💡First, I need to determine the percentage discount for each item using the formula:Percentage Discount = [(Original Price - Discounted Price) / Original Price] × 100For Item A:- Original Price = 150- Discounted Price = 135- Discount Amount = 150 - 135 = 15- Percentage Discount = (15 / 150) × 100 = 10%For Item B:- Original Price = 200- Discounted Price = 170- Discount Amount = 200 - 170 = 30- Percentage Discount = (30 / 200) × 100 = 15%For Item C:- Original Price = 120- Discounted Price = 102- Discount Amount = 120 - 102 = 18- Percentage Discount = (18 / 120) × 100 = 15%For Item D:- Original Price = 250- Discounted Price = 212.50- Discount Amount = 250 - 212.50 = 37.50- Percentage Discount = (37.50 / 250) × 100 = 15%After calculating the percentage discounts, I observe that Items B, C, and D all have the highest percentage discount of 15%, while Item A has a lower discount of 10%.Therefore, the items with the highest percentage discount are Items B, C, and D, each with a 15% discount.

💡To calculate the area of the shaded T-shaped region within the rectangle (ABCD), we can follow these steps:1. **Calculate the total area of the large rectangle (ABCD):** [ text{Area}_{ABCD} = 6 times 5 = 30 text{ square units} ]2. **Calculate the areas of the smaller rectangles:** - **Rectangle on the top left with dimensions (1 times 4):** [ text{Area}_{1} = 1 times 4 = 4 text{ square units} ] - **Rectangle on the bottom right with dimensions (1 times 4):** [ text{Area}_{2} = 1 times 4 = 4 text{ square units} ] - **Rectangle on the right with dimensions (1 times 3):** [ text{Area}_{3} = 1 times 3 = 3 text{ square units} ]3. **Calculate the area of the shaded T-shaped region:** [ text{Area}_{text{T-shape}} = text{Area}_{ABCD} - (text{Area}_{1} + text{Area}_{2} + text{Area}_{3}) = 30 - (4 + 4 + 3) = 19 text{ square units} ]Therefore, the area of the shaded T-shaped region is (boxed{19}) square units.

💡Okay, so I'm trying to figure out the length of this wire stretched between two poles. Let me start by visualizing the problem. There are two poles, one is 10 feet tall, and the other is 15 feet tall. They're 20 feet apart on flat ground. The wire is stretched from the top of the shorter pole to a point on the taller pole, and it makes a 30-degree angle with the horizontal line connecting the bases of the poles.Hmm, so I think I can model this as a right triangle. The horizontal distance between the poles is 20 feet, which would be one leg of the triangle. The vertical difference in height between the two poles is 15 minus 10, which is 5 feet. So, that's the other leg of the triangle. The wire itself would be the hypotenuse of this right triangle.But wait, the wire makes a 30-degree angle with the horizontal. That means I can use trigonometry here. Specifically, since I know the angle and the adjacent side (the horizontal distance), I can use the cosine function to find the length of the wire.Let me recall, cosine of an angle in a right triangle is equal to the adjacent side divided by the hypotenuse. So, cos(30°) = adjacent / hypotenuse. Here, the adjacent side is 20 feet, and the hypotenuse is the length of the wire, which I'll call 'w'.So, cos(30°) = 20 / w. I know that cos(30°) is √3 / 2. So, substituting that in, I get √3 / 2 = 20 / w.Now, I need to solve for 'w'. Let's rearrange the equation: w = 20 / (√3 / 2). Dividing by a fraction is the same as multiplying by its reciprocal, so this becomes w = 20 * (2 / √3) = 40 / √3.But wait, it's not good practice to have a square root in the denominator. I should rationalize the denominator. To do that, I'll multiply both the numerator and the denominator by √3. So, w = (40 * √3) / (√3 * √3) = (40√3) / 3.So, the length of the wire is 40√3 divided by 3 feet. Let me double-check my steps to make sure I didn't make a mistake. I used the cosine function correctly, substituted the known values, solved for 'w', and rationalized the denominator. It all seems to add up.I think that's it. The wire is (40√3)/3 feet long.

