💡First, I need to determine how many violet marbles Dan had initially. I know that Mary gave him 14 red marbles, and after receiving them, he has a total of 78 marbles.To find the number of violet marbles, I can subtract the number of red marbles from the total number of marbles.So, I subtract 14 red marbles from 78 total marbles:78 marbles (total) - 14 marbles (red) = 64 marbles (violet)This calculation shows that Dan initially had 64 violet marbles.

💡Alright, so I've got this geometry problem here with two lines and three planes, and I need to figure out which of the four propositions are true. Let me try to break this down step by step.First, let's list out the propositions again to keep them fresh in my mind:1. If line m is parallel to line l and m is perpendicular to plane α, then l is perpendicular to α.2. If line m is parallel to line l and m is parallel to plane α, then l is parallel to α.3. If plane α is perpendicular to plane β and line l is contained in α, then l is perpendicular to β.4. If the intersection of α and γ is line m, and the intersection of β and γ is line l, and α is parallel to β, then m is parallel to l.Okay, so I need to evaluate each of these one by one. Let's start with proposition 1.**Proposition 1:** If m is parallel to l and m is perpendicular to α, then l is perpendicular to α.Hmm, so if two lines are parallel, and one is perpendicular to a plane, does that mean the other is too? I think about what it means for a line to be perpendicular to a plane. A line is perpendicular to a plane if it's perpendicular to every line in that plane. So if m is perpendicular to α, then it's perpendicular to every line in α. Now, if l is parallel to m, then l should also be perpendicular to every line in α, right? Because parallel lines maintain the same angle with respect to any other line or plane. So, yeah, I think proposition 1 is true.**Proposition 2:** If m is parallel to l and m is parallel to α, then l is parallel to α.Alright, so if m is parallel to α, that means m doesn't intersect α and lies entirely outside of it, or it could be lying on α. But since m is parallel to l, does that mean l is also parallel to α? Wait, not necessarily. Because if m is parallel to α, l could be either parallel to α or it could lie on α. The problem states that l and m are non-coincident, so l can't be the same line as m. But if m is parallel to α, l could still lie on α, right? So l might not be strictly parallel to α; it could be lying within α. Therefore, proposition 2 isn't necessarily true. So I think proposition 2 is false.**Proposition 3:** If α is perpendicular to β and l is contained in α, then l is perpendicular to β.Okay, so α and β are perpendicular planes. If l is in α, does that make l perpendicular to β? Well, not necessarily. For example, imagine two perpendicular walls meeting at a corner. If I draw a line on one wall that's not along the corner, that line isn't necessarily perpendicular to the other wall. It might just intersect the other wall at some angle, not necessarily 90 degrees. So, l being in α doesn't guarantee it's perpendicular to β. Therefore, proposition 3 is false.**Proposition 4:** If α intersects γ along m, and β intersects γ along l, and α is parallel to β, then m is parallel to l.Alright, so α and β are parallel planes, and both intersect plane γ. The lines of intersection are m and l respectively. Since α and β are parallel, their intersections with γ should also be parallel. Because if two parallel planes are cut by a third plane, the lines of intersection are parallel. I remember this from some geometry concepts. So, yes, m should be parallel to l. Therefore, proposition 4 is true.So, summarizing:1. True2. False3. False4. TrueSo the true propositions are 1 and 4. Looking back at the options:A. 1, 2 B. 1, 3 C. 1, 4 D. 2, 4So the correct answer should be option C: 1, 4.**Final Answer**boxed{C}

