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Solve for 4w2-2 equals -7w?
4w - 2 = -7w11w - 2 = 011w = 2w = 2/11
What is the area of a poster that has a Length equals 2 times w and Width equals w plus 2?
Stating our known facts Let W = Width, and L = Length L = 2W, and W = W+2 ----What you given is a never ending loop, saying that W always equals itself + 2. I will assume that you meant to put W = L+2 instead. Letting L = 2W and W = L + 2, we can state that x (the total area) is defined asL W(substituted so we're only working for one term)x = (2W)(2W + 2) Simplifyingx = 4W2 + 4W Since we are left with 2 variables, it can be stated that there is not enough information to solve the problem.
A rectangle has an area of 124 cm2 has a length that is 4 times the width how long is the width also-round to the nearest tenth--answer options--5.6cm OR 22.3cm OR 11.1cm and 44.5cm?
124=4wxw 124/4=4w2/4 31=w2 5.56=w L=22.4 W=5.6
The length of a garden is 4 times its width the area of the garden is 256 sq feet how many feet of fence should you buy?
Let W = width of garden. Therefore 4W = length of garden. Area is length x width, ie W x 4W = 256 sq ft. Rewrite as: 4W2 = 256 W2 = 256/4 = 64, therefore W = 8ft, and length = 32ft. So total length of fence required is (2x8) + (2x32) = 80 ft. Happy building!
A rectangular swimming pool has a length that is four feet shorter than four times its width the surface area of the pool is 528 square ft find the dimensions of the pool?
length = (w*4)-4 area = 528ft2 W*L= area and since length is replaced with (w*4)-4 so W*((w*4)-4)=area so when you multiply w through you get 4w2 -4w -528 = 0 you can use the quadratic equation now to find the width then its easy to find length. Then you can test to find area.
Solve for 4w2-2 equals -7w?
4w - 2 = -7w11w - 2 = 011w = 2w = 2/11
How would you describe a rectangle?
(w+5)(4w+2) - 270 = (4w)(w) 4w2 + 22w + 10 - 270 = 4w2 22w = 260 w = 11.81818181 Length = 4w = 47.2727 or 47 3/11 Width = 11.8181 or 11 9/11
What is the area of a poster that has a Length equals 2 times w and Width equals w plus 2?
Stating our known facts Let W = Width, and L = Length L = 2W, and W = W+2 ----What you given is a never ending loop, saying that W always equals itself + 2. I will assume that you meant to put W = L+2 instead. Letting L = 2W and W = L + 2, we can state that x (the total area) is defined asL W(substituted so we're only working for one term)x = (2W)(2W + 2) Simplifyingx = 4W2 + 4W Since we are left with 2 variables, it can be stated that there is not enough information to solve the problem.
A rectangle has an area of 124 cm2 has a length that is 4 times the width how long is the width also-round to the nearest tenth--answer options--5.6cm OR 22.3cm OR 11.1cm and 44.5cm?
124=4wxw 124/4=4w2/4 31=w2 5.56=w L=22.4 W=5.6
The length of a garden is 4 times its width the area of the garden is 256 sq feet how many feet of fence should you buy?
Let W = width of garden. Therefore 4W = length of garden. Area is length x width, ie W x 4W = 256 sq ft. Rewrite as: 4W2 = 256 W2 = 256/4 = 64, therefore W = 8ft, and length = 32ft. So total length of fence required is (2x8) + (2x32) = 80 ft. Happy building!
A rectangular swimming pool has a length that is four feet shorter than four times its width the surface area of the pool is 528 square ft find the dimensions of the pool?
length = (w*4)-4 area = 528ft2 W*L= area and since length is replaced with (w*4)-4 so W*((w*4)-4)=area so when you multiply w through you get 4w2 -4w -528 = 0 you can use the quadratic equation now to find the width then its easy to find length. Then you can test to find area.
