Hello everyone! I'm asking for your help—I am struggling to solve our assignment. please help me answer the following questions below:1. The area of a concrete rectangular pathway is 350 m? and its perimeter pathwayis 90 m.Whatis the length of the pathway?2. A reamgularlothas an area of 240 m. What is the width of the lot if it requires 64 m offencing materials to enclose it?3. Thearea of a grdenis 160 m':.Suppose he length of the garden is 3 m more than twice itswidth. What is the length of the garden?4. The length of a tarpaulin is 3 ft. more than thrice its width and its area is 126 ft ?.What is the length of the tarpaulin?5. Mario and Kenneth work in a car wash station. The time that Mario takes in washinga caralone is 20 minutes less than the time that Kenneth takes in washing the same car. riainasthem work together in washing the car, it will take them 90 minutes How long willit talke each of them to wash the car?
Answer (1)
Answer:See below. Two could not be answered because the information was not sufficient (problems 1 and 2).Step-by-step explanation:1. The area of a concrete rectangular pathway is 350 m and its perimeter pathway is 90 m.What is the length of the pathway?The layout, as described, is unclear. what is a "perimeter" pathway relative to the "pathway?"2. A reamgularlothas an area of 240 m. What is the width of the lot if it requires 64 m of fencing materials to enclose it? What is a "reamgularlothas"? 3. The area of a garden is 160 m^2. Suppose the length of the garden is 3 m more than twice its width. What is the length of the garden?The garden described is in the shape of a rectangle. The area of a rectangle is: A = L*W, where L and W are Length and Width.We are told A = 160 m^2: so we can write: 1) 160m^2 = L*WWe also learn that L = 2W + 3m ["the length of the garden is 3m more than twice its width"]2) L = 2W + 3mLet's use the definition of L from (2) [2W + 3m] and use it in (1)1) 160m^2 = L*W 160m^2 = [2W + 3m]*W 160m^2 = 2W^2 + W*3m Reformat to: 2W^2 + (3m)W - 160m^2 = 0 Solve with the quadratic equation: W = 8.225m and W = - 9.725mWe can ignore the negative value. So the width (W) is 8.23 m.We want the length (L), so: 2) L = 2W + 3m 2) L = 2*(8.23m) + 3m L (length) = 19.46m4. The length of a tarpaulin is 3 ft. more than thrice its width and its area is 126 ft ?.What is the length of the tarpaulin?The tarp is a rectangle, and the area, A, is given by Length x Width, or A = L*WWe find that A = 126 ft^2 [The "^" was added to make this an area unit]Now we can write (1) 126 ft^2 = L*WWe learn that "length [L] of a tarpaulin is 3 ft. more than thrice its width [W]. We can write that as: (2) L = 3*W + 3 ftSubstitute (2) into (1): 126 ft^2 = L*W 126 ft^2 = (3*W + 3 ft)*W Simplify: 126 ft^2 = 3*W^2 + 3W ft Rearrange: 3*W^2 + 3W ft - 126 ft^2 = 0Solve with quadratic equation: X = 6 ft and x = -7 ftWe can ignore the minus 7 feet. The Width is 6 feet. Since from (1): 126 ft^2 = L*W 126 ft^2 = L*(6 ft) L = 21 feet5. Mario and Kenneth work in a car wash station. The time that Mario takes in washing a car alone is 20 minutes less than the time that Kenneth takes in washing the same car. If the two of them work together in washing the car, it will take them 90 minutes How long will it take each of them to wash the car?Let M and K stand for the rates Mario (M) and Kenneth) wash a car by themselves. The units would be cars/minute. The minutes required to wash 1 car are t for Kenneth and (t - 20) for MarioWe are told that We can express these as individual rates: Mario's rate is (1 car)/((t-20) min) Kenneth's rate is (1 car)/(t min)Working together, Kenneth and Mario can wash 1 car in 90 minutes.This means that: Mario's Rate + Kenneth's Rate = 1 car/90 minutes (1 car)/((t-20) min) + (1 car)/(t min) = (1 car/20 min)The common denominator is t(t-20)*20. Let's convert each term to the common denominator [I will drop the units for clarity, here]: Eventually, this reduces to t^2 - 60t + 400 = 0 [The common denominator is t(t-20)20]The quadratic solutions are 52.4 and 7.6 minutes. The 7.6 minute value for t is not possible, since t is the faster person's (Kenneth) time, and subtracting 20 from that would be a negative time.We are left with Kenneth's time of 52.4 minutes and Mario's time of (52.4 - 20) = 32.4 minutes.
