If the perimeter of the garden is (14x - 32) ft and has a side length of (x + 2) ft:
- the perimeter = 24 ft
- the area = 36 sq. ft
- if the perimeter of the garden is doubled, the perimeter of the new garden = 48 ft
- if the area of the garden is doubled, the area of the new garden = 72 sq. ft
<em><u>Recall</u></em>:
- A square has equal side lengths
- Perimeter of a square = 4(side length)
- Area of a square =

<em><u>Given:</u></em>
Perimeter of square (P) = 
Side length (s) = 
<em><u>First, let's find the </u></em><em><u>value of x</u></em><em><u> by creating an </u></em><em><u>equation </u></em><em><u>using the </u></em><em><u>perimeter </u></em><em><u>formula:</u></em>
- Perimeter of a square = 4(side length)


<em><u>Find how much fencing would be needed (</u></em><em><u>Perimeter </u></em><em><u>of the fence):</u></em>
- Perimeter of the fence =

Perimeter of the fence = 
<em><u>Find the </u></em><em><u>area </u></em><em><u>of the garden:</u></em>
- Area of the garden =

Area = 
Area = 
<u><em>Find the </em></u><u><em>perimeter </em></u><u><em>if the garden size is doubled:</em></u>
- Perimeter of the new garden = 2 x 24 = 48 ft
<em><u>Find the </u></em><em><u>area </u></em><em><u>if the garden size is doubled:</u></em>
- Perimeter of the new garden = 2 x 36 = 72 sq. ft
In summary, if the perimeter of the garden is (14x - 32) ft and has a side length of (x + 2) ft:
- the perimeter = 24 ft
- the area = 36 sq. ft
- if the perimeter of the garden is doubled, the perimeter of the new garden = 48 ft
- if the area of the garden is doubled, the area of the new garden = 72 sq. ft
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brainly.com/question/13511952
I Belive That There Can't Be 3 Numbers In A Rectangle Because At The Top You Get 24 And 6 And I Think It Is BxH One Half Times Base Times Height So 2xBxH .
6k+10.5=3k+12
-10.5 -10.5
6k=3k+1.5
-3k -3k
3k=1.5
Answer:
50
Step-by-step explanation:
35 +15 =50 15 dived by 2 = 7.5 x 2 =15 + 35 = 50
Answer:
6<em>i</em>
Step-by-step explanation:
sqrt(-121) - sqrt(-25)
11<em>i</em> - 5<em>i</em>
<u>6</u><u><em>i</em></u>