💡First, I need to determine how much money Tony spent in total. He started with 87 and had 61 left after his purchases. So, the total amount spent is 87 - 61 = 26.Next, I'll define the variables for the problem. Let ( C ) represent the number of pounds of cheese Tony bought. The cost of cheese is 7 per pound, so the total cost for cheese is ( 7C ) dollars. The cost of beef is 5 per pound, and Tony bought 1 pound of beef, so the total cost for beef is 5.The sum of the cost of cheese and beef equals the total amount spent:[7C + 5 = 26]To find the value of ( C ), I'll solve the equation step by step:[7C = 26 - 5][7C = 21][C = frac{21}{7} = 3]Therefore, Tony bought 3 pounds of cheese.

💡Okay, so I've got this problem here about two propositions, p and q, and I need to figure out the range of values for a based on whether these propositions are true or false. Let me try to break this down step by step.First, let's understand what each proposition is saying.Proposition p states that the function y = log_a(x + 2 - a) is an increasing function in the interval (0, +∞). Hmm, okay. So, logarithmic functions can be increasing or decreasing depending on the base. If the base a is greater than 1, the logarithmic function is increasing. If the base is between 0 and 1, it's decreasing. So, for y = log_a(x + 2 - a) to be increasing, a must be greater than 1.But wait, there's more to it. The argument of the logarithm, which is x + 2 - a, must be positive for all x in (0, +∞). That means x + 2 - a > 0 for all x > 0. The smallest value of x is approaching 0, so plugging that in, we get 0 + 2 - a > 0, which simplifies to 2 - a > 0, so a < 2. Therefore, for proposition p to be true, a must satisfy both a > 1 and a < 2. So, p is true when 1 < a < 2.Now, let's look at proposition q. It states that the function y = x² + 2ax + 1 has two zero points. That means the quadratic equation x² + 2ax + 1 = 0 has two real roots. For a quadratic equation ax² + bx + c = 0, the discriminant D = b² - 4ac must be positive for there to be two real roots.In this case, a is 1, b is 2a, and c is 1. So, the discriminant D = (2a)² - 4*1*1 = 4a² - 4. For two real roots, we need D > 0, so 4a² - 4 > 0. Dividing both sides by 4, we get a² - 1 > 0, which simplifies to a² > 1. Taking square roots, this means |a| > 1. But since a > 0 (given in the problem), this reduces to a > 1. So, proposition q is true when a > 1.Now, the problem says that the disjunction (p OR q) is true, and the conjunction (p AND q) is false. Let me recall what that means. If p OR q is true, at least one of p or q must be true. If p AND q is false, they cannot both be true at the same time.So, combining these two, it means exactly one of p or q is true, and the other is false. So, either p is true and q is false, or p is false and q is true.Let's consider the first case: p is true and q is false.From earlier, p is true when 1 < a < 2. For q to be false, since q is true when a > 1, q is false when a ≤ 1. But wait, p is true when a > 1, so if p is true, a must be greater than 1, but q is false only when a ≤ 1. There's no overlap here because a can't be both greater than 1 and less than or equal to 1 at the same time. So, this case leads to no solution.Now, the second case: p is false and q is true.p is false when a ≤ 1 or a ≥ 2. But q is true when a > 1. So, the overlap here is a > 1 and a ≥ 2, which simplifies to a ≥ 2. However, in the original problem, a ≠ 1, but since a > 1, that's already satisfied. So, the range here is a > 2.Wait, but let me double-check. If a is exactly 2, then for proposition p, the argument of the logarithm becomes x + 2 - 2 = x, so y = log_a(x). Since a = 2, which is greater than 1, log_2(x) is indeed increasing. So, p would be true when a = 2. But in this case, we're considering p being false, so a cannot be 2. Therefore, the range should be a > 2, not a ≥ 2.So, putting it all together, the only valid case is when p is false and q is true, which happens when a > 2.Let me just recap to make sure I didn't miss anything.- p is true when 1 < a < 2.- q is true when a > 1.- p OR q is true means at least one is true.- p AND q is false means they can't both be true.So, if a is between 1 and 2, both p and q are true, which violates the condition that p AND q is false. Therefore, a cannot be in (1, 2).If a is less than or equal to 1, p is false and q is also false because q requires a > 1. So, p OR q would be false, which contradicts the given condition that p OR q is true.Therefore, the only possible range is when a > 2, where p is false (since a > 2 makes the argument of the logarithm x + 2 - a, which for x > 0, could be positive or negative depending on a, but actually, for a > 2, when x approaches 0, x + 2 - a could be negative, making the logarithm undefined. Wait, hold on, I think I made a mistake here.Wait, earlier I concluded that for p to be true, a must be less than 2 because x + 2 - a > 0 for all x > 0. So, if a > 2, then when x approaches 0, x + 2 - a approaches 2 - a, which is negative. Therefore, the logarithm would not be defined for x near 0, meaning the function isn't increasing over the entire interval (0, +∞). So, p is false when a > 2 because the function isn't even defined properly in that interval.But wait, does that mean p is false for a > 2? Yes, because the function isn't defined for all x > 0. So, p is false when a ≥ 2.But q is true when a > 1. So, for a > 2, p is false and q is true, which satisfies the condition that p OR q is true and p AND q is false.Wait, but earlier I thought a = 2 makes p true, but actually, when a = 2, the argument becomes x, so y = log_2(x). But x is in (0, +∞), so it's defined, and since 2 > 1, it's increasing. So, p is true when a = 2. But according to the problem, p AND q is false, so if a = 2, p is true and q is true (since a > 1), which would make p AND q true, which contradicts the given condition. Therefore, a cannot be 2.So, the correct range is a > 2, not including 2.Therefore, the range of values for a is (2, +∞), which corresponds to option D.

