💡First, I need to determine how far the nearest edge of the picture is from the end of the wall. The wall is 22 feet wide, and the picture is 4 feet wide. Since the picture is centered on the wall, the space on both sides of the picture should be equal.To find the distance from the end of the wall to the nearest edge of the picture, I'll subtract the width of the picture from the total width of the wall and then divide the result by 2. This will give me the equal space on each side of the picture.So, 22 feet (total wall width) minus 4 feet (picture width) equals 18 feet. Dividing 18 feet by 2 gives me 9 feet. Therefore, the nearest edge of the picture is 9 feet away from the end of the wall.

💡First, I need to simplify the square root of the fraction ( frac{1}{3} ).To make the denominator a perfect square, I'll multiply both the numerator and the denominator by 3.This gives me ( sqrt{frac{3}{9}} ).Next, I can separate the square root of the numerator and the denominator, resulting in ( frac{sqrt{3}}{sqrt{9}} ).Since the square root of 9 is 3, the expression simplifies to ( frac{sqrt{3}}{3} ).Therefore, the simplified form of ( sqrt{frac{1}{3}} ) is ( boxed{frac{sqrt{3}}{3}} ).

💡Alright, so I've got this math problem here with four options, labeled A through D. Each option presents a different calculation involving exponents, and I need to figure out which one is correct. Let me take it step by step.Starting with **Option A**: It says ( a^{2} cdot a^{3} = a^{5} ). Hmm, I remember that when you multiply two exponents with the same base, you add the exponents. So, ( a^{2} cdot a^{3} ) should be ( a^{2+3} = a^{5} ). That seems right. So, A looks correct.Moving on to **Option B**: ( a^{2} + a^{3} = a^{5} ). Wait a minute, addition of exponents isn't as straightforward as multiplication. I don't think you can just add the exponents when you're adding terms. ( a^{2} ) and ( a^{3} ) are not like terms, so you can't combine them like that. So, B doesn't seem correct.Next is **Option C**: ( (ab^{2})^{3} = ab^{6} ). Okay, here we have a product raised to a power. I recall that when you raise a product to a power, you raise each factor to that power. So, ( (ab^{2})^{3} ) should be ( a^{3}b^{6} ). But the option says ( ab^{6} ), missing the exponent on ( a ). That means C is incorrect.Finally, **Option D**: ( a^{10} div a^{2} = a^{5} ). Dividing exponents with the same base means subtracting the exponents. So, ( a^{10} div a^{2} = a^{10-2} = a^{8} ). But the option says ( a^{5} ), which is wrong. So, D is also incorrect.After going through each option, it seems only **Option A** is correct.

💡First, I recognize that Bruce deposited one-fifth of his total birthday money into the bank, which amounts to 45.To find the total amount of money Bruce received for his birthday, I multiply the amount he deposited by 5:45 * 5 = 225Next, I know that Bruce's grandfather sent him 150. To determine how much his aunt sent him, I subtract the grandfather's contribution from the total birthday money:225 - 150 = 75Therefore, Bruce's aunt sent him 75.

