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lesya692 [45]
2 years ago
7

Evaluate the following equations

Mathematics
1 answer:
DochEvi [55]2 years ago
5 0

Step-by-step explanation:

A....

Solution ,

=  \frac{h - 3}{3}  = 3h \\

= h - 3 = 9h

= h - 3 - 9h = 0

= h - 9h = 3

=  - 8h = 3

Change the sign of both side

= 8h =  - 3

= h =  -  \frac{3}{8 }  \\

B....

Solution,

=  \frac{t}{2}  =  \frac{t + 1}{5}  \\

Multiply the both side by the least common multiple to eliminate the fraction...

= 5t = 2t + 2

= 5t - 2t = 2

= 3t = 2

= t =  \frac{2}{3}  \\

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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}}\\
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\\ \quad \\

\end{array}

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% right side info
\bullet \textit{ stretches or shrinks}\\
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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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