answer：First, I need to determine the total weight of the pie and the weight of the crust. The pie weighs 200 grams, and the crust accounts for 50 grams of that weight.Next, I'll calculate the weight of the pie that is not crust by subtracting the crust weight from the total weight. This gives me 200 grams minus 50 grams, which equals 150 grams.Then, to find the percentage of the pie that is not crust, I'll divide the non-crust weight by the total weight and multiply by 100. So, 150 grams divided by 200 grams equals 0.75, and multiplying by 100 gives me 75%.Therefore, 75% of the pie is not crust.

answer：Okay, so I have this problem where I need to prove that the product of factorials of some nonnegative integers ( K_1, K_2, ldots, K_n ) is greater than or equal to the factorial of the floor of ( K/n ) raised to the power of ( n ). Here, ( K ) is the sum of all these ( K_i )'s. First, I need to understand what exactly is being asked. We have these integers ( K_1 ) through ( K_n ), all nonnegative, and their sum is ( K ). The inequality we need to prove is:[K_1! cdot K_2! cdot ldots cdot K_n! geq left( leftlfloor frac{K}{n} rightrfloor! right)^n]So, essentially, the product of the factorials of each ( K_i ) should be at least as big as the factorial of the average value of the ( K_i )'s (rounded down) raised to the number of terms ( n ).I remember that factorials grow very rapidly, so the product of factorials might be quite large. Maybe I can use some inequality related to factorials or averages. The Arithmetic Mean-Geometric Mean (AM-GM) inequality comes to mind, but I'm not sure how to apply it directly here since we're dealing with factorials, not just numbers.Let me think about the AM-GM inequality. It states that for nonnegative real numbers, the arithmetic mean is greater than or equal to the geometric mean. Maybe I can relate this to the factorials somehow. If I take the logarithm of both sides, perhaps I can turn the product into a sum, which might make it easier to handle.Taking the natural logarithm of both sides of the inequality:[ln(K_1! cdot K_2! cdot ldots cdot K_n!) geq lnleft( left( leftlfloor frac{K}{n} rightrfloor! right)^n right)]Simplifying the right-hand side:[ln(K_1! cdot K_2! cdot ldots cdot K_n!) geq n cdot lnleft( leftlfloor frac{K}{n} rightrfloor! right)]Now, the left-hand side can be written as the sum of the logarithms:[sum_{i=1}^{n} ln(K_i!) geq n cdot lnleft( leftlfloor frac{K}{n} rightrfloor! right)]So, now the problem is transformed into showing that the average of the logarithms of the factorials is at least the logarithm of the factorial of the average. That is:[frac{1}{n} sum_{i=1}^{n} ln(K_i!) geq lnleft( leftlfloor frac{K}{n} rightrfloor! right)]This looks similar to Jensen's inequality, which applies to convex functions. The logarithm of a factorial is a convex function because the second derivative is positive. Let me recall that Jensen's inequality states that for a convex function ( f ):[frac{1}{n} sum_{i=1}^{n} f(x_i) geq fleft( frac{1}{n} sum_{i=1}^{n} x_i right)]In our case, ( f(x) = ln(x!) ), which is indeed convex. So, applying Jensen's inequality:[frac{1}{n} sum_{i=1}^{n} ln(K_i!) geq lnleft( left( frac{1}{n} sum_{i=1}^{n} K_i right)! right)]But wait, the right-hand side here is ( lnleft( left( frac{K}{n} right)! right) ), whereas in our transformed inequality, we have ( lnleft( leftlfloor frac{K}{n} rightrfloor! right) ). Since ( frac{K}{n} ) might not be an integer, we take the floor of it. I need to ensure that ( lnleft( left( frac{K}{n} right)! right) ) is greater than or equal to ( lnleft( leftlfloor frac{K}{n} rightrfloor! right) ). But actually, ( frac{K}{n} ) is greater than or equal to ( leftlfloor frac{K}{n} rightrfloor ), so ( left( frac{K}{n} right)! ) is greater than ( leftlfloor frac{K}{n} rightrfloor! ). However, since we're taking the logarithm, which is a monotonically increasing function, this implies:[lnleft( left( frac{K}{n} right)! right) geq lnleft( leftlfloor frac{K}{n} rightrfloor! right)]Therefore, the inequality from Jensen's gives us a stronger result than needed. So, combining these, we