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mezya [45]
3 years ago
7

The length of a rectangular garden is 8 feet longer than its width. If the garden's perimeter is 188 feet, what is the area of t

he garden in square feet?
Mathematics
1 answer:
NemiM [27]3 years ago
3 0

Answer: The area is 2,193 square feet (2,193 ft²)

Step-by-step explanation:The rectangular garden has its sides yet unknown but what we do know is that its length is 8 feet longer than its width. If therefore the width is W, then the length would be W + 8.

The perimeter is calculated as follows;

Perimeter = 2(L + W)

We can now substitute for the values given as follows;

188 = 2(W + 8 + W)

188 = 2(2W + 8)

188 = 4W + 16

Subtract 16 from both sides of the equation

172 = 4W

Divide both sides of the equation by 4

43 = W

If the width is 43, and the length is W + 8, then length equals 43 + 8 which gives us 51

Therefore the area is calculated as follows;

Area = L x W

Area = 51 x 43

Area = 2193

The area of the rectangular garden is 2,193 square feet

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4.43

Step-by-step explanation:

Given :The probability that the San Jose Sharks will win any given game is 0.3694 based on a 13-year win history of 382 wins out of 1,034 games played

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There are eight different jobs in a printer queue. Each job has a distinct tag which is a string of three upper case letters. Th
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Answer:

a. 40320 ways

b. 10080 ways

c. 25200 ways

d. 10080 ways

e. 10080 ways

Step-by-step explanation:

There are 8 different jobs in a printer queue.

a. They can be arranged in the queue in 8! ways.

No. of ways to arrange the 8 jobs = 8!

                                                        = 8*7*6*5*4*3*2*1

No. of ways to arrange the 8 jobs = 40320 ways

b. USU comes immediately before CDP. This means that these two jobs must be one after the other. They can be arranged in 2! ways. Consider both of them as one unit. The remaining 6 together with both these jobs can be arranged in 7! ways. So,

No. of ways to arrange the 8 jobs if USU comes immediately before CDP

= 2! * 7!

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c. First consider a gap of 1 space between the two jobs USU and CDP. One case can be that USU comes at the first place and CDP at the third place. The remaining 6 jobs can be arranged in 6! ways. Another case can be when USU comes at the second place and CDP at the fourth. This will go on until CDP is at the last place. So, we will have 5 such cases.

The no. of ways USU and CDP can be arranged with a gap of one space is:

6! * 6 = 4320

Then, with a gap of two spaces, USU can come at the first place and CDP at the fourth.  This will go on until CDP is at the last place and USU at the sixth. So there will be 5 cases. No. of ways the rest of the jobs can be arranged is 6! and the total no. of ways in which USU and CDP can be arranged with a space of two is: 5 * 6! = 3600

Then, with a gap of three spaces, USU will come at the first place and CDP at the fifth. We will have four such cases until CDP comes last. So, total no of ways to arrange the jobs with USU and CDP three spaces apart = 4 * 6!

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