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Viefleur [7K]
3 years ago
15

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
1 answer:
alexgriva [62]3 years ago
4 0

Answer:

(3x+3y+9)\ ft^{2}

Step-by-step explanation:

Let

x------> the length of the original play space

y-----> the width of the original play space

we know that

The area of the original play space is equal to

A=xy

The area of the new play space is equal to

An=(x+3)(y+3)\\An=xy+3x+3y+9

Find the difference

An-A=(xy+3x+3y+9)-(xy)\\An-A= (3x+3y+9)\ ft^{2}

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A number cube has faces numbered 1 through 6, and a coin has two sides, "heads" and "tails". The
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Answer:

The <em>required probability</em> is \dfrac{1}{12}.

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Let <em>A </em>be the event of rolling the number cube.

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Total number of possibilities of rolling the number cube and tossing the coin are <em>12 </em>here.

\{(1,H),(2,H),(3,H),(4,H),(5,H),(6,H),(1,T),(2,T),(3,T),(4,T),(5,T),(6,T)\}

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Which number is the same distance from 0 on a number line as 3/2 answers -3 -3/2 -2/3 2/3 3
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Henry has a fenced yard in the shape of a rectangle with a perimeter that measures 242 feet. If the width of the yard measures 5
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Answer: Length = 70 feet

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