💡First, I recognize that the longest chord in a circle is the diameter.Given that the longest chord is 22 units long, this means the diameter of the circle is 22 units.To find the radius, I divide the diameter by 2.So, the radius is 22 divided by 2, which equals 11 units.

💡Okay, so I have this system of equations with five variables and a parameter y. The equations are cyclic, meaning each equation relates consecutive variables, and the last one wraps around to the first. Let me write them down again to get a clear picture:1. ( x_5 + x_2 = y x_1 )2. ( x_1 + x_3 = y x_2 )3. ( x_2 + x_4 = y x_3 )4. ( x_3 + x_5 = y x_4 )5. ( x_4 + x_1 = y x_5 )Hmm, since all the equations are connected in a cycle, maybe I can express each variable in terms of the next one or something like that. Let me see.Starting with the first equation: ( x_5 + x_2 = y x_1 ). Maybe I can solve for ( x_5 ) here: ( x_5 = y x_1 - x_2 ). That might help substitute into another equation.Looking at the second equation: ( x_1 + x_3 = y x_2 ). If I solve for ( x_3 ), I get ( x_3 = y x_2 - x_1 ). Okay, so now I have expressions for ( x_5 ) and ( x_3 ) in terms of ( x_1 ) and ( x_2 ).Moving to the third equation: ( x_2 + x_4 = y x_3 ). I can substitute ( x_3 ) from the second equation here: ( x_2 + x_4 = y (y x_2 - x_1) ). Let's expand that: ( x_2 + x_4 = y^2 x_2 - y x_1 ). Then, solving for ( x_4 ): ( x_4 = y^2 x_2 - y x_1 - x_2 ). Simplify: ( x_4 = (y^2 - 1) x_2 - y x_1 ).Okay, so now I have ( x_4 ) in terms of ( x_1 ) and ( x_2 ). Let's move to the fourth equation: ( x_3 + x_5 = y x_4 ). I already have expressions for ( x_3 ) and ( x_5 ) from earlier. Let's substitute them in:( (y x_2 - x_1) + (y x_1 - x_2) = y x_4 ).Simplify the left side: ( y x_2 - x_1 + y x_1 - x_2 = (y x_2 - x_2) + (-x_1 + y x_1) = x_2 (y - 1) + x_1 (y - 1) = (x_1 + x_2)(y - 1) ).So, ( (x_1 + x_2)(y - 1) = y x_4 ). But from earlier, ( x_4 = (y^2 - 1) x_2 - y x_1 ). So substituting that in:( (x_1 + x_2)(y - 1) = y [(y^2 - 1) x_2 - y x_1] ).Let me expand the right side: ( y (y^2 - 1) x_2 - y^2 x_1 ).So, the equation becomes:( (x_1 + x_2)(y - 1) = y (y^2 - 1) x_2 - y^2 x_1 ).Let me bring all terms to one side:( (x_1 + x_2)(y - 1) - y (y^2 - 1) x_2 + y^2 x_1 = 0 ).Expanding the first term:( x_1 (y - 1) + x_2 (y - 1) - y (y^2 - 1) x_2 + y^2 x_1 = 0 ).Now, let's collect like terms for ( x_1 ) and ( x_2 ):For ( x_1 ):( x_1 (y - 1) + y^2 x_1 = x_1 (y - 1 + y^2) ).For ( x_2 ):( x_2 (y - 1) - y (y^2 - 1) x_2 = x_2 [ (y - 1) - y (y^2 - 1) ] ).Simplify each part:For ( x_1 ):( y^2 + y - 1 ).For ( x_2 ):First, factor ( y^2 - 1 ) as ( (y - 1)(y + 1) ):( (y - 1) - y (y - 1)(y + 1) = (y - 1)[1 - y(y + 1)] ).Simplify inside the brackets:( 1 - y^2 - y ).So, the coefficient for ( x_2 ) becomes:( (y - 1)(- y^2 - y + 1) ).Putting it all together, the equation is:( x_1 (y^2 + y - 1) + x_2 (y - 1)(- y^2 - y + 1) = 0 ).Hmm, this seems a bit complicated. Maybe I