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olga55 [171]
3 years ago
8

An egg that has been falling for x seconds has dropped at an average speed of 16x feet per second. If the egg is dropped from th

e top of a building, its total distance traveled is the product of the average rate times the time. Write a simplified expression to show the distance the egg has traveled after x seconds
Mathematics
1 answer:
ivanzaharov [21]3 years ago
5 0

average \: rate \: of \: fall = 16x \: feet \: per \: second \\ duration \: of \: fall \: (time) =  x \: seconds \\ distance \: of \: fall = 16x \: feet \: per \: second \times x \: seconds \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  =x(16x) \: feet \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   = 16 {x}^{2} \: feet \\  \: the \: egg \: traveled \: 16 {x}^{2} \: feet \: in \: x \: seconds.
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write a function g whose graph represents a translation 2 units to the right followed by a horizontal stretch by a factor or 2 o
Alenkinab [10]
\bf 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}}\\\\
f(x)=&{{  A}} \left|{{ B }}x+{{  C}}  \right|+{{  D}}
\\\\
--------------------\\\\

\bf \bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\
\left. \qquad   \right.  \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{  B}}\textit{ is negative}\\
\left. \qquad   \right.  \textit{reflection over the y-axis}

\bf \bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\left. \qquad  \right. if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\left. \qquad  \right.  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{  D}}\\
\left. \qquad  \right. if\ {{  D}}\textit{ is negative, downwards}\\\\
\left. \qquad  \right. if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{  B}}}

with that template in mind, let's see,
 
\bf f(x)=|x| \implies \begin{array}{lllccll}
f(x)=&1|&1x&+0|&+0\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}

so, to shift it to the right by 2 units, simply set C = 2.
to stretch it by 2, set A = 1/2.

the smaller A is, the wider it opens, the larger it is, the more it shrinks.
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3 years ago
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