💡First, I need to determine the number of one dollar bills Mary has in her wallet. She has a total of 60 bills consisting of one dollar, two dollar, and five dollar bills, and the total amount of money is 120.I'll start by defining variables:- Let ( x ) represent the number of one dollar bills.- Let ( y ) represent the number of two dollar bills.- Let ( z ) represent the number of five dollar bills.Based on the information provided, I can set up two equations:1. The total number of bills: ( x + y + z = 60 )2. The total amount of money: ( x + 2y + 5z = 120 )To solve this system of equations, I'll first solve the first equation for ( x ):[ x = 60 - y - z ]Next, I'll substitute this expression for ( x ) into the second equation:[ (60 - y - z) + 2y + 5z = 120 ]Simplifying this equation:[ 60 + y + 4z = 120 ][ y + 4z = 60 ]Now, I'll solve for ( y ):[ y = 60 - 4z ]Substituting this back into the expression for ( x ):[ x = 60 - (60 - 4z) - z ][ x = 60 - 60 + 4z - z ][ x = 3z ]To find specific values, I'll assume ( z = 10 ) (a reasonable whole number within the constraints):- ( z = 10 )- ( y = 60 - 4(10) = 20 )- ( x = 3(10) = 30 )However, this doesn't satisfy the total number of bills (30 + 20 + 10 = 60), but the total amount would be ( 30 + 40 + 50 = 120 ), which is correct. Therefore, Mary has 30 one dollar bills.

💡First, I need to recall the formula for calculating the area of a parallelogram. The formula is:Area = base × heightNext, I'll identify the given values from the problem. The base of the parallelogram is 12 cm, and the height is 18 cm.Now, I'll substitute these values into the formula:Area = 12 cm × 18 cmFinally, I'll perform the multiplication to find the area:Area = 216 cm²This means the area of the parallelogram is 216 square centimeters.

💡Okay, so I have this problem where a point P(m, n) is moving on the circle defined by the equation x² + y² = 2. I need to find the trajectory equation of another point Q, which has coordinates (m + n, 2mn). Then, I have to determine what shape this trajectory is.First, let me understand what the circle x² + y² = 2 looks like. It's a circle centered at the origin with a radius of √2. So, any point P(m, n) on this circle satisfies m² + n² = 2.Now, point Q has coordinates (m + n, 2mn). So, the x-coordinate of Q is m + n, and the y-coordinate is 2mn. I need to find a relationship between x and y for point Q, eliminating the parameters m and n.One approach is to express m and n in terms of trigonometric functions since they lie on a circle. So, I can write m = √2 cosθ and n = √2 sinθ, where θ is the parameter representing the angle made with the positive x-axis.Substituting these into Q's coordinates:x = m + n = √2 cosθ + √2 sinθy = 2mn = 2*(√2 cosθ)*(√2 sinθ)Simplify y:y = 2*(√2 * √2)*(cosθ sinθ) = 2*(2)*(cosθ sinθ) = 4 cosθ sinθI remember that 2 cosθ sinθ is equal to sin(2θ), so 4 cosθ sinθ is 2 sin(2θ). Therefore, y = 2 sin(2θ).Now, let's look at x:x = √2 cosθ + √2 sinθ = √2 (cosθ + sinθ)I can factor out √2, but maybe it's better to express this in terms of a single trigonometric function. I recall that a cosθ + b sinθ can be written as R cos(θ - φ), where R = √(a² + b²) and tanφ = b/a.In this case, a = 1 and b = 1, so R = √(1 + 1) = √2, and tanφ = 1/1 = 1, so φ = π/4.Therefore, x = √2 * √2 cos(θ - π/4) = 2 cos(θ - π/4)Wait, hold on. Let me check that again.x = √2 (cosθ + sinθ) = √2 * [√2 cos(θ - π/4)] because cosθ + sinθ = √2 cos(θ - π/4). So, x = √2 * √2 cos(θ - π/4) = 2 cos(θ - π/4)Yes, that's correct. So, x = 2 cos(θ - π/4). Therefore, the range of x is from -2 to 2 because cosine varies between -1 and 1.Similarly, y = 2 sin(2θ). The range of y is from -2 to 2 as well because sine varies between -1 and 1.Now, I need to eliminate θ from the equations for x and y to get a relation between x and y.Let me express θ in terms of x or y. Let's see.From x = 2 cos(θ - π/4), we can write cos(θ - π/4) = x/2.Similarly, from y = 2 sin(2θ), we can write sin(2θ) = y/2.Now, let's try to express sin(2θ) in terms of cos(θ - π/4). Maybe using some trigonometric identities.I know that 2θ can be written as 2(θ - π/4 + π/4) = 2(θ - π/4) + π/2.So, sin(2θ) = sin[2(θ - π/4) + π/2]Using the sine addition formula:sin(A + B) = sinA cosB + cosA sinBSo, sin[2(θ - π/4) + π/2] = sin[2(θ - π/4)] cos(π/2) + cos[2(θ - π/4)] sin(π/2)But cos(π/2) = 0 and sin(π/2) = 1, so this simplifies to:sin[2(θ - π/4)]*0 + cos[2(θ - π/4)]*1 = cos[2(θ - π/4)]Therefore, sin(2θ) = cos[2(θ - π/4)]So, y/2 = cos[2(θ - π/4)]But from x = 2 cos(θ - π/4), we have cos(θ - π/4) = x/2.Let me denote φ = θ - π/4, so cosφ = x/2.Then, cos[2φ] = 2 cos²φ - 1 = 2(x/2)² - 1 = 2(x²/4) - 1 = x²/2 - 1But we also have y/2 = cos[2φ], so y/2 = x²/2 - 1Multiply both sides by 2:y = x² - 2So, the equation of the trajectory is y = x² - 2.Now, considering the ranges of x and y, since x ranges from -2 to 2 and y ranges from -2 to 2, the trajectory is a parabola opening upwards, but only the portion where x is between -2 and 2 and y is between -2 and 2.So, it's a segment of the parabola y = x² - 2.

💡Okay, so I have this equation to solve: (frac{4x^2 - 3x + 2}{x + 2} = 4x - 5). Hmm, it's a rational equation, meaning there's a fraction involved. I remember that to solve equations like this, one common method is cross-multiplication. But before I jump into that, I should make sure I understand the equation properly.First, let me rewrite the equation to visualize it better:[frac{4x^2 - 3x + 2}{x + 2} = 4x - 5]So, the left side is a rational expression, and the right side is a linear expression. My goal is to find the value(s) of (x) that satisfy this equation.Since there's a denominator (x + 2), I should note that (x) cannot be (-2) because that would make the denominator zero, which is undefined. So, (x neq -2) is an important restriction to keep in mind.Now, moving on to solving the equation. As I thought earlier, cross-multiplication might be a good approach here. That means I'll multiply both sides of the equation by (x + 2) to eliminate the denominator. Let's do that:Multiply both sides by (x + 2):[4x^2 - 3x + 2 = (4x - 5)(x + 2)]Okay, so now I have a polynomial equation. The next step is to expand the right-hand side so that I can combine like terms and solve for (x).Let me expand ((4x - 5)(x + 2)). I'll use the distributive property (FOIL method) for this:First, multiply (4x) by (x): (4x cdot x = 4x^2)Then, (4x) multiplied by (2): (4x cdot 2 = 8x)Next, (-5) multiplied by (x): (-5 cdot x = -5x)Finally, (-5) multiplied by (2): (-5 cdot 2 = -10)Now, let's combine these terms:[4x^2 + 8x - 5x - 10]Combine like terms (8x - 5x):[4x^2 + 3x - 10]So, the right-hand side simplifies to (4x^2 + 3x - 10).Now, let's rewrite the equation with this expansion:[4x^2 - 3x + 2 = 4x^2 + 3x - 10]Hmm, now I have a quadratic equation on both sides. To solve for (x), I should get all terms on one side so that I can set the equation equal to zero.First, I can subtract (4x^2) from both sides to eliminate the quadratic term:[4x^2 - 3x + 2 - 4x^2 = 4x^2 + 3x - 10 - 4x^2]Simplifying both sides:Left side: (-3x + 2)Right side: (3x - 10)So now the equation is:[-3x + 2 = 3x - 10]Alright, now I have a linear equation. Let's solve for (x). I'll start by getting all the (x) terms on one side and the constants on the other.First, I'll add (3x) to both sides to move the (x) terms to the right:[-3x + 2 + 3x = 3x - 10 + 3x]Simplifying both sides:Left side: (2)Right side: (6x - 10)So now the equation is:[2 = 6x - 10]Next, I'll isolate the term with (x) by adding (10) to both sides:[2 + 10 = 6x - 10 + 10]Simplifying:Left side: (12)Right side: (6x)So now we have:[12 = 6x]To solve for (x), I'll divide both sides by (6):[frac{12}{6} = frac{6x}{6}]Simplifying:[2 = x]Or, written as:[x = 2]Wait a minute, I should check if this solution is valid. Remember earlier, I noted that (x neq -2) because it would make the original denominator zero. Since (x = 2) doesn't violate that condition, it's a valid solution.Just to be thorough, I'll plug (x = 2) back into the original equation to make sure both sides are equal.Left side:[frac{4(2)^2 - 3(2) + 2}{2 + 2} = frac{4(4) - 6 + 2}{4} = frac{16 - 6 + 2}{4} = frac{12}{4} = 3]Right side:[4(2) - 5 = 8 - 5 = 3]Both sides equal 3, so (x = 2) is indeed the correct solution.I think that's it. I don't see any mistakes in my steps, and the solution checks out. So, the solution is (x = 2).**Final Answer**The solution is (boxed{2}).