💡Okay, so I have this function ( f(x) = x^4 + a x^3 + b x^2 + a x + 1 ) where ( a ) and ( b ) are real numbers. The problem says that this function has at least one root, and I need to find the minimum value of ( a^2 - b ). Hmm, okay, let me try to figure this out step by step.First, since the function has at least one root, let's say ( m ) is a root. That means when I plug ( m ) into the function, it equals zero. So, I can write:[ f(m) = m^4 + a m^3 + b m^2 + a m + 1 = 0 ]Hmm, that's a quartic equation. Quartic equations can be tricky, but maybe I can manipulate this equation to find a relationship between ( a ), ( b ), and ( m ). Let me try dividing both sides of the equation by ( m^2 ) to see if that simplifies things:[ frac{m^4}{m^2} + a frac{m^3}{m^2} + b frac{m^2}{m^2} + a frac{m}{m^2} + frac{1}{m^2} = 0 ]Simplifying each term, this becomes:[ m^2 + a m + b + frac{a}{m} + frac{1}{m^2} = 0 ]Okay, so that's:[ m^2 + frac{1}{m^2} + a left( m + frac{1}{m} right) + b = 0 ]I notice that ( m^2 + frac{1}{m^2} ) can be written as ( left( m + frac{1}{m} right)^2 - 2 ). Let me use that identity:[ left( m + frac{1}{m} right)^2 - 2 + a left( m + frac{1}{m} right) + b = 0 ]Let me set ( t = m + frac{1}{m} ) to simplify the equation further. Then, the equation becomes:[ t^2 - 2 + a t + b = 0 ]So, solving for ( b ):[ b = -t^2 - a t + 2 ]Now, the expression I need to minimize is ( a^2 - b ). Let's substitute the expression for ( b ) into this:[ a^2 - b = a^2 - (-t^2 - a t + 2) ][ = a^2 + t^2 + a t - 2 ]Hmm, so ( a^2 - b = a^2 + a t + t^2 - 2 ). This looks like a quadratic in terms of ( a ). Maybe I can complete the square to find the minimum value.Let me rewrite the expression:[ a^2 + a t + t^2 - 2 ]Completing the square for the terms involving ( a ):[ a^2 + a t = left(a + frac{t}{2}right)^2 - frac{t^2}{4} ]So substituting back:[ left(a + frac{t}{2}right)^2 - frac{t^2}{4} + t^2 - 2 ][ = left(a + frac{t}{2}right)^2 + frac{3 t^2}{4} - 2 ]Okay, so now I have:[ a^2 - b = left(a + frac{t}{2}right)^2 + frac{3 t^2}{4} - 2 ]Since squares are always non-negative, the term ( left(a + frac{t}{2}right)^2 ) is greater than or equal to zero. Therefore, the minimum value of ( a^2 - b ) occurs when this term is zero, which happens when ( a = -frac{t}{2} ).So, substituting ( a = -frac{t}{2} ) into the expression:[ a^2 - b = 0 + frac{3 t^2}{4} - 2 ][ = frac{3 t^2}{4} - 2 ]Now, I need to find the minimum value of ( frac{3 t^2}{4} - 2 ). Since ( t = m + frac{1}{m} ), let's analyze the possible values of ( t ).For real numbers ( m ) (since ( m ) is a root and ( a, b ) are real), ( m ) can be positive or negative. Let's consider ( m > 0 ) first. By the AM-GM inequality, ( m + frac{1}{m} geq 2 ). Similarly, if ( m < 0 ), let ( m = -n ) where ( n > 0 ), then ( m + frac{1}{m} = -n - frac{1}{n} leq -2 ).So, ( t ) can be any real number such that ( |t| geq 2 ). Therefore, ( t^2 geq 4 ).Plugging this back into the expression ( frac{3 t^2}{4} - 2 ):Since ( t^2 geq 4 ), the minimum value of ( frac{3 t^2}{4} - 2 ) occurs when ( t^2 = 4 ):[ frac{3 times 4}{4} - 2 = 3 - 2 = 1 ]Therefore, the minimum value of ( a^2 - b ) is 1.Wait, let me just double-check my steps to make sure I didn't make any mistakes. I started by expressing ( b ) in terms of ( t ) and ( a ), then substituted into ( a^2 - b ) and completed the square. That seems correct. Then, I found that the minimum occurs when ( a = -frac{t}{2} ) and ( t^2 = 4 ), leading to the minimum value of 1. Yeah, that seems solid.So, I think the minimum value of ( a^2 - b ) is indeed 1.**Final Answer**The minimum value of ( a^2 - b ) is boxed{1}.

💡First, I need to calculate the area of Rectangle A by multiplying its length and breadth.Next, I will calculate the area of Rectangle B in the same way by multiplying its length and breadth.After obtaining both areas, I will form a ratio of the area of Rectangle A to the area of Rectangle B.Finally, I will simplify this ratio to its lowest terms to find the simplest form of the ratio.

💡Okay, so I have this problem here about mixing cashews and Brazil nuts to make a 50-pound mixture. The cashews cost 6.75 per pound, and the Brazil nuts cost 5.00 per pound. I need to figure out what the selling price per pound of the mixture should be. Alright, let's break this down step by step.First, I need to understand how much of each type of nut is in the mixture. The problem says there are 20 pounds of cashews. Since the total mixture is 50 pounds, that means the rest must be Brazil nuts. So, 50 pounds minus 20 pounds equals 30 pounds of Brazil nuts. Got it.Now, I need to calculate the total cost of the cashews and the Brazil nuts separately. For the cashews, it's 6.75 per pound, and we have 20 pounds. So, I'll multiply 20 by 6.75 to find the total cost for the cashews. Let me do that: 20 times 6 is 120, and 20 times 0.75 is 15. So, 120 plus 15 equals 135. Okay, so the cashews cost 135 in total.Next, for the Brazil nuts, they cost 5.00 per pound, and we have 30 pounds of them. So, I'll multiply 30 by 5 to find the total cost for the Brazil nuts. That's straightforward: 30 times 5 equals 150. So, the Brazil nuts cost 150 in total.Now, to find the total cost of the entire mixture, I need to add the cost of the cashews and the cost of the Brazil nuts together. That would be 135 plus 150. Let me add those up: 135 plus 150 equals 285. So, the total cost for the 50-pound mixture is 285.Finally, to find the selling price per pound of the mixture, I need to divide the total cost by the total weight of the mixture. That means I'll take 285 and divide it by 50 pounds. Let me do that division: 285 divided by 50. Hmm, 50 goes into 285 five times because 5 times 50 is 250, and that leaves a remainder of 35. Then, 35 divided by 50 is 0.7. So, altogether, that's 5.70 per pound.Wait, let me double-check my calculations to make sure I didn't make any mistakes. For the cashews: 20 pounds times 6.75 per pound is indeed 135. For the Brazil nuts: 30 pounds times 5.00 per pound is 150. Adding those together gives 285, and dividing that by 50 pounds gives 5.70 per pound. Yeah, that seems correct.I think I've got it. The selling price per pound of the mixture should be 5.70.

