💡First, I need to determine how many apples, oranges, and bananas Mary initially selected. I'll denote the number of apples as ( A ), oranges as ( O ), and bananas as ( B ). Mary selected a total of 12 pieces of fruit, so:[A + O + B = 12]The average price of the 12 pieces of fruit is 55 cents, so the total cost is:[12 times 55 = 660 text{ cents}]The cost equation based on the prices of each fruit is:[40A + 60O + 80B = 660]Next, I want to find out how many oranges Mary must put back to achieve an average price of 50 cents for the remaining fruit. Let ( x ) be the number of oranges she puts back. The new total number of pieces of fruit will be ( 12 - x ), and the new total cost will be ( 660 - 60x ) cents.The equation for the new average price is:[frac{660 - 60x}{12 - x} = 50]Multiplying both sides by ( 12 - x ) to eliminate the fraction:[660 - 60x = 50(12 - x)]Expanding the right side:[660 - 60x = 600 - 50x]Solving for ( x ):[60x - 50x = 660 - 600][10x = 60][x = 6]Therefore, Mary must put back 6 oranges to achieve the desired average price.

💡Okay, so I've got this problem here with two parts, (I) and (II). Let me try to tackle them one by one.Starting with part (I): I need to find the intervals where the function ( f(x) = sin(2x - frac{pi}{6}) + 2cos^2 x - 1 ) is monotonically increasing. Hmm, okay. I remember that to find where a function is increasing or decreasing, I need to look at its derivative. So, first, I should probably simplify the function ( f(x) ) as much as possible before taking the derivative.Looking at ( f(x) ), I see two terms: ( sin(2x - frac{pi}{6}) ) and ( 2cos^2 x - 1 ). Wait, ( 2cos^2 x - 1 ) is a familiar expression—it's the double-angle identity for cosine. Specifically, ( cos(2x) = 2cos^2 x - 1 ). So, that simplifies the second term to ( cos(2x) ).So now, ( f(x) = sin(2x - frac{pi}{6}) + cos(2x) ). Hmm, maybe I can combine these two sine and cosine terms into a single sine or cosine function using some trigonometric identities. Let me recall that ( sin(A - B) = sin A cos B - cos A sin B ). So, expanding ( sin(2x - frac{pi}{6}) ), I get:( sin(2x)cos(frac{pi}{6}) - cos(2x)sin(frac{pi}{6}) ).We know that ( cos(frac{pi}{6}) = frac{sqrt{3}}{2} ) and ( sin(frac{pi}{6}) = frac{1}{2} ). So, substituting these values in, we have:( sin(2x) cdot frac{sqrt{3}}{2} - cos(2x) cdot frac{1}{2} ).So, ( f(x) = frac{sqrt{3}}{2}sin(2x) - frac{1}{2}cos(2x) + cos(2x) ).Wait, the last two terms are ( -frac{1}{2}cos(2x) + cos(2x) ), which simplifies to ( frac{1}{2}cos(2x) ).So now, ( f(x) = frac{sqrt{3}}{2}sin(2x) + frac{1}{2}cos(2x) ).Hmm, this looks like the form ( Asin(2x) + Bcos(2x) ), which can be written as a single sine function with a phase shift. The general identity is ( Asintheta + Bcostheta = Csin(theta + phi) ), where ( C = sqrt{A^2 + B^2} ) and ( phi = arctan(frac{B}{A}) ) or something like that.Let me compute ( C ) first. Here, ( A = frac{sqrt{3}}{2} ) and ( B = frac{1}{2} ). So,( C = sqrt{ (frac{sqrt{3}}{2})^2 + (frac{1}{2})^2 } = sqrt{ frac{3}{4} + frac{1}{4} } = sqrt{1} = 1 ).Okay, so ( C = 1 ). Now, to find the phase shift ( phi ), we can use ( tanphi = frac{B}{A} = frac{frac{1}{2}}{frac{sqrt{3}}{2}} = frac{1}{sqrt{3}} ). So, ( phi = arctan(frac{1}{sqrt{3}}) = frac{pi}{6} ).Therefore, ( f(x) = sin(2x + frac{pi}{6}) ). That's a much simpler expression!Now, to find where ( f(x) ) is increasing, I need to find where its derivative is positive. Let's compute the derivative:( f'(x) = frac{d}{dx} sin(2x + frac{pi}{6}) = 2cos(2x + frac{pi}{6}) ).So, ( f'(x) = 2cos(2x + frac{pi}{6}) ). We need to find the intervals where ( f'(x) > 0 ), which is equivalent to ( cos(2x + frac{pi}{6}) > 0 ).The cosine function is positive in the intervals ( (-frac{pi}{2} + 2kpi, frac{pi}{2} + 2kpi) ) for any integer ( k ). So, let's set up the inequality:( -frac{pi}{2} + 2kpi < 2x + frac{pi}{6} < frac{pi}{2} + 2kpi ).Let me solve for ( x ):Subtract ( frac{pi}{6} ) from all parts:( -frac{pi}{2} - frac{pi}{6} + 2kpi < 2x < frac{pi}{2} - frac{pi}{6} + 2kpi ).Simplify the left side:( -frac{2pi}{3} + 2kpi < 2x < frac{pi}{3} + 2kpi ).Divide all parts by 2:( -frac{pi}{3} + kpi < x < frac{pi}{6} + kpi ).So, the function ( f(x) ) is increasing on the intervals ( (kpi - frac{pi}{3}, kpi + frac{pi}{6}) ) for all integers ( k ).Wait, but the question says "intervals of monotonic increase", so I should present these as closed intervals? Because at the endpoints, the derivative is zero, which is neither increasing nor decreasing. So, maybe it's better to write them as closed intervals where the function is non-decreasing. Hmm, but in calculus, usually, when we talk about increasing, we mean strictly increasing, so maybe open intervals. But sometimes, people include the endpoints where the derivative is zero as part of the increasing interval. I think in this case, since the derivative is zero at those points, it's more precise to write them as closed intervals.So, the intervals are ( [kpi - frac{pi}{3}, kpi + frac{pi}{6}] ) for all integers ( k ).Alright, that's part (I) done.Moving on to part (II). This seems more involved. Let me read it again:In triangle ABC, the sides opposite to angles A, B, and C are a, b, and c respectively. It is known that ( f(A) = frac{1}{2} ), and b, a, c form an arithmetic sequence, and ( overrightarrow{AB} cdot overrightarrow{AC} = 9 ). Find the value of a.Okay, so we have triangle ABC with sides a, b, c opposite angles A, B, C respectively.Given:1. ( f(A) = frac{1}{2} ). From part (I), we know that ( f(x) = sin(2x + frac{pi}{6}) ). So, ( sin(2A + frac{pi}{6}) = frac{1}{2} ).2. The sides b, a, c form an arithmetic sequence. So, the sequence