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natali 33 [55]
3 years ago
15

Let c represent how much it costs for one person to go to a baseball game. How can we represent the total for 6 people to go to

the game? A) 6c. B) c-6. C) c/6. D) c+6.
PLZ HELP ASAP
Mathematics
2 answers:
svetlana [45]3 years ago
6 0

Answer:

It's A

Step-by-step explanation:

A is 6 times c it's simple math

Paladinen [302]3 years ago
5 0

d)c+6

because all of them dont make since. d is representing it costs c for 6 people to get into the baseball game

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\fb \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{  A}}sin({{  B}}x+{{  C}})+{{  D}}
\\\\
f(x)=&{{  A}}cos({{  B}}x+{{  C}})+{{  D}}\\\\
f(x)=&{{  A}}tan({{  B}}x+{{  C}})+{{  D}}
\end{array}
\\\\
-------------------

\bf \bullet \textit{ stretches or shrinks}\\
\left. \qquad   \right. \textit{horizontally by amplitude } |{{  A}}|\\\\
\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}

\bf \bullet \textit{function period or frequency}\\
\left. \qquad  \right. \frac{2\pi }{{{  B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
\left. \qquad  \right. \frac{\pi }{{{  B}}}\ for\ tan(\theta),\ cot(\theta)

now, with that template in mind,

\bf \stackrel{parent~function}{y=sin(\theta )}\qquad \qquad y=\stackrel{A}{\frac{1}{2}}sin\left(\stackrel{B}{\frac{1}{2}}\theta   \right)
\\\\\\
Amplitude\implies \frac{1}{2}
\\\\\\
Period\implies \cfrac{2\pi }{B}\implies \cfrac{2\pi }{\frac{1}{2}}\implies 4\pi

which is pretty much the same sin(θ) function, but squished by 1/2 and elongated up to 4π, check the picture below.


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