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Bas_tet [7]
3 years ago
8

Area and explanation please

Mathematics
2 answers:
Xelga [282]3 years ago
7 0
FIrst you would have to cut this up into a square and a triangle. Since you want all the square's sides to be equal, in this case 8 units, you subtract 8 from 15 to get 7. When you cut it up, the square's side would be 8 units while the triangles side would be 7 inches, since 8+7=15. So if you take just the square itself, it would have an area of 64 units squared, or 64 un.^2, because all sides are 8 units and 8 x 8 = 64. Moving onto the triangle, since we cut it off, the width would be 7 units. It still has the same length though so you would have to multiple 7x8 to get the area, which is 56. But since it's a triangle, you also have to divide 56 by two since a triangle is half of a square. So the area of the triangle would turn out to be 28 units squared, or 28 un.^2. Lol, sorry this is long but bear with me, I'm almost done. Now that we have the area for both sides, we have to add them together to get the area of the shape altogether. The area of the square, 64, plus the area of the triangle, 28, equals 92. The total area of this shape is 92 square units (92 un. ^2). 

Hope this helps :)
BigorU [14]3 years ago
4 0
The answer is 92. ...
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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 \\
f(x)=&{{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}\\\\
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\end{array}\qquad


\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}

with that template in mind, let's see

\bf \begin{array}{lllcclll}
f(x)=&1log(&1x&+3)&+2\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}
\\\\\\
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