💡Alright, so I have this problem where I need to find the units digit of ( R_{12345} ) where ( R_n = frac{1}{2}(a^n + b^n) ) and ( a = 3 + 2sqrt{2} ), ( b = 3 - 2sqrt{2} ). The answer choices are given as (A) 1, (B) 3, (C) 5, (D) 7, (E) 9.First, I need to understand what ( R_n ) represents. It looks like it's the average of ( a^n ) and ( b^n ). Since ( a ) and ( b ) are conjugates (because they have the form ( 3 + 2sqrt{2} ) and ( 3 - 2sqrt{2} )), their sum and product are rational numbers. Specifically, ( a + b = 6 ) and ( ab = 1 ). This might come in handy later.I remember that for sequences defined by such expressions, especially when dealing with conjugate pairs, there's often a recurrence relation that can be used to compute terms without having to calculate large exponents each time. Let me try to derive that.Starting with ( R_{n+1} ) and ( R_{n-1} ), I can write:[R_{n+1} = frac{1}{2}(a^{n+1} + b^{n+1})][R_{n-1} = frac{1}{2}(a^{n-1} + b^{n-1})]I also know that ( a ) and ( b ) satisfy the quadratic equation ( x^2 - 6x + 1 = 0 ) because ( a + b = 6 ) and ( ab = 1 ). This suggests that ( a^n ) and ( b^n ) satisfy a linear recurrence relation. Let me see if I can express ( R_{n+1} ) in terms of ( R_n ) and ( R_{n-1} ).Multiplying both sides of the recurrence equation ( x^2 = 6x - 1 ) by ( x^{n-1} ), we get:[x^{n+1} = 6x^n - x^{n-1}]Applying this to both ( a ) and ( b ):[a^{n+1} = 6a^n - a^{n-1}][b^{n+1} = 6b^n - b^{n-1}]Adding these two equations:[a^{n+1} + b^{n+1} = 6(a^n + b^n) - (a^{n-1} + b^{n-1})]Dividing both sides by 2:[R_{n+1} = 6R_n - R_{n-1}]So, the recurrence relation is ( R_{n+1} = 6R_n - R_{n-1} ). That's a second-order linear recurrence relation with constant coefficients. Good, now I can use this to compute terms of the sequence.Next, I need the initial conditions to start the recurrence. Let's compute ( R_0 ) and ( R_1 ):- ( R_0 = frac{1}{2}(a^0 + b^0) = frac{1}{2}(1 + 1) = 1 )- ( R_1 = frac{1}{2}(a + b) = frac{1}{2}(6) = 3 )So, ( R_0 = 1 ) and ( R_1 = 3 ). Now, using the recurrence relation, I can compute subsequent terms.But since we're interested in the units digit of ( R_{12345} ), computing all terms up to 12345 is impractical. Instead, I can look for a pattern in the units digits of ( R_n ) modulo 10.Let me compute the units digits of the first few terms:- ( R_0 = 1 ) → units digit 1- ( R_1 = 3 ) → units digit 3- ( R_2 = 6R_1 - R_0 = 6*3 - 1 = 18 - 1 = 17 ) → units digit 7- ( R_3 = 6R_2 - R_1 = 6*17 - 3 = 102 - 3 = 99 ) → units digit 9- ( R_4 = 6R_3 - R_2 = 6*99 - 17 = 594 - 17 = 577 ) → units digit 7- ( R_5 = 6R_4 - R_3 = 6*577 - 99 = 3462 - 99 = 3363 ) → units digit 3- ( R_6 = 6R_5 - R_4 = 6*3363 - 577 = 20178 - 577 = 19601 ) → units digit 1- ( R_7 = 6R_6 - R_5 = 6*19601 - 3363 = 117606 - 3363 = 114243 ) → units digit 3Wait, looking at the units digits:- ( R_0 ): 1- ( R_1 ): 3- ( R_2 ): 7- ( R_3 ): 9- ( R_4 ): 7- ( R_5 ): 3- ( R_6 ): 1- ( R_7 ): 3I notice that starting from ( R_0 ), the units digits are: 1, 3, 7, 9, 7, 3, 1, 3, ... It seems like after ( R_6 ), the units digit cycles back to 1, and then repeats the same pattern: 1, 3, 7, 9, 7, 3, 1, 3, etc. So, the cycle length is 6.To confirm, let's compute a few more terms:- ( R_7 ): 3 (as above)- ( R_8 = 6R_7 - R_6 = 6*114243 - 19601 = 685458 - 19601 = 665857 ) → units digit 7- ( R_9 = 6R_8 - R_7 = 6*665857 - 114243 = 3995142 - 114243 = 3880899 ) → units digit 9- ( R_{10} = 6R_9 - R_8 = 6*3880899 - 665857 = 23285394 - 665857 = 22619537 ) → units digit 7- ( R_{11} = 6R_{10} - R_9 = 6*22619537 - 3880899 = 135717222 - 3880899 = 131836323 ) → units digit 3- ( R_{12} = 6R_{11} - R_{10} = 6*131836323 - 22619537 = 791017938 - 22619537 = 768398401 ) → units digit 1Yes, the pattern repeats every 6 terms: 1, 3, 7, 9, 7, 3, then back to 1, and so on. So, the cycle length is 6.Now, to find the units digit of ( R_{12345} ), I need to determine where 12345 falls in this 6-term cycle. That is, I need to compute ( 12345 mod 6 ).Calculating ( 12345 div 6 ):6 * 2057 = 1234212345 - 12342 = 3So, ( 12345 mod 6 = 3 ). That means ( R_{12345} ) has the same units digit as ( R_3 ).Looking back at the units digits:- ( R_0 ): 1- ( R_1 ): 3- ( R_2 ): 7- ( R_3 ): 9Therefore, the units digit of ( R_{12345} ) is 9.**Final Answer**boxed{E}

