💡Alright, so I have this problem about a cone. The total surface area is given as πS square units. When the lateral surface is unfolded, it forms a sector with a 60-degree angle. I need to find the volume of the cone. Hmm, okay, let me think step by step.First, I remember that the lateral surface area of a cone, when unfolded, becomes a sector of a circle. The radius of this sector is equal to the slant height (l) of the cone. The arc length of this sector should be equal to the circumference of the base of the cone. That makes sense because when you roll the sector back into a cone, the arc becomes the circumference.So, the circumference of the base of the cone is 2πR, where R is the radius of the base. The sector has a central angle of 60 degrees, which is π/3 radians. The length of the arc of the sector can also be calculated using the formula for arc length: θl, where θ is in radians and l is the radius of the sector (which is the slant height of the cone).Putting that together, the arc length is θl = (π/3)l. This should equal the circumference of the base, which is 2πR. So, I can write the equation:(π/3)l = 2πRSimplifying this, I can divide both sides by π:(1/3)l = 2RThen multiply both sides by 3:l = 6ROkay, so the slant height l is six times the radius R. Got that.Next, I remember that in a cone, the slant height l, the radius R, and the height H form a right triangle. So, by the Pythagorean theorem:l² = R² + H²We already know that l = 6R, so substituting that in:(6R)² = R² + H²Which simplifies to:36R² = R² + H²Subtract R² from both sides:35R² = H²Taking the square root of both sides:H = R√35Alright, so the height H is R times the square root of 35.Now, the total surface area of the cone is given as πS. The total surface area of a cone is the sum of the base area and the lateral surface area. The base area is πR², and the lateral surface area is πRl.So, total surface area:πS = πR² + πRlWe can factor out πR:πS = πR(R + l)But we already know that l = 6R, so substituting that in:πS = πR(R + 6R) = πR(7R) = 7πR²So, πS = 7πR²Divide both sides by π:S = 7R²Then, solving for R²:R² = S/7So, R = √(S/7)Now, we can find H since H = R√35:H = √(S/7) * √35 = √(35S/7) = √(5S)So, H = √(5S)Now, to find the volume of the cone. The formula for the volume is:V = (1/3)πR²HWe have R² = S/7 and H = √(5S). Plugging these into the volume formula:V = (1/3)π(S/7)(√(5S))Simplify this:V = (1/3)(π)(S/7)(√5 * √S) = (π S √5 √S) / 21Since √S * √S = S, but here it's √5 * √S, so it's √(5S). Wait, no, actually, √5 is a constant, and √S is separate. So, it's π S √5 √S / 21, which can be written as π S √(5S) / 21.Wait, hold on, √5 * √S is √(5S), right? Because √a * √b = √(ab). So, yes, √5 * √S = √(5S). So, the volume is:V = (π S √(5S)) / 21So, that's the volume of the cone.Let me just double-check my steps to make sure I didn't make any mistakes.1. Found that l = 6R from the sector arc length.2. Used Pythagorean theorem to find H = R√35.3. Calculated total surface area and solved for R² = S/7.4. Substituted R and H into the volume formula.Everything seems to check out. So, I think that's the correct volume.**Final Answer**The volume of the cone is boxed{dfrac{pi S sqrt{5S}}{21}}.