made a mistake in expanding. Let me check.Wait, perhaps there's a better approach. Since the system is cyclic, maybe I can express each variable in terms of ( x_1 ) and find a relationship.Let me try expressing each variable step by step:From equation 1: ( x_5 = y x_1 - x_2 ).From equation 2: ( x_3 = y x_2 - x_1 ).From equation 3: ( x_4 = y x_3 - x_2 ). Substitute ( x_3 ): ( x_4 = y (y x_2 - x_1) - x_2 = y^2 x_2 - y x_1 - x_2 = (y^2 - 1) x_2 - y x_1 ).From equation 4: ( x_5 = y x_4 - x_3 ). Substitute ( x_4 ) and ( x_3 ):( x_5 = y [(y^2 - 1) x_2 - y x_1] - (y x_2 - x_1) ).Expand:( x_5 = y (y^2 - 1) x_2 - y^2 x_1 - y x_2 + x_1 ).Combine like terms:For ( x_2 ): ( y (y^2 - 1) x_2 - y x_2 = y (y^2 - 1 - 1) x_2 = y (y^2 - 2) x_2 ).For ( x_1 ): ( - y^2 x_1 + x_1 = ( - y^2 + 1 ) x_1 ).So, ( x_5 = y (y^2 - 2) x_2 + (1 - y^2) x_1 ).But from equation 1, we have ( x_5 = y x_1 - x_2 ). So, set them equal:( y x_1 - x_2 = y (y^2 - 2) x_2 + (1 - y^2) x_1 ).Bring all terms to one side:( y x_1 - x_2 - y (y^2 - 2) x_2 - (1 - y^2) x_1 = 0 ).Factor ( x_1 ) and ( x_2 ):For ( x_1 ): ( y - (1 - y^2) = y - 1 + y^2 = y^2 + y - 1 ).For ( x_2 ): ( -1 - y (y^2 - 2) = -1 - y^3 + 2 y = - y^3 + 2 y - 1 ).So, the equation becomes:( (y^2 + y - 1) x_1 + (- y^3 + 2 y - 1) x_2 = 0 ).Hmm, this is a relationship between ( x_1 ) and ( x_2 ). Let me write it as:( (y^2 + y - 1) x_1 = (y^3 - 2 y + 1) x_2 ).So, ( x_1 = frac{y^3 - 2 y + 1}{y^2 + y - 1} x_2 ).Let me simplify the fraction:Factor numerator: ( y^3 - 2 y + 1 ). Maybe factor by grouping or rational roots. Let's try y=1: 1 - 2 + 1 = 0. So, y=1 is a root. So, factor as (y - 1)(y^2 + y - 1).Yes, because:( (y - 1)(y^2 + y - 1) = y^3 + y^2 - y - y^2 - y + 1 = y^3 - 2 y + 1 ).Perfect. So, numerator is ( (y - 1)(y^2 + y - 1) ).Denominator is ( y^2 + y - 1 ).So, ( x_1 = frac{(y - 1)(y^2 + y - 1)}{y^2 + y - 1} x_2 = (y - 1) x_2 ).So, ( x_1 = (y - 1) x_2 ).That's a nice relationship. So, ( x_1 ) is proportional to ( x_2 ).Now, let's use this in our earlier expressions.From equation 2: ( x_3 = y x_2 - x_1 = y x_2 - (y - 1) x_2 = (y - (y - 1)) x_2 = 1 x_2 ).So, ( x_3 = x_2 ).From equation 3: ( x_4 = (y^2 - 1) x_2 - y x_1 = (y^2 - 1) x_2 - y (y - 1) x_2 ).Simplify:( (y^2 - 1 - y (y - 1)) x_2 = (y^2 - 1 - y^2 + y) x_2 = ( -1 + y ) x_2 = (y - 1) x_2 ).So, ( x_4 = (y - 1) x_2 ).From equation 1: ( x_5 = y x_1 - x_2 = y (y - 1) x_2 - x_2 = (y^2 - y - 1) x_2 ).So, now we have all variables in terms of ( x_2 ):- ( x_1 = (y - 1) x_2 )- ( x_3 = x_2 )- ( x_4 = (y - 1) x_2 )- ( x_5 = (y^2 - y - 1) x_2 )Now, let's use equation 5: ( x_4 + x_1 = y x_5 ).Substitute the expressions:( (y - 1) x_2 + (y - 1) x_2 = y (y^2 - y - 1) x_2 ).Simplify left side:( 2 (y - 1) x_2 = y (y^2 - y - 1) x_2 ).Assuming ( x_2 neq 0 ) (since we're looking for non-trivial solutions), we can divide both sides by ( x_2 ):( 2 (y - 1) = y (y^2 - y - 1) ).Expand the right side:( 2 y - 2 = y^3 - y^2 - y ).Bring