answer：First, I need to determine Gillian's usual monthly phone bill. Let's denote this amount as ( x ).If the monthly bill increases by 10%, the new monthly bill becomes ( 1.10x ).Given that the total phone bill for the entire next year after the increase is 660, I can set up the equation:[12 times 1.10x = 660]To find ( x ), I'll first divide both sides of the equation by 12:[1.10x = frac{660}{12} = 55]Next, I'll solve for ( x ) by dividing both sides by 1.10:[x = frac{55}{1.10} = 50]Therefore, Gillian's usual monthly phone bill is 50.

answer：First, I need to determine the total area of the sides of the box that Nathan wants to line with velvet. The box has two long sides and two short sides.For the long sides, each measures 8 inches by 6 inches. Therefore, the area of one long side is:8 inches * 6 inches = 48 square inchesSince there are two long sides, the total area for the long sides is:2 * 48 square inches = 96 square inchesNext, for the short sides, each measures 5 inches by 6 inches. The area of one short side is:5 inches * 6 inches = 30 square inchesWith two short sides, the total area for the short sides is:2 * 30 square inches = 60 square inchesAdding the areas of the long and short sides together gives:96 square inches + 60 square inches = 156 square inchesNathan needs a total of 236 square inches of velvet. To find the combined area of the top and bottom of the box, I subtract the area of the sides from the total area required:236 square inches - 156 square inches = 80 square inchesSince the top and bottom are equal in area, I divide the remaining area by 2 to find the area of one of them:80 square inches / 2 = 40 square inchesTherefore, the area of the top and bottom of the box is 40 square inches each.

answer：Okay, so I have this equation here: the square root of (x plus 16) equals 5. Hmm, I need to find the value of x that makes this true. Alright, let's break this down step by step.First, I remember that when you have a square root equation like this, a good strategy is to get rid of the square root to make it easier to solve for x. How do I do that? Oh right, I can square both sides of the equation. That should eliminate the square root on the left side.So, if I square both sides, what happens? On the left side, squaring the square root of something should just give me that something. So, squaring the left side, sqrt(x + 16), would give me x + 16. On the right side, squaring 5 gives me 25. So now my equation is x + 16 = 25.Alright, now I have a simpler equation: x plus 16 equals 25. To solve for x, I need to isolate it. That means I should get rid of the 16 that's being added to x. How do I do that? I subtract 16 from both sides of the equation. Subtracting 16 from the left side, x + 16 minus 16 is just x. On the right side, 25 minus 16 is 9. So, x equals 9. Wait, let me double-check that. If I plug x = 9 back into the original equation, does it hold true? Let's see. The original equation is sqrt(x + 16) = 5. Plugging in 9, we get sqrt(9 + 16) which is sqrt(25), and that's 5. Yep, that works out. So, x = 9 is the correct solution.I think that's it. I squared both sides to eliminate the square root, solved for x by subtracting 16, and then checked my answer by plugging it back into the original equation. Everything seems to make sense, and the check confirms that x = 9 is indeed the right answer.

