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4 years ago

A playground is constructed with the dimensions and shapes shown below

Mathematics

1 answer:

Alchen [17]4 years ago

6 0

check the picture below.

so the playground is really just 3 rectangles and one triangle, now the triangle has a base of 8 and a height of 6. We can simply get the area of each figure, sum them up and that's the area of the playground.

$\bf \stackrel{\textit{area of the triangle}}{\cfrac{1}{2}(8)(6)}~~+~~\stackrel{\textit{rectangle}}{(8\cdot 8)}~~+~~\stackrel{\textit{rectangle}}{(6\cdot 6)}~~+~~\stackrel{\textit{rectangle}}{(9\cdot 4)}
\\\\\\
24+64+36+36\implies 160$

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The maximum grain yield for corn is achieved by planting at a density of 37,000 plants per

Firlakuza [10]

According to the question, there are 37000 plants per hectare, the total corn plan required is 475,665.75 corn plants

<h3>Determining the area of a trapezium?</h3>

The trapezium is a 2 -dimensional plane shape with the area calculated as shown below:

The formula for calculating the area of a trapezoid is expressed as:

Area = 0.5(a + b)h

Given the following parameters

a = 500 ft
b = 900ft
height h = 800ft

Substitute the given parameters
Area = 1/2 (500 + 900) * 800
Area = 1/2 (1400) * 800
Area = 560000 square feet

Since 1 acre is equivalent to 43560 square feet, hence the total acre required is expressed as:

Total acre = 560000/43560
Total acre = 12.856 acres

<h3>Convert the acres to total corn plant</h3>

Also since there are 37000 plants per hectare, the total corn plan required is 475,665.75 corn plants

Learn more on area of trapezoid here: brainly.com/question/1463152

6 0

2 years ago

(a) Find the first 3 terms in ascending powers of x of the binomial expansion of (2+x/2)^6

Firdavs [7]

Answer:(2+x/2)^6=64(1+3x/2+15x^2/16+5x^3/16+....) (2.05)^6=74.2203

Step-by-step explanation:

3 0

3 years ago

What is the area of this composite figure ? <br> 21 yd<br> 21 yd<br> 10 yd

butalik [34]

Answer:

area is 630

we can multiply 2 1 by 21= 63 by 10=630

8 0

3 years ago

Uestion

Stella [2.4K]

Check the picture below, so the park looks more or less like so, with the paths in red, so let's find those midpoints.

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad K(\stackrel{x_2}{1}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 1 -3}{2}~~~ ,~~~ \cfrac{ 3 +1}{2} \right) \implies \left(\cfrac{ -2 }{2}~~~ ,~~~ \cfrac{ 4 }{2} \right)\implies JK=(-1~~,~~2) \\\\[-0.35em] ~\dotfill$

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ L(\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 +5}{2}~~~ ,~~~ \cfrac{ -3 -1}{2} \right) \implies \left(\cfrac{ 4 }{2}~~~ ,~~~ \cfrac{ -4 }{2} \right)\implies LM=(2~~,~~-2) \\\\[-0.35em] ~\dotfill$

$~~~~~~~~~~~~\textit{distance between 2 points} \\\\ JK(\stackrel{x_1}{-1}~,~\stackrel{y_1}{2})\qquad LM(\stackrel{x_2}{2}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ JKLM=\sqrt{(~~2 - (-1)~~)^2 + (~~-2 - 2~~)^2} \\\\\\ JKLM=\sqrt{(2 +1)^2 + (-2 - 2)^2} \implies JKLM=\sqrt{( 3 )^2 + ( -4 )^2} \\\\\\ JKLM=\sqrt{ 9 + 16 } \implies JKLM=\sqrt{ 25 }\implies \boxed{JKLM=5}$

now, let's check the other path, JM and KL

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 -3}{2}~~~ ,~~~ \cfrac{ -3 +1}{2} \right) \implies \left(\cfrac{ -4 }{2}~~~ ,~~~ \cfrac{ -2 }{2} \right)\implies JM=(-2~~,~~-1) \\\\[-0.35em] ~\dotfill$

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ K(\stackrel{x_1}{1}~,~\stackrel{y_1}{3})\qquad L(\stackrel{x_2}{5}~,~\stackrel{y_2}{-1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 5 +1}{2}~~~ ,~~~ \cfrac{ -1 +3}{2} \right) \implies \left(\cfrac{ 6 }{2}~~~ ,~~~ \cfrac{ 2 }{2} \right)\implies KL=(3~~,~~1) \\\\[-0.35em] ~\dotfill$

$~~~~~~~~~~~~\textit{distance between 2 points} \\\\ JM(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-1})\qquad KL(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ JMKL=\sqrt{(~~3 - (-2)~~)^2 + (~~1 - (-1)~~)^2} \\\\\\ JMKL=\sqrt{(3 +2)^2 + (1 +1)^2} \implies JMKL=\sqrt{( 5 )^2 + ( 2 )^2} \\\\\\ JMKL=\sqrt{ 25 + 4 } \implies \boxed{JMKL=\sqrt{ 29 }}$