is b, a, c, which is arithmetic. That means the difference between consecutive terms is constant. So, ( a - b = c - a ), which simplifies to ( 2a = b + c ).3. The dot product ( overrightarrow{AB} cdot overrightarrow{AC} = 9 ). Hmm, the dot product of vectors AB and AC. I need to recall how to compute that. The dot product of two vectors is equal to the product of their magnitudes times the cosine of the angle between them. So, ( overrightarrow{AB} cdot overrightarrow{AC} = |AB||AC|cos A ). But in triangle ABC, |AB| is the length of side c, |AC| is the length of side b, and the angle between AB and AC is angle A. So, this dot product is ( bc cos A = 9 ).So, summarizing the given information:1. ( sin(2A + frac{pi}{6}) = frac{1}{2} ).2. ( 2a = b + c ).3. ( bc cos A = 9 ).We need to find the value of a.Let me tackle each piece step by step.First, let's find angle A from the equation ( sin(2A + frac{pi}{6}) = frac{1}{2} ).We know that ( sin theta = frac{1}{2} ) has solutions ( theta = frac{pi}{6} + 2kpi ) or ( theta = frac{5pi}{6} + 2kpi ) for integers k.So, ( 2A + frac{pi}{6} = frac{pi}{6} + 2kpi ) or ( 2A + frac{pi}{6} = frac{5pi}{6} + 2kpi ).Let me solve for A in both cases.Case 1: ( 2A + frac{pi}{6} = frac{pi}{6} + 2kpi ).Subtract ( frac{pi}{6} ) from both sides:( 2A = 2kpi ).Divide by 2:( A = kpi ).But in a triangle, angles are between 0 and ( pi ), so possible values are ( A = 0 ) or ( A = pi ). But both are degenerate triangles, so we discard these solutions.Case 2: ( 2A + frac{pi}{6} = frac{5pi}{6} + 2kpi ).Subtract ( frac{pi}{6} ):( 2A = frac{4pi}{6} + 2kpi = frac{2pi}{3} + 2kpi ).Divide by 2:( A = frac{pi}{3} + kpi ).Again, since A is an angle in a triangle, it must be between 0 and ( pi ). So, possible solutions are ( A = frac{pi}{3} ) or ( A = frac{pi}{3} + pi = frac{4pi}{3} ). But ( frac{4pi}{3} ) is more than ( pi ), so it's invalid. Thus, the only valid solution is ( A = frac{pi}{3} ).So, angle A is ( 60^circ ) or ( frac{pi}{3} ) radians.Alright, moving on. We have ( 2a = b + c ). So, sides b, a, c are in arithmetic progression with a as the middle term.Also, ( bc cos A = 9 ). Since ( A = frac{pi}{3} ), ( cos A = frac{1}{2} ). So, substituting, we get:( bc cdot frac{1}{2} = 9 ) => ( bc = 18 ).So, ( bc = 18 ).We need to find a. Let's see what other relationships we can use. Since we have a triangle, the Law of Cosines might be helpful. The Law of Cosines states that:( a^2 = b^2 + c^2 - 2bc cos A ).We know ( A = frac{pi}{3} ), so ( cos A = frac{1}{2} ). So, substituting:( a^2 = b^2 + c^2 - 2bc cdot frac{1}{2} = b^2 + c^2 - bc ).We also know that ( 2a = b + c ). Let me denote ( b + c = 2a ). So, perhaps I can express ( b^2 + c^2 ) in terms of ( (b + c)^2 ).Recall that ( (b + c)^2 = b^2 + 2bc + c^2 ). So, ( b^2 + c^2 = (b + c)^2 - 2bc ).Substituting into the expression