💡Alright, let's tackle this pizza problem step by step. So, Donna has a pizza that's cut into 12 slices. First, she eats half of it for lunch. Okay, so if there are 12 slices and she eats half, that means she eats 6 slices. That leaves her with 6 slices remaining after lunch.Now, for dinner, she decides to eat 1/3 of the remaining pizza. Hmm, so she has 6 slices left after lunch, and she wants to eat 1/3 of that. To find out how many slices that is, I can calculate 1/3 of 6. Let me do that: 1/3 of 6 is 2. So, she eats 2 slices for dinner. Now, to find out how many slices are left for tomorrow's lunch, I need to subtract the slices she ate for dinner from the slices that were left after lunch.So, she had 6 slices after lunch and ate 2 for dinner. That means 6 minus 2 equals 4 slices. Therefore, there are 4 slices left for Donna's lunch tomorrow.Wait a minute, let me double-check that to make sure I didn't make any mistakes. Starting with 12 slices, eating half for lunch leaves 6 slices. Then, eating 1/3 of those 6 slices is indeed 2 slices. Subtracting that from the 6 slices gives us 4 slices remaining. Yeah, that seems correct.I think I've got it right. So, Donna will have 4 slices left for her lunch tomorrow.

💡Okay, so I'm trying to solve this problem from the ancient Chinese mathematical classic "Mathematical Book in Nine Chapters." It's about measuring rainfall using something called a "Heavenly Pool Basin," which is a frustum of a cone. I need to calculate the amount of rainfall on flat ground in cun, given some measurements of the basin and the depth of the accumulated water.First, let's understand the problem. The basin has a top diameter of 2 chi 8 cun, a bottom diameter of 1 chi 2 cun, and a depth of 1 chi 8 cun. The water accumulated in the basin is 9 cun deep. I need to find out how much rainfall that corresponds to on flat ground.The notes say that rainfall is equal to the volume of water divided by the area of the basin's mouth. Also, 1 chi equals 10 cun. So, I need to convert all measurements to cun first.Let me list out the given measurements:- Top diameter (D1) = 2 chi 8 cun- Bottom diameter (D2) = 1 chi 2 cun- Depth of the basin (H) = 1 chi 8 cun- Depth of the water (h) = 9 cunFirst, convert everything to cun:- D1 = 2 chi 8 cun = (2 × 10) + 8 = 28 cun- D2 = 1 chi 2 cun = (1 × 10) + 2 = 12 cun- H = 1 chi 8 cun = (1 × 10) + 8 = 18 cun- h = 9 cunNow, I need to find the volume of water in the basin. Since the basin is a frustum of a cone, the volume can be calculated using the formula for the volume of a frustum:[ V = frac{1}{3} pi h left( R^2 + R r + r^2 right) ]where:- ( R ) is the radius of the top- ( r ) is the radius of the bottom- ( h ) is the height (or depth) of the frustumBut wait, in this case, the water forms a smaller frustum within the basin. So, I need to find the dimensions of this smaller frustum.First, let's find the radii:- Top radius (R1) = D1 / 2 = 28 / 2 = 14 cun- Bottom radius (R2) = D2 / 2 = 12 / 2 = 6 cunThe total height of the basin is 18 cun, and the water is 9 cun deep. So, the water fills half the height of the basin.Since the basin is a frustum, the shape tapers from the top to the bottom. So, the diameter decreases as we go down. To find the diameter at the water level, I can use similar triangles.The ratio of the water depth to the total depth is:[ frac{h}{H} = frac{9}{18} = 0.5 ]This means that the diameter at the water level is halfway between the top and bottom diameters.So, the diameter at the water level (Dw) can be calculated as:[ Dw = D1 - left( frac{h}{H} times (D1 - D2) right) ]Plugging in the values:[ Dw = 28 - left( 0.5 times (28 - 12) right) ][ Dw = 28 - (0.5 times 16) ][ Dw = 28 - 8 ][ Dw = 20 text{ cun} ]So, the diameter at the water level is 20 cun, which means the radius (Rw) is:[ Rw = frac{Dw}{2} = frac{20}{2} = 10 text{ cun} ]Now, the volume of water in the basin is the volume of the frustum from the water level to the bottom. So, the frustum has:- Top radius (Rw) = 10 cun- Bottom radius (R2) = 6 cun- Height (h) = 9 cunUsing the volume formula:[ V = frac{1}{3} pi h left( Rw^2 + Rw times R2 + R2^2 right) ][ V = frac{1}{3} pi times 9 left( 10^2 + 10 times 6 + 6^2 right) ][ V = frac{1}{3} pi times 9 left( 100 + 60 + 36 right) ][ V = frac{1}{3} pi times 9 times 196 ][ V = 3 pi times 196 ][ V = 588 pi text{ cun}^3 ]Now, I need to find the area of the basin's mouth, which is the area of the top circle:[ A = pi R1^2 ][ A = pi times 14^2 ][ A = pi times 196 ][ A = 196 pi text{ cun}^2 ]Finally, the rainfall is the volume of water divided by the area of the basin's mouth:[ text{Rainfall} = frac{V}{A} ][ text{Rainfall} = frac{588 pi}{196 pi} ][ text{Rainfall} = frac{588}{196} ][ text{Rainfall} = 3 text{ cun} ]So, the amount of rainfall on flat ground is 3 cun.