💡Okay, so I have this problem where two real numbers are chosen independently at random from the interval ([-30, 15]). I need to find the probability that the product of these two numbers is greater than zero. Hmm, okay, let's break this down step by step.First, I remember that the product of two numbers is positive if both numbers are positive or both are negative. So, the product will be greater than zero in two cases: either both numbers are positive or both are negative. That makes sense because multiplying two negatives gives a positive, and multiplying two positives also gives a positive.Now, the interval given is ([-30, 15]). Let me visualize this interval. It starts at -30 and goes up to 15. So, the total length of the interval is (15 - (-30) = 45). That means the interval is 45 units long.Next, I need to figure out the lengths of the negative and positive parts of this interval. The negative part is from -30 to 0, which is 30 units long. The positive part is from 0 to 15, which is 15 units long. So, the negative interval is twice as long as the positive interval.Since the numbers are chosen independently and uniformly at random, the probability of picking a negative number is the length of the negative interval divided by the total length. That would be ( frac{30}{45} = frac{2}{3} ). Similarly, the probability of picking a positive number is ( frac{15}{45} = frac{1}{3} ).Okay, so now I have the probabilities of picking a negative or a positive number. To find the probability that the product is positive, I need to consider both cases where both numbers are negative or both are positive.Let me calculate the probability that both numbers are negative. Since the choices are independent, I can multiply the probabilities. So, that would be ( frac{2}{3} times frac{2}{3} = frac{4}{9} ).Similarly, the probability that both numbers are positive is ( frac{1}{3} times frac{1}{3} = frac{1}{9} ).Now, to get the total probability that the product is positive, I add these two probabilities together. So, ( frac{4}{9} + frac{1}{9} = frac{5}{9} ).Wait, let me double-check that. The negative interval is twice as long as the positive interval, so the probability of picking a negative number is higher. Therefore, the probability of both numbers being negative should be higher than both being positive, which it is ((frac{4}{9}) vs. (frac{1}{9})). Adding them together gives (frac{5}{9}), which seems reasonable.Is there another way to think about this? Maybe using areas? Since we're dealing with two numbers, we can represent all possible outcomes as points in a coordinate system where one axis is the first number and the other axis is the second number. The total area would then represent all possible pairs, and the area where the product is positive would be the regions where both numbers are negative or both are positive.The total area of the square would be the length of the interval squared, which is (45 times 45 = 2025). The area where both numbers are negative would be a square with side length 30, so (30 times 30 = 900). The area where both numbers are positive would be a square with side length 15, so (15 times 15 = 225). Adding these together gives (900 + 225 = 1125). Then, the probability would be ( frac{1125}{2025} ).Simplifying ( frac{1125}{2025} ), both numerator and denominator are divisible by 225. (1125 ÷ 225 = 5) and (2025 ÷ 225 = 9). So, that simplifies to ( frac{5}{9} ), which matches what I got earlier.Okay, so that seems consistent. Both methods give me the same result, so I feel more confident that ( frac{5}{9} ) is the correct probability.Wait, just to make sure, let me consider the cases where one number is negative and the other is positive. The product would be negative in those cases, right? So, the probability of the product being negative would be the remaining probability. Let's check that.The probability of one number being negative and the other positive can be calculated in two ways: first number negative and second positive, or first positive and second negative. So, that would be ( frac{2}{3} times frac{1}{3} + frac{1}{3} times frac{2}{3} = frac{2}{9} + frac{2}{9} = frac{4}{9} ).Adding the probability of the product being positive (( frac{5}{9} )) and the probability of the product being negative (( frac{4}{9} )) gives ( frac{5}{9} + frac{4}{9} = 1 ), which makes sense because these are the only two possibilities (excluding the case where one or both numbers are zero, but since we're dealing with real numbers, the probability of picking exactly zero is zero, so it doesn't affect the result).Therefore, all the probabilities add up correctly, and I'm confident that the probability of the product being greater than zero is ( frac{5}{9} ).**Final Answer**boxed{D}