💡Okay, so I've got this math problem here, and I need to simplify it. Let me take a deep breath and look at it step by step. The expression is:frac{2(a+(a+1)+(a+2)+ldots+2a)}{a^{2}+3a+2}+frac{6left(a^{1/2}+b^{1/2}right)}{(a-b)^{0.6}(a+2)}:left(left(a^{1/2}-b^{1/2}right)(a-b)^{-2/5}right)^{-1}Alright, that's a bit intimidating, but let's break it down into smaller parts. I'll start with the first fraction:frac{2(a+(a+1)+(a+2)+ldots+2a)}{a^{2}+3a+2}Hmm, the numerator is a sum of consecutive integers from ( a ) to ( 2a ). I remember that the sum of an arithmetic series can be calculated using the formula ( S = frac{n}{2}(first term + last term) ), where ( n ) is the number of terms.So, let's figure out how many terms there are. Starting at ( a ) and ending at ( 2a ), with each term increasing by 1. The number of terms ( n ) is ( 2a - a + 1 = a + 1 ).Now, applying the sum formula:S = frac{a + 1}{2} times (a + 2a) = frac{a + 1}{2} times 3a = frac{3a(a + 1)}{2}So, the numerator becomes:2 times frac{3a(a + 1)}{2} = 3a(a + 1)Now, the denominator is ( a^2 + 3a + 2 ). Let me factor that:a^2 + 3a + 2 = (a + 1)(a + 2)So, the first fraction simplifies to:frac{3a(a + 1)}{(a + 1)(a + 2)} = frac{3a}{a + 2}Alright, that wasn't too bad. Now, moving on to the second part of the expression:frac{6left(a^{1/2}+b^{1/2}right)}{(a-b)^{0.6}(a+2)}:left(left(a^{1/2}-b^{1/2}right)(a-b)^{-2/5}right)^{-1}Hmm, the colon here probably means division. So, it's like dividing the first fraction by the second part. Let me rewrite it:frac{6left(sqrt{a} + sqrt{b}right)}{(a - b)^{0.6}(a + 2)} div left( left( sqrt{a} - sqrt{b} right) (a - b)^{-2/5} right)^{-1}Dividing by something is the same as multiplying by its reciprocal. So, flipping the denominator:frac{6left(sqrt{a} + sqrt{b}right)}{(a - b)^{0.6}(a + 2)} times left( sqrt{a} - sqrt{b} right) (a - b)^{2/5}Now, let's simplify the exponents. ( (a - b)^{0.6} ) is the same as ( (a - b)^{3/5} ) because 0.6 is 3/5. So, we have:frac{6left(sqrt{a} + sqrt{b}right)}{(a - b)^{3/5}(a + 2)} times left( sqrt{a} - sqrt{b} right) (a - b)^{2/5}Multiplying the terms with the same base, ( (a - b)^{3/5} times (a - b)^{2/5} = (a - b)^{5/5} = (a - b)^1 = (a - b) ).So now, the expression becomes:frac{6left(sqrt{a} + sqrt{b}right)left(sqrt{a} - sqrt{b}right)}{(a - b)(a + 2)}I remember that ( (sqrt{a} + sqrt{b})(sqrt{a} - sqrt{b}) = a - b ). So, substituting that in:frac{6(a - b)}{(a - b)(a + 2)} = frac{6}{a + 2}Great, so the second part simplifies to ( frac{6}{a + 2} ).Now, putting it all together, the original expression becomes:frac{3a}{a + 2} + frac{6}{a + 2}Since both terms have the same denominator, I can combine them:frac{3a + 6}{a + 2} = frac{3(a + 2)}{a + 2} = 3Wait, that simplifies nicely to just 3. So, regardless of the values of ( a ) and ( b ) (as long as they satisfy the initial conditions), the entire expression simplifies to 3.Let me just double-check my steps to make sure I didn't make any mistakes. Starting from the sum, factoring, simplifying exponents, and then combining everything. It all seems to add up. I think I'm confident that the final answer is 3.**Final Answer**boxed{3}

💡First, I need to determine Jungkook's total score for Korean, math, and science. Since his average score is 92 points across these three subjects, I can calculate the total score by multiplying the average by the number of subjects.Next, I need to find out the new average score after including the English test. The average increased by 2 points, making the new average 94 points. With the inclusion of the English test, there are now four subjects. I will calculate the new total score by multiplying the new average by the total number of subjects.Finally, to find Jungkook's English score, I will subtract the original total score (for Korean, math, and science) from the new total score (which includes English). This difference will give me Jungkook's score in English.