answer：Alright, so I've got this geometry problem here. Let me try to parse it step by step. The problem says:"Given that point ( P ) lies on the line ( 3x + 4y + 8 = 0 ), and ( PA ) and ( PB ) are the two tangents drawn from ( P ) to the circle ( x^2 + y^2 - 2x - 2y + 1 = 0 ). Let ( A ) and ( B ) be the points of tangency, and ( C ) be the center of the circle. Find the minimum possible area of the quadrilateral ( PACB )."Okay, so I need to find the minimum area of quadrilateral ( PACB ) where ( P ) is on the given line, and ( PA ) and ( PB ) are tangents to the given circle. Let me break this down.First, I should probably figure out the center and radius of the given circle. The equation is ( x^2 + y^2 - 2x - 2y + 1 = 0 ). I remember that the general form of a circle is ( x^2 + y^2 + Dx + Ey + F = 0 ), and the center is at ( (-D/2, -E/2) ) with radius ( sqrt{(D/2)^2 + (E/2)^2 - F} ).So, for this circle, ( D = -2 ), ( E = -2 ), and ( F = 1 ). Therefore, the center ( C ) is at ( (-(-2)/2, -(-2)/2) = (1, 1) ). The radius ( r ) is ( sqrt{(1)^2 + (1)^2 - 1} = sqrt{1 + 1 - 1} = sqrt{1} = 1 ). Got that down: center at (1,1), radius 1.Next, point ( P ) lies on the line ( 3x + 4y + 8 = 0 ). I need to consider all possible points ( P ) on this line and find the one that minimizes the area of quadrilateral ( PACB ).Quadrilateral ( PACB ) is formed by points ( P ), ( A ), ( C ), and ( B ). Since ( PA ) and ( PB ) are tangents from ( P ) to the circle, points ( A ) and ( B ) lie on the circle, and ( PA = PB ) because the lengths of tangents from a common external point to a circle are equal.So, quadrilateral ( PACB ) is a kite because two pairs of adjacent sides are equal (( PA = PB ) and ( AC = BC ), since both are radii of the circle). The area of a kite can be calculated in a couple of ways, but since we have right angles at ( A ) and ( B ) (because the radius is perpendicular to the tangent at the point of contact), the quadrilateral can be split into two congruent right triangles: ( PAC ) and ( PBC ).Therefore, the area of ( PACB ) is twice the area of triangle ( PAC ). The area of triangle ( PAC ) is ( frac{1}{2} times PA times AC ). Since ( AC ) is the radius, which is 1, and ( PA ) is the length of the tangent from ( P ) to the circle.So, the area of quadrilateral ( PACB ) is ( 2 times frac{1}{2} times PA times 1 = PA times 1 = PA ). Therefore, to minimize the area of ( PACB ), I need to minimize the length of the tangent ( PA ).Wait, is that right? If the area is equal to ( PA ), then minimizing ( PA ) would minimize the area. Hmm, let me think. Alternatively, maybe I should express the area in terms of other variables.Alternatively, the area can be thought of as the product of the lengths of the diagonals multiplied by the sine of the angle between them divided by two. But in this case, since it's a kite with two right angles, maybe the first approach is better.But let me double-check. If quadrilateral ( PACB ) is a kite with two right triangles, each with legs ( PA ) and ( AC ), then each triangle has area ( frac{1}{2} times PA times AC ). So, total area is ( 2 times frac{1}{2} times PA times AC = PA times AC ). Since ( AC = 1 ), the area is indeed ( PA times 1 = PA ). So, minimizing the area is equivalent to minimizing ( PA ).Therefore, I need to find the point ( P ) on the line ( 3x + 4y + 8 = 0 ) such that the length of the tangent from ( P ) to the circle is minimized.I remember that the length of the tangent from a point ( (x_1, y_1) ) to a circle with center ( (h, k) ) and radius ( r ) is given by ( sqrt{(x_1 - h)^2 + (y_1 - k)^2 - r^2} ). So, in this case, the length ( PA ) is ( sqrt{(x_P - 1)^2 + (y_P - 1)^2 - 1^2} ).Therefore, to minimize ( PA ), I need to minimize ( sqrt{(x_P - 1)^2 + (y_P - 1)^2 - 1} ). Since the square root is a monotonically increasing function, minimizing the expression inside the square root will minimize the entire expression. So, I can instead minimize ( (x_P - 1)^2 + (y_P - 1)^2 - 1 ).But ( (x_P - 1)^2 + (y_P - 1)^2 ) is the square of the distance from ( P ) to the center ( C(1,1) ). So, to minimize ( PA ), I need to minimize the distance from ( P ) to ( C ), because ( PA = sqrt{PC^2 - r^2} ). So, the smaller ( PC ) is, the smaller ( PA ) will be.Therefore, the minimal ( PA ) occurs when ( PC ) is minimized. So, I need to find the point ( P ) on the line ( 3x + 4y + 8 = 0 ) that is closest to the center ( C(1,1) ). The minimal distance from ( C ) to the line is the perpendicular distance.I remember the formula for the distance from a point ( (x_0, y_0) ) to the line ( ax + by + c = 0 ) is ( frac{|ax_0 + by_0 + c|}{sqrt{a^2 + b^2}} ).So, plugging in the values, the distance ( d ) from ( C(1,1) ) to the line ( 3x + 4y + 8 = 0 ) is:( d = frac{|3(1) + 4(1) + 8|}{sqrt{3^2 + 4^2}} = frac{|3 + 4 + 8|}{5} = frac{15}{5} = 3 ).So, the minimal distance from ( C ) to the line is 3 units. Therefore, the minimal ( PC ) is 3, and the minimal ( PA ) is ( sqrt{PC^2 - r^2} = sqrt{3^2 - 1^2} = sqrt{9 - 1} = sqrt{8} = 2sqrt{2} ).Therefore, the minimal area of quadrilateral ( PACB ) is equal to ( PA times 1 = 2sqrt{2} times 1 = 2sqrt{2} ).Wait, hold on. Earlier, I thought the area was equal to ( PA ), but actually, the area is ( 2 times frac{1}{2} times PA times AC ). So, that is ( PA times AC ). Since ( AC = 1 ), it's ( PA times 1 = PA ). So, yes, the area is equal to ( PA ). Therefore, the minimal area is ( 2sqrt{2} ).But let me double-check my reasoning. If ( PA ) is minimized, then the area is minimized. But is there another way the area could be smaller? For example, if the angle between ( PA ) and ( PC ) changes, does that affect the area? Wait, no, because the area is fixed once ( PA ) is fixed since ( AC ) is fixed at 1.Alternatively, maybe I can parametrize point ( P ) on the line and express the area in terms of coordinates, then find the minimum. Let me try that approach to confirm.Let me parametrize the line ( 3x + 4y + 8 = 0 ). I can express ( y ) in terms of ( x ): ( y = (-3x - 8)/4 ). So, any point ( P ) on the line can be written as ( (x, (-3x - 8)/4 ) ).Now, the distance from ( P(x, (-3x - 8)/4 ) ) to the center ( C(1,1) ) is:( PC = sqrt{(x - 1)^2 + left( frac{-3x - 8}{4} - 1 right)^2 } )Simplify the y-component:( frac{-3x - 8}{4} - 1 = frac{-3x - 8 - 4}{4} = frac{-3x - 12}{4} = frac{-3(x + 4)}{4} )So, ( PC = sqrt{(x - 1)^2 + left( frac{-3(x + 4)}{4} right)^2 } )Simplify further:( PC = sqrt{(x - 1)^2 + left( frac{9(x + 4)^2}{16} right) } )Now, the length of the tangent ( PA ) is ( sqrt{PC^2 - r^2} = sqrt{PC^2 - 1} ). So, the area of quadrilateral ( PACB ) is ( PA times 1 = sqrt{PC^2 - 