💡Alright, so I've got this problem here about sets M and N, and I need to find their union. Let me try to understand what each set represents and then figure out how to combine them.First, set M is defined as {x | -1 < x < 2}. Okay, that means M includes all real numbers x that are greater than -1 and less than 2. So, if I imagine a number line, M would cover everything from just above -1 up to just below 2. It's an open interval, so the endpoints aren't included.Now, set N is a bit trickier. It's defined as {x | y = sqrt(x - 1)}. Hmm, so this is the set of all x values for which the square root of (x - 1) is defined. I remember that the expression inside a square root has to be non-negative because you can't take the square root of a negative number in the set of real numbers. So, for y to be a real number, (x - 1) must be greater than or equal to zero.Let me write that down: x - 1 >= 0. Solving for x, I add 1 to both sides, which gives me x >= 1. So, set N includes all real numbers x that are greater than or equal to 1. That means N starts at 1 and goes to infinity, right? So, N is the interval [1, ∞).Now, I need to find the union of M and N, which is M ∪ N. The union of two sets includes all elements that are in either set or in both. So, I need to combine the intervals of M and N.Set M is (-1, 2), and set N is [1, ∞). Let me visualize this on a number line. M starts just above -1 and goes up to just below 2. N starts at 1 and goes all the way to the right, infinitely. So, where do these two intervals overlap?Looking at the number line, M covers from -1 to 2, and N covers from 1 to infinity. The overlapping part is from 1 to 2. But since we're taking the union, we need to include all parts of both intervals. So, from -1 to 2 is covered by M, and from 1 to infinity is covered by N. Combining these, the union would start at -1 (but not including -1) and go all the way to infinity.Wait, let me make sure. M is up to 2, and N starts at 1 and goes beyond. So, from -1 to 2, M covers that, and N covers from 1 onwards. So, the union would be from -1 to infinity. But hold on, M doesn't go beyond 2, but N does. So, the union would be from -1 to infinity, but M only goes up to 2. So, from -1 to 2, it's covered by M, and from 2 onwards, it's covered by N.But wait, N starts at 1, so from 1 to 2, both M and N overlap. So, the union is from -1 to infinity. But looking at the options, let me see:A: {x | x > -1}B: {x | 1 ≤ x < 2}C: {x | -1 < x < 2}D: {x | x ≥ 0}Hmm, option A is {x | x > -1}, which would be (-1, ∞). But wait, does the union go all the way to infinity? Because N does, but M stops at 2. So, does the union include everything from -1 onwards?Wait, M is up to 2, and N starts at 1 and goes to infinity. So, the union would be from -1 to infinity because N covers everything beyond 1, and M covers from -1 to 2. So, together, they cover from -1 to infinity.But let me double-check. If I take all x such that x > -1, that would include everything from just above -1 to infinity. But M only goes up to 2, and N covers from 1 onwards. So, from -1 to 1, only M covers that. From 1 to 2, both M and N cover it. From 2 onwards, only N covers it.So, combining all these, the union is indeed all x where x > -1, which is option A.But wait, let me make sure I'm not missing anything. Is there any restriction beyond what's given? M is open at -1 and 2, and N is closed at 1. So, in the union, the point at 1 is included because N includes it. But since M is open at 2, the union at 2 is still open. However, since N continues beyond 2, the union isn't restricted at 2; it just continues on.So, yes, the union is all real numbers greater than -1, which is option A.

