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3 years ago

Find the area of the large rectangle by finding the area of the two small rectangles and adding them.

Mathematics

2 answers:

ASHA 777 [7]3 years ago

5 0

Answer:

Area of the given figure is 21 square units

Step-by-step explanation:

Area of the rectangle formula is length times width

Area of the top rectangle

Length = 5 and width = 3

So area of top rectangle =5 times 3 is 15 square units

Area of bottom rectangle

length =2 and width =3

So area of bottom rectangle is 2 times 3 is 6 square units

Total area = $15+6= 21$ square units

jekas [21]3 years ago

4 0

The total area is 21 (15+6=21)

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Write a formula for function f in terms of function g.

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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3 years ago