for ( a^2 ):( a^2 = (b + c)^2 - 2bc - bc = (b + c)^2 - 3bc ).But ( b + c = 2a ), so:( a^2 = (2a)^2 - 3bc ).Compute ( (2a)^2 = 4a^2 ). So,( a^2 = 4a^2 - 3bc ).Rearranging:( 0 = 3a^2 - 3bc ).Divide both sides by 3:( 0 = a^2 - bc ).So, ( a^2 = bc ).But earlier, we found that ( bc = 18 ). Therefore,( a^2 = 18 ).Taking the square root:( a = sqrt{18} = 3sqrt{2} ).So, the value of a is ( 3sqrt{2} ).Wait, let me double-check the steps to make sure I didn't make any mistakes.1. Found angle A correctly: ( A = frac{pi}{3} ).2. From the arithmetic sequence, ( 2a = b + c ).3. From the dot product, ( bc cos A = 9 ), which gave ( bc = 18 ).4. Applied Law of Cosines: ( a^2 = b^2 + c^2 - bc ).5. Expressed ( b^2 + c^2 ) as ( (b + c)^2 - 2bc ), which is ( (2a)^2 - 2bc ).6. Substituted into the equation: ( a^2 = 4a^2 - 2bc - bc = 4a^2 - 3bc ).7. Plugged ( bc = 18 ): ( a^2 = 4a^2 - 54 ).Wait, hold on, in step 6, I think I made a miscalculation.Wait, let's go back.Law of Cosines: ( a^2 = b^2 + c^2 - 2bc cos A ).Since ( cos A = frac{1}{2} ), this becomes:( a^2 = b^2 + c^2 - bc ).Then, ( b^2 + c^2 = (b + c)^2 - 2bc ).So, ( a^2 = (b + c)^2 - 2bc - bc = (b + c)^2 - 3bc ).Since ( b + c = 2a ), ( (b + c)^2 = 4a^2 ).Thus, ( a^2 = 4a^2 - 3bc ).Bringing terms together:( 0 = 3a^2 - 3bc ).Divide by 3:( 0 = a^2 - bc ).Thus, ( a^2 = bc ).But ( bc = 18 ), so ( a^2 = 18 ), hence ( a = 3sqrt{2} ).Wait, so my initial calculation was correct. So, the value of a is indeed ( 3sqrt{2} ).Just to make sure, let's recap:We had:1. ( f(A) = frac{1}{2} ) leading to ( A = frac{pi}{3} ).2. Arithmetic sequence: ( 2a = b + c ).3. Dot product: ( bc = 18 ).4. Law of Cosines: ( a^2 = b^2 + c^2 - bc ).5. Express ( b^2 + c^2 ) in terms of ( (b + c)^2 ), leading to ( a^2 = 4a^2 - 3bc ).6. Substituted ( bc = 18 ), solved for ( a ), got ( a = 3sqrt{2} ).Everything seems consistent. So, I think that's the correct answer.**Final Answer**Ⅰ. The intervals of monotonic increase for ( f(x) ) are (boxed{[kpi - frac{pi}{3}, kpi + frac{pi}{6}]}) for all integers ( k ).Ⅱ. The value of ( a ) is (boxed{3sqrt{2}}).

💡To determine how many multiples of 5 exist between 105 and 500, I will follow these steps:1. **Identify the first multiple of 5 greater than 105**: The smallest multiple of 5 after 105 is 110.2. **Identify the last multiple of 5 less than or equal to 500**: The largest multiple of 5 before or equal to 500 is 500 itself.3. **List the multiples of 5 between 110 and 500**: These multiples form the sequence: 110, 115, 120, ..., 495, 500.4. **Simplify the sequence by dividing each term by 5**: This gives: 22, 23, 24, ..., 99, 100.5. **Calculate the number of terms in the simplified sequence**: The sequence starts at 22 and ends at 100. The total number of terms is (100 - 22 + 1 = 79).Therefore, there are 79 multiples of 5 between 105 and 500.