💡Okay, so I have this problem here: A circle has an area of (4pi) square units. I need to find the length of the circle's diameter. Hmm, let me think about how to approach this.First, I remember that the area of a circle is calculated using the formula (A = pi r^2), where (A) is the area and (r) is the radius of the circle. Since the problem gives me the area, which is (4pi), I can set up the equation like this:[pi r^2 = 4pi]Alright, now I need to solve for (r). To do that, I can start by dividing both sides of the equation by (pi) to simplify things. Let's see:[frac{pi r^2}{pi} = frac{4pi}{pi}]This simplifies to:[r^2 = 4]Okay, so now I have (r^2 = 4). To find the value of (r), I need to take the square root of both sides. Taking the square root of 4 gives me 2, right? So,[r = sqrt{4} = 2]Wait a second, is that all? I think so. The radius (r) is 2 units. But the question asks for the diameter, not the radius. I remember that the diameter (d) of a circle is twice the radius. So, I can calculate the diameter by multiplying the radius by 2:[d = 2r = 2 times 2 = 4]So, the diameter of the circle is 4 units. Let me just double-check my steps to make sure I didn't make any mistakes. Starting with the area formula, plugging in the given area, solving for the radius, and then finding the diameter by doubling the radius. Everything seems to add up. Yeah, I think that's correct.

💡Okay, so I have this problem here: Given the function ( f(x) = sin omega x cos omega x - sin^2 omega x ) where ( omega > 0 ). It says that the function ( f(x) ) is monotonically decreasing on the interval ( left( frac{pi}{2}, pi right) ). I need to find the range of the real number ( omega ). The options are A: ( (0, frac{1}{2}] ), B: ( (0, frac{1}{4}] ), C: ( [frac{1}{4}, frac{5}{8}] ), and D: ( [frac{1}{2}, frac{5}{4}] ).Alright, let me try to figure this out step by step. First, I should probably simplify the function ( f(x) ) to make it easier to analyze. The function is given as ( sin omega x cos omega x - sin^2 omega x ). Hmm, I remember some trigonometric identities that might help here.I recall that ( sin 2theta = 2 sin theta cos theta ), so maybe I can rewrite ( sin omega x cos omega x ) using that identity. Let me try that:( sin omega x cos omega x = frac{1}{2} sin 2omega x ).Okay, so that simplifies the first term. Now, the second term is ( sin^2 omega x ). I remember another identity: ( sin^2 theta = frac{1 - cos 2theta}{2} ). Let me apply that here:( sin^2 omega x = frac{1 - cos 2omega x}{2} ).So substituting these back into the original function, we have:( f(x) = frac{1}{2} sin 2omega x - frac{1 - cos 2omega x}{2} ).Let me simplify this further:( f(x) = frac{1}{2} sin 2omega x - frac{1}{2} + frac{1}{2} cos 2omega x ).Combine the constants:( f(x) = frac{1}{2} (sin 2omega x + cos 2omega x) - frac{1}{2} ).Hmm, this looks a bit more manageable. Maybe I can write this as a single sine function using the amplitude-phase form. I remember that ( A sin theta + B cos theta = C sin(theta + phi) ) where ( C = sqrt{A^2 + B^2} ) and ( phi = arctan left( frac{B}{A} right) ) or something like that.In this case, ( A = frac{1}{2} ) and ( B = frac{1}{2} ), so ( C = sqrt{ (frac{1}{2})^2 + (frac{1}{2})^2 } = sqrt{ frac{1}{4} + frac{1}{4} } = sqrt{ frac{1}{2} } = frac{sqrt{2}}{2} ).And the phase shift ( phi ) is ( arctan