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Alekssandra [29.7K]
3 years ago
6

Ron Finley is a gardener who decided a few years ago to plant several gardens in South Central Los Angeles. He wants his next ga

rden to be a rectangle where the length is three times the width. Additionally, it must have a perimeter of 24 yards because of space constraints. What are the length and the width of the garden?
Mathematics
1 answer:
yaroslaw [1]3 years ago
3 0

Well, we can be sure that whatever the width is, we can call it ' W '. Then, from information in the question, the length of the garden is ' 3W '.

Now, the perimeter of a rectangle is (length + width + length + width). Using the fancy algebra labels I just gave them, that's (3W + W + 3W + W). And now I can go through that, add up all the Ws, and get a total of 8W for the perimeter.

But he question tells us that the perimeter is 24 yards, so 8W = 24 yds.

Divide each side of that equation by 8, and we discover that W = 3 yds. And if THAT's true, then 3W = 9 yds. Bada bing ! We have the dimensions of the garden.

It's 3 yards wide and 9 yards long.

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Hoochie [10]

The correct answer is C (7, 9)

Firstly we know that each point is 6 away from the other in terms of x and in terms of y. Now we also know that for every 6, we will be one away from point B and five away from point A. We know this because the ratio is AB 5:1, meaning that the 5 is on the A side (they both come first).

So, we can just add 5 to each of the A value numbers to get point P.

A = (2, 4)

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P = (7, 9)

6 0
3 years ago
Find the mid segment
Airida [17]

Answer:

11

Step-by-step explanation:

13 + 9 = 22

22/2 = 11

7 0
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marta [7]

Answer:

Step-by-step explanation:

2x+y=14+y

x=7

7+13y=35-y

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7 0
3 years ago
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How many different rectangles can you draw, that has an area of 28 cm sq. ? pls help me
jolli1 [7]

Answer:

You can draw three rectangles:

1- 28 * 1.

2- 14*2

3- 7*4

Step-by-step explanation:

We have a total area of 28. If we use only integer numbers, we can find all the divisors of 28. The possible rectangles will be organized with them, taking into account that the product of them, which means the area should be 28.

28 = 1*2*2*7

We can organize them as follows:

R1: 28 = 1*28

R2: 28 = 2*14

R3: 28 = 4*7

Finally, we can conclude that there are only three possibilities

1- 28 by 1.

2- 14 by 2

3- 7 by4

The perimeters will be:

Perimeter 1 = 2x1 + 2x28 = 58

Perimeter 2 = 2x2 + 2x14 = 32

Perimeter 3 = 2x4+2x7 = 22

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There would be 203,759 total registered doctors that year.
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