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lisov135 [29]
4 years ago
11

What is the solution set of 2x2+x=15?

Mathematics
2 answers:
vodomira [7]4 years ago
7 0
4+x=15
---> x=15-4
---> x=11
Alexeev081 [22]4 years ago
5 0

Answer:

5/2,-3

Step-by-step explanation:

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Write an equation of cosine function with amplitude 2 and a period 4pi
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\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{  A}}cos({{  B}}x+{{  C}})+{{  D}}\\
f(x)=&{{  A}}sin({{  B}}x+{{  C}})+{{  D}}\\
f(x)=&{{  A}}tan({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\

\end{array}

\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{  A}}|\\\\
\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{function period}\\
\qquad \frac{2\pi }{{{  B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\
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\end{array}

now.. hmmm let's see, keeping in mind the template above

so, amplitude of 2, A=2, thus |A| = ±2, so, you can use either

and period of 4π, well, that simply means that 

\bf \cfrac{2\pi }{B}=4\pi \implies \cfrac{2\pi }{4\pi }=B\implies \cfrac{1}{2}=B\\\\
-----------------------------\\\\

\begin{array}{llll}
f(\theta)=&\pm 2cos&\left(\frac{1}{2}\theta  \right)\\
&\ \uparrow &\ \uparrow \\
&A&B
\end{array}
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