1} ).Therefore, to minimize the area, I need to minimize ( sqrt{PC^2 - 1} ), which is equivalent to minimizing ( PC^2 - 1 ). So, let's compute ( PC^2 ):( PC^2 = (x - 1)^2 + frac{9(x + 4)^2}{16} )Let me compute this:First, expand ( (x - 1)^2 = x^2 - 2x + 1 )Next, expand ( (x + 4)^2 = x^2 + 8x + 16 ), so ( frac{9(x + 4)^2}{16} = frac{9x^2 + 72x + 144}{16} )Therefore, ( PC^2 = x^2 - 2x + 1 + frac{9x^2 + 72x + 144}{16} )To combine these, let's express everything over 16:( PC^2 = frac{16x^2 - 32x + 16}{16} + frac{9x^2 + 72x + 144}{16} )Combine the numerators:( 16x^2 - 32x + 16 + 9x^2 + 72x + 144 = (16x^2 + 9x^2) + (-32x + 72x) + (16 + 144) = 25x^2 + 40x + 160 )Therefore, ( PC^2 = frac{25x^2 + 40x + 160}{16} )So, ( PC^2 - 1 = frac{25x^2 + 40x + 160}{16} - 1 = frac{25x^2 + 40x + 160 - 16}{16} = frac{25x^2 + 40x + 144}{16} )Therefore, the area squared is ( (PA)^2 = PC^2 - 1 = frac{25x^2 + 40x + 144}{16} ). Wait, no, actually, the area is ( PA ), which is ( sqrt{PC^2 - 1} ). So, the area squared is ( PC^2 - 1 ).But to find the minimal area, I can instead minimize ( PC^2 - 1 ), which is a quadratic function in ( x ). Let me write that:( f(x) = frac{25x^2 + 40x + 144}{16} )To find the minimum, since the coefficient of ( x^2 ) is positive, the function has a minimum at its vertex. The x-coordinate of the vertex is at ( x = -b/(2a) ), where ( a = 25/16 ) and ( b = 40/16 = 5/2 ).Wait, actually, let me write ( f(x) = frac{25}{16}x^2 + frac{40}{16}x + frac{144}{16} = frac{25}{16}x^2 + frac{5}{2}x + 9 ).So, ( a = 25/16 ), ( b = 5/2 ). Therefore, the vertex is at ( x = -b/(2a) = -(5/2)/(2*(25/16)) = -(5/2)/(25/8) = -(5/2)*(8/25) = -(40/50) = -4/5 ).So, the minimal value occurs at ( x = -4/5 ). Let me compute ( f(-4/5) ):( f(-4/5) = frac{25}{16}*(-4/5)^2 + frac{5}{2}*(-4/5) + 9 )Compute each term:First term: ( frac{25}{16}*(16/25) = 1 )Second term: ( frac{5}{2}*(-4/5) = -2 )Third term: 9So, adding them up: ( 1 - 2 + 9 = 8 )Therefore, the minimal value of ( f(x) = PC^2 - 1 ) is 8, so the minimal area is ( sqrt{8} = 2sqrt{2} ).Wait, that's consistent with my earlier result. So, that's good.Alternatively, I could have used calculus. Take the derivative of ( f(x) ) with respect to ( x ), set it to zero, and solve for ( x ). But since it's a quadratic, the vertex formula suffices.So, both methods give me the same minimal area of ( 2sqrt{2} ).Just to recap:1. Found the center and radius of the circle: ( C(1,1) ) and ( r = 1 ).2. Recognized that the area of quadrilateral ( PACB ) is equal to ( PA ) because it's composed of two right triangles each with area ( frac{1}{2} times PA times 1 ).3. Noted that minimizing ( PA ) would minimize the area.4. Realized that ( PA = sqrt{PC^2 - r^2} ), so minimizing ( PA ) is equivalent to minimizing ( PC ).5. Calculated the minimal distance from ( C ) to the line ( 3x + 4y + 8 = 0 ), which is 3.6. Therefore, minimal ( PA = sqrt{3^2 - 1^2} = 2sqrt{2} ).7. Alternatively, parametrized ( P ) on the line, expressed ( PC^2 ) in terms of ( x ), found the minimal value, and confirmed the minimal area is ( 2sqrt{2} ).I think that's thorough. I don't see any mistakes in my reasoning, and both approaches confirm the same result. So, I'm confident that the minimal area is ( 2sqrt{2} ).**Final Answer**The minimum possible area of the quadrilateral ( PACB ) is boxed{2sqrt{2}}.