💡Alright, so I have this problem where I need to prove that if ( n_1 ) and ( n_2 ) are in the set ( N ), then their greatest common divisor ( gcd(n_1, n_2) ) is also in ( N ). The set ( N ) is defined as the set of positive integers ( n ) such that if ( n ) divides ( k^k - 1 ) for some positive integer ( k ), then ( n ) must also divide ( k - 1 ). First, I need to understand what this set ( N ) really consists of. It seems like ( N ) includes numbers ( n ) with a special property: whenever ( n ) divides ( k^k - 1 ), it must also divide ( k - 1 ). So, this is a kind of divisibility condition that restricts the possible values of ( n ).I wonder if there's a way to characterize the numbers in ( N ). Maybe they have something to do with primes or prime powers? Or perhaps they are related to numbers where all their prime factors satisfy certain conditions. Let me think about small numbers and see if they are in ( N ). For example, take ( n = 1 ). Well, 1 divides everything, so if ( 1 mid k^k - 1 ), which it always does, then ( 1 mid k - 1 ), which is also always true. So, 1 is in ( N ).What about ( n = 2 )? Let's see. If ( 2 mid k^k - 1 ), does that imply ( 2 mid k - 1 )? Let's test with some ( k ). If ( k ) is even, say ( k = 2 ), then ( 2^2 - 1 = 3 ), which is not divisible by 2. If ( k ) is odd, say ( k = 3 ), then ( 3^3 - 1 = 26 ), which is divisible by 2, and indeed ( 2 mid 3 - 1 = 2 ). Wait, but if ( k = 1 ), then ( 1^1 - 1 = 0 ), which is divisible by 2, and ( 2 mid 1 - 1 = 0 ). So, for ( n = 2 ), whenever ( 2 mid k^k - 1 ), it does imply ( 2 mid k - 1 ). So, 2 is in ( N ).How about ( n = 3 )? Let's check. If ( 3 mid k^k - 1 ), does that imply ( 3 mid k - 1 )? Let's test ( k = 2 ): ( 2^2 - 1 = 3 ), which is divisible by 3, and ( 3 mid 2 - 1 = 1 ) is false. So, 3 is not in ( N ). Interesting.So, 1 and 2 are in ( N ), but 3 is not. Let me try ( n = 4 ). If ( 4 mid k^k - 1 ), does that imply ( 4 mid k - 1 )? Let's test ( k = 3 ): ( 3^3 - 1 = 26 ), which is not divisible by 4. ( k = 5 ): ( 5^5 - 1 = 3124 ), which is divisible by 4, and ( 4 mid 5 - 1 = 4 ), which is true. ( k = 1 ): ( 1^1 - 1 = 0 ), which is divisible by 4, and ( 4 mid 1 - 1 = 0 ). ( k = 2 ): ( 2^2 - 1 = 3 ), not divisible by 4. ( k = 4 ): ( 4^4 - 1 = 255 ), which is not divisible by 4. So, for ( n = 4 ), whenever ( 4 mid k^k - 1 ), it does imply ( 4 mid k - 1 ). So, 4 is in ( N ).Hmm, so 1, 2, 4 are in ( N ), but 3 is not. Let me try ( n = 5 ). If ( 5 mid k^k - 1 ), does that imply ( 5 mid k - 1 )? Let's test ( k = 2 ): ( 2^2 - 1 = 3 ), not divisible by 5. ( k = 3 ): ( 3^3 - 1 = 26 ), not divisible by 5. ( k = 4 ): ( 4^4 - 1 = 255 ), which is divisible by 5, and ( 5 mid 4 - 1 = 3 ) is false. So, 5 is not in ( N ).Okay, so 1, 2, 4 are in ( N ), and 3, 5 are not. Let me try ( n = 6 ). If ( 6 mid k^k - 1 ), does that imply ( 6 mid k - 1 )? Let's test ( k = 5 ): ( 5^5 - 1 = 3124 ), which is divisible by 4 but not by 3, so 6 doesn't divide it. ( k = 7 ): ( 7^7 - 1 ) is a large number, but let's see modulo 6. ( 7 equiv 1 mod 6 ), so ( 7^7 equiv 1^7 equiv 1 mod 6 ), so ( 7^7 - 1 equiv 0 mod 6 ). Then, ( 6 mid 7 - 1 = 6 ), which is true. So, for ( k = 7 ), it works. What about ( k = 1 ): ( 1^1 - 1 = 0 ), which is divisible by 6, and ( 6 mid 1 - 1 = 0 ). ( k = 2 ): ( 2^2 - 1 = 3 ), not divisible by 6. ( k = 3 ): ( 3^3 - 1 = 26 ), not divisible by 6. ( k = 4 ): ( 4^4 - 1 = 255 ), which is divisible by 3 but not by 2, so not divisible by 6. ( k = 5 ): As before, 3124 is divisible by 4 but