💡Alright, so I have this algebra problem here, and I need to figure out the value of "a" such that when I calculate the expression ((x^{2} + ax + 5) cdot (-2x) - 6x^{2}), there's no (x^{2}) term in the result. The options given are A: -3, B: -1/3, C: 0, and D: 3. Hmm, okay, let's break this down step by step.First, I think I need to expand the expression ((x^{2} + ax + 5) cdot (-2x)). To do that, I'll distribute the (-2x) across each term inside the parentheses. So, let's see:1. Multiply (x^{2}) by (-2x): That should give me (-2x^{3}).2. Multiply (ax) by (-2x): That would be (-2ax^{2}).3. Multiply 5 by (-2x): That results in (-10x).So, after distributing, the expression becomes (-2x^{3} - 2ax^{2} - 10x). Now, I need to subtract (6x^{2}) from this result. Let me write that out:[-2x^{3} - 2ax^{2} - 10x - 6x^{2}]Okay, now I need to combine like terms. The (x^{3}) term is just (-2x^{3}), and there's no other (x^{3}) term to combine with. For the (x^{2}) terms, I have (-2ax^{2}) and (-6x^{2}). Combining those gives me (-2ax^{2} - 6x^{2}), which can be factored as (-2a - 6) times (x^{2}). Then, the (x) term is just (-10x), and there are no other (x) terms to combine with.So, putting it all together, the expression simplifies to:[-2x^{3} + (-2a - 6)x^{2} - 10x]Now, the problem states that the result does not contain the term (x^{2}). That means the coefficient of (x^{2}) must be zero. So, I can set up the equation:[-2a - 6 = 0]Solving for "a":1. Add 6 to both sides: (-2a = 6)2. Divide both sides by -2: (a = -3)So, the value of "a" should be -3. Let me double-check to make sure I didn't make any mistakes.Starting with the original expression:[(x^{2} + ax + 5) cdot (-2x) - 6x^{2}]Substituting (a = -3):[(x^{2} - 3x + 5) cdot (-2x) - 6x^{2}]Expanding the first part:[-2x^{3} + 6x^{2} - 10x]Subtracting (6x^{2}):[-2x^{3} + 6x^{2} - 10x - 6x^{2} = -2x^{3} + 0x^{2} - 10x]Yes, the (x^{2}) term cancels out, leaving (-2x^{3} - 10x). So, there's no (x^{2}) term, which confirms that (a = -3) is the correct answer.I think that's solid. I don't see any errors in my calculations, and the substitution checks out. So, I'm confident that the answer is A: -3.

💡First, I recognize that Joanna has 8. Her brother has a certain multiple of her money, which I'll denote as ( x ). Therefore, her brother has ( 8x ) dollars.Her sister has half as much money as Joanna, which means her sister has ( frac{8}{2} = 4 ) dollars.The total amount of money that Joanna, her brother, and her sister have together is 36. I can set up the equation:[8 + 8x + 4 = 36]Combining the known amounts:[8x + 12 = 36]Subtracting 12 from both sides:[8x = 24]Dividing both sides by 8:[x = 3]This means the brother has 3 times as much money as Joanna. Therefore, the ratio of the brother's money to Joanna's money is ( 3:1 ).