💡Okay, so I have this problem here: In triangle ABC, if sin A sin B < cos A cos B, then what is the shape of triangle ABC? Hmm, interesting. I need to figure out whether it's acute, right, or obtuse. Let me think about how to approach this.First, I remember that in a triangle, the sum of angles A, B, and C is π radians, or 180 degrees. So, A + B + C = π. That might come in handy later.The given inequality is sin A sin B < cos A cos B. Hmm, that looks a bit like the formula for cosine of a sum. Let me recall: cos(A + B) = cos A cos B - sin A sin B. Oh! So, if I rearrange that formula, I get cos(A + B) = cos A cos B - sin A sin B, which means that cos(A + B) = cos A cos B - sin A sin B.Wait, so if I move things around, I can write sin A sin B = cos A cos B - cos(A + B). Hmm, but the given inequality is sin A sin B < cos A cos B. So, substituting from the formula, that would mean cos A cos B - cos(A + B) < cos A cos B. If I subtract cos A cos B from both sides, I get -cos(A + B) < 0. Multiplying both sides by -1 (and remembering to flip the inequality sign), I get cos(A + B) > 0.So, cos(A + B) > 0. That tells me something about the angle A + B. Since cosine is positive in the first and fourth quadrants, but since A and B are angles in a triangle, they must each be between 0 and π, so A + B must be between 0 and 2π. But in a triangle, A + B must be less than π because C is also positive. So, A + B is between 0 and π.Therefore, cos(A + B) > 0 implies that A + B is in the first quadrant, meaning A + B < π/2. Wait, no, that's not right. Cosine is positive in the first and fourth quadrants, but since A + B is between 0 and π, cosine is positive when A + B is between 0 and π/2, and negative when A + B is between π/2 and π.Wait, so if cos(A + B) > 0, then A + B must be less than π/2? But that seems too restrictive because in a triangle, A + B = π - C. So, if A + B < π/2, then π - C < π/2, which would mean that C > π/2. So, angle C is greater than π/2, which is 90 degrees, meaning it's obtuse.Wait, hold on, let me double-check that. If A + B < π/2, then C = π - (A + B) > π - π/2 = π/2. So yes, angle C is greater than π/2, which makes triangle ABC an obtuse triangle.But wait, I thought earlier that if cos(A + B) > 0, then A + B is less than π/2. But is that necessarily true? Let me think about the cosine function. Cosine is positive in the first and fourth quadrants, but since A + B is between 0 and π, as I said before, cosine is positive only when A + B is between 0 and π/2. Because once A + B exceeds π/2, cosine becomes negative. So, yes, cos(A + B) > 0 implies A + B < π/2, which in turn implies that angle C is greater than π/2.Therefore, triangle ABC is obtuse.Wait, but let me make sure I didn't make a mistake in the algebra. Starting from sin A sin B < cos A cos B. Then, moving everything to one side: cos A cos B - sin A sin B > 0. But cos A cos B - sin A sin B is equal to cos(A + B). So, cos(A + B) > 0. So, that's correct.Therefore, since cos(A + B) > 0, A + B is less than π/2, so angle C is greater than π/2, making the triangle obtuse.I think that's solid. So, the shape is an obtuse triangle.**Final Answer**The shape of triangle ABC is boxed{text{Obtuse Triangle}}.

💡First, I need to determine the possible length of the third side of a triangle when two sides are given as 4 and 10. I'll use the triangle inequality theorem, which states that the sum of the lengths of any two sides must be greater than the length of the remaining side, and the difference of the lengths of any two sides must be less than the length of the remaining side.Let's denote the length of the third side as ( x ). According to the triangle inequality theorem, I have two inequalities:1. ( x + 4 > 10 )2. ( x + 10 > 4 )3. ( |10 - 4| < x )Solving these inequalities:1. From ( x + 4 > 10 ), subtracting 4 from both sides gives ( x > 6 ).2. The second inequality, ( x + 10 > 4 ), simplifies to ( x > -6 ), which is always true since ( x ) is a positive length.3. From ( |10 - 4| < x ), simplifying the absolute value gives ( 6 < x ).Combining ( 6 < x ) with the fact that ( x ) must also be less than the sum of the other two sides (( 10 + 4 = 14 )), we get ( 6 < x < 14 ).Now, evaluating the given options:- ( A: 5 ) does not satisfy ( 6 < x < 14 ).- ( B: 6 ) does not satisfy ( 6 < x < 14 ) because ( x ) must be strictly greater than 6.- ( C: 11 ) satisfies ( 6 < x < 14 ).- ( D: 16 ) does not satisfy ( 6 < x < 14 ).Therefore, the only possible length for the third side that satisfies the conditions is 11.

💡First, identify the number of apples the chef had before making pies, which is 46 apples.Next, determine the number of apples left after making pies, which is 14 apples.To find the difference between the number of apples before and after making pies, subtract the number of apples left from the original number of apples.So, 46 apples minus 14 apples equals 32 apples.Therefore, the difference is 32 apples.