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3 years ago

Please help!!!!!!!!!

Mathematics

1 answer:

malfutka [58]3 years ago

3 0

If given the circumference you can solve for the radius which will be the height of the parallelogram. also with the circumference you can divide by 2 to find the base of the parallelogram. so all you have to do now is times the radius by half the circumference to find the area

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

Contact [7]

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.

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

$3^o\\x\leq5\leq10\\\\\left\{\begin{array}{ccc}x +5 \ \textgreater \ 10\\ x^2+5^2 \ \textgreater \ 10^2\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \ 5\\ x^2 \ \textgreater \ 75\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \ 5\\ x \ \textgreater \ \sqrt{75}\approx8.7\end{array}\right\\\boxed{x\in\O}\\\\Answer:\boxed{x=9\ or\ x=11}\to your\ answer:\boxed{\boxed{x=11}}$

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7. Marco is making beaded bracelets. Each

stiks02 [169]

One way to find the least common multiple of two numbers is to first list the prime factors of each number.

8 = 2 x 2 x 2

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2: three occurrences

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So, our LCM should be

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