💡Okay, so I have this problem: For how many non-negative real values of ( x ) is ( sqrt{196 - sqrt[3]{x}} ) an integer? Hmm, let me try to figure this out step by step.First, I know that the square root of something is an integer. That means the expression inside the square root, which is ( 196 - sqrt[3]{x} ), must be a perfect square. Because if the square root is an integer, say ( n ), then ( n^2 ) is the inside expression.So, let me write that down: ( sqrt{196 - sqrt[3]{x}} = n ), where ( n ) is a non-negative integer. Squaring both sides, I get ( 196 - sqrt[3]{x} = n^2 ). Now, I can solve for ( sqrt[3]{x} ). Subtracting ( n^2 ) from both sides, I have ( sqrt[3]{x} = 196 - n^2 ). To find ( x ), I need to cube both sides of this equation, right? So, ( x = (196 - n^2)^3 ).Okay, so for each integer value of ( n ), I can find a corresponding ( x ). But I need to make sure that ( x ) is a non-negative real number, which it will be as long as ( 196 - n^2 ) is non-negative because the cube of a real number is real, and if ( 196 - n^2 ) is negative, ( x ) would still be real, but since we're dealing with non-negative ( x ), maybe we need to ensure ( 196 - n^2 ) is non-negative? Wait, actually, ( sqrt[3]{x} ) can be negative if ( x ) is negative, but the problem specifies non-negative ( x ), so ( sqrt[3]{x} ) must also be non-negative. Therefore, ( 196 - n^2 ) must be greater than or equal to zero.So, ( 196 - n^2 geq 0 ). That means ( n^2 leq 196 ). Taking square roots on both sides, ( n leq sqrt{196} ). Since ( sqrt{196} = 14 ), ( n ) must be less than or equal to 14. But ( n ) is a non-negative integer, so ( n ) can be 0, 1, 2, ..., up to 14.Therefore, the possible integer values for ( n ) are 0, 1, 2, ..., 14. That's 15 different values in total. For each of these ( n ), there's a corresponding ( x ) given by ( x = (196 - n^2)^3 ).Let me double-check to make sure I didn't miss anything. If ( n = 0 ), then ( x = (196 - 0)^3 = 196^3 ), which is a valid non-negative real number. If ( n = 14 ), then ( x = (196 - 14^2)^3 = (196 - 196)^3 = 0^3 = 0 ), which is also valid. For each integer ( n ) in between, ( 196 - n^2 ) is positive, so ( x ) is positive as well.Is there any chance that different ( n ) could lead to the same ( x )? Let me see. Suppose ( n_1 ) and ( n_2 ) are two different integers such that ( (196 - n_1^2)^3 = (196 - n_2^2)^3 ). Then, ( 196 - n_1^2 = 196 - n_2^2 ), which implies ( n_1^2 = n_2^2 ). Since ( n_1 ) and ( n_2 ) are non-negative integers, this would mean ( n_1 = n_2 ). So, each ( n ) gives a unique ( x ).Therefore, the number of non-negative real values of ( x ) is equal to the number of integers ( n ) from 0 to 14, inclusive. That's 15 values.Wait, just to make sure, let me count them: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14. Yep, that's 15 numbers.So, I think the answer is 15.**Final Answer**The number of non-negative real values of ( x ) is boxed{15}.

💡Alright, let's tackle this problem step by step. So, we have a parabola C defined by the equation ( y^2 = 4x ), and its focus is F. There's also a line l given by ( y = frac{2}{3}(x + 2) ) that intersects the parabola at points M and N. We need to find the value of the dot product ( overrightarrow{FM} cdot overrightarrow{FN} ).First, I remember that for a parabola in the form ( y^2 = 4ax ), the focus is at ( (a, 0) ). In this case, comparing ( y^2 = 4x ) with the standard form, we see that ( 4a = 4 ), so ( a = 1 ). Therefore, the focus F is at ( (1, 0) ).Next, let's look at the line ( l: y = frac{2}{3}(x + 2) ). This line intersects the parabola at points M and N. To find these points of intersection, I need to solve the system of equations:1. ( y^2 = 4x )2. ( y = frac{2}{3}(x + 2) )I can substitute the expression for y from the second equation into the first equation. Let's do that:( left( frac{2}{3}(x + 2) right)^2 = 4x )Expanding the left side:( frac{4}{9}(x + 2)^2 = 4x )Multiply both sides by 9 to eliminate the denominator:( 4(x + 2)^2 = 36x )Divide both sides by 4:( (x + 2)^2 = 9x )Expand ( (x + 2)^2 ):( x^2 + 4x + 4 = 9x )Bring all terms to one side:( x^2 + 4x + 4 - 9x = 0 )Simplify:( x^2 - 5x + 4 = 0 )Now, factor the quadratic equation:( (x - 1)(x - 4) = 0 )So, the solutions are ( x = 1 ) and ( x = 4 ).Now, let's find the corresponding y-values for these x-values using the equation of the line ( y = frac{2}{3}(x + 2) ).For ( x = 1 ):( y = frac{2}{3}(1 + 2) = frac{2}{3}(3) = 2 )So, one point of intersection is ( M(1, 2) ).For ( x = 4 ):( y = frac{2}{3}(4 + 2) = frac{2}{3}(6) = 4 )So, the other point