💡First, I need to determine the total distance the train travels while passing the bridge. This distance is the sum of the length of the train and the length of the bridge.Next, I'll calculate the total distance by adding the length of the train (400 meters) to the length of the bridge (800 meters), which gives a total distance of 1200 meters.Now, I'll use the formula for speed, which is distance divided by time. The train takes 72 seconds to cover 1200 meters. So, the speed in meters per second (m/s) is 1200 meters divided by 72 seconds, resulting in approximately 16.67 m/s.To convert this speed to kilometers per hour (km/h), I'll multiply the speed in m/s by the conversion factor of 3.6. Therefore, 16.67 m/s multiplied by 3.6 equals 60 km/h.Thus, the speed of the train is 60 kilometers per hour.

💡Okay, so I have this geometry problem here. Let me try to visualize it first. There's a figure with points A, B, C, and D. AB is parallel to DC, which probably means that lines AB and DC never meet and are always the same distance apart. Then, angle ACB is a right angle, so triangle ACB is a right-angled triangle. Also, AC equals CB, which tells me that triangle ACB is an isosceles right-angled triangle. That means the two legs AC and CB are equal, and the angles opposite them should each be 45 degrees. So, angle BAC and angle ABC are both 45 degrees.Next, it says AB equals BD. So, the length of segment AB is the same as the length of segment BD. I need to figure out the value of angle CBD, which is denoted as b degrees. Hmm, okay. Let me try to draw this figure step by step in my mind.First, I imagine triangle ACB with right angle at C. Since AC equals CB, it's an isosceles right triangle, so angles at A and B are each 45 degrees. Now, AB is parallel to DC. So, DC must be a line segment that's parallel to AB. Since AB is one side of the triangle, DC is probably extending from point D somewhere else in the figure.Given that AB equals BD, point D must be somewhere such that when you connect B to D, the length BD is equal to AB. So, BD is equal in length to AB, which is the hypotenuse of the isosceles right triangle ACB. That might mean that triangle ABD is also an isosceles triangle with AB equal to BD.Wait, but AB is parallel to DC. So, DC is parallel to AB, which is the hypotenuse of triangle ACB. Maybe DC is another side of a parallelogram? Or perhaps it's part of another triangle. I need to figure out how these points are connected.Let me try to sketch this mentally. Points A, B, and C form a right-angled isosceles triangle with the right angle at C. AB is the hypotenuse. Now, AB is parallel to DC, so DC must be a line segment starting from point D and going in the same direction as AB. Since AB is parallel to DC, the slope of AB should be the same as the slope of DC if we consider coordinate geometry.But maybe I don't need coordinates. Let me think about the angles. Since AB is parallel to DC, the angles formed by a transversal should be equal. For example, if I draw a line from C to D, that's a transversal cutting the parallel lines AB and DC. So, the corresponding angles should be equal. That might help me find some angle relationships.Given that AC equals CB, triangle ACB is isosceles, so angles at A and B are 45 degrees each. Now, angle ACB is 90 degrees. So, in triangle ACB, we have all the angles accounted for: 45, 45, and 90 degrees.Now, AB equals BD. So, BD is equal to AB. Since AB is the hypotenuse of triangle ACB, BD must be equal in length to that hypotenuse. So, point D is somewhere such that when you connect B to D, the length is equal