💡Okay, so I have this function f(x) = ln(ax + 1) + x³ - x² - ax, and I need to find the range of the real number a such that f(x) is increasing on the interval [2, +∞). Hmm, okay. Let me try to break this down step by step.First, I remember that if a function is increasing on an interval, its derivative must be non-negative throughout that interval. So, I should start by finding the derivative of f(x). Let's compute f'(x).The function f(x) has four terms: ln(ax + 1), x³, -x², and -ax. Let's differentiate each term with respect to x.1. The derivative of ln(ax + 1) with respect to x is (a)/(ax + 1). That's using the chain rule, right? The derivative of ln(u) is (1/u) * u', and here u = ax + 1, so u' = a.2. The derivative of x³ is 3x².3. The derivative of -x² is -2x.4. The derivative of -ax is -a.So, putting it all together, the derivative f'(x) is:f'(x) = (a)/(ax + 1) + 3x² - 2x - a.Hmm, okay. So, f'(x) = (a)/(ax + 1) + 3x² - 2x - a.Now, since f(x) is increasing on [2, +∞), f'(x) must be greater than or equal to zero for all x in [2, +∞). So, I need to ensure that f'(x) ≥ 0 for x ≥ 2.Let me write that down:f'(x) = (a)/(ax + 1) + 3x² - 2x - a ≥ 0 for all x ≥ 2.Hmm, this looks a bit complicated. Maybe I can simplify this expression. Let me see.First, let's combine the terms involving a. So, we have (a)/(ax + 1) - a. Let's factor out a:a [1/(ax + 1) - 1] + 3x² - 2x.Simplify the expression inside the brackets:1/(ax + 1) - 1 = [1 - (ax + 1)] / (ax + 1) = (-ax) / (ax + 1).So, substituting back, we have:a * (-ax)/(ax + 1) + 3x² - 2x.Which simplifies to:(-a²x)/(ax + 1) + 3x² - 2x.Hmm, okay. So, f'(x) = (-a²x)/(ax + 1) + 3x² - 2x.This still looks a bit messy, but maybe I can combine these terms into a single fraction. Let's try that.First, let's write all terms over the common denominator (ax + 1):[ -a²x + (3x² - 2x)(ax + 1) ] / (ax + 1) ≥ 0.Let me expand the numerator:- a²x + (3x² - 2x)(ax + 1).First, expand (3x² - 2x)(ax + 1):= 3x² * ax + 3x² * 1 - 2x * ax - 2x * 1= 3a x³ + 3x² - 2a x² - 2x.So, the numerator becomes:- a²x + 3a x³ + 3x² - 2a x² - 2x.Let me combine like terms:- a²x - 2x + 3a x³ + (3x² - 2a x²).Factor where possible:For the x terms: -x(a² + 2).For the x² terms: x²(3 - 2a).And the x³ term: 3a x³.So, numerator is:3a x³ + (3 - 2a)x² - (a² + 2)x.So, putting it all together, f'(x) = [3a x³ + (3 - 2a)x² - (a² + 2)x] / (ax + 1).Hmm, okay. So, f'(x) is equal to that fraction. Since we're looking for f'(x) ≥ 0 for x ≥ 2, and the denominator is ax + 1.Now, since x is in [2, +∞), and a is a real number, we need to consider the sign of the denominator. If a is positive, then ax + 1 is positive for all x ≥ 2. If a is negative, ax + 1 could potentially be zero or negative for some x ≥ 2, which would cause issues with the logarithm in the original function as well.Wait, actually, the original function has ln(ax + 1). For ln(ax + 1) to be defined, ax + 1 must be positive. So, for x ≥ 2, ax + 1 > 0. So, if a is positive, then ax + 1 is definitely positive for x ≥ 2. If a is negative, we need to ensure that ax + 1 > 0 for x ≥ 2.So, if a is negative, then ax + 1 > 0 implies that x < -1/a. But since x is in [2, +∞), this would require that -1/a > 2, which implies that a > -1/2. Wait, but a is negative, so this is a bit tricky.Wait, let me think again. If a is negative, then ax + 1 > 0 for x ≥ 2 implies that:ax + 1 > 0 => x < (-1)/a.But since x is in [2, +∞), this inequality would require that (-1)/a > 2, because x has to be less than (-1)/a for ax + 1 to be positive, but x is also greater than or equal to 2. So, to have ax + 1 > 0 for all x ≥ 2, we need (-1)/a > 2, which implies that a > -1/2. But since a is negative, this means that a must be in (-1/2, 0).But wait, if a is negative, say a = -1/4, then ax + 1 = (-1/4)x + 1. For x = 2, this is (-1/4)*2 + 1 = -1/2 + 1 = 1/2 > 0. For x = 4, it's (-1/4)*4 + 1 = -1 + 1 = 0. So, at x = 4, it's zero, which is not allowed because ln(0) is undefined. So, actually, we need ax + 1 > 0 for all x ≥ 2, which would require that a ≥ 0.Wait, because if a is negative, then as x increases, ax + 1 decreases, so eventually, it will become zero or negative, which is not allowed. Therefore, a must be non-negative. So, a ≥ 0.Therefore, in the denominator, ax + 1 is positive for all x ≥ 2, as long as a ≥ 0.So, since the denominator is positive, the sign of f'(x) is determined by the numerator. Therefore, for f'(x) ≥ 0, we need the numerator:3a x³ + (3 - 2a)x² - (a² + 2)x ≥ 0 for all x ≥ 2.So, let's denote the numerator as N(x):N(x) = 3a x³ + (3 - 2a)x² - (a² + 2)x.We need N(x) ≥ 0 for all x ≥ 2.Hmm, okay. So, perhaps we can factor N(x) or analyze its behavior.Let me factor out an x:N(x) = x [3a x² + (3 - 2a)x - (a² + 2)].So, N(x) = x [3a x² + (3 - 2a)x - (a² + 2)].Since x ≥ 2, x is positive, so the sign of N(x) is determined by the quadratic in the brackets:Q(x) = 3a x² + (3 - 2a)x - (a² + 2).So, we need Q(x) ≥ 0 for all x ≥ 2.Therefore, the problem reduces to finding a such that Q(x) = 3a x² + (3 - 2a)x - (a² + 2) ≥ 0 for all x ≥ 2.Now, Q(x) is a quadratic function in x. Let's analyze it.First, since the coefficient of x² is 3a, which is non-negative because a ≥ 0. So, if a > 0, Q(x) is a quadratic opening upwards. If a = 0, Q(x) becomes a linear function.Let me check the case when a = 0 separately.Case 1: a = 0.Then, Q(x) = 3*0*x² + (3 - 0)x - (0 + 2) = 0 + 3x - 2 = 3x - 2.So, Q(x) = 3x - 2. We need Q(x) ≥ 0 for x ≥ 2.At x = 2, Q(2) = 3*2 - 2 = 6 - 2 = 4 ≥ 0.Since Q(x) is linear with a positive slope, it will be increasing. Therefore, for x ≥ 2, Q(x) ≥ 4 ≥ 0. So, a = 0 is acceptable.Case 2: a > 0.In this case, Q(x) is a quadratic opening upwards. So, it will have a minimum at its vertex. To ensure that Q(x) ≥ 0 for all x ≥ 2, we need to ensure that the minimum of Q(x) on [2, +∞) is non-negative.The vertex of a quadratic ax² + bx + c is at x = -b/(2a). In our case, Q(x) = 3a x² + (3 - 2a)x - (a² + 2). So, the vertex is at:x_v = -(3 - 2a)/(2*3a) = (2a - 3)/(6a).Hmm, let's compute x_v:x_v = (2a - 3)/(6a) = (2a)/(6a) - 3/(6a) = (1/3) - (1)/(2a).So, x_v = 1/3 - 1/(2a).Now, since a > 0, 1/(2a) is positive, so x_v = 1/3 - positive number. So, x_v < 1/3.Therefore, the vertex is at x_v < 1/3, which is less than 2. So, on the interval [2, +∞), the quadratic Q(x) is increasing because it's opening upwards and the vertex is to the left of x = 2.Therefore, the minimum value of Q(x) on [2, +∞) occurs at x = 2.Therefore, to ensure Q(x) ≥ 0 for all x ≥ 2, it suffices to ensure that Q(2) ≥ 0.So, let's compute Q(2):Q(2) = 3a*(2)^2 + (3 - 2a)*(2) - (a² + 2).Compute each term:3a*(4) = 12a.(3 - 2a)*2 = 6 - 4a.-(a² + 2) = -a² - 2.So, Q(2) = 12a + 6 - 4a - a² - 2.Combine like terms:12a - 4a = 8a.6 - 2 = 4.So, Q(2) = -a² + 8a + 4.We need Q(2) ≥ 0:-a² + 8a + 4 ≥ 0.Multiply both sides by -1 (remember to reverse the inequality):a² - 8a - 4 ≤ 0.Now, solve the quadratic inequality a² - 8a - 4 ≤ 0.First, find the roots of the equation a² - 8a - 4 = 0.Using the quadratic formula:a = [8 ± sqrt(64 + 16)] / 2 = [8 ± sqrt(80)] / 2 = [8 ± 4*sqrt(5)] / 2 = 4 ± 2*sqrt(5).So, the roots are a = 4 + 2√5 and a = 4 - 2√5.Since the coefficient of a² is positive, the quadratic opens upwards. Therefore, the inequality a² - 8a - 4 ≤ 0 holds between the roots:4 - 2√5 ≤ a ≤ 4 + 2√5.But remember, we are in the case where a > 0. So, we need to consider the intersection of this interval with a > 0.Compute 4 - 2√5:√5 ≈ 2.236, so 2√5 ≈ 4.472.Thus, 4 - 4.472 ≈ -0.472.So, 4 - 2√5 ≈ -0.472, which is negative. Since a > 0, the lower bound is 0. So, the acceptable interval for a is:0 < a ≤ 4 + 2√5.But wait, since a = 0 is also acceptable (as we saw in Case 1), the overall range of a is:0 ≤ a ≤ 4 + 2√5.Therefore, the range of the real number a is [0, 4 + 2√5].Let me just double-check my steps to make sure I didn't make any mistakes.1. Found the derivative f'(x) correctly.2. Simplified f'(x) to N(x)/(ax + 1), where N(x) is a cubic polynomial.3. Factored out x to get N(x) = x * Q(x), where Q(x) is quadratic.4. Noted that since x ≥ 2, x is positive, so the sign of N(x) is determined by Q(x).5. Analyzed Q(x) for a = 0 and a > 0.6. For a > 0, found that the vertex of Q(x) is at x_v < 1/3, so on [2, +∞), Q(x) is increasing, so minimum at x = 2.7. Computed Q(2) = -a² + 8a + 4 ≥ 0, leading to a² - 8a - 4 ≤ 0.8. Solved the quadratic inequality, found roots at 4 ± 2√5, and since a > 0, took the interval [0, 4 + 2√5].Seems solid. I think that's the correct range for a.**Final Answer**The range of the real number ( a ) is (boxed{[0, 4 + 2sqrt{5}]}).

💡First, I need to determine the number of children at the event. There are 12 adults and an unknown number of children. The event planner wants to set up tables with the same combination of adults and children, and no one should be left out. The maximum number of tables that can be set up is 4.Since there are 12 adults and 4 tables, I can divide the number of adults by the number of tables to find out how many adults will be at each table:12 adults ÷ 4 tables = 3 adults per table.Next, I need to determine how many children will be at each table. Let's denote the number of children as C. Since the number of children must also be evenly distributed across the 4 tables, C must be divisible by 4.Therefore, the number of children at the event must be a multiple of 4. Without additional information, I can conclude that the number of children is a multiple of 4.