answer：To determine the maximum value of the function ( f(x) = x^2 + 2x + 1 ) on the interval ([-2, 2]), I will follow these steps:First, I'll find the critical points by taking the derivative of the function and setting it equal to zero. The derivative ( f'(x) = 2x + 2 ). Setting this equal to zero gives ( 2x + 2 = 0 ), which simplifies to ( x = -1 ).Next, I'll evaluate the function at the critical point ( x = -1 ) and at the endpoints of the interval, ( x = -2 ) and ( x = 2 ). Calculating ( f(-2) ):[f(-2) = (-2)^2 + 2(-2) + 1 = 4 - 4 + 1 = 1]Calculating ( f(-1) ):[f(-1) = (-1)^2 + 2(-1) + 1 = 1 - 2 + 1 = 0]Calculating ( f(2) ):[f(2) = 2^2 + 2(2) + 1 = 4 + 4 + 1 = 9]Finally, by comparing the values ( f(-2) = 1 ), ( f(-1) = 0 ), and ( f(2) = 9 ), I conclude that the maximum value of the function on the interval ([-2, 2]) is 9.

answer：First, I'll calculate the number of shirts Kendra wears during the weekdays. She wears one shirt each day for school, which totals 5 shirts. Additionally, she changes into a different shirt for an after-school club three days a week, adding 3 more shirts. This gives a total of 8 shirts for the weekdays.Next, I'll consider her weekend activities. On Saturday, she wears one shirt all day, and on Sunday, she wears a different shirt to church and another shirt for the rest of the day. This adds 3 shirts for the weekend.Adding the weekday and weekend shirts together, Kendra needs a total of 11 shirts. Since she wants to do laundry only once every two weeks, she needs to have enough shirts for the entire two-week period without needing to wash them in between. Therefore, she needs 11 shirts in total.

answer：Okay, so I have this problem: I need to find how many numbers less than 1000 are divisible by 4 and don't contain any of the digits 6, 7, 8, 9, or 0. Hmm, that sounds a bit tricky, but let me break it down step by step.First, let's understand the constraints. The numbers have to be less than 1000, so we're dealing with one-digit, two-digit, and three-digit numbers. Also, they must be divisible by 4. Additionally, none of the digits can be 6, 7, 8, 9, or 0. So, the allowed digits are 1, 2, 3, 4, and 5. That gives me 5 choices for each digit.Wait, but hold on. The number can't have 0, so for numbers with multiple digits, the first digit can't be 0 anyway, but since we're excluding 0 entirely, all digits must be from 1 to 5. That simplifies things a bit.Now, let's think about the divisibility by 4. I remember that a number is divisible by 4 if the number formed by its last two digits is divisible by 4. So, for example, 124 is divisible by 4 because 24 is divisible by 4. This rule applies regardless of how many digits the number has, as long as we're considering the last two digits.So, for one-digit numbers, since they only have one digit, the rule simplifies. A single-digit number is divisible by 4 if it's 4. Because 4 divided by 4 is 1, which is an integer. So, 4 is the only single-digit number that fits.For two-digit numbers, the entire number is considered, so we need to find all two-digit numbers where both digits are from 1 to 5, and the number itself is divisible by 4. Similarly, for three-digit numbers, the last two digits must form a number divisible by 4, and all three digits must be from 1 to 5.Alright, let's tackle each case separately: one-digit, two-digit, and three-digit numbers.Starting with one-digit numbers. As I thought, only 4 is allowed. So that's 1 number.Moving on to two-digit numbers. Each digit can be 1, 2, 3, 4, or 5. So, the first digit has 5 choices, and the second digit has 5 choices, making 25 possible two-digit numbers. But we need to find how many of these are divisible by 4.To find that, I can list all two-digit numbers with digits from 1 to 5 and check which ones are divisible by 4. Alternatively, I can find a pattern or a formula.Let me think. The