💡First, calculate the total number of classes attended by all students daily. With 1500 students each attending 6 classes, the total number of classes is 1500 multiplied by 6, which equals 9000 classes.Next, determine the number of unique classes by dividing the total number of classes by the number of students per class. Dividing 9000 by 25 students per class gives 360 unique classes.Finally, calculate the number of teachers needed by dividing the number of unique classes by the number of classes each teacher can manage. Dividing 360 by 3 classes per teacher results in 120 teachers required.

💡First, I need to determine how much money Gwen spent. She started with 7 and has 5 left after spending some amount.To find the amount she spent, I can subtract the remaining money from the initial amount she received.So, 7 minus 5 equals 2.Therefore, Gwen spent 2.

💡Okay, so I'm trying to figure out the period of this function ( f(x) ) that satisfies the equation ( f(x+2) + f(x-2) = f(x) ). Hmm, periodic functions repeat their values at regular intervals, called periods. So, I need to find the smallest positive number ( T ) such that ( f(x + T) = f(x) ) for all ( x ).Let me start by writing down the given equation:( f(x+2) + f(x-2) = f(x) )I think I can use this equation to express ( f(x+2) ) in terms of ( f(x) ) and ( f(x-2) ). Let me rearrange the equation:( f(x+2) = f(x) - f(x-2) )Okay, so now I have ( f(x+2) ) expressed in terms of ( f(x) ) and ( f(x-2) ). Maybe I can use this recursively to find expressions for ( f(x+4) ), ( f(x+6) ), and so on.Let's try to find ( f(x+4) ). To do that, I'll substitute ( x+2 ) into the equation for ( f(x+2) ):( f((x+2)+2) = f(x+2) - f((x+2)-2) )Simplifying that:( f(x+4) = f(x+2) - f(x) )But I already have an expression for ( f(x+2) ), which is ( f(x) - f(x-2) ). Let's substitute that in:( f(x+4) = (f(x) - f(x-2)) - f(x) )Simplifying:( f(x+4) = -f(x-2) )Interesting, so ( f(x+4) = -f(x-2) ). Maybe I can use this to find ( f(x+6) ). Let's try substituting ( x+4 ) into the original equation:( f((x+4)+2) + f((x+4)-2) = f(x+4) )Simplifying:( f(x+6) + f(x+2) = f(x+4) )But from earlier, ( f(x+4) = -f(x-2) ) and ( f(x+2) = f(x) - f(x-2) ). Let's substitute these in:( f(x+6) + (f(x) - f(x-2)) = -f(x-2) )Simplify:( f(x+6) + f(x) - f(x-2) = -f(x-2) )Subtract ( f(x) - f(x-2) ) from both sides:( f(x+6) = -f(x-2) - f(x) + f(x-2) )Wait, that simplifies to:( f(x+6) = -f(x) )So, ( f(x+6) = -f(x) ). Hmm, that's a useful relation. If I apply this again, shifting ( x ) by 6, I get:( f(x+12) = -f(x+6) )But since ( f(x+6) = -f(x) ), substituting that in:( f(x+12) = -(-f(x)) = f(x) )So, ( f(x+12) = f(x) ). That means the function repeats every 12 units. Therefore, the period of the function is 12.Let me double-check to make sure I didn't make any mistakes. Starting from the original equation, I expressed ( f(x+2) ) in terms of ( f(x) ) and ( f(x-2) ), then found ( f(x+4) ) in terms of ( f(x-2) ), and so on. Each step seems consistent, and the final result shows that shifting ( x ) by 12 brings the