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iren [92.7K]
3 years ago
11

The cube with side 2 is cut from the corner of rectangular prism with dimensions 4×3×5. Find the volume and total surface area o

f the new object.
Mathematics
1 answer:
True [87]3 years ago
5 0

Answer:

Volume: 52 Units Squared

Surface Area: 94 units.

Step-by-step explanation:

The volume is relatively simple to find. Just subtract the original volume by the 2x2x2 cube's volume. The original volume is 60. The cube's volume is 2x2x2 which is 8. 60-8=52.

The surface area is harder to find. Try to envision the corner of the rectangle being cut out. We see that each side of the cube has a surface area of 2x2 which is 4. In the picture, we see that three sides of the rectangle has been partially removed. But since each side of the cube has an equal surface area, it is safe to minus 3 of the sides that has been partially removed by 3. However, since that it is the corner, the "dent" that the cube made in the rectangle also needed to be counted. As we said, each of the sides of a cube has a surface area of 4, so since that the dent has 3 sides, we see that the surface area of the dent is 4x3 which is 12. Now we need to count the unaffected sides of the rectangle. There are three of them. Just multiply the edges to find the surface area of each side. Add all of the values up: 11+16+12+8+15+12+20=94 units.

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emmasim [6.3K]
\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 \\
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}}}
\end{array}


so hmm if you notice, your f(x) there, looks just like g(x), but is "shifted horizontally" by about 9 units to the right

that simply means, based on that template above, that C/B = -9, and we can simply make B = 1  and C = -9, and we get C/B = -9/1 which is -9

thus   \bf \begin{array}{llcll}
f(x)=g(&1x&-9)\\
&\uparrow &\uparrow \\
&B&C
\end{array}\iff f(x)=g(x-9)
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Westkost [7]
<span>{x | x ≤ -8} hope this helps</span>
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Answer:

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Step-by-step explanation:

You need to figure out which information is necessary for this problem and which is not.

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