💡Okay, so I'm trying to figure out this problem where Alice, Bob, Charlie, and Dave are hiking and trying to determine how far away the nearest town is. Each of them made a statement about the distance, but none of their statements were true. I need to figure out the possible values for the distance, d, based on their false statements.First, let me list out their statements:- Alice said, "We are at least 8 miles away." So, if her statement is false, that means the distance is less than 8 miles. In mathematical terms, that would be d < 8.- Bob stated, "We are at most 7 miles away." If this is false, then the distance must be greater than 7 miles. So, d > 7.- Charlie said, "The nearest town is at most 6 miles away." If this is false, then the distance must be greater than 6 miles. So, d > 6.- Dave mentioned, "We are exactly 9 miles away." If this is false, then the distance is not equal to 9 miles. So, d ≠ 9.Now, I need to consider all these conditions together and find the values of d that satisfy all of them.Starting with Alice's statement: d < 8. So, the distance is less than 8 miles.Bob's statement: d > 7. So, the distance is greater than 7 miles.Charlie's statement: d > 6. But since Bob's statement already requires d > 7, which is a stronger condition than d > 6, we can ignore Charlie's statement because if d is greater than 7, it's automatically greater than 6.Dave's statement: d ≠ 9. Since we already have d < 8 from Alice, d can't be 9 anyway, so this doesn't add any new information.Putting it all together, we have:- From Alice: d < 8- From Bob: d > 7So, combining these two, we get 7 < d < 8.Looking at the answer choices:A. (6,7)B. (7,8)C. (7,9)D. (8,∞)E. (9,∞)The interval 7 < d < 8 corresponds to option B, which is (7,8).I should double-check to make sure none of the other options could also be valid. Let's see:Option A: (6,7) – But Bob's statement being false requires d > 7, so this interval doesn't fit.Option C: (7,9) – This includes values up to 9, but Alice's statement being false requires d < 8, so this interval is too wide.Option D: (8,∞) – Alice's statement being false requires d < 8, so this interval doesn't fit.Option E: (9,∞) – Again, Alice's statement being false requires d < 8, so this doesn't fit either.Therefore, the only interval that satisfies all the conditions is (7,8), which is option B.

💡Alright, so I've got this problem here where I need to compare the sizes of three different expressions: a, b, and c. Let me write them down again to make sure I have them right.a is equal to 1.5 raised to the power of 0.3. Okay, so that's like 1.5 to the 0.3. Then, b is the logarithm base 7 of 6, which I think is written as log₇6. And finally, c is the tangent of 300 degrees, which is tan(300°). Alright, so I need to figure out which of these is the largest, which is in the middle, and which is the smallest. Let's take them one by one.Starting with a: 1.5^0.3. Hmm, 1.5 is greater than 1, and when you raise a number greater than 1 to a positive power, it's going to be greater than 1. So, a is definitely greater than 1. But how much greater? Maybe I can approximate it or use logarithms to get a better sense.Next, b is log base 7 of 6. So, logarithms can be tricky, but I remember that log base a of b is the exponent you need to raise a to get b. Since 6 is less than 7, log₇6 should be less than 1, right? Because 7^1 is 7, and 6 is less than 7, so the exponent needed to get 6 from 7 must be less than 1. So, b is between 0 and 1.Lastly, c is tan(300°). Okay, tangent of an angle. I know that 300 degrees is in the fourth quadrant, where tangent is negative because sine is negative and cosine is positive in the fourth quadrant. Also, 300 degrees is equivalent to -60 degrees, so tan(300°) is the same as tan(-60°), which is -tan(60°). And tan(60°) is √3, so tan(300°) is -√3, which is approximately -1.732. So, c is negative.Putting it all together:- a is greater than 1- b is between 0 and 1- c is negativeSo, in terms of size, c is the smallest because it's negative, then b is next because it's between 0 and 1, and a is the largest because it's greater than 1.Wait, let me double-check to make sure I didn't make any mistakes. For a, 1.5^0.3. Maybe I can calculate it more precisely. Let's see, 1.5^0.3 is the same as e^(0.3 * ln(1.5)). Let me compute ln(1.5) first. I know that ln(1) is 0, ln(e) is 1, and ln(2) is about 0.693. So, ln(1.5) is somewhere between 0 and 0.693. Let me approximate it as 0.405. So, 0.3 * 0.405 is approximately 0.1215. Then, e^0.1215 is approximately 1.129. So, a is roughly 1.129.For b, log base 7 of 6. I can use the change of base formula: log₇6 = ln6 / ln7. Let's compute ln6 and ln7. I know that ln6 is about 1.792 and ln7 is about 1.946. So, log₇6 ≈ 1.792 / 1.946 ≈ 0.921. So, b is approximately 0.921.And c is tan(300°), which we already determined is -√3 ≈ -1.732.So, plugging in the approximate values:- a ≈ 1.129- b ≈ 0.921- c ≈ -1.732So, ordering them from smallest to largest: c < b < a.Wait, but I initially thought b was between 0 and 1, which it is, and a is greater than 1, so that still holds. So, c is the smallest, then b, then a.I think that's correct. But just to be thorough, let me check if there's any chance that a could be less than b or something. Well, a is about 1.129, and b is about 0.921, so no, a is definitely larger than b. And c is negative, so it's definitely smaller than both.So, the conclusion is that c is less than b, which is less than a.**Final Answer**The size relationship is boxed{c < b < a}.