to AB, and DC is parallel to AB.I think I need to consider triangle BCD. Since DC is parallel to AB, and AB equals BD, maybe triangle BCD has some special properties. Let me try to figure out the angles in triangle BCD.First, angle CBD is the angle we need to find, which is angle b. So, in triangle BCD, we have angle at B, which is angle CBD, angle at C, which might be related to angle ACB, and angle at D.Wait, angle ACB is 90 degrees, but that's in triangle ACB. How does that relate to triangle BCD? Maybe through some angle relationships because of the parallel lines.Since AB is parallel to DC, and CB is a transversal, the angle at C in triangle BCD might be equal to angle at A in triangle ACB. Because of the parallel lines, the alternate interior angles should be equal. So, angle BCD might be equal to angle BAC, which is 45 degrees.But wait, angle BCD is at point C, between points B, C, and D. If DC is parallel to AB, then the angle between CB and DC should be equal to the angle between CB and AB. But CB is connected to both A and D. Hmm, maybe I need to think about corresponding angles.Alternatively, maybe I can use the fact that AB equals BD and triangle ABD is isosceles. If AB equals BD, then triangle ABD has two equal sides, so the base angles are equal. That would mean angle at A and angle at D are equal. But AB is parallel to DC, so maybe angle at D is related to angle at C.This is getting a bit confusing. Maybe I should assign coordinates to the points to make it more concrete. Let's place point C at the origin (0,0). Since triangle ACB is a right-angled isosceles triangle with AC equal to CB, let's say AC is along the x-axis and CB is along the y-axis. So, point A is at (a, 0) and point B is at (0, a) for some positive value a.Then, AB is the hypotenuse from (a,0) to (0,a). The length of AB is sqrt((a)^2 + (a)^2) = a*sqrt(2). Now, AB is parallel to DC. So, DC must have the same slope as AB. The slope of AB is (a - 0)/(0 - a) = -1. So, DC must also have a slope of -1.Since DC is parallel to AB and has a slope of -1, and point C is at (0,0), point D must lie somewhere along the line y = -x + c for some constant c. But we need to determine the coordinates of D such that BD equals AB.Point B is at (0,a). So, BD is the distance from (0,a) to point D. Let's denote point D as (d, -d + c). But since DC is parallel to AB, and AB goes from (a,0) to (0,a), which is a line with slope -1, DC must also have slope -1. So, if point C is at (0,0), then point D must lie somewhere on the line y = -x.Wait, because if DC is parallel to AB, which has slope -1, and point C is at (0,0), then DC must lie along the line y = -x. So, point D must be at some point (d, -d). That simplifies things.So, point D is at (d, -d). Now, BD is the distance from point B (0,a) to point D (d, -d). The length of BD is sqrt((d - 0)^2 + (-d - a)^2) = sqrt(d^2 + ( -d - a)^2).We are told that AB equals BD. The length of AB is a*sqrt(2), as calculated earlier. So, we have:sqrt(d^2 + (-d - a)^2) = a*sqrt(2)Let's square both sides to eliminate the square roots:d^2 + (-d - a)^2 = (a*sqrt(2))^2Simplify the left side:d^2 + (d + a)^2 = 2a^2Expand (d + a)^2:d^2 + d^2 + 2ad + a^2 = 2a^2Combine like terms:2d^2 + 2ad + a^2 = 2a^2Subtract 2a^2 from both sides:2d^2 + 2ad + a^2 - 2a^2 = 0Simplify:2d^2 + 2ad - a^2 = 0Divide the entire equation by 2 to simplify:d^2 + ad - (a^2)/2 = 0Now, we have a quadratic equation in terms of d:d^2 + ad - (a^2)/2 = 0Let's solve for d using the quadratic formula. The quadratic is of the form:d^2 + (a)d - (a^2)/2 = 0So, coefficients are:A = 1, B = a, C = -a^2/2The