💡Okay, so I have this problem here about propositions and negations. It says: For the proposition p: For all real numbers x, x squared plus x plus one is less than zero. Then, not p is what? Hmm, let me try to figure this out step by step.First, I remember that in logic, the negation of a universal statement is an existential statement. So, if p is saying "for all x, something is true," then not p would be "there exists an x for which that something is not true." That makes sense because if it's not true for every x, then there must be at least one x where it fails.So, p is ∀x ∈ ℝ, x² + x + 1 < 0. To negate this, I should switch the universal quantifier to an existential quantifier and negate the inner statement. That means ¬p would be ∃x ∈ ℝ such that x² + x + 1 is not less than zero. In other words, x² + x + 1 ≥ 0.Wait, let me make sure I'm doing this right. The original statement p is saying that for every real number x, when you plug it into the quadratic equation x² + x + 1, the result is negative. So, p is claiming that this quadratic is always negative for all real x.But I know that quadratic equations can open upwards or downwards depending on the coefficient of x². In this case, the coefficient is positive (1), so the parabola opens upwards. That means it has a minimum point. If the minimum value of the quadratic is above the x-axis, then the quadratic is always positive. If it touches or crosses the x-axis, then it's not always positive.So, maybe I should check if x² + x + 1 is ever negative. To do that, I can find its discriminant. The discriminant of a quadratic ax² + bx + c is b² - 4ac. Here, a is 1, b is 1, and c is 1. So, discriminant D = 1² - 4*1*1 = 1 - 4 = -3.Since the discriminant is negative, the quadratic doesn't cross the x-axis. It means that the quadratic is always positive because it opens upwards and never touches the x-axis. So, x² + x + 1 is always greater than zero for all real x.Wait, but the original proposition p says that x² + x + 1 is less than zero for all x. But we just saw that it's actually always greater than zero. So, p is a false statement because it's claiming something that isn't true.But the question is about the negation of p, not whether p is true or false. So, regardless of the truth value, the negation of p should be ∃x ∈ ℝ such that x² + x + 1 ≥ 0. That makes sense because if p is saying it's always negative, then the opposite is that there exists at least one x where it's not negative, meaning it's either zero or positive.But since we already know that x² + x + 1 is always positive, this negation is actually a true statement. However, the question is just asking for the form of ¬p, not whether it's true or false.So, to recap: p is a universal statement claiming that for all real x, x² + x + 1 is negative. The negation of that is an existential statement claiming that there exists at least one real x for which x² + x + 1 is not negative, i.e., it's greater than or equal to zero.I think that's it. So, ¬p is ∃x ∈ ℝ, x² + x + 1 ≥ 0.

💡First, I need to determine how much Edward spent in total on the toy cars and the race track. He started with 17.80 and has 8.00 left, so the total amount spent is 17.80 minus 8.00, which equals 9.80.Next, I know that the race track cost 6.00. To find out how much was spent on the toy cars alone, I subtract the cost of the race track from the total amount spent: 9.80 minus 6.00 equals 3.80.Since Edward bought 4 toy cars, I can find the cost of each toy car by dividing the total amount spent on the toy cars by the number of toy cars. So, 3.80 divided by 4 equals 0.95.Therefore, each toy car costs 0.95.

