How many triangles can be drawn with side lengths 3ft, 4ft, and 5ft

Mathematics

1 answer:

AleksandrR [38]3 years ago

6 0

Answer:

Step-by-step explanation:

The way to figure out if it is a triangle or not is by using the triangle inequality theorem. It says that the sum of 2 sides of the triangle must be larger than the third side. Therefore, there are 3 triangles that we can make with the side lengths of 3ft, 4ft, and 5ft.

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<h3>The number of marbles in the bag?</h3>

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brainly.com/question/251701

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$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

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% left side templates
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\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}
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y=&2(&1x&-2)^2&-4\\
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