all terms to one side:( y^3 - y^2 - y - 2 y + 2 = 0 ).Simplify:( y^3 - y^2 - 3 y + 2 = 0 ).Now, let's solve this cubic equation: ( y^3 - y^2 - 3 y + 2 = 0 ).Try rational roots: possible roots are ±1, ±2.Test y=1: 1 - 1 - 3 + 2 = -1 ≠ 0.Test y=2: 8 - 4 - 6 + 2 = 0. Yes, y=2 is a root.So, factor out (y - 2):Using polynomial division or synthetic division:Divide ( y^3 - y^2 - 3 y + 2 ) by (y - 2):Coefficients: 1 | -1 | -3 | 2Bring down 1.Multiply by 2: 1*2=2. Add to -1: 1.Multiply by 2: 1*2=2. Add to -3: -1.Multiply by 2: -1*2=-2. Add to 2: 0.So, the cubic factors as (y - 2)(y^2 + y - 1).Set equal to zero: (y - 2)(y^2 + y - 1) = 0.Solutions: y=2, and y = [ -1 ± sqrt(1 + 4) ] / 2 = [ -1 ± sqrt(5) ] / 2.So, y can be 2, or the golden ratio and its conjugate.So, for y=2, and y = [ -1 ± sqrt(5) ] / 2, we have non-trivial solutions.Now, let's analyze each case.**Case 1: y = 2**From earlier, we have:- ( x_1 = (2 - 1) x_2 = x_2 )- ( x_3 = x_2 )- ( x_4 = (2 - 1) x_2 = x_2 )- ( x_5 = (4 - 2 - 1) x_2 = 1 x_2 )So, all variables are equal: ( x_1 = x_2 = x_3 = x_4 = x_5 = x_2 ). Let me denote ( x_2 = t ), so all variables are t. Thus, the solution is any scalar multiple of (1,1,1,1,1).**Case 2: y = [ -1 + sqrt(5) ] / 2 and y = [ -1 - sqrt(5) ] / 2**Let me denote y as the roots of ( y^2 + y - 1 = 0 ). So, for these y, the coefficient ( y^2 + y - 1 = 0 ).From earlier, we have:- ( x_1 = (y - 1) x_2 )- ( x_3 = x_2 )- ( x_4 = (y - 1) x_2 )- ( x_5 = (y^2 - y - 1) x_2 )But since ( y^2 + y - 1 = 0 ), we can express ( y^2 = 1 - y ). So, substitute into ( x_5 ):( x_5 = (1 - y - y - 1) x_2 = (-2 y) x_2 ).Wait, let me check:( y^2 - y - 1 = (1 - y) - y - 1 = 1 - y - y - 1 = -2 y ).Yes, correct. So, ( x_5 = -2 y x_2 ).So, now, we have:- ( x_1 = (y - 1) x_2 )- ( x_3 = x_2 )- ( x_4 = (y - 1) x_2 )- ( x_5 = -2 y x_2 )Now, let's check equation 5: ( x_4 + x_1 = y x_5 ).Substitute:( (y - 1) x_2 + (y - 1) x_2 = y (-2 y x_2) ).Simplify left side: ( 2 (y - 1) x_2 ).Right side: ( -2 y^2 x_2 ).But since ( y^2 = 1 - y ), substitute:Right side: ( -2 (1 - y) x_2 = (-2 + 2 y) x_2 ).Left side: ( 2 (y - 1) x_2 = (2 y - 2) x_2 ).Which is equal to the right side. So, it checks out.Therefore, for these y, the solutions are proportional to the vector:( (x_1, x_2, x_3, x_4, x_5) = (y - 1, 1, 1, y - 1, -2 y) x_2 ).So, if we let ( x_2 = t ), then:- ( x_1 = (y - 1) t )- ( x_2 = t )- ( x_3 = t )- ( x_4 = (y - 1) t )- ( x_5 = -2 y t )This gives us a non-trivial solution for these specific y values.**Conclusion:**The system has non-trivial solutions only when y is 2 or one of the roots of ( y^2 + y - 1 = 0 ), which are ( y = frac{-1 pm sqrt{5}}{2} ).For y=2, all variables are equal, so the solution is any scalar multiple of (1,1,1,1,1).For ( y = frac{-1 pm sqrt{5}}{2} ), the solutions are scalar multiples of the vector ( (y - 1, 1, 1, y - 1, -2 y) ).Otherwise, the only solution is the trivial one where all variables are zero.

