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lisabon 2012 [21]
3 years ago
15

At India’s arcade 2/3 of the games are racing games. Among the racing games, 1/2 are motorcycle racing games. What fraction of t

he games at India’s arcade are motorcycle racing games
Mathematics
1 answer:
OleMash [197]3 years ago
7 0

Answer:

\frac{1}{3}

Step-by-step explanation:

Motorcycle racing games = \frac{1}{2} × \frac{2}{3} (Directly multiply)

                                           = \frac{2}{6} (Simplify)

                                           = \frac{1}{3} (Tada!)

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Answer:

Area of triangle is 9.88 units^2

Step-by-step explanation:

We need to find the area of triangle

Given E(5,1), F(0,4), D(0,8)

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Area\,\,of\,\,triangle =\sqrt{s(s-a)(s-b)s-c)} \\where\,\, s = \frac{a+b+c}{2}

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Length of side DE = a = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

Length of side DE = a = =\sqrt{(5-0)^2+(1-8)^2}\\=\sqrt{(5)^2+(-7)^2}\\=\sqrt{25+49}\\=\sqrt{74}\\=8.60

Length of side EF = b = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

Length of side EF = b = =\sqrt{(0-5)^2+(4-1)^2}\\=\sqrt{(-5)^2+(3)^2}\\=\sqrt{25+9}\\=\sqrt{34}\\=5.8

Length of side FD = c = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

Length of side FD = c = =\sqrt{(0-0)^2+(8-4)^2}\\=\sqrt{(0)^2+(4)^2}\\=\sqrt{0+16}\\=\sqrt{16}\\=4

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s = a+b+c/2

s= 8.6+5.8+4/2

s= 9.2

Area of triangle==\sqrt{s(s-a)(s-b)s-c)}\\=\sqrt{9.2(9.2-8.6)(9.2-5.8)(9.2-4)}\\=\sqrt{9.2(0.6)(3.4)(5.2)}\\=\sqrt{97.5936}\\=9.88

So, area of triangle is 9.88 units^2

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