the form ( (n, n + d) ), where ( n ) is the smaller number in the pair. But since the numbers go up to 60, ( n + d ) must not exceed 60. That means ( d ) has to be chosen such that when you add it to any number in the pair, you don't go beyond 60.Now, if we're pairing all 60 cards, we need to make sure that the entire set from 1 to 60 can be divided into pairs with this common difference ( d ). This means that the way we pair them should cover all numbers without overlap or missing numbers.Let me think about how pairing works. If we start with the smallest number, 1, it should pair with ( 1 + d ). Then the next number, 2, should pair with ( 2 + d ), and so on. But this can only continue until we reach a point where adding ( d ) would exceed 60.Wait, but if we keep pairing like this, we might end up with some numbers left unpaired or overlapping. So, maybe there's a better way to think about this. Perhaps the entire set can be divided into multiple blocks where each block contains pairs with the same difference ( d ).For example, if ( d = 1 ), then each pair would be consecutive numbers: (1,2), (3,4), (5,6), etc. But in this case, the modulus difference is 1, which is valid. Similarly, if ( d = 2 ), the pairs would be (1,3), (2,4), (5,7), (6,8), etc.But wait, if ( d = 2 ), the first pair is (1,3), then (2,4), but then the next pair would have to be (5,7), skipping 5 and 6. Hmm, does that cover all numbers? Let's see: 1,3; 2,4; 5,7; 6,8; 9,11; 10,12; and so on. It seems like it does cover all numbers, but in this case, the modulus difference is 2.But how do we ensure that this works for all numbers up to 60? Maybe ( d ) has to be such that when we pair the numbers, we don't leave any gaps or exceed 60.Let me formalize this a bit. If we have a common difference ( d ), then each pair is ( (k, k + d) ). To pair all 60 numbers, we need to ensure that every number from 1 to 60 is included in exactly one pair. So, the pairing process must cover the entire set without overlap or omission.This suggests that the set {1, 2, ..., 60} can be partitioned into pairs where each pair has a difference of ( d ). For this to be possible, the number 60 must be divisible by ( 2d ), because each pair consists of two numbers, and we need to cover all 60 numbers.Wait, let me think about that again. If each pair has a difference of ( d ), then each pair is two numbers apart by ( d ). So, the total number of such pairs would be ( frac{60}{2} = 30 ) pairs. But if we're partitioning the set into pairs with a specific difference, the way these pairs are formed might require that the entire set is divided into blocks where each block can be paired with a difference of ( d ).Alternatively, maybe the entire set can be divided into multiple sequences where each sequence is an arithmetic progression with common difference ( d ), and each sequence has two elements. So, each pair is an arithmetic progression of length 2 with common difference ( d ).But for this to cover the entire set, the number of such sequences must be such that they tile the entire set without overlapping. This would require that the total number of elements, 60, is divisible by the number of elements in each sequence, which is 2. But since 60 is divisible by 2, that's fine, but we also need to ensure that the way we partition the set into these sequences works for the given ( d ).Wait, maybe another approach. If we fix a difference ( d ), then the pairs would be (1, 1+d), (2, 2+d), ..., up to some point. But this would only work if 1+d, 2+d, etc., are all within 1 to 60. So, the maximum number we can pair is 60 - d. Therefore, the number of such pairs would be 60 - d, but since we need 30 pairs (because 60/2 = 30), we have:Number of pairs = 60 - d = 30Wait, that can't be right because 60 - d = 30 implies d = 30. But that would mean each pair is (1,31), (2,32), ..., (30,60). That works, but that's just one possible way with d=30.But we need to consider all possible d's such that the entire set can be partitioned into pairs with difference d. So, maybe d has to be such that the set can be divided into multiple such pairings.Wait, perhaps I need to think in terms of modular arithmetic. If we consider the numbers modulo d, then each number can be represented as ( k ) where ( k ) is from 0 to d-1. Then, for each residue class modulo d, we can pair the numbers in that class with the next residue class.But I'm not sure if that's the right approach. Maybe another way: if we have a difference d, then the pairs are (1,1+d), (2,2+d), ..., (d, 2d), then (d+1, d+1+d), etc. But this would require that 2d divides 60, because each block of 2d numbers can be paired in d pairs with difference d.So, 60 must be divisible by 2d, meaning that 2d is a divisor of 60. Therefore, d must be a divisor of 30, because 60 divided by 2 is 30.So, the possible values of d are the divisors of 30. The divisors of 30 are: 1, 2, 3, 5, 6, 10, 15, 30.Therefore, there are 8 possible values of d, each corresponding to a way of pairing the cards with the same modulus difference.But wait, let me verify this with an example. Let's take d=1. Then, the pairs would be (1,2), (3,4), ..., (59,60). That works, and it's a valid pairing.For d=2, the pairs would be (1,3), (2,4), (5,7), (6,8), ..., (57,59), (58,60). That also works.Similarly, for d=3, the pairs would be (1,4), (2,5), (3,6), (7,10), (8,11), (9,12), ..., up to (58,61) but 61 is beyond 60, so we need to adjust. Wait, no, actually, if d=3, then the pairs would be (1,4), (2,5), (3,6), (7,10), (8,11), (9,12), ..., (58,61) is not possible, so actually, the last pair would be (57,60). So, it still works because 60 is divisible by 3*2=6, so 60/6=10 blocks, each block of 6 numbers can be paired into 3 pairs with difference 3.Wait, maybe my earlier reasoning was correct. Since 60 must be divisible by 2d, so d must be a divisor of 30. Therefore, the number of ways is equal to the number of divisors of 30, which are 8.So, the answer should be 8.