💡Alright, so I have this problem here about points in a plane. Let me try to understand it step by step. The problem says:Let ( n ) and ( k ) be positive integers, and ( S ) be a set of ( n ) points in a plane, satisfying:1. No three points in ( S ) are collinear;2. For each point ( P ) in ( S ), there are at least ( k ) points in ( S ) that are equidistant from ( P ).We need to prove that ( k < frac{1}{2} + sqrt{2n} ).Hmm, okay. So, we have a set of points with no three on a straight line, which is a common condition to avoid trivial cases. The second condition is interesting: for each point, there are at least ( k ) other points that are equidistant from it. So, for every point ( P ), there's a circle centered at ( P ) that passes through at least ( k ) other points in the set.I think the strategy here is to use some combinatorial geometry. Maybe we can count something related to circles and their intersections. Let me try to break it down.First, let's consider each point ( P_i ) in the set ( S ). For each ( P_i ), there's a circle ( C_i ) centered at ( P_i ) that contains at least ( k ) other points from ( S ). So, each circle ( C_i ) has at least ( k ) points on it.Now, let's think about how many times each point lies on these circles. For each point ( P_j ), how many circles ( C_i ) pass through ( P_j )? Let's denote this number as ( e_j ). So, ( e_j ) is the number of circles centered at other points that pass through ( P_j ).Since each circle ( C_i ) has at least ( k ) points, the total number of incidences between points and circles is at least ( nk ). In other words, the sum of all ( e_j ) over all points ( P_j ) is at least ( nk ).Mathematically, that's:[sum_{j=1}^n e_j geq nk]Now, let's think about how many pairs of circles intersect at each point. If two circles ( C_i ) and ( C_j ) intersect at a point ( P ), then ( P ) lies on both circles. But since no three points are collinear, each pair of circles can intersect in at most two points. However, in our case, since all circles are centered at points in ( S ), and we're only considering intersections at points in ( S ), each pair of circles can intersect at at most two points in ( S ).Wait, actually, since no three points are collinear, two circles can intersect at at most two points, but those points have to be in ( S ). So, for each pair of circles ( C_i ) and ( C_j ), the number of intersection points in ( S ) is at most two.Therefore, the total number of intersection points across all pairs of circles is at most ( 2 times binom{n}{2} = n(n - 1) ).But another way to count the number of intersection points is to consider, for each point ( P_j ), how many pairs of circles pass through it. For each ( P_j ), the number of pairs of circles passing through it is ( binom{e_j}{2} ). So, the total number of intersection points is:[sum_{j=1}^n binom{e_j}{2}]Since each intersection point is counted once for each pair of circles passing through it, and we know that the total number of intersection points is at most ( n(n - 1) ), we have:[sum_{j=1}^n binom{e_j}{2} leq n(n - 1)]Now, let's express ( binom{e_j}{2} ) in terms of ( e_j ):[binom{e_j}{2} = frac{e_j(e_j - 1)}{2}]So, the inequality becomes:[sum_{j=1}^n frac{e_j(e_j - 1)}{2} leq n(n - 1)]Multiplying both sides by 2:[sum_{j=1}^n e_j(e_j - 1) leq 2n(n - 1)]Expanding the left side:[sum_{j=1}^n e_j^2 - sum_{j=1}^n e_j leq 2n(n - 1)]We already know from earlier that:[sum_{j=1}^n e_j geq nk]So, substituting that into the