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____ [38]
3 years ago
13

You are installing a new pre-constructed fence in front of your house. each fence section measures 4 ½ feet wide, and each end w

ill also have a decorative piece that measures 1 ¾ feet wide.if the space for the fence is 30 feet wide, what is the most number of 4 ½ foot fence sections you could install?
Mathematics
2 answers:
Tatiana [17]3 years ago
7 0

Answer:

4 sections can be installed.

Step-by-step explanation:

Fence space for the fence in front of the house = 30 feet

Each fence section measures = 4.5 feet wide

let the numbers of sections installed = x

Then measure of fence covered by the sections = 4.5x feet

At each end of the section a piece of decorative piece measure= 1.75 feet wide

If x sections are installed then number of decorative pieces required will be same as x.

Therefore, length of decorative pieces = 1.75x feet

Now the total length of fence = Length of sections installed + Length of decorative pieces

4.5x + 1.75x = 30

6.25x = 30

x = \frac{30}{6.25}

x = 4.8

Since number of sections can not be in the fraction or decimals so maximum 4 sections can be installed.

irga5000 [103]3 years ago
4 0
Since, each side of the fence should have the 1 3/4 feet wide decorative piece, the width of the fence should be subtracted by 2 pieces (1 3/4 feet)
                                 = 26.5 feet
Then, divide the remaining width by 4.5 feet and the answer is 5.88. Thus, approximately 5 pieces of the 4 1/2 feet should be installed. 
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The simplified form of the difference quotient of equation 9x^2+5x-1 . in the form 9x + 9h + 5.

In question for equation 9x^2+5x-1, simplified to difference quotient in the form of Ax+Bh+C where A, B, C are integers

<h3>What is equation?</h3>

equation is the relationship between variable and represented as  9x^2+5x-1  is example of polynomial equation.

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Learn more about equation here:
brainly.com/question/10413253

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