two-digit numbers range from 11 to 55, with both digits from 1 to 5. So, the smallest is 11, the largest is 55.I know that a number is divisible by 4 if the number formed by its last two digits is divisible by 4. Since these are two-digit numbers, the entire number needs to be divisible by 4.So, let's list all two-digit numbers with digits 1-5 and check divisibility by 4.Starting with 11: 11 divided by 4 is 2.75, not an integer. So, not divisible.12: 12 / 4 = 3, which is an integer. So, 12 is good.13: 13 / 4 = 3.25, not divisible.14: 14 / 4 = 3.5, not divisible.15: 15 / 4 = 3.75, not divisible.21: 21 / 4 = 5.25, not divisible.22: 22 / 4 = 5.5, not divisible.23: 23 / 4 = 5.75, not divisible.24: 24 / 4 = 6, divisible. So, 24 is good.25: 25 / 4 = 6.25, not divisible.31: 31 / 4 = 7.75, not divisible.32: 32 / 4 = 8, divisible. So, 32 is good.33: 33 / 4 = 8.25, not divisible.34: 34 / 4 = 8.5, not divisible.35: 35 / 4 = 8.75, not divisible.41: 41 / 4 = 10.25, not divisible.42: 42 / 4 = 10.5, not divisible.43: 43 / 4 = 10.75, not divisible.44: 44 / 4 = 11, divisible. So, 44 is good.45: 45 / 4 = 11.25, not divisible.51: 51 / 4 = 12.75, not divisible.52: 52 / 4 = 13, divisible. So, 52 is good.53: 53 / 4 = 13.25, not divisible.54: 54 / 4 = 13.5, not divisible.55: 55 / 4 = 13.75, not divisible.So, the two-digit numbers that fit are: 12, 24, 32, 44, 52. That's 5 numbers.Wait, let me count again: 12, 24, 32, 44, 52. Yep, 5 numbers.So, two-digit numbers contribute 5.Now, moving on to three-digit numbers. This is a bit more complex because we have to consider the last two digits for divisibility by 4, and all three digits must be from 1 to 5.So, let's think about this. For a three-digit number, the hundreds digit can be 1-5, the tens digit can be 1-5, and the units digit can be 1-5. But the number formed by the tens and units digits must be divisible by 4.So, if I can find all possible two-digit endings (from 11 to 55) that are divisible by 4, then for each such ending, the hundreds digit can be any of 1-5, giving 5 possibilities for each valid ending.Therefore, first, I need to find how many two-digit numbers (from 11 to 55, with digits 1-5) are divisible by 4. Then, multiply that count by 5 to get the total three-digit numbers.Wait, but hold on. Earlier, for two-digit numbers, we found 5 numbers: 12, 24, 32, 44, 52. So, does that mean there are 5 possible endings for the three-digit numbers?Yes, exactly. So, for each of these 5 endings, the hundreds digit can be 1, 2, 3, 4, or 5. So, 5 endings * 5 hundreds digits = 25 three-digit numbers.But let me verify this because sometimes when dealing with three-digit numbers, the hundreds digit can sometimes affect the divisibility, but in this case, since we're only concerned with the last two digits, the hundreds digit doesn't matter. So, as long as the last two digits form a number divisible by 4, the entire number is divisible by 4.Therefore, yes, 25 three-digit numbers.Wait, but hold on again. Let me make sure that the two-digit endings we found earlier are indeed all the possible endings. Because in the two-digit case, we found 5 numbers, but in the three-digit case, the last two digits can be any of those 5, regardless of the hundreds digit.But let me think: are there any other two-digit numbers with digits 1-5 that are divisible by 4 besides the 5 we found? Let's check again.Earlier, I listed all two-digit numbers from 11 to 55 with digits 1-5 and found 12, 24, 32, 44, 52. Let me double-check if I missed any.Starting from 11:11: No12: Yes13: No14: No15: No21: No22: No23: No24: Yes25: No31: No32: Yes33: No34: No35: No41: No42: No43: No44: Yes45: No51: No52: Yes53: No54: No55: NoSo, yes, only 12, 24, 32, 44, 52. So, 5 two-digit endings.Therefore, for three-digit numbers, each of these 5 endings can be combined with 5 hundreds digits, giving 25 numbers.So, three-digit