answer:1. **Position and Distance Setup**: Place A at (0,0) and B at (8,0). We need to find C = (x, y) satisfying specific perimeter and area conditions. 2. **Area Condition Setup**: [ text{Area} = frac{1}{2} left| 8y right| = 4|y| ] Setting this equal to 80, we solve: [ 4|y| = 80 implies |y| = 20 implies y = 20 text{ or } y = -20 ] 3. **Perimeter Condition Setup**: [ AB = 8, quad AC = sqrt{x^2 + y^2}, quad BC = sqrt{(x-8)^2 + y^2} ] Using y = 20 or y = -20, then: [ AC = BC = sqrt{x^2 + 400} ] The perimeter is: [ P = 8 + 2sqrt{x^2 + 400} ] Setting this equal to 40: [ 8 + 2sqrt{x^2 + 400} = 40 implies 2sqrt{x^2 + 400} = 32 implies sqrt{x^2 + 400} = 16 implies x^2 + 400 = 256 implies x^2 = -144 ] Since x^2 = -144 is impossible, no real solutions for x exist. 4. **Conclusion**: There are no points C such that triangle ABC has a perimeter of 40 units and an area of 80 square units. Therefore, the answer is 0. The final answer is boxed{textbf{(A) }0}

answer:Given that the quadrilateral provided can be cut by two straight lines into 6 pieces, we consider a specific case of a rectangle inscribed in an isosceles right triangle. 1. Assume that the sides of the rectangle are related by the ratio (5:2). 2. Suppose we have an isosceles right triangle (triangle ABC) with the hypotenuse (AB) equal to 45 units. 3. The right triangle is inscribed in such a way that two vertices of the rectangle lie on the hypotenuse (AB), and the other two vertices lie on the legs (BC) and (AC). Let's denote the vertices of the rectangle as (M, N, K, L) such that: - (M) and (N) are on the hypotenuse (AB), - (K) is on (BC), and - (L) is on (AC). Assume the segments: [ AM = ML = NK = NB = 2x ] [ MN = KL = 5x ] Given that the sum of the segments ( AM + MN + NB) equals the hypotenuse (AB): Now, we set up the equation: [ AM + MN + NB = 45 ] [ 2x + 5x + 2x = 45 ] 2. Simplify the equation to find (x): [ 9x = 45 ] [ x = frac{45}{9} ] [ x = 5 ] 3. From the value of (x), calculate the lengths of the sides of the rectangle: [ LM = KN = 2x = 2 cdot 5 = 10 ] [ MN = KL = 5x = 5 cdot 5 = 25 ] Thus, the sides of the rectangle are: [ text{Length} = 25 ] [ text{Width} = 10 ] 4. If considering another scenario where (MN) might be the smaller side: Here, the same calculations show enough symmetry, reaffirming if needed through an illustrative diagram. # Conclusion: Therefore, the lengths of the sides of the rectangle inscribed in the given triangle are: [ boxed{25text{ and }10} ] For a different configuration involving another ratio or setup: [ boxed{18.75text{ and }7.5} ]

answer:Let's break down the tasks and their respective times: 1. Assembling ingredients: 1 hour 2. Baking cakes in each oven: 1.5 hours (under normal circumstances) 3. Decorating each cake: 1 hour 4. Handling two special orders: 2 * 0.5 hours = 1 hour Now, let's calculate the adjusted baking times for each oven due to the malfunctions: 1. First oven: 1.5 hours * 2 = 3 hours 2. Second oven: 1.5 hours * 3 = 4.5 hours 3. Third oven: 1.5 hours * 4 = 6 hours Since Matthew can bake in all three ovens simultaneously, the total baking time will be determined by the oven that takes the longest, which is the third oven at 6 hours. Now, let's add up all the times: 1. Assembling ingredients: 1 hour 2. Baking cakes (longest oven time): 6 hours 3. Decorating cakes: 1 hour (this can be done while the last batch is baking, so it doesn't add to the total time) 4. Handling special orders: 1 hour Total time = Assembling ingredients + Baking cakes + Handling special orders Total time = 1 hour + 6 hours + 1 hour Total time = 8 hours So, it took Matthew boxed{8} hours to complete all his tasks on the day the ovens failed and he had to handle two special orders.

answer:When the graph of f(x) = cos(ωx) is shifted to the left by frac{pi}{3omega} units, the resulting function is g(x) = cos(ω(x + frac{pi}{3omega})) = cos(ωx + frac{pi}{3}). From 2kπ - π ≤ ωx + frac{pi}{3} ≤ 2kπ, k ∈ ℤ, we obtain frac{2kpi}{omega} - frac{4pi}{3omega} ≤ x ≤ frac{2kpi}{omega} - frac{pi}{3omega}, k ∈ ℤ. Thus, the intervals where the function is monotonically increasing are [frac{2kpi}{omega} - frac{4pi}{3omega}, frac{2kpi}{omega} - frac{pi}{3omega}], k ∈ ℤ. If y = g(x) is monotonically increasing in the interval (frac{2pi}{3}, frac{4pi}{3}), then we have: begin{cases} frac{2kpi}{omega} - frac{4pi}{3omega} leq frac{2pi}{3} frac{2kpi}{omega} - frac{pi}{3omega} geq frac{4pi}{3} end{cases}, which simplifies to begin{cases} omega geq 3k - 2 omega leq frac{3k}{2} - frac{1}{4} end{cases}. Since ω > 0, when k = 1, we have begin{cases} omega geq 1 omega leq frac{5}{4} end{cases}, which means 1 ≤ ω ≤ frac{5}{4}. When k = 2, we have begin{cases} omega geq 4 omega leq frac{11}{4} end{cases}, which yields no solution. Hence, the range of ω is [1, frac{5}{4}]. Therefore, the correct answer is boxed{D. [1, frac{5}{4}]}.