inequality:[sum_{j=1}^n e_j^2 - nk leq 2n(n - 1)]Which simplifies to:[sum_{j=1}^n e_j^2 leq 2n(n - 1) + nk]Now, we can apply the Cauchy-Schwarz inequality to relate ( sum e_j^2 ) and ( sum e_j ). The Cauchy-Schwarz inequality states that:[left( sum_{j=1}^n e_j right)^2 leq n sum_{j=1}^n e_j^2]Substituting our earlier result:[(nk)^2 leq n sum_{j=1}^n e_j^2]Which simplifies to:[n^2k^2 leq n sum_{j=1}^n e_j^2]Dividing both sides by ( n ):[nk^2 leq sum_{j=1}^n e_j^2]But from earlier, we have:[sum_{j=1}^n e_j^2 leq 2n(n - 1) + nk]Combining these two results:[nk^2 leq 2n(n - 1) + nk]Dividing both sides by ( n ) (since ( n ) is positive):[k^2 leq 2(n - 1) + k]Rearranging terms:[k^2 - k - 2(n - 1) leq 0]This is a quadratic inequality in terms of ( k ). Let's solve for ( k ):[k^2 - k - 2(n - 1) leq 0]Using the quadratic formula, the roots of the equation ( k^2 - k - 2(n - 1) = 0 ) are:[k = frac{1 pm sqrt{1 + 8(n - 1)}}{2}]Since ( k ) is a positive integer, we only consider the positive root:[k leq frac{1 + sqrt{1 + 8(n - 1)}}{2}]Simplifying inside the square root:[sqrt{1 + 8(n - 1)} = sqrt{8n - 7}]But to make it simpler, notice that:[sqrt{8n - 7} < sqrt{8n}]So,[k < frac{1 + sqrt{8n}}{2}]Simplifying further:[frac{1 + sqrt{8n}}{2} = frac{1}{2} + frac{sqrt{8n}}{2} = frac{1}{2} + sqrt{2n}]Therefore, we have:[k < frac{1}{2} + sqrt{2n}]Which is what we needed to prove.**Final Answer**boxed{k < frac{1}{2} + sqrt{2n}}

💡Okay, so I have this problem here about three numbers. The mean of these numbers is 8 more than the least number and 20 less than the greatest number. Also, the median of the three numbers is 10. I need to find their sum. Hmm, let me try to break this down step by step.First, let's denote the three numbers as ( a ), ( b ), and ( c ). Since the median is 10, that means when we arrange the numbers in order, the middle one is 10. So, without loss of generality, let's assume ( a leq b leq c ). Therefore, ( b = 10 ).Now, the mean of the three numbers is given by ( frac{a + b + c}{3} ). According to the problem, this mean is 8 more than the least number, which is ( a ). So, we can write the equation:[frac{a + b + c}{3} = a + 8]Similarly, the mean is also 20 less than the greatest number, which is ( c ). So, another equation is:[frac{a + b + c}{3} = c - 20]Since both expressions equal the mean, they must equal each other. Therefore:[a + 8 = c - 20]Let me solve this equation for ( c ):[c = a + 28]Okay, so ( c ) is 28 more than ( a ). Now, let's substitute ( b = 10 ) and ( c = a + 28 ) back into the equation for the mean. Let's use the first equation:[frac{a + 10 + (a + 28)}{3} = a + 8]Simplify the numerator:[frac{2a + 38}{3} = a + 8]Multiply both sides by 3 to eliminate the denominator:[2a + 38 = 3a + 24]Now, subtract ( 2a ) from both sides:[38 = a + 24]Subtract 24 from both sides:[a = 14]So, the smallest number ( a ) is 14. Then, since ( c = a + 28 ), we have:[c = 14 + 28 = 42]Now, we have all three numbers: ( a = 14 ), ( b = 10 ), and ( c = 42 ). Let's check if these satisfy the original conditions.First, the median is indeed 10, which matches. Next, let's calculate the mean:[frac{14 + 10 + 42}{3} = frac{66}{3} = 22]Now, check if the mean is 8 more than the least number:[14 + 8 = 22]Yes, that's correct. And check if the mean is 20 less than the greatest number:[42 - 20 = 22]That also works. So, the numbers are 14, 10, and 42, and their sum is:[14 + 10 + 42 = 66]Therefore, the sum of the three numbers is 66.**Final Answer**boxed{C}