numbers contribute 25.Now, adding up all the cases:One-digit: 1Two-digit: 5Three-digit: 25Total numbers: 1 + 5 + 25 = 31.Wait, hold on. Let me make sure I didn't miss anything.First, one-digit: only 4. Correct.Two-digit: 12, 24, 32, 44, 52. That's 5. Correct.Three-digit: For each of the 5 two-digit endings, 5 hundreds digits, so 25. Correct.So, total is 31.But wait, let me think again. Is 00 considered? No, because numbers can't contain 0. So, numbers like 04 or 00 aren't allowed. So, our count is correct.But let me think about another approach to verify.Another way is to consider that all numbers less than 1000, with digits only from 1-5, and divisible by 4.So, for each number of digits (1, 2, 3), count how many numbers meet the criteria.We did that.Alternatively, we can model this as a counting problem with constraints.For one-digit numbers: only 4. So, 1.For two-digit numbers: first digit 1-5, second digit 1-5, and the number divisible by 4. We found 5.For three-digit numbers: first digit 1-5, second digit 1-5, third digit 1-5, and the last two digits form a number divisible by 4. Since the last two digits can be 12, 24, 32, 44, 52, each with 5 possibilities for the first digit, so 5*5=25.So, total is 1 + 5 + 25 = 31.Wait, but let me think about the three-digit numbers again. Is there a possibility that some of the two-digit endings could result in the same three-digit number? No, because each three-digit number is uniquely determined by its hundreds digit and the two-digit ending.So, 25 is correct.Alternatively, perhaps I can think in terms of generating functions or combinatorics, but that might be overcomplicating.Alternatively, let's think about how many numbers less than 1000 with digits 1-5 are divisible by 4.Total numbers with digits 1-5: For one-digit: 5, two-digit: 25, three-digit: 125. So, total 155 numbers.Out of these, how many are divisible by 4?Since numbers are equally distributed modulo 4, approximately 1/4 of them should be divisible by 4. 155 / 4 is approximately 38.75, but our count is 31, which is less. Hmm, that discrepancy makes me think maybe my initial approach is wrong.Wait, but numbers aren't necessarily equally distributed modulo 4 because the digits are restricted. So, the distribution might not be uniform.Therefore, the initial approach is more accurate.Alternatively, maybe I can compute the total number of such numbers and then compute how many are divisible by 4.But since the numbers are not uniformly distributed modulo 4, that might not help.Alternatively, perhaps I can model the problem as a finite automaton or use dynamic programming to count numbers with digits 1-5 and divisible by 4.But that might be more complex.Alternatively, perhaps I can think of the numbers as strings of digits and compute how many such strings are divisible by 4.But since the divisibility by 4 depends only on the last two digits, as we established earlier, the count is equal to the number of valid last two digits multiplied by the number of possibilities for the preceding digits.So, for one-digit numbers: only 4.For two-digit numbers: as we found, 5.For three-digit numbers: 5 (two-digit endings) * 5 (hundreds digit) = 25.So, total 1 + 5 + 25 = 31.Therefore, I think 31 is the correct answer.But let me double-check my two-digit numbers again because sometimes it's easy to miss or double-count.Two-digit numbers with digits 1-5:12: 12 / 4 = 3, yes.24: 24 / 4 = 6, yes.32: 32 / 4 = 8, yes.44: 44 / 4 = 11, yes.52: 52 / 4 = 13, yes.So, 5 numbers.Yes, that's correct.Therefore, I think my answer is correct: 31 numbers.**Final Answer**The total number of such numbers is boxed{31}.