💡First, I recognize that the 27th day of the month is a Tuesday. To find the day of the week for the 1st day, I can work backwards week by week.Starting from the 27th, subtracting 7 days brings me to the 20th, which would also be a Tuesday. Subtracting another 7 days, I reach the 13th, still a Tuesday. Continuing this process, I find that the 6th is also a Tuesday.Now, counting backwards from the 6th, the 5th would be a Monday, the 4th a Sunday, the 3rd a Saturday, the 2nd a Friday, and finally, the 1st would be a Thursday.Therefore, the first day of the month was a Thursday.

💡Okay, so I have this inequality to prove: (left(1 + frac{a}{b}right)^n + left(1 + frac{b}{a}right)^n geq 2^{n+1}), where (a) and (b) are positive real numbers and (n) is a natural number. Hmm, let me think about how to approach this.First, I remember that inequalities often involve some standard techniques or known inequalities. Maybe I can use the AM-GM inequality here? The AM-GM inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. That might be useful since we have two terms here that seem symmetric in (a) and (b).Let me denote (x = frac{a}{b}) and (y = frac{b}{a}). Then, notice that (x times y = 1), so they are reciprocals of each other. That might simplify things.So, substituting, the expression becomes (left(1 + xright)^n + left(1 + yright)^n). Since (y = frac{1}{x}), this is (left(1 + xright)^n + left(1 + frac{1}{x}right)^n).Now, applying the AM-GM inequality to these two terms: (frac{left(1 + xright)^n + left(1 + frac{1}{x}right)^n}{2} geq sqrt{left(1 + xright)^n left(1 + frac{1}{x}right)^n}).Simplifying the right-hand side, we get (sqrt{left[(1 + x)(1 + frac{1}{x})right]^n}). Let's compute ((1 + x)(1 + frac{1}{x})):[(1 + x)left(1 + frac{1}{x}right) = 1 + frac{1}{x} + x + 1 = 2 + x + frac{1}{x}]So, the expression becomes (sqrt{(2 + x + frac{1}{x})^n}). Now, I need to find a lower bound for (2 + x + frac{1}{x}). Wait, I remember that for any positive real number (x), (x + frac{1}{x} geq 2) by AM-GM. So, (2 + x + frac{1}{x} geq 2 + 2 = 4). Therefore, (2 + x + frac{1}{x} geq 4), which implies that ((2 + x + frac{1}{x})^n geq 4^n). Taking the square root of both sides, we get (sqrt{(2 + x + frac{1}{x})^n} geq sqrt{4^n} = 2^n).So, putting it all together, from the AM-GM inequality:[frac{left(1 + xright)^n + left(1 + frac{1}{x}right)^n}{2} geq 2^n]Multiplying both sides by 2:[left(1 + xright)^n + left(1 + frac{1}{x}right)^n geq 2^{n+1}]Substituting back (x = frac{a}{b}), we get:[left(1 + frac{a}{b}right)^n + left(1 + frac{b}{a}right)^n geq 2^{n+1}]So, that seems to work out. I think I followed the right steps, using substitution and applying AM-GM twice: once directly on the terms and then on the expression inside the square root. It all ties together nicely to give the desired inequality.Just to double-check, let me test with specific values. Let’s take (a = b = 1) and (n = 1). Then the left-hand side is ((1 + 1)^1 + (1 + 1)^1 = 2 + 2 = 4), and the right-hand side is (2^{1+1} = 4). So, equality holds here, which is consistent with AM-GM, as equality occurs when the terms are equal.Another test: (a = 2), (b = 1), (n = 2). Then, (left(1 + 2right)^2 + left(1 + frac{1}{2}right)^2 = 9 + left(frac{3}{2}right)^2 = 9 + 2.25 = 11.25). The right-hand side is (2^{2+1} = 8). So, 11.25 ≥ 8, which is true.What if (a) is much larger than (b)? Let’s say (a = 100), (b = 1), (n = 1). Then, (left(1 + 100right) + left(1 + frac{1}{100}right) = 101 + 1.01 = 102.01), and the right-hand side is (2^{1+1} = 4). So, 102.01 ≥ 4, which is true.Seems like the inequality holds in these cases, so I think my proof is correct.**Final Answer**The inequality is proven, and the result is boxed{2^{n+1}}.