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pashok25 [27]
3 years ago
14

farmer joe is enclosing a rectangular area on his arm for his chicken with 200 feet of fencing that he recently acquired in a tr

ade. two equal lengths of fencing, of unknown length x ft, will run perpendicular to the side of the barn, and a single length of fencing of unknown length (200-2x) ft will run parallel to the side of the barn. to the nearest square foot, what is the maximum possible area that joe can enclose with his 200ft of fencing
Mathematics
1 answer:
Llana [10]3 years ago
8 0
For this case, the area is given by:
 A = x * (200-2x)
 Rewriting:
 A = 200x-2x ^ 2
 Deriving the expression we have:
 A '= 200-4x
 Equaling zero we have:
 200-4x = 0
 We clear x:
 4x = 200
 x = 200/4
 x = 50 feet
 Then, the maximum area is:
 A (50) = 50 * (200-2 * 50)
 A (50) = 5000 feet ^ 2
 Answer:
 
the maximum possible area that can be enclosed with his 200ft of fencing is:
 A
 (50) = 5000 feet ^ 2
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Answer:

B.

Step-by-step explanation:

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Then the time taken in all levels of game = (7\frac{1}{3}) l<em> </em>

Hence, the total time ( in minutes ) = Set up time + time in all level

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We have 60-minute of free trial of the game,

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2 years ago
Which two tables represent the same function?
topjm [15]

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The 1st and the 5th tables represent the same function

Step-by-step explanation:

* Lets explain how to solve the problem

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- By using cross multiplication

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∴ 2y - 16 = -x + 4 ⇒ add x and 16 for two sides

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# (x1 , y1) = (4 , 5) and (x2 , y2) = (6 , 4)

∵ x1 = 4 , x2 = 6 and y1 = 5 , y2 = 4

∴ \frac{y-5}{x-4}=\frac{4-5}{6-4}

∴ \frac{y-5}{x-4}=\frac{-1}{2}

- By using cross multiplication

∴ 2(y - 5) = -1(x - 4) ⇒ simplify

∴ 2y - 10 = -x + 4 ⇒ add x and 10 for two sides

∴ x + 2y = 14 ⇒ (2)

# (x1 , y1) = (2 , 8) and (x2 , y2) = (8 , 5)

∵ x1 = 2 , x2 = 8 and y1 = 8 , y2 = 5

∴ \frac{y-8}{x-2}=\frac{5-8}{8-2}

∴ \frac{y-8}{x-2}=\frac{-3}{6}=====\frac{y-8}{x-2}=\frac{-1}{2}

- By using cross multiplication

∴ 2(y - 8) = -1(x - 2) ⇒ simplify

∴ 2y - 16 = -x + 2 ⇒ add x and 16 for two sides

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# (x1 , y1) = (2 , 10) and (x2 , y2) = (6 , 14)

∵ x1 = 2 , x2 = 6 and y1 = 10 , y2 = 14

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∴ \frac{y-10}{x-2}=\frac{4}{4}======\frac{y-10}{x-2}=1

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∴ (y - 10) = (x - 2)

∴ y - 10 = x - 2 ⇒ add 2 and subtract y in the two sides

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2 years ago
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3 years ago
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Tanya [424]

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3 years ago
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