function back to its original value. So, I think 12 is indeed the period.I wonder if there's a smaller period. Let's see. If the period were 6, then ( f(x+6) = f(x) ). But from earlier, we have ( f(x+6) = -f(x) ). So, if ( f(x+6) = f(x) ), then ( f(x) = -f(x) ), which implies ( f(x) = 0 ) for all ( x ). But the problem doesn't specify that the function is zero, just that it's periodic. So, unless the function is identically zero, 6 can't be the period. Therefore, 12 is the smallest period.Another way to think about it is to consider the characteristic equation of the recurrence relation. The given equation is a linear recurrence relation with constant coefficients. The characteristic equation would be ( r^2 + 1 = r ), which simplifies to ( r^2 - r + 1 = 0 ). Solving this quadratic equation:( r = [1 pm sqrt{1 - 4}]/2 = [1 pm isqrt{3}]/2 )These roots are complex numbers with magnitude 1 and angles ( pi/3 ) and ( -pi/3 ). The general solution to the recurrence is then a combination of terms like ( e^{ipi/3 x} ) and ( e^{-ipi/3 x} ), which correspond to periodic functions with period ( 2pi / (pi/3) ) = 6 ). Wait, that seems conflicting with my earlier conclusion of period 12.Hmm, maybe I'm mixing up the periods here. The characteristic roots have an angle of ( pi/3 ), so the period of the corresponding exponential functions would be ( 2pi / (pi/3) ) = 6 ). But in the functional equation, we're dealing with shifts of 2 units, not 1. So, perhaps the period in terms of the function's argument is different.Let me think. If the characteristic roots have a period of 6 in terms of the variable ( x ), but our shifts are in steps of 2, then the overall period of the function might be related to 6 divided by the step size. Wait, that might not be the right way to think about it.Alternatively, considering the functional equation, each shift is by 2 units. So, to get back to the original function, we need to shift by a multiple of 2 that results in a full period. From the earlier steps, we saw that shifting by 12 units brings the function back to itself. So, even though the characteristic roots suggest a period of 6, the actual period of the function considering the step size of 2 is 12.I think that makes sense because the functional equation relates values separated by 2 units, so the period has to be a multiple of 2. Since the characteristic roots suggest a period of 6, but our shifts are in steps of 2, the least common multiple of 6 and 2 is 6, but from our earlier derivation, we saw that 6 leads to ( f(x+6) = -f(x) ), which isn't a period unless the function is zero. Therefore, the next multiple would be 12, which satisfies ( f(x+12) = f(x) ).So, putting it all together, the period of the function is 12.

💡First, substitute (x = 2) into the equation (x^{2y} = 16), resulting in (2^{2y} = 16).Next, express 16 as a power of 2: (16 = 2^4). This gives the equation (2^{2y} = 2^4).Since the bases are the same, set the exponents equal to each other: (2y = 4).Finally, solve for (y) by dividing both sides by 2: (y = frac{4}{2} = 2).