left( frac{B}{A} right) = arctan(1) = frac{pi}{4} ).So, ( sin 2omega x + cos 2omega x = sqrt{2} sin left( 2omega x + frac{pi}{4} right) ).Wait, but in our function, we have ( frac{1}{2} (sin 2omega x + cos 2omega x) ), so that would be ( frac{sqrt{2}}{2} sin left( 2omega x + frac{pi}{4} right) ).Therefore, the function simplifies to:( f(x) = frac{sqrt{2}}{2} sin left( 2omega x + frac{pi}{4} right) - frac{1}{2} ).Alright, so now we have ( f(x) ) expressed as a sine function with some amplitude and phase shift, minus a constant. Since we're interested in the monotonicity of ( f(x) ), we should probably look at its derivative.Let me compute the derivative ( f'(x) ):( f'(x) = frac{sqrt{2}}{2} cdot 2omega cos left( 2omega x + frac{pi}{4} right) ).Simplifying:( f'(x) = sqrt{2} omega cos left( 2omega x + frac{pi}{4} right) ).For ( f(x) ) to be monotonically decreasing on ( left( frac{pi}{2}, pi right) ), the derivative ( f'(x) ) must be less than or equal to zero on that interval. So, we have:( sqrt{2} omega cos left( 2omega x + frac{pi}{4} right) leq 0 ) for all ( x in left( frac{pi}{2}, pi right) ).Since ( sqrt{2} omega ) is positive (because ( omega > 0 )), this inequality simplifies to:( cos left( 2omega x + frac{pi}{4} right) leq 0 ) for all ( x in left( frac{pi}{2}, pi right) ).So, we need ( cos theta leq 0 ) where ( theta = 2omega x + frac{pi}{4} ) for ( x in left( frac{pi}{2}, pi right) ).The cosine function is less than or equal to zero in the intervals ( [frac{pi}{2} + 2kpi, frac{3pi}{2} + 2kpi] ) for integers ( k ). Therefore, we need:( frac{pi}{2} + 2kpi leq 2omega x + frac{pi}{4} leq frac{3pi}{2} + 2kpi ) for all ( x in left( frac{pi}{2}, pi right) ).Let me rewrite this inequality:( frac{pi}{2} + 2kpi leq 2omega x + frac{pi}{4} leq frac{3pi}{2} + 2kpi ).Subtract ( frac{pi}{4} ) from all parts:( frac{pi}{2} - frac{pi}{4} + 2kpi leq 2omega x leq frac{3pi}{2} - frac{pi}{4} + 2kpi ).Simplify:( frac{pi}{4} + 2kpi leq 2omega x leq frac{5pi}{4} + 2kpi ).Divide all parts by 2:( frac{pi}{8} + kpi leq omega x leq frac{5pi}{8} + kpi ).Now, since ( x ) is in ( left( frac{pi}{2}, pi right) ), let's express the inequalities in terms of ( omega ).First, let's consider the lower bound:( omega x geq frac{pi}{8} + kpi ).Since ( x > frac{pi}{2} ), the smallest ( omega x ) can be is ( omega cdot frac{pi}{2} ). So,( omega cdot frac{pi}{2} geq frac{pi}{8} + kpi ).Similarly, the upper bound:( omega x leq frac{5pi}{8} + kpi ).Since ( x < pi ), the largest ( omega x ) can be is ( omega cdot pi ). So,( omega cdot pi leq frac{5pi}{8} + kpi ).Let me write these two inequalities:1. ( frac{pi}{2} omega geq frac{pi}{8} + kpi )2. ( pi omega leq frac{5pi}{8} + kpi )Let's simplify both inequalities by dividing through by ( pi ):1. ( frac{1}{2} omega geq frac{1}{8} + k )2. ( omega leq frac{5}{8} + k )Now, let's solve for ( omega ):From the first inequality:( omega geq frac{1}{4} + 2k ).From the second inequality:( omega leq frac{5}{8} + k ).So, combining these, we have:( frac{1}{4} + 2k leq omega leq frac{5}{8} + k ).Now, we need to find integer values of ( k ) such that this inequality is possible. Let's consider different integer values for ( k ).First, let's try ( k = 0 ):Then, the inequality becomes:( frac{1}{4} leq omega leq frac{5}{8} ).That's a valid range for ( omega ). Let's check if this works.Next, let's