answer：First, I need to understand the given sets ( M ) and ( N ). Set ( M ) is explicitly provided as ( {1, 2, 3, 4, 5, 6} ). Set ( N ) is defined by the inequality ( -2 < x < 5 ) where ( x ) is an integer.Next, I'll determine the elements of set ( N ) by finding all integer values that satisfy the inequality ( -2 < x < 5 ). This gives me ( x = -1, 0, 1, 2, 3, 4 ), so ( N = {-1, 0, 1, 2, 3, 4} ).Now, to find the intersection ( M cap N ), I'll identify the common elements between sets ( M ) and ( N ). Comparing the two sets:- ( M = {1, 2, 3, 4, 5, 6} )- ( N = {-1, 0, 1, 2, 3, 4} )The common elements are ( 1, 2, 3, ) and ( 4 ). Therefore, the intersection ( M cap N ) is ( {1, 2, 3, 4} ).

answer：Alright, so I have this trigonometry problem here. It says that sin α cos β equals -1/2, and I need to find the range of values for cos α sin β. The options are given as A, B, C, D, each with different intervals. Hmm, okay, let me try to figure this out step by step.First, I remember that there are some trigonometric identities that relate products of sine and cosine terms. Maybe I can use those to express cos α sin β in terms of something else. Let me recall... Oh, right! There are sum and difference formulas for sine. Specifically, sin(α + β) equals sin α cos β plus cos α sin β. Similarly, sin(α - β) equals sin α cos β minus cos α sin β.So, if I write those down:1. sin(α + β) = sin α cos β + cos α sin β2. sin(α - β) = sin α cos β - cos α sin βGiven that sin α cos β is -1/2, I can substitute that into both equations. Let's do that.Substituting into the first equation:sin(α + β) = (-1/2) + cos α sin βAnd into the second equation:sin(α - β) = (-1/2) - cos α sin βNow, I know that the sine function has a range between -1 and 1. So, sin(α + β) must be between -1 and 1, and the same goes for sin(α - β). That gives me two inequalities:1. -1 ≤ (-1/2) + cos α sin β ≤ 12. -1 ≤ (-1/2) - cos α sin β ≤ 1Let me solve each of these inequalities for cos α sin β.Starting with the first inequality:-1 ≤ (-1/2) + cos α sin β ≤ 1I can subtract (-1/2) from all parts to isolate cos α sin β:-1 + 1/2 ≤ cos α sin β ≤ 1 + 1/2Which simplifies to:-1/2 ≤ cos α sin β ≤ 3/2Okay, so from the first inequality, cos α sin β is between -1/2 and 3/2.Now, moving on to the second inequality:-1 ≤ (-1/2) - cos α sin β ≤ 1Again, I'll subtract (-1/2) from all parts:-1 + 1/2 ≤ -cos α sin β ≤ 1 + 1/2Which simplifies to:-1/2 ≤ -cos α sin β ≤ 3/2Hmm, now I have -1/2 ≤ -cos α sin β ≤ 3/2. To solve for cos α sin β, I need to multiply each part by -1. But remember, when you multiply or divide an inequality by a negative number, the inequality signs flip.So, multiplying by -1:1/2 ≥ cos α sin β ≥ -3/2Which can be rewritten as:-3/2 ≤ cos α sin β ≤ 1/2Alright, so from the second inequality, cos α sin β is between -3/2 and 1/2.Now, I have two ranges for cos α sin β:1. From the first inequality: -1/2 ≤ cos α sin β ≤ 3/22. From the second inequality: -3/2 ≤ cos α sin β ≤ 1/2To find the overall range of cos α sin β, I need to take the intersection of these two intervals. The intersection is where both conditions are satisfied.Looking at the lower bounds: the first interval starts at -1/2, and the second starts at -3/2. So, the higher lower bound is -1/2.Looking at the upper bounds: the first interval ends at 3/2, and the second ends at 1/2. So, the lower upper bound is 1/2.Therefore, the intersection is:-1/2 ≤ cos α sin β ≤ 1/2So, the range of values for cos α sin β is from -1/2 to 1/2.Let me just double-check my steps to make sure I didn't make a mistake. I used the sum and difference identities correctly, substituted the given value, applied the range of sine function, solved the inequalities, and then took the intersection. Everything seems to add up. I think this is the correct range.**Final Answer**boxed{D}