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11111nata11111 [884]
3 years ago
10

The play space is 10 ft to 12 ft. Logan is getting a second dog and wants to increase the length of the play space by 3 feet and

the width by 3 feet. What will be the difference in the area, in square feet, between the original play space and the new play space
Mathematics
2 answers:
Serga [27]3 years ago
5 0

Answer: 75\ ft^2

Step-by-step explanation:

Given

The dimension of playspace is 10\ ft\times 12\ ft

If the length and width is increased by 3 ft

New length is  13\ ft

New width is 15\ ft

So, the dimension becomes 13\ ft\times 15\ ft

The difference between the areas of the playspace is

\Rightarrow 13\times 15-10\times 12\\\Rightarrow 195-120=75\ ft^2

Aleks04 [339]3 years ago
4 0

Answer: 75 square feet

Step-by-step explanation:

The original dimensions are 10 ft by 12 ft, so the area is (10 f)(15 f)

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a quantity with an initial valuve of 390 decays continuously at a rate of 5% per decade what is the value of the quantity after
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4695.90

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