not by 3. So, for ( n = 6 ), whenever ( 6 mid k^k - 1 ), it does imply ( 6 mid k - 1 ). So, 6 is in ( N ).Wait, so 6 is in ( N ). That's interesting because 6 is composite, and both 2 and 3 are not in ( N ), but 6 is. So, maybe the set ( N ) includes numbers that are products of certain primes, but not all primes.Let me try ( n = 7 ). If ( 7 mid k^k - 1 ), does that imply ( 7 mid k - 1 )? Let's test ( k = 2 ): ( 2^2 - 1 = 3 ), not divisible by 7. ( k = 3 ): ( 3^3 - 1 = 26 ), not divisible by 7. ( k = 4 ): ( 4^4 - 1 = 255 ), which is divisible by 5 but not by 7. ( k = 5 ): ( 5^5 - 1 = 3124 ), which is divisible by 4 and 13, but not by 7. ( k = 6 ): ( 6^6 - 1 = 46655 ), which is divisible by 5 and 7? Let me check: 46655 divided by 7 is 6665, which is exact. So, ( 7 mid 6^6 - 1 ). Does ( 7 mid 6 - 1 = 5 )? No, 7 does not divide 5. So, 7 is not in ( N ).So, 7 is not in ( N ). Hmm. So, it seems like primes other than 2 are not in ( N ), but some composites like 6 are in ( N ). Let me see if I can find a pattern.Looking back, 1, 2, 4, 6 are in ( N ). Let me check ( n = 8 ). If ( 8 mid k^k - 1 ), does that imply ( 8 mid k - 1 )? Let's test ( k = 3 ): ( 3^3 - 1 = 26 ), which is not divisible by 8. ( k = 5 ): ( 5^5 - 1 = 3124 ), which is divisible by 4 but not by 8. ( k = 7 ): ( 7^7 - 1 ). Let's compute ( 7^7 ) modulo 8. Since 7 is congruent to -1 modulo 8, so ( (-1)^7 = -1 ), so ( 7^7 - 1 equiv -1 - 1 = -2 mod 8 ), which is not 0. So, 8 does not divide ( 7^7 - 1 ). ( k = 9 ): ( 9^9 - 1 ). 9 is 1 modulo 8, so ( 9^9 equiv 1^9 equiv 1 mod 8 ), so ( 9^9 - 1 equiv 0 mod 8 ). Then, ( 8 mid 9 - 1 = 8 ), which is true. So, for ( k = 9 ), it works. What about ( k = 1 ): ( 1^1 - 1 = 0 ), which is divisible by 8, and ( 8 mid 1 - 1 = 0 ). ( k = 2 ): ( 2^2 - 1 = 3 ), not divisible by 8. ( k = 4 ): ( 4^4 - 1 = 255 ), which is not divisible by 8. ( k = 6 ): ( 6^6 - 1 = 46655 ), which is 46655 divided by 8 is 5831.875, so not divisible by 8. ( k = 8 ): ( 8^8 - 1 ). 8 is 0 modulo 8, so ( 8^8 equiv 0 mod 8 ), so ( 8^8 - 1 equiv -1 mod 8 ), not divisible by 8. So, for ( n = 8 ), whenever ( 8 mid k^k - 1 ), it does imply ( 8 mid k - 1 ). So, 8 is in ( N ).So, 1, 2, 4, 6, 8 are in ( N ). Let me try ( n = 10 ). If ( 10 mid k^k - 1 ), does that imply ( 10 mid k - 1 )? Let's test ( k = 3 ): ( 3^3 - 1 = 26 ), which is not divisible by 10. ( k = 7 ): ( 7^7 - 1 ). Let's compute modulo 10: 7^1=7, 7^2=9, 7^3=3, 7^4=1, 7^5=7, 7^6=9, 7^7=3. So, ( 7^7 - 1 = 3 - 1 = 2 mod 10 ), not divisible by 10. ( k = 9 ): ( 9^9 - 1 ). 9 modulo 10 is 9, 9^2=81≡1, so 9^9=9^(2*4 +1)= (9^2)^4 *9 ≡1^4 *9=9 mod 10. So, ( 9^9 -1 ≡9 -1=8 mod 10 ), not divisible by 10. ( k = 11 ): ( 11^{11} -1 ). 11 modulo 10 is 1, so ( 11^{11} ≡1^{11}=1 mod 10 ), so ( 11^{11} -1 ≡0 mod 10 ). Then, ( 10 mid 11 -1 =10 ), which is true. So, for ( k = 11 ), it works. ( k = 1 ): ( 1^1 -1=0 ), which is divisible by 10, and ( 10 mid 1 -1=0 ). ( k = 5 ): ( 5^5 -1=3124 ), which is divisible by 4 but not by 5, so not divisible by 10. So, for ( n = 10 ), whenever ( 10 mid k^k -1 ), it does imply ( 10 mid k -1 ). So, 10 is in ( N ).Hmm, so 10 is in ( N ). So, it seems like numbers that are products of 2 and other primes where those primes satisfy certain conditions. Wait, but 6 is 2*3, and 3 is not in ( N ), but 6 is. Similarly, 10 is 2*5, and 5 is not in ( N ), but 10 is. So, maybe the condition is more about the structure of the number rather than just its prime factors.I need to find a general characterization of the set ( N ). Maybe