have:[frac{1}{n} sum_{i=1}^{n} ln(K_i!) geq lnleft( left( frac{K}{n} right)! right) geq lnleft( leftlfloor frac{K}{n} rightrfloor! right)]Which implies:[frac{1}{n} sum_{i=1}^{n} ln(K_i!) geq lnleft( leftlfloor frac{K}{n} rightrfloor! right)]Exponentiating both sides to remove the logarithm, we get back to the original inequality:[K_1! cdot K_2! cdot ldots cdot K_n! geq left( leftlfloor frac{K}{n} rightrfloor! right)^n]So, by applying Jensen's inequality to the convex function ( ln(x!) ), we've established the required inequality. But wait, I should double-check if ( ln(x!) ) is indeed convex. The second derivative of ( ln(x!) ) is the second derivative of the sum ( sum_{k=1}^{x} ln(k) ). The first derivative is ( ln(x) ), and the second derivative is ( frac{1}{x} ), which is positive for ( x > 0 ). So yes, ( ln(x!) ) is convex, which justifies the use of Jensen's inequality.Another way to think about this is by considering the multiplicative version of the AM-GM inequality. Since factorials are multiplicative, maybe there's a way to relate the product of factorials to the factorial of the average. However, I think the approach using Jensen's inequality is more straightforward here.I also wonder if there's a combinatorial interpretation. For example, the product ( K_1! K_2! cdots K_n! ) could represent the number of ways to arrange objects in different groups, while ( leftlfloor frac{K}{n} rightrfloor!^n ) might represent arranging them in a more uniform way. Since arranging objects in groups with more variability tends to give more permutations, this could explain why the product is larger.But I think the analytical approach using Jensen's inequality is sufficient for the proof. It directly applies the convexity property to establish the required inequality.**Final Answer**boxed{K_1!K_2!cdots K_n! ge leftlfloor frac{K}{n} rightrfloor!^n}

answer：Okay, so I have this hyperbola equation: (frac{x^2}{a^2} - frac{y^2}{b^2} = 1), where (a > 0) and (b > 0). The foci are (F_1) and (F_2), which are on the left and right sides respectively. I know that for a hyperbola, the foci are located at ((pm c, 0)), where (c^2 = a^2 + b^2). So, (F_1) is at ((-c, 0)) and (F_2) is at ((c, 0)).Point (A) is on one of the asymptotes of the hyperbola. The asymptotes for this hyperbola are the lines (y = pm frac{b}{a}x). So, point (A) lies somewhere on either (y = frac{b}{a}x) or (y = -frac{b}{a}x).The problem states that (AF_2) is perpendicular to (F_1F_2). Let me visualize this: (F_1F_2) is the line segment connecting the two foci, which lies along the x-axis from ((-c, 0)) to ((c, 0)). So, (F_1F_2) is a horizontal line. If (AF_2) is perpendicular to this, then (AF_2) must be a vertical line.Wait, if (AF_2) is vertical, that means the line from (A) to (F_2) is vertical. Since (F_2) is at ((c, 0)), a vertical line through (F_2) would have the equation (x = c). Therefore, point (A) must lie somewhere on this vertical line (x = c).But point (A) is also on one of the asymptotes. So, let's find the coordinates of point (A). Since (A) is on both the asymptote and the vertical line (x = c), we can substitute (x = c) into the equation of the asymptote.Let's take the positive asymptote (y = frac{b}{a}x) first. Substituting (x = c), we get (y = frac{b}{a}c). So, point (A) is at ((c, frac{bc}{a})). Similarly, if we take the negative asymptote (y = -frac{b}{a}x), substituting (x = c) gives (y = -frac{bc}{a}), so point (A) would be at ((c, -frac{bc}{a})).But since the problem doesn't specify which asymptote, we can consider both possibilities. However, since the hyperbola is symmetric, the result should be the same regardless of which asymptote we choose. So, for simplicity, let's take (A = (c, frac{bc}{a})).Next, we need to find the distance from the origin (O) to the line (AF_1). The origin is at ((0, 0)), and (F_1) is at ((-c, 0)). So, line (AF_1) connects point (A = (c, frac{bc}{a})) to (F_1 = (-c, 0)).To find the distance from the origin to this line, we can use the formula for the distance from a point to a line. First, we need the equation of line (AF_1).Let's find the slope of line (AF_1). The slope (m) is given by:[m = frac{y_2 - y_1}{x_2 - x_1} = frac{0 - frac{bc}{a}}{-c - c} = frac{-frac{bc}{a}}{-2c} = frac{frac{bc}{a}}{2c} = frac{b}{2a}]So, the slope of line (AF_1) is (frac{b}{2a}).Now, using the point-slope form of the equation of a line, we can write the equation of (AF_1). Let's use point (F_1 = (-c, 0)):[y - 0 = frac{b}{2a}(x - (-c)) implies y = frac{b}{2a}(x + c)]So, the equation of line (AF_1) is (y = frac{b}{2a}x + frac{bc}{2a}).We can rewrite this in standard form (Ax + By + C = 0):[frac{b}{2a}x - y + frac{bc}{2a} = 0]Multiplying both sides by (2a) to eliminate the denominators:[b x - 2a y + bc = 0]So, the standard form is (b x - 2a y + bc = 0).Now, the distance (d) from the origin ((0, 0)) to this line is given by the formula:[d = frac{|b(0) - 2a(0) + bc|}{sqrt{b^2 + (-2a)^2}} = frac{|bc|}{sqrt{b^2 + 4a^2}} = frac{bc}{sqrt{b^2 + 4a^2}}]The problem states that this distance is (frac{1}{3}) of (|OF_1|). Let's find (|OF_1|). Since (F_1) is at ((-c, 0)), the distance from the origin to (F_1) is:[|OF_1| = sqrt{(-c - 0)^2 + (0 - 0)^2} = sqrt{c^2} = c]So, the distance from the origin to line (AF_1) is (frac{1}{3}c). Therefore, we have:[frac{bc}{sqrt{b^2 + 4a^2}} = frac{1}{3}c]We can cancel (c) from both sides (assuming (c neq 0), which it isn't because (a) and (b) are positive):[frac{b}{sqrt{b^2 + 4a^2}} = frac{1}{3}]Let's solve for (a) in terms of (b). Multiply both sides by (sqrt{b^2 + 4a^2}):[b = frac{1}{3} sqrt{b^2 + 4a^2}]Multiply both sides by 3:[3b = sqrt{b^2 + 4a^2}]Square both sides:[9b^2 = b^2 + 4a^2]Subtract (b^2) from both sides:[8b^2 = 4a^2]Divide both sides by 4:[2b^2 = a^2]So, (a^2 = 2b^2), which implies (a = bsqrt{2}).Now, recall that the slope of the asymptotes is (pm frac{b}{a}). Since we found (a = bsqrt{2}), the slope is:[frac{b}{a} = frac{b}{bsqrt{2}} = frac{1}{sqrt{2}} = frac{sqrt{2}}{2}]Therefore, the slopes of the asymptotes are (frac{sqrt{2}}{2}) and (-frac{sqrt{2}}{2}).Looking back at the options:A: (sqrt{5}) or (-sqrt{5})B: (sqrt{2}) or (-sqrt{2})C: 1 or -1D: (frac{sqrt{2}}{2}) or (-frac{sqrt{2}}{2})So, the correct answer is D.**Final Answer**boxed{D}

answer：Okay, so I have this geometry problem here, and I need to figure out the distance from the center of a sphere to the plane of a right triangle. The sphere has a radius of 10, and the triangle has sides 8, 15, and 17. Each side of the triangle is tangent to the sphere. Hmm, okay, let me try to visualize this.First, I know that a right triangle with sides 8, 15, and 17 is a Pythagorean triple because 8² + 15² = 64 + 225 = 289, which is 17². So, that checks out. Now, the triangle is situated in 3D space such that each side is tangent to the sphere. That means the sphere is tangent to all three sides of the triangle.I remember that when a sphere is tangent to all sides of a polygon, especially a triangle, the sphere's center must be at a certain distance from the plane of the triangle. I think this distance can be found using the sphere's radius and the inradius of the triangle. Wait, the inradius? Let me recall.The inradius of a triangle is the radius of the largest circle that fits inside the triangle and is tangent to all three sides. In this case, since the sphere is tangent to all three sides, the cross-section of the sphere with the plane of the triangle should be the incircle of the triangle. So, the inradius of the triangle is equal to the radius of the sphere's cross-section in that plane.But wait, the sphere has a radius of 10, and the inradius of the triangle is something else. So, if I can find the inradius of the triangle, I can relate it to the sphere's radius to find the distance from the center of the sphere to the plane of the triangle.Let me calculate the inradius of the triangle. The formula for the inradius (r) of a triangle is:r = A / swhere A is the area of the triangle, and s is the semi-perimeter.First, let's find the area of the triangle. Since it's a right triangle, the area is (base * height) / 2. The legs are 8 and 15, so:A = (8 * 15) / 2 = 60Now, the semi-perimeter (s) is half the sum of all sides:s = (8 + 15 + 17) / 2 = (40) / 2 = 20So, the inradius is:r = 60 / 20 = 3Okay, so the inradius