Two identical square pyramids were joined at their bases to form the composite figure below.

Mathematics

2 answers:

lapo4ka [179]3 years ago

6 0

Answer:

It's A. I took the test

Step-by-step explanation:

gulaghasi [49]3 years ago

3 0

The answer is A!

Hope that helps

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Answer:

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$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\

\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\\\
\bullet \textit{ horizontal or "phase" shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\
\end{array}$

$\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{function period}\\
\qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\
\qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}$

now.. with that template above in mind, let's see yours

$\bf \begin{array}{lllllll}
f(x)=&3tan(&4x+&\pi )\\
f(x)=&3tan(&4x+&1\pi )&+0\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}$

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