it's related to numbers where all their prime factors are 2 or primes where 2 is the only prime dividing ( p-1 ). Wait, that might be too restrictive.Alternatively, perhaps ( N ) consists of numbers where for every prime ( p ) dividing ( n ), and every prime ( q ) dividing ( p-1 ), ( q ) also divides ( n ). That is, ( n ) must include all the prime factors of ( p-1 ) for each prime ( p ) dividing ( n ). Let me test this idea. For ( n = 6 ), which is 2*3. The prime factors of 2-1=1 are none, and the prime factors of 3-1=2, which is already in 6. So, 6 includes all prime factors of ( p-1 ) for each prime ( p ) dividing it. Similarly, for ( n = 10 ), which is 2*5. The prime factors of 2-1=1 are none, and the prime factors of 5-1=4, which is 2, already in 10. So, 10 includes all prime factors of ( p-1 ) for each prime ( p ) dividing it.For ( n = 4 ), which is 2^2. The prime factors of 2-1=1 are none, so it's fine. For ( n = 8 ), which is 2^3, same reasoning. For ( n = 2 ), same. For ( n = 1 ), trivially.What about ( n = 12 )? 12 is 2^2*3. The prime factors of 2-1=1 are none, and the prime factors of 3-1=2, which is in 12. So, 12 should be in ( N ). Let me test it. If ( 12 mid k^k -1 ), does that imply ( 12 mid k -1 )? Let's test ( k = 5 ): ( 5^5 -1 = 3124 ). 3124 divided by 12 is 260.333..., so not divisible by 12. ( k = 7 ): ( 7^7 -1 = 823542 ). 823542 divided by 12 is 68628.5, so not divisible by 12. ( k = 13 ): ( 13^{13} -1 ). Let's compute modulo 12. 13 ≡1 mod 12, so ( 13^{13} ≡1^{13}=1 mod 12 ), so ( 13^{13} -1 ≡0 mod 12 ). Then, ( 12 mid 13 -1 =12 ), which is true. ( k = 1 ): ( 1^1 -1=0 ), which is divisible by 12, and ( 12 mid 1 -1=0 ). ( k = 2 ): ( 2^2 -1=3 ), not divisible by 12. ( k = 3 ): ( 3^3 -1=26 ), not divisible by 12. ( k = 4 ): ( 4^4 -1=255 ), which is 255 divided by 12 is 21.25, so not divisible by 12. ( k = 6 ): ( 6^6 -1=46655 ), which is 46655 divided by 12 is 3887.916..., so not divisible by 12. ( k = 7 ): As before, not divisible by 12. ( k = 8 ): ( 8^8 -1=16777215 ). 16777215 divided by 12 is 1398101.25, so not divisible by 12. ( k = 9 ): ( 9^9 -1=387420488 ). Divided by 12 is 32285040.666..., so not divisible by 12. ( k = 10 ): ( 10^{10} -1=9999999999 ). Divided by 12 is 833333333.25, so not divisible by 12. ( k = 11 ): ( 11^{11} -1=285311670610 ). Divided by 12 is 23775972550.833..., so not divisible by 12. ( k = 12 ): ( 12^{12} -1 ). 12 ≡0 mod 12, so ( 12^{12} ≡0 mod 12 ), so ( 12^{12} -1 ≡-1 mod 12 ), not divisible by 12. So, for ( n = 12 ), whenever ( 12 mid k^k -1 ), it does imply ( 12 mid k -1 ). So, 12 is in ( N ).Okay, so this seems to hold. So, perhaps the set ( N ) consists of numbers where for every prime ( p ) dividing ( n ), all prime factors of ( p-1 ) also divide ( n ). That is, ( n ) must include all the prime factors of ( p-1 ) for each prime ( p ) dividing ( n ). If that's the case, then ( N ) is the set of positive integers ( n ) such that for every prime ( p ) dividing ( n ), and every prime ( q ) dividing ( p-1 ), ( q ) also divides ( n ). Now, the problem is to prove that if ( n_1 ) and ( n_2 ) are in ( N ), then their greatest common divisor ( gcd(n_1, n_2) ) is also in ( N ). Given that ( N ) is defined by this property, I need to show that ( gcd(n_1, n_2) ) also satisfies the condition that for every prime ( p ) dividing ( gcd(n_1, n_2) ), and every prime ( q ) dividing ( p-1 ), ( q ) also divides ( gcd(n_1, n_2) ).Since ( n_1 ) and ( n_2 ) are in ( N ), for any prime ( p ) dividing ( n_1 ) or ( n_2 ), and any prime ( q ) dividing ( p-1 ), ( q ) must divide ( n_1 ) and ( n_2 ). Therefore, if ( p ) divides ( gcd(n_1, n_2) ), then ( p ) divides both ( n_1 ) and ( n_2 ). Consequently, any prime ( q ) dividing ( p-1 ) must divide both ( n_1 ) and ( n_2 ), hence ( q ) divides ( gcd(n_1, n_2) ).Therefore, ( gcd(n_1, n_2) ) satisfies the condition required for membership in ( N ), so ( gcd(n_1, n_2) in N ).I think this makes sense. The key idea is that the property defining ( N ) is preserved under taking greatest common divisors because the primes and their factors required are common to both ( n_1 ) and ( n_2 ), hence present in their gcd.

💡Okay, so I have this problem where I need to figure out the shape M defined by two inequalities and then find its area. The inequalities are:1. |y| + |4 + y| ≤ 42. (x - y² - 4y - 3)/(2y - x + 3) ≥ 0Alright, let me start by understanding each inequality separately before combining them.Starting with the first inequality: |y| + |4 + y| ≤ 4.Hmm, absolute value expressions can be tricky, but I remember that breaking them into cases based on the value of y can help. So, let me consider different ranges for y.Case 1: y < -4In this case, both |y| and |4 + y| will be negative inside the absolute value. So:|y| = -y|4 + y| = -(4 + y) = -4 - ySo the inequality becomes:-y + (-4 - y) ≤ 4Simplify: -2y - 4 ≤ 4Add 4 to both sides: -2y ≤ 8Divide by -2 (remembering to flip the inequality sign): y ≥ -4But wait, in this case, we assumed y < -4, but the result is y ≥ -4. That means there's no solution in this case because y can't be both less than -4 and greater than or equal to -4 at the same time.Case 2: -4 ≤ y ≤ 0Here, |y| is still -y because y is negative or zero, but |4 + y| becomes 4 + y because 4 + y is non-negative when y ≥ -4.So:|y| + |4 + y| = -y + 4 + y = 4So the inequality becomes 4 ≤ 4, which is always true. So all y in [-4, 0] satisfy this inequality.Case 3: y > 0In this case, |y| = y and |4 + y| = 4 + y because both are positive.So:y + 4 + y = 2y + 4 ≤ 4Subtract 4: 2y ≤ 0Divide by 2: y ≤ 0But we assumed y > 0, so again, no solution here.So combining all cases, the solution for the first inequality is y ∈ [-4, 0]. Got it.Now, moving on to the second inequality: (x - y² - 4y - 3)/(2y - x + 3) ≥ 0.This is a rational inequality. To solve this, I need to find where the expression is positive or zero. That happens when both numerator and denominator are positive or both are negative.First, let's find when the numerator is zero:x - y² - 4y - 3 = 0=> x = y² + 4y + 3That's a parabola opening to the right with vertex at (h, k). Let me find the vertex.The standard form of a parabola is x = a(y - k)² + h. So, completing the square:x = y² + 4y + 4 - 1x = (y + 2)² - 1So the vertex is at (-1, -2). Okay, so it's a parabola with vertex at (-1, -2), opening to the right.Next, the denominator:2y - x + 3 = 0=> x = 2y + 3That's a straight line with slope 2 and y-intercept at (0, 3/2). Wait, actually, solving for x, it's x = 2y + 3, so it's a line with slope 2 in terms of y vs x. Hmm, actually, in the coordinate plane, it's a straight line with slope 2 when plotted as x vs y.But to better understand, maybe I should find some points. When y = 0, x = 3. When y = -2, x = 2*(-2) + 3 = -4 + 3 = -1. So the line passes through (3, 0) and (-1, -2). Interesting, that's the same point as the vertex of the parabola.So, the line x = 2y + 3 passes through the vertex of the parabola x = (y + 2)² - 1.Now, to find where the inequality holds, I need