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zheka24 [161]
3 years ago
8

a square green rug has a blue square in the center. the side length of the blue square is x inches. the width of the green band

that surrounds the blue square is 6 inches. what is the area of the green band
Mathematics
2 answers:
andreev551 [17]3 years ago
6 0

Answer:

12(x+3) square inches.

Step-by-step explanation:

Given:

A square green rug has a blue square in the center.

The side length of the blue square is x inches as shown in the figure:

The width of the green band that surrounds the blue square is 6 inches.

Question asked:

What is the area of the green band ?

Solution:

<u>Side length of blue square = </u>x<u />

First of all we will calculate area of blue square.

As we know:

Area of square=(Side)^{2}

                       =x^{2}

Thus, area of blue square = x^{2}

Now, we will calculate area of square green rug:

Side length of green rug = Side length of blue square + width which surrounds the blue square

Side length of green rug = x+6

Thus,  area of square green rug = =(x+6)^{2} \\

<em><u>Now, we will find area of the green band ( shaded region with green color as shown in the figure)</u></em>

Area of the green band = Area of square green rug -  Area of blue square        

                                       =(x+6)^{2} -x^{2} \\=(x+6+x)(x+6-x)\               [a^{2} -b^{2} =(a+b)(a-b)]

                                       =(2x+6)6\\=12x+36\\

                                       =12(x+3)\ taking\ 12\  as\ common

Therefore, the area of the green band is 12(x+3) square inches.

Download docx
Musya8 [376]3 years ago
6 0

Answer is

32x-256

most people put the wrong answer

Step-by-step explanation:

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Answer:

ans=13.59%

Step-by-step explanation:

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In our problem, we have that:

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Pr(42 \leq X \leq 64) = 0.6827

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-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have Pr(31 \leq X \leq 75) = 0.9545 = 0.9545 subtracted by Pr(42 \leq X \leq 64) = 0.6827 is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.

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The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.

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4 years ago
Jeff drove 5,784.5 miles in 12 days. To the nearest mile, what was the rate of miles Jeff drove each day? A. 482 B. 488 C. 500 D
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<u>Answer:</u>

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Also, Distance can be calculated by the formula

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Answer: 13x -8y

Hey there!

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