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dimulka [17.4K]
3 years ago
8

A rectangular floor space in a community center is 40 ft long and 32 ft wide. A volunteer is setting up chairs on the floor spac

e. Each chair needs an area of approximately 4 ft by 5 ft.
Which estimate best approximates the number of chairs that can be set up on the floor space?
Mathematics
2 answers:
Sever21 [200]3 years ago
7 0
To approximate the number of chairs that can be set up in floor space, we just have to divide the total floor area by the area that each chair needs. That is,
                                                 n = (40 ft x 32 ft) / (4 ft x 5 ft)
                                                   n = 64
Therefore, there are 64 chairs that can be set up in the floor. 
Verizon [17]3 years ago
6 0

Answer:

Actually, its 60

Step-by-step explanation:

I just took the quiz.

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3)


let's use (100, 36.5) and (300, 106.5)


\bf (\stackrel{x_1}{100}~,~\stackrel{y_1}{36.5})\qquad  (\stackrel{x_2}{300}~,~\stackrel{y_2}{106.5}) \\\\\\ slope =  m\implies  \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{106.5-36.5}{300-100}\implies \cfrac{70}{200}\implies \cfrac{7}{20} \\\\\\ \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-36.5=\cfrac{7}{20}(x-100) \\\\\\ y-36.5=\cfrac{7}{20}x-35\implies y=\cfrac{7}{20}x+1.5


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and for this one let's pick say (35, 88) and (45, 82)


\bf (\stackrel{x_1}{35}~,~\stackrel{y_1}{88})\qquad  (\stackrel{x_2}{45}~,~\stackrel{y_2}{82}) \\\\\\ slope =  m\implies  \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{82-88}{45-35}\implies \cfrac{-6}{10}\implies -\cfrac{3}{5} \\\\\\ \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-88=-\cfrac{3}{5}(x-35) \\\\\\ y-88=-\cfrac{3}{5}x+21\implies y=-\cfrac{3}{8}x+109

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