PLZ HELP ONLY ONE QUESTION!!!!!! 33 POINTS

Mathematics

1 answer:

Elden [556K]3 years ago

3 0

The volume would be V = (side of base)^2 * (height). The height would be 3. To see this, make yourself a model if need be. Assuming that the dimensions of the "square piece of cardboard" are x by x, then the length (or the width) of a side of the square bottom would be x-3-3, or x - 6.

So: V = (x-6)^2*3 = (area of base)*(depth of box)
= 3(x-6)^2 (answer)

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Andreyy89

$\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 \\
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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}}}
\end{array}$

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

$\bf \begin{array}{llll}
g(x)=&4ln(x)\implies &2(2)ln(x)\\
&\uparrow &\uparrow \\
&A&A
\end{array}$

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Answer:

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