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mart [117]
3 years ago
14

A rectangular piece of paper with length 35 cm and width 14 cm has two semicircles cut out of it, as shown below.

Mathematics
1 answer:
Ede4ka [16]3 years ago
3 0

Answer: 336.14 cm²

Step-by-step explanation:

To find the area of the rectangle after being cut, we want to find the area of the two semicircles and subtract it from the area of the rectangle. The area of the rectangle is just base times height, or 35cm times 14cm = 490cm² . Since there are two semicircles with the same diameter, we can just solve for the area of a circle and subtract it. To find the area of the circle, we need the radius, which we get by dividing the diameter by 2. After that, we calculate the radius to be 7cm, squared and multiplied by 3.14 (area of a circle) to get 153.86 cm². Subtract the areas, and we get 490 - 153.86 = 336.14 cm²

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Answer:

y \leq 6 + \frac{3x}{4}

Step-by-step explanation:

3x - 4y \geq 24

-3x                 -3x

(-4y \geq 24 - 3x ) / -4

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An 8 foot square floor is to be covered with square tiles measuring 8 inches on each side. If each tile costs 50 cents, how much
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9514 1404 393

Answer:

  (c) $72

Step-by-step explanation:

Each tile is 8/12 ft = 2/3 ft on a side. Then 8/(2/3) = 12 tiles will fit along each edge of the square area to be tiled. That is ...

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3 years ago
A farmer with 950 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to o
Gelneren [198K]

Answer:

22,562.5 ft²

Step-by-step explanation:

The largest possible area will be obtained when half the fencing is used for the long side of the 4 pens, so the dimension in that direction is (950 ft)/(2·2) = 237.5 ft.

The other half of the fencing will be used for the 2 ends and 3 partitions, each of which will be (950 ft)/(2·5) = 95 ft.

Then the overall area of the 4 pens is ...

... (237.5 ft)(95 ft) = 22,562.5 ft²

_____

<em>General Solution</em>

Suppose L is the length of fence available, and x is the length of the long side of the enclosed area. For n pens, the enclosed area will be ...

... A = x(L-2x)/(n+1)

For constant values of A, L, n, this describes a downward-opening parabola with zeros at x=0 and x=L/2. The vertex of the parabola (point of maximum area) will be halfway between these zeros, at x = (0 + L/2)/2 = L/4. That is, half the available fence is used in each of the orthogonal directions.

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... A = x(L -(3+1)x)/(4+1)

and, once again, we find that half the fence is used in each of the orthogonal directions when we maximize the overall area.

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