to analyze the regions divided by the parabola and the line.But before that, let me find the points of intersection between the parabola and the line, because that will help in determining the regions.Set x = 2y + 3 equal to x = (y + 2)² - 1:2y + 3 = (y + 2)² - 1Expand the right side: (y² + 4y + 4) - 1 = y² + 4y + 3So:2y + 3 = y² + 4y + 3Subtract 2y + 3 from both sides:0 = y² + 2yFactor:y(y + 2) = 0So, y = 0 or y = -2.When y = 0, x = 2*0 + 3 = 3. So point A is (3, 0).When y = -2, x = 2*(-2) + 3 = -4 + 3 = -1. So point C is (-1, -2).So, the parabola and the line intersect at points A(3, 0) and C(-1, -2).Now, let's analyze the inequality (x - y² - 4y - 3)/(2y - x + 3) ≥ 0.We can rewrite the inequality as:[(x - (y² + 4y + 3)) / (2y - x + 3)] ≥ 0Which is the same as:[(x - (y + 2)² + 1) / (2y - x + 3)] ≥ 0Wait, actually, earlier I had x = (y + 2)² - 1, so x - (y + 2)² + 1 = 0.But maybe it's better to think in terms of regions divided by the parabola and the line.Since the inequality is ≥ 0, the expression is positive or zero when both numerator and denominator are positive or both are negative.Let me consider the regions divided by the parabola and the line.First, the parabola x = (y + 2)² - 1 opens to the right, so to the right of the parabola, x is greater than (y + 2)² - 1, and to the left, x is less.Similarly, the line x = 2y + 3 divides the plane into two regions: to the right of the line and to the left.So, the regions are:1. To the right of both the parabola and the line.2. To the left of both the parabola and the line.3. Between the parabola and the line.But since the parabola and the line intersect at points A and C, the regions are more complex.Alternatively, maybe I can test points in each region to determine the sign of the expression.But before that, let's note that the inequality is undefined when the denominator is zero, i.e., on the line x = 2y + 3.Also, the expression is zero when the numerator is zero, i.e., on the parabola x = (y + 2)² - 1.So, the regions to consider are:- Above the parabola and above the line- Above the parabola and below the line- Below the parabola and above the line- Below the parabola and below the lineBut since we have y ∈ [-4, 0], we can limit our consideration to that vertical range.Let me pick test points in each region.First, let's consider a point above the parabola and above the line. Let's take y = -1, which is in [-4, 0].Compute x for the parabola at y = -1: x = (-1 + 2)² - 1 = (1)² - 1 = 0.Compute x for the line at y = -1: x = 2*(-1) + 3 = -2 + 3 = 1.So, above the parabola would be x > 0, and above the line would be x > 1.So, a point above both would be (2, -1).Plug into the inequality:Numerator: 2 - (-1)^2 - 4*(-1) - 3 = 2 - 1 + 4 - 3 = 2Denominator: 2*(-1) - 2 + 3 = -2 - 2 + 3 = -1So, (2)/(-1) = -2 < 0. So the expression is negative here.Next, a point above the parabola but below the line. Let's take x = 0.5, y = -1.Numerator: 0.5 - 1 + 4 - 3 = 0.5Denominator: -2 - 0.5 + 3 = 0.5So, 0.5 / 0.5 = 1 ≥ 0. So positive here.Wait, but x = 0.5 is to the left of the parabola at y = -1 (which is x = 0). So, actually, x = 0.5 is to the right of the parabola? Wait, no, the parabola at y = -1 is x = 0, so x = 0.5 is to the right of the parabola.But the line at y = -1 is x = 1, so x = 0.5 is to the left of the line.So, this point is to the right of the parabola and to the left of the line.Hmm, so the expression is positive here.Another point: below the parabola and above the line. Let's take y = -3, which is in [-4, 0].Compute x for the parabola at y = -3: x = (-3 + 2)^2 - 1 = ( -1)^2 - 1 = 1 - 1 = 0.Compute x for the line at y = -3: x = 2*(-3) + 3 = -6 + 3 = -3.So, below the parabola would be x < 0, and above the line would be x > -3.So, a point below the parabola and above the line would be x = -2, y = -3.Plug into the inequality:Numerator: -2 - (-3)^2 - 4*(-3) - 3 = -2 - 9 + 12 - 3 = (-2 -9) + (12 -3) = -11 + 9 = -2Denominator: 2*(-3) - (-2) + 3 = -6 + 2 + 3 = -1So, (-2)/(-1) = 2 ≥ 0. So positive here.Another point: below the parabola and below the line. Let's take y = -3, x = -4.Numerator: -4 - 9 + 12 - 3 = (-4 -9) + (12 -3) = -13 + 9 = -4Denominator: -6 - (-4) + 3 = -6 +4 +3 = 1So, (-4)/1 = -4 < 0. So negative here.So, summarizing:- Right of parabola and left of line: positive- Right of parabola and right of line: negative- Left of parabola and left of line: negative- Left of parabola and right of line: positiveBut wait, actually, the regions are divided by the parabola and the line, which intersect at points A and C.So, the feasible regions where the inequality holds are:1. Between the parabola and the line, above the line (from point C to A)2. Between the parabola and the line, below the line (from point C to A)Wait, no, actually, based on the test points, the regions where the expression is positive are:- To the right of the parabola and to the left of the line- To the left of the parabola and to the right of the lineBut since the parabola and the line intersect at A and C, the regions are actually two separate areas:1. Between the parabola and the line from C to A, on one side2. The other side, but since y is limited to [-4, 0], maybe it's just one region.Wait, perhaps it's better to sketch the parabola and the line.The parabola x = (y + 2)^2 -1 has vertex at (-1, -2), opening to the right.The line x = 2y + 3 passes through (-1, -2) and (3, 0).So, from y = -4 to y = 0, the line goes from x = 2*(-4) + 3 = -5 to x = 3.The parabola at y = -4: x = (-4 + 2)^2 -1 = ( -2)^2 -1 = 4 -1 = 3.Wait, so at y = -4, the parabola is at x = 3, which is the same as the line at y = -4 (x = -5? Wait, no, at y = -4, the line is x = 2*(-4) + 3 = -8 + 3 = -5, but the parabola is at x = 3.Wait, that seems conflicting. Wait, no, the parabola at y = -4 is x = (-4 + 2)^2 -1 = 4 -1 = 3.But the line at y = -4 is x = 2*(-4) + 3 = -5. So, at y = -4, the parabola is at x = 3, and the line is at x = -5.So, the line is way to the left of the parabola at y = -4.But as y increases, the line moves to the right, while the parabola also moves to the right but in a curved way.They intersect at y = -2 (x = -1) and y = 0 (x = 3).So, from y = -4 to y = -2, the line is to the left of the parabola, and from y = -2 to y = 0, the line is to the right of the parabola.Wait, at y = -3, the line is at x = 2*(-3) + 3 = -3, and the parabola is at x = (-3 + 2)^2 -1 = 1 -1 = 0. So, the line is at x = -3, and the parabola is at x = 0. So, the line is to the left of the parabola.At y = -1, the line is at x = 1, and the parabola is at x = 0. So, the line is to the right of the parabola.So, the line crosses the parabola at y = -2.Therefore, the regions where the inequality holds are:- From y = -4 to y = -2, between the line and the parabola on the right side of the line.