answer：Okay, so I have this problem about a regular quadrilateral pyramid, which is P-ABCD. I need to find the ratio of two volumes: V_{G-PAD} over V_{G-PAB}, where G is the centroid of triangle PBC. Hmm, okay, let me break this down step by step.First, let me visualize the pyramid. A regular quadrilateral pyramid means that the base is a square, right? So ABCD is a square, and P is the apex. All the edges from the apex to the base are equal in length, so PA = PB = PC = PD. That should help with some symmetry arguments.Now, G is the centroid of triangle PBC. The centroid is the intersection point of the medians, so it's the average of the coordinates of the three vertices. That might be useful if I assign coordinates to the pyramid.Maybe assigning coordinates will make this easier. Let me set up a coordinate system. Let's place the square base ABCD on the xy-plane with its center at the origin. So, let me denote the coordinates as follows:- Let’s say point A is at (a, a, 0)- Point B is at (-a, a, 0)- Point C is at (-a, -a, 0)- Point D is at (a, -a, 0)- The apex P is at (0, 0, h), where h is the height of the pyramid.Wait, actually, since it's a regular pyramid, all the edges PA, PB, PC, PD are equal. So, the distance from P to each base vertex is the same. Let me compute that distance.The distance from P(0,0,h) to A(a,a,0) is sqrt[(a)^2 + (a)^2 + (h)^2] = sqrt(2a² + h²). Similarly, the distance from P to B, C, D will be the same. So, that's good.Now, I need to find the centroid G of triangle PBC. Let me find the coordinates of G. The centroid is the average of the coordinates of P, B, and C.Coordinates of P: (0, 0, h)Coordinates of B: (-a, a, 0)Coordinates of C: (-a, -a, 0)So, centroid G has coordinates:x-coordinate: (0 + (-a) + (-a))/3 = (-2a)/3y-coordinate: (0 + a + (-a))/3 = 0z-coordinate: (h + 0 + 0)/3 = h/3So, G is at (-2a/3, 0, h/3). Got that.Now, I need to find the volumes of two tetrahedrons: G-PAD and G-PAB.Let me recall that the volume of a tetrahedron can be found using the scalar triple product formula. If I have four points, say O, A, B, C, then the volume is |(OA · (OB × OC))| / 6.So, for each tetrahedron, I can choose a point as the origin and compute the vectors accordingly.Let me first find V_{G-PAD}. The tetrahedron is G-P-A-D.Similarly, V_{G-PAB} is the tetrahedron G-P-A-B.Wait, actually, let me make sure: the notation V_{G-PAD} means the volume of the tetrahedron with vertices G, P, A, D. Similarly, V_{G-PAB} is the volume with vertices G, P, A, B.Yes, that's correct.So, let me compute these volumes.First, let me write down the coordinates of all the points involved.Points:- G: (-2a/3, 0, h/3)- P: (0, 0, h)- A: (a, a, 0)- D: (a, -a, 0)- B: (-a, a, 0)So, for V_{G-PAD}, the four points are G, P, A, D.Similarly, for V_{G-PAB}, the four points are G, P, A, B.Let me compute V_{G-PAD} first.To compute the volume, I can use the scalar triple product. Let me choose point G as the origin for this computation. So, I need vectors from G to P, G to A, and G to D.Wait, actually, no. The scalar triple product is based on vectors from a common point. So, if I take G as the origin, then vectors GP, GA, GD would be:GP = P - G = (0 - (-2a/3), 0 - 0, h - h/3) = (2a/3, 0, 2h/3)GA = A - G = (a - (-2a/3), a - 0, 0 - h/3) = (5a/3, a, -h/3)GD = D - G = (a - (-2a/3), -a - 0, 0 - h/3) = (5a/3, -a, -h/3)So, the volume V_{G-PAD} is |(GP · (GA × GD))| / 6.Similarly, for V_{G-PAB}, the vectors from G would be GP, GA, and GB.Let me compute GA × GD first.Compute GA × GD:GA = (5a/3, a, -h/3)GD = (5a/3, -a, -h/3)Cross product GA × GD is:|i     j     k||5a/3  a   -h/3||5a/3 -a   -h/3|So, determinant:i * [a*(-h/3) - (-h/3)*(-a)] - j * [5a/3*(-h/3) - (-h/3)*5a/3] + k * [5a/3*(-a) - a*5a/3]Compute each component:i component: [ -a h / 3 - (h a / 3) ] = (-a h / 3 - a h / 3) = (-2 a h / 3)j component: - [ (-5a h / 9) - (-5a h / 9) ] = - [ (-5a h / 9 + 5a h / 9) ] = - [0] = 0k component: [ (-5a² / 3) - (5a² / 3) ] = (-10a² / 3)So, GA × GD = (-2 a h / 3, 0, -10 a² / 3)Now, compute GP · (GA × GD):GP = (2a/3, 0, 2h/3)Dot product:(2a/3)*(-2 a h / 3) + 0*0 + (2h/3)*(-10 a² / 3)Compute each term:First term: (2a/3)*(-2 a h / 3) = (-4 a² h) / 9Second term: 0Third term: (2h/3)*(-10 a² / 3) = (-20 a² h) / 9Total: (-4 a² h / 9) + (-20 a² h / 9) = (-24 a² h) / 9 = (-8 a² h) / 3So, the absolute value is | -8 a² h / 3 | = 8 a² h / 3Therefore, V_{G-PAD} = (8 a² h / 3) / 6 = (8 a² h) / 18 = (4 a² h) / 9Okay, so V_{G-PAD} is (4 a² h)/9.Now, let me compute V_{G-PAB}.Similarly, the tetrahedron G-P-A-B.Again, I can use the scalar triple product. Let me take G as the origin again.Vectors from G:GP = (2a/3, 0, 2h/3) as before.GA = (5a/3, a, -h/3) as before.GB = B - G = (-a - (-2a/3), a - 0, 0 - h/3) = (-a + 2a/3, a, -h/3) = (-a/3, a, -h/3)So, vectors are GP, GA, GB.Compute the scalar triple product GP · (GA × GB)First, compute GA × GB.GA = (5a/3, a, -h/3)GB = (-a/3, a, -h/3)Cross product GA × GB:|i     j     k||5a/3  a   -h/3||-a/3  a   -h/3|Compute determinant:i * [a*(-h/3) - (-h/3)*a] - j * [5a/3*(-h/3) - (-h/3)*(-a/3)] + k * [5a/3*a - a*(-a/3)]Compute each component:i component: [ -a h / 3 - (-a h / 3) ] = (-a h / 3 + a h / 3) = 0j component: - [ (-5a h / 9) - (a h / 9) ] = - [ (-6a h / 9) ] = - [ (-2a h / 3) ] = 2a h / 3k component: [5a² / 3 - (-a² / 3)] = (5a² / 3 + a² / 3) = 6a² / 3 = 2a²So, GA × GB = (0, 2a h / 3, 2a²)Now, compute GP · (GA × GB):GP = (2a/3, 0, 2h/3)Dot product:(2a/3)*0 + 0*(2a h / 3) + (2h/3)*(2a²)Compute each term:First term: 0Second term: 0Third term: (2h/3)*(2a²) = 4a² h / 3So, the scalar triple product is 4a² h / 3Therefore, V_{G-PAB} = |4a² h / 3| / 6 = (4a² h / 3) / 6 = (4a² h) / 18 = (2a² h) / 9So, V_{G-PAB} is (2a² h)/9.Now, the ratio V_{G-PAD} / V_{G-PAB} is (4a² h / 9) / (2a² h / 9) = (4/9)/(2/9) = 4/2 = 2.Wait, that seems straightforward, but let me verify my calculations because sometimes signs can mess things up.Wait, in the cross product GA × GD, I had (-2 a h / 3, 0, -10 a² / 3). Then, GP · (GA × GD) was (-8 a² h)/3, so absolute value 8 a² h / 3, divided by 6 gives 4 a² h / 9.Similarly, for GA × GB, I had (0, 2a h / 3, 2a²). Then, GP · (GA × GB) was 4a² h / 3, absolute value same, divided by 6 gives 2a² h / 9.So, the ratio is indeed 2.Hmm, that seems correct. But let me think if there's another way to approach this, maybe using ratios of areas or something else.Alternatively, since G is the centroid of triangle PBC, it divides the medians in a 2:1 ratio. Maybe I can use some properties of centroids and volumes.But since I already got the ratio as 2 using coordinates, and the calculations seem consistent, I think that's correct.So, the value of V_{G-PAD}/V_{G-PAB} is 2.**Final Answer**The value of frac{V_{G-PAD}}{V_{G-PAB}} is boxed{2}.

answer：First, I know that Jayden has 20 pencils.Since Jayden has twice as many pencils as Marcus, I can find out how many pencils Marcus has by dividing Jayden's pencils by 2. So, Marcus has 10 pencils.Dana has 15 more pencils than Jayden. Adding 15 to Jayden's 20 pencils, Dana has 35 pencils.To find out how many more pencils Dana has than Marcus, I subtract the number of pencils Marcus has from the number Dana has. That means Dana has 25 more pencils than Marcus.

answer：First, I need to evaluate each term in the expression separately.For the first term, (lg 25), since the logarithm base 10 of 25 can be expressed as (log_{10} 25), and knowing that (25 = 5^2), I can rewrite this as (2 log_{10} 5). Using the approximate value of (log_{10} 5 approx 0.69897), this term becomes approximately (2 times 0.69897 = 1.39794).Next, for the second term, (log_3 27), I recognize that (27 = 3^3). Therefore, (log_3 27 = 3).For the third term, (lg 4), which is (log_{10} 4), I can express 4 as (2^2), so this becomes (2 log_{10} 2). Using the approximate value of (log_{10} 2 approx 0.30103), this term is approximately (2 times 0.30103 = 0.60206).Finally, I add all the evaluated terms together: (1.39794 + 3 + 0.60206 = 5).