💡Okay, so I have this problem about a cosine function, and I need to find some values related to it. Let me try to understand what's being asked here.First, the function is given as ( y = 2 cos(omega x + theta) ). I know that in a cosine function, the amplitude is the coefficient in front of the cosine, which here is 2. The period of a cosine function is usually ( 2pi ), but here it's modified by ( omega ). The phase shift is given by ( theta ), which affects where the cosine wave starts on the x-axis.The problem states that the graph intersects the y-axis at the point ( (0, sqrt{3}) ). That means when ( x = 0 ), ( y = sqrt{3} ). So, plugging ( x = 0 ) into the function, we get ( y = 2 cos(theta) = sqrt{3} ). Hmm, so ( cos(theta) = sqrt{3}/2 ). I remember that ( cos(pi/6) = sqrt{3}/2 ), so ( theta ) must be ( pi/6 ). That seems straightforward.Next, the problem mentions that the minimum positive period of the function is ( pi ). The period of a cosine function ( cos(omega x + theta) ) is ( 2pi / omega ). So, if the period is ( pi ), then ( 2pi / omega = pi ), which means ( omega = 2 ). Okay, so that gives me ( omega = 2 ) and ( theta = pi/6 ). That takes care of part 1.Now, moving on to part 2. We have a point ( A ) at ( (pi/2, 0) ) and a point ( P ) on the graph of the function. The midpoint ( Q ) of segment ( PA ) is given as ( (x_0, y_0) ), and we know that ( y_0 = sqrt{3}/2 ) and ( x_0 ) is between ( pi/2 ) and ( pi ). We need to find ( x_0 ).Let me visualize this. Point ( A ) is on the x-axis at ( (pi/2, 0) ), and point ( P ) is somewhere on the cosine curve. The midpoint ( Q ) is halfway between ( P ) and ( A ). Since ( Q ) has a y-coordinate of ( sqrt{3}/2 ), that tells me something about the y-coordinate of ( P ).If ( Q ) is the midpoint, then its coordinates are the average of the coordinates of ( P ) and ( A ). So, if ( P ) is ( (x, y) ), then:[x_0 = frac{x + pi/2}{2}][y_0 = frac{y + 0}{2} = frac{y}{2}]Given that ( y_0 = sqrt{3}/2 ), then ( y = sqrt{3} ). So, point ( P ) must have a y-coordinate of ( sqrt{3} ). Since ( P ) lies on the function ( y = 2 cos(2x + pi/6) ), we can set up the equation:[sqrt{3} = 2 cos(2x + pi/6)]Dividing both sides by 2:[cos(2x + pi/6) = sqrt{3}/2]I know that ( cos(pi/6) = sqrt{3}/2 ) and ( cos(11pi/6) = sqrt{3}/2 ) as well, since cosine is positive in the first and fourth quadrants. So, the general solutions for ( 2x + pi/6 ) are:[2x + pi/6 = pi/6 + 2pi k quad text{or} quad 2x + pi/6 = -pi/6 + 2pi k]Where ( k ) is any integer. Let me solve for ( x ) in both cases.First case:[2x + pi/6 = pi/6 + 2pi k]Subtract ( pi/6 ) from both sides:[2x = 2pi k]Divide by 2:[x = pi k]Second case:[2x + pi/6 = -pi/6 + 2pi k]Subtract ( pi/6 ) from both sides:[2x = -pi/3 + 2pi k]Divide by 2:[x = -pi/6 + pi k]So, the solutions for ( x ) are ( x = pi k ) and ( x = -pi/6 + pi k ).Now, we need to find ( x ) such that ( x_0 ) is between ( pi/2 ) and ( pi ). Remember that ( x_0 = frac{x + pi/2}{2} ). So, let's express ( x ) in terms of ( x_0 ):[x = 2x_0 - pi/2]So, substituting back into our solutions:First case:[2x_0 - pi/2 = pi k][2x_0 = pi k + pi/2][x_0 = frac{pi k}{2} + pi/4]Second case:[2x_0 - pi/2 = -pi/6 + pi k][2x_0 = -pi/6 + pi k + pi/2][2x_0 = pi k + pi/3][x_0 = frac{pi k}{2} + pi/6]Now, we need ( x_0 ) to be in the interval ( [pi/2, pi] ). Let's test integer values of ( k ) to see which solutions fall into this interval.Starting with the first case:( x_0 = frac{pi k}{2} + pi/4 )Let’s try ( k = 1 ):[x_0 = frac{pi}{2} + pi/4 = 3pi/4 approx 2.356]Which is between ( pi/2 approx 1.571 ) and ( pi approx 3.142 ). So, this is a valid solution.Next, ( k = 2 ):[x_0 = pi + pi/4 = 5pi/4 approx 3.927]Which is greater than ( pi ), so it's outside our interval.