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ololo11 [35]
3 years ago
14

A rectangular park is 70 meters wide and 100 meters long.

Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
6 0

9514 1404 393

Answer:

  60 m wide and 110 m long

Step-by-step explanation:

The more the shape deviates from a square, the smaller the area will be. The area can be reduced by making the park longer and narrower. The sum of length and width must remain the same for the perimeter to be unchanged.

Given park: 70 m × 100 m = 7000 m²

Smaller park: 60 m × 110 m = 6600 m² -- less area than 7000 m²

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d=rt

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3 years ago
The ammunition storage room has 10 feet between the floor and the ceiling. Each box of ammunition is 10 feet tall. Each crate of
KIM [24]

Answer:

the maximum number of crates that can be stacked between the floor and ceiling

\frac{height \hspace{0.1cm} of  \hspace{0.1cm} ammunition \hspace{0.1cm} box}{height \hspace{0.1cm} of  \hspace{0.1cm} each \hspace{0.1cm} crate}  = \frac{10 \hspace{0.1cm}  feet}{12 \hspace{0.1cm}  inches}  =  \frac{10 \times 12  \hspace{0.1cm} inches }{12  \hspace{0.1cm} inches}  = 10 \hspace{0.1cm}  crates

where 1 foot  = 12 inches. SO the answer is that a maximum of 10 crates can be stacked from floor to ceiling.

Step-by-step explanation:

i) the maximum number of crates that can be stacked between the floor and ceiling

\frac{height \hspace{0.1cm} of  \hspace{0.1cm} ammunition \hspace{0.1cm} box}{height \hspace{0.1cm} of  \hspace{0.1cm} each \hspace{0.1cm} crate}  = \frac{10 \hspace{0.1cm}  feet}{12 \hspace{0.1cm}  inches}  =  \frac{10 \times 12  \hspace{0.1cm} inches }{12  \hspace{0.1cm} inches}  = 10 \hspace{0.1cm}  crates

where 1 foot  = 12 inches

3 0
3 years ago
Explain in detail how to add and regroup 28+5
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You need to get the Variables alone

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EXPLANATION

The line of reflection from WXY to W'X'Y' is the line y=-x as shown in the following graph:

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