How so you round 7,895,164 to the nearest 1,000

Mathematics

1 answer:

lana [24]3 years ago

5 0

Answer: 7,895,164 rounded to the nearest 1,000 is 7,895,000

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In the figure to the right, you are given a square ABCD. Each of its sides is divided into two segments, such that AE = BF = CG

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Proving that EFGH is a square: points E,F,G,H
separate the sides of the square ABCD into two lines. if AE=BF=CG=DH then
AH=DG=EB=FC and it shows that if we connect the points E,F,G,H to each other with line we will have a square

HAE~EBF~FCG~GDH by rule SAS

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3 years ago

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Find the area of the triangle ABC with the coordinates of A(10, 15) B(15, 15) C(30, 9).

lions [1.4K]

Check the picture below. so, that'd be the triangle's sides hmmm so let's use Heron's Area formula for it.

$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{15}~,~\stackrel{y_2}{15}) ~\hfill a=\sqrt{[ 15- 10]^2 + [ 15- 5]^2} \\\\\\ ~\hfill \boxed{a=\sqrt{125}} \\\\\\ (\stackrel{x_1}{15}~,~\stackrel{y_1}{15})\qquad (\stackrel{x_2}{30}~,~\stackrel{y_2}{9}) ~\hfill b=\sqrt{[ 30- 15]^2 + [ 9- 15]^2} \\\\\\ ~\hfill \boxed{b=\sqrt{261}}$

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

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