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navik [9.2K]
2 years ago
7

I don’t understand this or what it means? I would like it if you showed work and explained why, thx so much!

Mathematics
2 answers:
Alexeev081 [22]2 years ago
6 0
Area of the rectangle: {base times height}
4cm x 8cm = 32cm^2

Area of the triangle: {base times height divided by two}
(4cm x 11cm) : 2 = 22cm^2

Total area = 22cm^2 + 32cm^2 = 54cm^2
Firlakuza [10]2 years ago
6 0

Answer:

54cm^2

Step-by-step explanation:

You need to calculate area of the given figure. To solve this there are two figures together, triangle and rectangle.

first find area of rectangle given length = 8 cm and breadth = 4cm

therefore area of rectangle  = l * b = 8 * 4 = 32 cm^2

To find area of triangle:

The triangle itself is divided into two right angled triangle.

Now base of one right angled triangle will be 4/2 = 2cm

Area of rt. angled trangle = 1/2 * b*h = 1/2 * 2 * 11 = 11cm^2

Therefore area of whole triangle = 11*2 = 22cm^2 (since there are two rt.angled triangle)

Total area of given figure = 32 + 22 = 54 cm^2

(Mark answer as brainliest if you like this answer)

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slamgirl [31]
\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}}}
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now, with that template in mind, let's see

3 units to the right, that means C/B = -3 so hmm C = -3 and B = 1 will do, -3/1 = -3

vertical stretch by 2, so A = 2
reflected over the x-axis, so that means is flipped upside-down, so A = -2 then

and shifted down by 3, do D  = -3

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