try ( k = 1 ):Then, the inequality becomes:( frac{1}{4} + 2(1) = frac{9}{4} leq omega leq frac{5}{8} + 1 = frac{13}{8} ).But ( frac{9}{4} = 2.25 ) and ( frac{13}{8} = 1.625 ), which is not possible because ( 2.25 > 1.625 ). So, no solution for ( k = 1 ).Similarly, for ( k = -1 ):( frac{1}{4} + 2(-1) = frac{1}{4} - 2 = -frac{7}{4} leq omega leq frac{5}{8} + (-1) = frac{5}{8} - 1 = -frac{3}{8} ).But ( omega > 0 ), so this range is invalid.Therefore, the only valid integer value for ( k ) is ( k = 0 ), giving us ( frac{1}{4} leq omega leq frac{5}{8} ).So, the range of ( omega ) is ( [frac{1}{4}, frac{5}{8}] ), which corresponds to option C.Wait, let me double-check if this makes sense. If ( omega ) is in that range, then ( 2omega x + frac{pi}{4} ) should lie in the interval where cosine is non-positive, i.e., between ( frac{pi}{2} ) and ( frac{3pi}{2} ).Let's take ( omega = frac{1}{4} ). Then, for ( x = frac{pi}{2} ):( 2 cdot frac{1}{4} cdot frac{pi}{2} + frac{pi}{4} = frac{pi}{4} + frac{pi}{4} = frac{pi}{2} ).And for ( x = pi ):( 2 cdot frac{1}{4} cdot pi + frac{pi}{4} = frac{pi}{2} + frac{pi}{4} = frac{3pi}{4} ).So, the argument of cosine goes from ( frac{pi}{2} ) to ( frac{3pi}{4} ), which is within the interval where cosine is negative or zero. Hence, the derivative is non-positive, so ( f(x) ) is decreasing.Similarly, for ( omega = frac{5}{8} ):At ( x = frac{pi}{2} ):( 2 cdot frac{5}{8} cdot frac{pi}{2} + frac{pi}{4} = frac{5pi}{8} + frac{pi}{4} = frac{7pi}{8} ).At ( x = pi ):( 2 cdot frac{5}{8} cdot pi + frac{pi}{4} = frac{5pi}{4} + frac{pi}{4} = frac{6pi}{4} = frac{3pi}{2} ).So, the argument goes from ( frac{7pi}{8} ) to ( frac{3pi}{2} ), which is still within the interval where cosine is negative or zero. Therefore, the derivative remains non-positive, and ( f(x) ) is decreasing.If ( omega ) were larger than ( frac{5}{8} ), say ( omega = frac{3}{4} ), then at ( x = pi ):( 2 cdot frac{3}{4} cdot pi + frac{pi}{4} = frac{3pi}{2} + frac{pi}{4} = frac{7pi}{4} ).But ( frac{7pi}{4} ) is in the fourth quadrant where cosine is positive, so the derivative would be positive there, meaning ( f(x) ) would be increasing at that point, which contradicts the requirement that ( f(x) ) is decreasing on the entire interval.Similarly, if ( omega ) were smaller than ( frac{1}{4} ), say ( omega = frac{1}{8} ), then at ( x = frac{pi}{2} ):( 2 cdot frac{1}{8} cdot frac{pi}{2} + frac{pi}{4} = frac{pi}{8} + frac{pi}{4} = frac{3pi}{8} ).Which is less than ( frac{pi}{2} ), so cosine is positive there, making the derivative positive, which again contradicts the requirement.Therefore, the range ( [frac{1}{4}, frac{5}{8}] ) is indeed the correct range for ( omega ) to ensure that ( f(x) ) is monotonically decreasing on ( left( frac{pi}{2}, pi right) ).**Final Answer**boxed{C}

💡First, I need to determine the running speed. Since it takes 30 minutes to run 4 miles, the speed can be calculated by dividing the distance by the time.Next, I'll find the distance to the library, which is exactly halfway between the house and the park. This means the library is 2 miles away from the house.Using the calculated speed, I'll determine the time it takes to run from the house to the library.Finally, I'll calculate the total time by adding the time to reach the library and the time to go from the library to the park.