of intersection is ( N(4, 4) ).Now, we have points M(1, 2) and N(4, 4), and the focus F(1, 0). We need to find the vectors ( overrightarrow{FM} ) and ( overrightarrow{FN} ) and then compute their dot product.First, let's find ( overrightarrow{FM} ). This vector goes from F(1, 0) to M(1, 2). So, the components are:( overrightarrow{FM} = (1 - 1, 2 - 0) = (0, 2) )Next, let's find ( overrightarrow{FN} ). This vector goes from F(1, 0) to N(4, 4). So, the components are:( overrightarrow{FN} = (4 - 1, 4 - 0) = (3, 4) )Now, we need to compute the dot product of these two vectors. The dot product of two vectors ( (a, b) ) and ( (c, d) ) is given by ( ac + bd ).So, let's compute:( overrightarrow{FM} cdot overrightarrow{FN} = (0)(3) + (2)(4) = 0 + 8 = 8 )Therefore, the value of ( overrightarrow{FM} cdot overrightarrow{FN} ) is 8.**Final Answer**boxed{8}

💡Alright, so I've got this problem here about arranging acts in a performance. Let me try to understand it step by step. Okay, there are originally 4 scheduled acts, and someone wants to add 2 more comedy skits. The catch is that the original sequence of those 4 acts has to stay the same. So, I guess the order of the first 4 acts can't be changed, but we can insert these 2 new skits somewhere in the lineup. The question is asking how many different ways we can arrange these 6 acts under these conditions.Hmm, so I need to figure out how many different arrangements are possible. The options given are A: 20, B: 25, C: 30, D: 32. I need to figure out which one is correct.Let me think about how to approach this. Since the original 4 acts must remain in their original order, but we can insert the 2 new skits anywhere, including at the beginning or end. So, it's like we're adding 2 new elements into a sequence of 4, without changing the order of the original 4.I remember something about permutations with restrictions. Maybe it's similar to arranging items where some are identical or have to stay in a certain order. But in this case, the original acts are distinct, and the new skits are also distinct, I assume.Wait, no, actually, the problem doesn't specify whether the skits are identical or different. Hmm, that's important. If the skits are identical, the number of arrangements would be different than if they're distinct.Let me check the problem statement again. It says "add 2 more comedy skits." It doesn't specify if they're different or the same. Hmm, in most cases, unless specified, I think they're considered distinct. So, I'll proceed under the assumption that the two skits are different.So, we have 4 original acts, let's call them A, B, C, D, in that order, and we need to insert two new skits, let's say X and Y, into the lineup. The original order of A, B, C, D must remain unchanged.So, how do we calculate the number of ways to insert X and Y into the sequence?I think this is a problem of inserting two elements into a sequence of four, maintaining the original order. The number of ways to do this is similar to choosing positions for the new elements among the existing ones.In combinatorics, when you have n items and you want to insert k new items, the number of ways is given by the combination formula C(n + k, k). But wait, is that correct?Wait, no, that's when you're choosing positions without considering the order of the new items. But since the new skits are distinct, the order matters. So, it's actually a permutation problem.Let me think again. If we have 4 original acts, there are 5 possible gaps where we can insert the new skits: before A, between A and B, between B and C, between C and D, and after D. So, 5 gaps.We need to insert 2 skits into these 5 gaps. Since the skits are distinct, the order in which we insert them matters.So, for the first skit, we have 5 choices of gaps. For the second skit, since we've already inserted one skit, there are now 6 gaps (because inserting a skit creates a new gap). So, for the second skit, we have 6 choices.Therefore, the total number of arrangements would be 5 * 6 = 30.Wait, but hold on. Is that correct? Because if we insert both skits into the same gap, they would be next to each other, but since they're distinct, their order matters. So, inserting X before Y in the same gap is different from inserting Y before X.But in the calculation above, when we choose 5 gaps for the first skit and 6 for the second, we are already accounting for the order because choosing gap 1 for X and gap 2 for Y is different from choosing gap 2 for X and gap 1 for Y.Wait, no, actually, if we insert both skits into the same gap, the