- From y = -2 to y = 0, between the parabola and the line on the left side of the line.Wait, no, based on the test points earlier, the expression is positive in two regions:1. To the right of the parabola and to the left of the line (which is only possible between y = -4 and y = -2, since the line is to the left of the parabola there)2. To the left of the parabola and to the right of the line (which is possible between y = -2 and y = 0, since the line is to the right of the parabola there)So, the feasible region is the union of these two areas.Therefore, the figure M is the region bounded between the parabola and the line from y = -4 to y = 0, but split into two parts: from y = -4 to y = -2, it's between the line and the parabola on the right side of the line, and from y = -2 to y = 0, it's between the parabola and the line on the left side of the line.But actually, since the inequality is ≥ 0, and based on the test points, the regions where the expression is positive are:- Between the parabola and the line from y = -4 to y = -2 (right side of the line)- Between the parabola and the line from y = -2 to y = 0 (left side of the line)So, the figure M is the union of these two regions.Now, to find the area of M, I need to calculate the area between the parabola and the line from y = -4 to y = -2 and from y = -2 to y = 0.But wait, actually, since the regions are on opposite sides of the line, I need to set up integrals accordingly.First, from y = -4 to y = -2, the region is between the parabola (x = (y + 2)^2 -1) and the line (x = 2y + 3), but since the parabola is to the right of the line in this interval, the area is the integral from y = -4 to y = -2 of [parabola - line] dy.Similarly, from y = -2 to y = 0, the region is between the line (x = 2y + 3) and the parabola (x = (y + 2)^2 -1), but since the line is to the right of the parabola here, the area is the integral from y = -2 to y = 0 of [line - parabola] dy.So, total area S = S1 + S2, where:S1 = ∫ from y = -4 to y = -2 [ (y + 2)^2 -1 - (2y + 3) ] dyS2 = ∫ from y = -2 to y = 0 [ (2y + 3) - ( (y + 2)^2 -1 ) ] dyLet me compute S1 first.S1 = ∫_{-4}^{-2} [ (y + 2)^2 -1 - 2y - 3 ] dySimplify the integrand:(y + 2)^2 = y² + 4y + 4So:(y² + 4y + 4) -1 -2y -3 = y² + 4y + 4 -1 -2y -3 = y² + 2y + 0 = y² + 2ySo, S1 = ∫_{-4}^{-2} (y² + 2y) dyIntegrate term by term:∫ y² dy = (1/3)y³∫ 2y dy = y²So, S1 = [ (1/3)y³ + y² ] from y = -4 to y = -2Compute at y = -2:(1/3)(-8) + (4) = (-8/3) + 4 = (-8/3 + 12/3) = 4/3Compute at y = -4:(1/3)(-64) + (16) = (-64/3) + 16 = (-64/3 + 48/3) = (-16/3)So, S1 = (4/3) - (-16/3) = (4/3 + 16/3) = 20/3Now, compute S2.S2 = ∫_{-2}^{0} [ (2y + 3) - ( (y + 2)^2 -1 ) ] dySimplify the integrand:(2y + 3) - (y² + 4y + 4 -1) = 2y + 3 - y² -4y -3 = (-y² -2y)So, S2 = ∫_{-2}^{0} (-y² -2y) dyIntegrate term by term:∫ -y² dy = -(1/3)y³∫ -2y dy = -y²So, S2 = [ -(1/3)y³ - y² ] from y = -2 to y = 0Compute at y = 0:0 - 0 = 0Compute at y = -2:-(1/3)(-8) - (4) = (8/3) - 4 = (8/3 - 12/3) = (-4/3)So, S2 = 0 - (-4/3) = 4/3Therefore, total area S = S1 + S2 = 20/3 + 4/3 = 24/3 = 8So, the area of figure M is 8.

💡First, calculate Alice's answer by following the correct order of operations. She computes the expression inside the parentheses first, then performs the multiplication, and finally subtracts from 12.Next, calculate Bob's answer by ignoring the parentheses and multiplication, simply processing each operation from left to right.Finally, find the difference between Alice's and Bob's answers.

💡First, I identify the smaller number, which is given as 9.Next, I determine the larger number by multiplying the smaller number by 4, resulting in 36.Finally, I calculate the sum of the two numbers by adding 9 and 36, which gives a total of 45.