( k = 0 ):[x_0 = 0 + pi/4 = pi/4 approx 0.785]Which is less than ( pi/2 ), so also outside.Now, the second case:( x_0 = frac{pi k}{2} + pi/6 )Let’s try ( k = 1 ):[x_0 = frac{pi}{2} + pi/6 = 2pi/3 approx 2.094]Which is within ( [pi/2, pi] ).( k = 2 ):[x_0 = pi + pi/6 = 7pi/6 approx 3.665]Which is greater than ( pi ), so invalid.( k = 0 ):[x_0 = 0 + pi/6 = pi/6 approx 0.524]Which is less than ( pi/2 ), so invalid.So, from both cases, the valid solutions within the interval are ( x_0 = 3pi/4 ) and ( x_0 = 2pi/3 ).Wait, but I need to make sure that these ( x_0 ) values correspond to valid ( x ) values on the function. Let me check:For ( x_0 = 3pi/4 ):[x = 2x_0 - pi/2 = 2*(3pi/4) - pi/2 = 3pi/2 - pi/2 = pi]So, ( x = pi ). Plugging into the function:[y = 2 cos(2pi + pi/6) = 2 cos(13pi/6) = 2*(sqrt{3}/2) = sqrt{3}]Which is correct.For ( x_0 = 2pi/3 ):[x = 2x_0 - pi/2 = 2*(2pi/3) - pi/2 = 4pi/3 - pi/2 = (8pi/6 - 3pi/6) = 5pi/6]Plugging into the function:[y = 2 cos(2*(5pi/6) + pi/6) = 2 cos(10pi/6 + pi/6) = 2 cos(11pi/6) = 2*(sqrt{3}/2) = sqrt{3}]Which is also correct.So, both solutions are valid. Therefore, the possible values of ( x_0 ) are ( 2pi/3 ) and ( 3pi/4 ).**Final Answer**1. The values are ( theta = boxed{dfrac{pi}{6}} ) and ( omega = boxed{2} ).2. The value of ( x_0 ) is ( boxed{dfrac{2pi}{3}} ) or ( boxed{dfrac{3pi}{4}} ).

💡First, I need to determine the original area of the towel by multiplying its length and breadth.After bleaching, the towel loses 20% of its length and 10% of its breadth. This means the new length is 80% of the original length, and the new breadth is 90% of the original breadth.Next, I'll calculate the new area of the towel using the reduced length and breadth.To find the percentage decrease in area, I'll subtract the new area from the original area and then divide by the original area. Finally, I'll multiply by 100 to get the percentage.

💡First, calculate the total amount of money Whitney received from her mom by multiplying the number of 20 bills by 20.Next, determine the cost of the posters by multiplying the number of posters by the cost per poster.Then, calculate the cost of the notebooks by multiplying the number of notebooks by the cost per notebook.Add the costs of the posters and notebooks to find the total amount spent on these items.Subtract the total amount spent on posters and notebooks from the total money Whitney had to find out how much money she spent on bookmarks.Finally, divide the amount spent on bookmarks by the cost per bookmark to find the number of bookmarks Whitney bought.

💡First, I need to determine the dimensions of the mirror inside the frame. Since the frame has a uniform width of 15 cm around the rectangle, I will subtract twice this width from both the length and the width of the outer perimeter to account for the frame on both sides.Next, I'll calculate the inner dimensions by subtracting 30 cm (15 cm from each side) from both the length and the width of the outer frame.Finally, I'll compute the area of the mirror by multiplying the inner length and inner width.