number of gaps doesn't increase. So, if we insert both skits into the same gap, there are 5 choices for the gap, and then 2! ways to arrange the skits within that gap.Alternatively, if we insert them into different gaps, then we have 5 choices for the first skit and 4 remaining choices for the second skit, but since the order matters, it's 5 * 4 * 2! ?Wait, I'm getting confused. Let me try a different approach.The total number of ways to insert 2 distinct skits into 5 gaps is equal to the number of ways to choose positions for the skits, considering that they can be in the same gap or different gaps, and that the order matters.This is similar to arranging 2 distinct items into 5 distinct boxes, where each box can hold any number of items, and the order within the boxes matters.In combinatorics, this is calculated as P(n + k - 1, k), where n is the number of gaps and k is the number of items. So, P(5 + 2 - 1, 2) = P(6, 2) = 6 * 5 = 30.Yes, that matches my earlier calculation. So, the total number of arrangements is 30.But wait, let me double-check. If the skits are identical, the number would be C(5 + 2 - 1, 2) = C(6, 2) = 15. But since they're distinct, it's 30.Looking back at the options, C is 30, so that should be the answer.But just to make sure, let's think of a smaller example. Suppose we have 2 original acts, A and B, and we want to insert 1 skit, X. The number of arrangements should be 3: XAB, AXB, ABX.If we have 2 skits, X and Y, the number of arrangements would be:- Both skits before A: XYAB, YXAB- One skit before A and one between A and B: XAYB, YABX- One skit before A and one after B: XABY, YABX- One skit between A and B and one after B: AXYB, AYXB- Both skits after B: ABXY, ABYXWait, that's 2 + 2 + 2 + 2 + 2 = 10 arrangements.But according to the formula, n = 2 original acts, so 3 gaps. Inserting 2 distinct skits: P(3 + 2 -1, 2) = P(4, 2) = 4 * 3 = 12. But in reality, we have 10 arrangements. Hmm, discrepancy here.Wait, maybe my formula is incorrect. Let me recount the arrangements:1. XYAB2. YXAB3. XAYB4. YABX5. XABY6. YABX (duplicate)7. AXYB8. AYXB9. ABXY10. ABYXWait, I think I duplicated YABX. So, actually, there are 9 unique arrangements.But according to the formula, it should be 12. Hmm, that's a problem.Wait, maybe my initial assumption about the formula is wrong. Let me think differently.When inserting k distinct items into n gaps, the number of ways is n^k, because for each item, you have n choices of gaps. But in our case, n = 5 gaps, k = 2 skits, so 5^2 = 25.But earlier, I thought it was 30. Hmm, now I'm confused.Wait, no, because when inserting the first skit, you have 5 choices, and then inserting the second skit, you have 6 choices (since inserting the first skit creates a new gap). So, it's 5 * 6 = 30.But in the smaller example, with n = 2 acts, 3 gaps, inserting 2 skits, it would be 3 * 4 = 12, but we only have 9 unique arrangements.So, there's a discrepancy here. Maybe the formula isn't directly applicable.Alternatively, perhaps the correct way is to think of it as arranging the 6 acts where the original 4 are in a fixed order, and the 2 skits can be placed anywhere.This is similar to permutations of multiset. The formula for permutations of a multiset is:Total permutations = (n + k)! / (n! * k!)But in this case, the original 4 acts are distinct and must stay in order, and the 2 skits are distinct.Wait, actually, the problem is similar to arranging 6 items where 4 are of one type (must stay in order) and 2 are of another type (distinct). But I'm not sure.Alternatively, think of it as choosing positions for the skits among the 6 slots, and then arranging the skits in those positions.There are C(6, 2) ways to choose positions for the skits, and then 2! ways to arrange the skits in those positions. So, total arrangements = C(6, 2) * 2! = 15 * 2 = 30.Yes, that makes sense. So, the total number of arrangements is 30.In the smaller example with 2 acts and 2 skits, it would be C(4, 2) * 2! = 6 * 2 = 12, but earlier I only found 9 arrangements. Hmm, maybe I missed some arrangements.Wait, let's recount with 2 acts, A and B, and 2 skits, X and Y.The possible arrangements are:1. X Y A B2. X A Y B3. X A B Y4. Y X A B5. Y A X B6. Y A B X7. A X Y B8. A X B Y9. A Y X B10. A Y B X11. A B X Y12. A B Y XOh, I see, I missed some arrangements earlier. So, there are indeed 12 arrangements, matching the formula. So, the formula works.Therefore, in the original problem, with 4 acts and 2 skits, the number of arrangements is C(6, 2) * 2! = 15 * 2 = 30.So, the answer should be C: 30.