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slava [35]
3 years ago
10

The excavation for a house and the trucks to carry away the material have the dimensions shown. About how many level truck loads

are necessary to remove all the dirt? The truck is 5.0 for the length 2.0 for the height 3.0 for the side base The house is 4.0 for the height 14.5 for the length and 11.0 for the side base 100 points !!!!! and brainlist
Mathematics
1 answer:
Inessa05 [86]3 years ago
8 0

Answer:

<h3>This means that about 21 level truck loads are necessary to remove all dirts.</h3>

Step-by-step explanation:

To get the number of level truck loads needed to remove all the dirts in the house, first we need to calculate volume of the truck and the house.

Volume of truck = Length * Breadth * Height

If the truck is 5.0 for the length 2.0 for the height 3.0 for the side base;

Volume of the truck = 5*2*3

Volume of truck = 30

For the house;

Volume of house = Length * Breadth * Height

If the house is 4.0 for the height 14.5 for the length and 11.0 for the side base, then

Volume of the house = 4*14.5*11 = 638

Number of level truck loads needed to remove all the dirts = Volume of house/Volume of the truck = 638/30 ≈ 21.3

This means that about 21 level truck loads are necessary to remove all dirts.

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---------------

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Find the area of the triangle ABC with the coordinates of A(10, 15) B(15, 15) C(30, 9).
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~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{15}~,~\stackrel{y_2}{15}) ~\hfill a=\sqrt{[ 15- 10]^2 + [ 15- 5]^2} \\\\\\ ~\hfill \boxed{a=\sqrt{125}} \\\\\\ (\stackrel{x_1}{15}~,~\stackrel{y_1}{15})\qquad (\stackrel{x_2}{30}~,~\stackrel{y_2}{9}) ~\hfill b=\sqrt{[ 30- 15]^2 + [ 9- 15]^2} \\\\\\ ~\hfill \boxed{b=\sqrt{261}}

(\stackrel{x_1}{30}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{5}) ~\hfill c=\sqrt{[ 10- 30]^2 + [ 5- 9]^2} \\\\\\ ~\hfill \boxed{c=\sqrt{416}} \\\\[-0.35em] ~\dotfill

\qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{125}\\ b=\sqrt{261}\\ c=\sqrt{416}\\ s\approx 23.87 \end{cases} \\\\\\ A\approx\sqrt{23.87(23.87-\sqrt{125})(23.87-\sqrt{261})(23.87-\sqrt{416})}\implies \boxed{A\approx 90}

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