of the triangle is 3. That means the cross-section of the sphere with the plane of the triangle is a circle with radius 3. But the sphere itself has a radius of 10. So, how does this relate to the distance from the center of the sphere to the plane?I think it's related to the Pythagorean theorem. If we consider the sphere's center, the plane of the triangle, and the inradius, we can form a right triangle where one leg is the distance from the center to the plane (let's call this distance d), another leg is the inradius (3), and the hypotenuse is the sphere's radius (10).So, according to the Pythagorean theorem:d² + r² = R²Where:- d is the distance from the center to the plane,- r is the inradius (3),- R is the sphere's radius (10).Plugging in the values:d² + 3² = 10²d² + 9 = 100d² = 100 - 9d² = 91d = √91So, the distance from the center of the sphere to the plane of the triangle is √91.Let me just double-check my reasoning. The sphere is tangent to all three sides of the triangle, so the cross-section in the plane of the triangle is indeed the incircle with radius 3. The sphere's radius is 10, so the distance from the center to the plane is the other leg of the right triangle formed by 10, 3, and d. Using Pythagoras, I get d = √91. That seems right.I don't think I made any mistakes here. The key was recognizing that the sphere's cross-section in the triangle's plane is the incircle, and then applying the Pythagorean theorem to relate the sphere's radius, the inradius, and the distance from the center to the plane.**Final Answer**boxed{D}

answer：First, I need to find the slopes of the two lines to determine the value of ( a ) that makes them perpendicular.For the first line, ( 2x + y + 2 = 0 ), I'll rearrange it into slope-intercept form:[y = -2x - 2]This shows that the slope (( m_1 )) of the first line is (-2).Next, for the second line, ( ax + 4y - 2 = 0 ), I'll also rearrange it into slope-intercept form:[4y = -ax + 2 y = -frac{a}{4}x + frac{1}{2}]This gives the slope (( m_2 )) of the second line as (-frac{a}{4}).Since the lines are perpendicular, the product of their slopes should be (-1):[m_1 times m_2 = -1 (-2) times left(-frac{a}{4}right) = -1 frac{a}{2} = -1 a = -2]Now that I have ( a = -2 ), I can find the intersection point by solving the system of equations:[2x + y + 2 = 0 -2x + 4y - 2 = 0]From the first equation:[y = -2x - 2]Substitute ( y ) into the second equation:[-2x + 4(-2x - 2) - 2 = 0 -2x - 8x - 8 - 2 = 0 -10x - 10 = 0 -10x = 10 x = -1]Substitute ( x = -1 ) back into the equation for ( y ):[y = -2(-1) - 2 = 2 - 2 = 0]Therefore, the intersection point of the two perpendicular lines is ((-1, 0)).

answer：Alright, so I have this problem where I need to plot two quadratic equations on the same set of axes and find where they intersect. The original equation is ( y = ax^2 + bx + c ), and the second equation is obtained by replacing ( x ) with ( 2x ) in the original equation. I'm told that ( b ) is not zero and ( c ) is not zero either. The question is asking where these two graphs intersect, and the options are ( x = 0 ), ( x = -frac{b}{3a} ), both, or neither.First, I need to write down both equations clearly. The original equation is straightforward: ( y = ax^2 + bx + c ). Now, replacing ( x ) with ( 2x ) in this equation should give me the second equation. Let me do that step by step to make sure I don't make a mistake.So, replacing ( x ) with ( 2x ) in ( ax^2 ) would give ( a(2x)^2 ). Calculating that, ( (2x)^2 = 4x^2 ), so that term becomes ( 4ax^2 ). Next, replacing ( x ) with ( 2x ) in the ( bx ) term gives ( b(2x) = 2bx ). The constant term ( c ) remains the same because there's no ( x ) in it. So, putting it all together, the second equation is ( y = 4ax^2 + 2bx + c ).Now, I need to find the points where these two graphs intersect. To do that, I should set the two equations equal to each other since at the points of intersection, the ( y )-values will be the same for the same ( x )-value. So, I set ( ax^2 + bx + c = 4ax^2 + 2bx + c ).Let me write that equation down:( ax^2 + bx + c = 4ax^2 + 2bx + c )Now, I need to solve for ( x ). Let me subtract ( ax^2 + bx + c ) from both sides to get all terms on one side:( 0 = 4ax^2 + 2bx + c - ax^2 - bx - c )Simplifying the right side:First, combine like terms:- ( 4ax^2 - ax^2 = 3ax^2 )- ( 2bx - bx = bx )- ( c - c = 0 )So, the equation simplifies to:( 0 = 3ax^2 + bx )I can factor out an ( x ) from the right side:( 0 = x(3a x + b) )So, this gives me two possible solutions:1. ( x = 0 )2. ( 3a x + b = 0 ) which leads to ( x = -frac{b}{3a} )So, it seems like there are two points of intersection: one at ( x = 0 ) and another at ( x = -frac{b}{3a} ).But wait, let me double-check to make sure I didn't make any mistakes in my algebra. Starting from the beginning:Original equation: ( y = ax^2 + bx + c )After replacing ( x ) with ( 2x ): ( y = a(2x)^2 + b(2x) + c = 4ax^2 + 2bx + c )Setting them equal: ( ax^2 + bx + c = 4ax^2 + 2bx + c )Subtracting ( ax^2 + bx + c ) from both sides:( 0 = 3ax^2 + bx )Factoring out ( x ):( 0 = x(3a x + b) )So, yes, that seems correct. Therefore, the solutions are ( x = 0 ) and ( x = -frac{b}{3a} ).But just to be thorough, maybe I should plug these ( x )-values back into the original equations to find the corresponding ( y )-values and ensure that they are indeed the same for both equations.First, let's check ( x = 0 ):Plugging into the original equation: ( y = a(0)^2 + b(0) + c = c )Plugging into the second equation: ( y = 4a(0)^2 + 2b(0) + c = c )So, both give ( y = c ), which means the point ( (0, c) ) is indeed a point of intersection.Now, let's check ( x = -frac{b}{3a} ):Plugging into the original equation:( y = aleft(-frac{b}{3a}right)^2 + bleft(-frac{b}{3a}right) + c )Calculating each term:First term: ( aleft(frac{b^2}{9a^2}right) = frac{b^2}{9a} )Second term: ( -frac{b^2}{3a} )Third term: ( c )So, combining these:( y = frac{b^2}{9a} - frac{b^2}{3a} + c )To combine the fractions, I need a common denominator, which is ( 9a ):( y = frac{b^2}{9a} - frac{3b^2}{9a} + c = -frac{2b^2}{9a} + c )Now, plugging ( x = -frac{b}{3a} ) into the second equation:( y = 4aleft(-frac{b}{3a}right)^2 + 2bleft(-frac{b}{3a}right) + c )Calculating each term:First term: ( 4aleft(frac{b^2}{9a^2}right) = frac{4b^2}{9a} )Second term: ( -frac{2b^2}{3a} )Third term: ( c )Combining these:( y = frac{4b^2}{9a} - frac{2b^2}{3a} + c )Again, finding a common denominator for the fractions:( y = frac{4b^2}{9a} - frac{6b^2}{9a} + c = -frac{2b^2}{9a} + c )So, both equations give the same ( y )-value at ( x = -frac{b}{3a} ), which confirms that this is indeed another point of intersection.Therefore, the two graphs intersect at ( x = 0 ) and ( x = -frac{b}{3a} ). So, the correct answer should include both options A and B.But just to make sure I haven't overlooked anything, let me think about the nature of quadratic equations. Quadratic equations can intersect at up to two points. In this case, since both are parabolas, they can intersect at two points, one point, or no points depending on their shapes and positions.In our case, since we've found two distinct solutions for ( x ), it means the two parabolas intersect at two distinct points. Therefore, both ( x = 0 ) and ( x = -frac{b}{3a} ) are valid points of intersection.I also want to consider if there's any possibility that these solutions might not be valid under certain conditions. For example, if ( a = 0 ), the equation would no longer be quadratic, but the problem states ( b neq 0 ) and ( c neq 0 ), but doesn't specify anything about ( a ). However, since the equation is given as ( y = ax^2 + bx + c ), I assume ( a neq 0 ) because otherwise, it wouldn't be a quadratic equation.Also, since ( b neq 0 ), the solution ( x = -frac{b}{3a} ) is a valid, non-zero value. If ( b = 0 ), then the second solution would be undefined, but since ( b neq 0 ), we're safe.Another thing to consider is whether these points are distinct. If ( -frac{b}{3a} = 0 ), then both solutions would be the same. But since ( b neq 0 ), ( -frac{b}{3a} ) cannot be zero, so the two points of intersection are indeed distinct.Therefore, after carefully going through the algebra and verifying the solutions, I'm confident that the two graphs intersect at both ( x = 0 ) and ( x = -frac{b}{3a} ).