quadratic formula is:d = [-B ± sqrt(B^2 - 4AC)] / (2A)Plugging in the values:d = [-a ± sqrt(a^2 - 4*1*(-a^2/2))]/2Simplify inside the square root:sqrt(a^2 - 4*(1)*(-a^2/2)) = sqrt(a^2 + 2a^2) = sqrt(3a^2) = a*sqrt(3)So, we have:d = [-a ± a*sqrt(3)] / 2This gives two solutions:d = [-a + a*sqrt(3)] / 2 and d = [-a - a*sqrt(3)] / 2Simplify:d = a(-1 + sqrt(3))/2 and d = a(-1 - sqrt(3))/2Since we're dealing with coordinates, we need to consider the position of point D. If we take d = a(-1 + sqrt(3))/2, that would place D in the fourth quadrant because x-coordinate is positive (since sqrt(3) ≈ 1.732, so -1 + 1.732 ≈ 0.732) and y-coordinate is -d, which would be negative. Alternatively, d = a(-1 - sqrt(3))/2 would place D in the third quadrant because both x and y would be negative.But in the original figure, AB is parallel to DC, and since AB is from (a,0) to (0,a), which is in the first quadrant, DC should be extending in a similar direction but from point C at (0,0). So, point D is likely in the fourth quadrant, making DC go from (0,0) to (d, -d) where d is positive. Therefore, we take the positive solution:d = a(-1 + sqrt(3))/2So, point D is at (d, -d) where d = a(-1 + sqrt(3))/2.Now, we need to find angle CBD, which is angle at point B between points C, B, and D.Point C is at (0,0), point B is at (0,a), and point D is at (d, -d). So, angle CBD is the angle between vectors BC and BD.Vector BC is from B to C: (0 - 0, 0 - a) = (0, -a)Vector BD is from B to D: (d - 0, -d - a) = (d, -d - a)To find angle between vectors BC and BD, we can use the dot product formula:cos(theta) = (BC . BD) / (|BC| |BD|)First, compute the dot product BC . BD:(0)(d) + (-a)(-d - a) = 0 + a(d + a) = a(d + a)Compute |BC|:|BC| = sqrt(0^2 + (-a)^2) = sqrt(a^2) = aCompute |BD|:|BD| = sqrt(d^2 + (-d - a)^2) = sqrt(d^2 + (d + a)^2)But we already know that |BD| = AB = a*sqrt(2)So, |BD| = a*sqrt(2)Therefore, cos(theta) = [a(d + a)] / [a * a*sqrt(2)] = (d + a) / (a*sqrt(2))Simplify:cos(theta) = (d + a)/(a*sqrt(2)) = (d/a + 1)/sqrt(2)We have d = a(-1 + sqrt(3))/2, so d/a = (-1 + sqrt(3))/2Therefore:cos(theta) = [(-1 + sqrt(3))/2 + 1]/sqrt(2) = [(-1 + sqrt(3) + 2)/2]/sqrt(2) = [(1 + sqrt(3))/2]/sqrt(2)Simplify numerator:(1 + sqrt(3))/2 divided by sqrt(2) is equal to (1 + sqrt(3))/(2*sqrt(2))We can rationalize the denominator:(1 + sqrt(3))/(2*sqrt(2)) * (sqrt(2)/sqrt(2)) = (sqrt(2) + sqrt(6))/4So, cos(theta) = (sqrt(2) + sqrt(6))/4Now, we need to find theta such that cos(theta) = (sqrt(2) + sqrt(6))/4I recall that cos(15°) = (sqrt(6) + sqrt(2))/4Yes, because cos(15°) can be derived using the cosine of difference formula:cos(45° - 30°) = cos45 cos30 + sin45 sin30 = (sqrt(2)/2)(sqrt(3)/2) + (sqrt(2)/2)(1/2) = sqrt(6)/4 + sqrt(2)/4 = (sqrt(6) + sqrt(2))/4So, cos(theta) = cos(15°), which means theta = 15°Therefore, angle CBD is 15 degrees. So, b = 15.Wait, let me double-check my steps to make sure I didn't make any mistakes.1. Assigned coordinates correctly: C at (0,0), A at (a,0), B at (0,a). That seems fine.2. Found slope of AB is -1, so DC must also have slope -1. Therefore, D lies on y = -x. That makes sense.3. Expressed D as (d, -d). Correct.4. Calculated BD distance and set it equal to AB. Correct.5. Solved quadratic equation for d, got two solutions, chose the positive one for D in fourth quadrant. That seems reasonable.6. Used vectors BC and BD to compute angle CBD. Correct approach.7. Calculated dot product, magnitudes, and found cos(theta) = (sqrt(2) + sqrt(6))/4, which is cos(15°). Therefore, theta = 15°. That seems correct.I think all the steps are valid. So, the value of b is 15 degrees.