It says to find the unknown side length. If necessary, write the answer in simplest radical form.

Mathematics

1 answer:

fomenos2 years ago

3 0

You'd use the Pythagorean theorem for this since it's a right triangle.
$a^2+b^2=c^2$
So plug in your given numbers to get: $9^2+40^2=x^2$
Then solve:
$81+1600=x^2$
$1681=x^2$
$\sqrt{1681}= \sqrt{x^2}$
$41=x$

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

\begin{array}{rllll} 
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \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 \\
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\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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now, with that template in mind, let's take a peek at this function

$\bf \begin{array}{lllcclll}
y=&2(&1x&-2)^2&-4\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\\\\
-----------------------------\\\\
A\cdot B=2\impliedby \textit{shrunk by a factor of 2, of half-size}\\\\
\cfrac{C}{B}= \cfrac{-2}{1}\implies -2\impliedby \textit{horizontal right shift of 2 units}\\\\
D=-4\impliedby \textit{vertical down shift of 4 units}$

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