Please help me solve this by today I'm in a hurry

Mathematics

1 answer:

Ghella [55]3 years ago

3 0

Answer:

The 3rd one

Step-by-step explanation:

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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.

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

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

8 0

3 years ago

Read 2 more answers

How are real numbers used to slove problems

Marina86 [1]

Answer:Real numbers are used in measurements of continuously varying quantities such as size and time, in contrast to the natural numbers 1, 2, 3, …, arising from counting. ... The real numbers include the positive and negative integers and fractions (or rational numbers) and also the irrational numbers.

Step-by-step explanation:

3 0

3 years ago

Which expression is equivalent to the following complex fraction?

marta [7]

Answer:

$\dfrac{-4x+7}{2(x-2)}$

Step-by-step explanation:

The first step is to combine the parts of the numerator and denominator into one rational expression each. Those will have the same denominator, so their ratio is the ratio of their numerators.

$\dfrac{\dfrac{3}{x-1}-4}{2-\dfrac{2}{x-1}}=\dfrac{\left(\dfrac{3-4(x-1)}{x-1}\right)}{\left(\dfrac{2(x-1)-2}{x-1}\right)}=\dfrac{3-4(x-1)}{2(x-1)-2}\\\\=\dfrac{3-4x+4}{2x-2-2}=\boxed{\dfrac{-4x+7}{2(x-2)}}$