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DerKrebs [107]
3 years ago
13

"Find an integer, x, such that 5, 10, and x represent the lengths of the sides of an acute triangle.

Mathematics
2 answers:
Contact [7]3 years ago
8 0
A, b, c - the lengths of the sides of the triangle
and a ≤ b ≤ c
then:
a + b > c and if the triangle is an acute triangle then a² + b² > c².

1^o\\5\leq10\leq x\\\\\left\{\begin{array}{ccc}5+10 \ \textgreater \  x\\5^2+10^2 \ \textgreater \  x^2\end{array}\right\to\left\{\begin{array}{ccc}x \ \textless \  15\\ x^2 \ \textless \  125\end{array}\right\to\left\{\begin{array}{ccc}x \ \textless \  15\\ x \ \textless \  \sqrt{125}\approx11.1\end{array}\right\\\\\boxed{x=11}\\\\2^o\\5\leq x\leq10\\\\\left\{\begin{array}{ccc}5+x \ \textgreater \  10\\ 5^2+x^2 \ \textgreater \  10^2\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \  5\\ x^2 \ \textgreater \  75\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \  5\\ x \ \textgreater \  \sqrt{75}\approx8.7\end{array}\right\\\boxed{x=9}

3^o\\x\leq5\leq10\\\\\left\{\begin{array}{ccc}x +5 \ \textgreater \  10\\ x^2+5^2 \ \textgreater \  10^2\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \ 5\\ x^2 \ \textgreater \ 75\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \ 5\\ x \ \textgreater \ \sqrt{75}\approx8.7\end{array}\right\\\boxed{x\in\O}\\\\Answer:\boxed{x=9\ or\ x=11}\to your\ answer:\boxed{\boxed{x=11}}
lawyer [7]3 years ago
4 0
If given the two sides of a triangle, the third side, x, must be greater than the difference between the given two sides.

The third side, x must also be less than the sum of the given two sides.

Given  5, and 10.

The third side, x > (10 - 5)            x > 5

The third side, x < (10 + 5)            x < 15

x > 5  and    x < 15

5 < x < 15

The third side x, is between 5 and 15.  It could be any 6, 7, 8, 9, 10, 11,...., 14

But since the question stated that the triangle is acute, x can be like 11.

You can use Cosine Rule to check the angles of triangle, 5, 10, 11. You would discover all the angles are acute.

Answer is option D.
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Remy drinks 214 cups of water every 145 hours.
Tcecarenko [31]

Answer:

1  69/145 cups

Step-by-step explanation:

We can use ratio's to solve this problem.  We will put water over time

214 cups           x cups

--------------- = --------------

145 hours        1 hours


Using cross products

214 *1 =145 x

Divide both sides by 125

214/145 = 145x/145

214/145 = x

145 goes into 215 1 time with 69 left over

1 69/145


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