Question

@jim_thompson5910

Answer 1

need to restart my computer, isnt this one pi as well?

Answer 2

\[\Large \sin\left(2x\right)\]
is the same as
\[\Large 1*\sin\left(2x-0\right)+0\]


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Notice how the color-coded terms match up
\[\Large {\color{red}{1}}*\sin\left({\color{blue}{2}}x-{\color{green}{0}}\right)+{\color{purple}{0}}\]
\[\Large {\color{red}{A}}*\sin\left({\color{blue}{B}}x-{\color{green}{C}}\right)+{\color{purple}{D}}\]
where...
|A| = amplitude
2pi/B = period
C/B = phase shift
y = D is the equation of the midline

Answer 3

so we see
A = 1
B = 2
C = 0
D = 0

Answer 4

`isnt this one pi as well?`

T = 2pi/B
T = 2pi/2
T = pi

yep the period is pi

Answer 5

yes!

Answer 6

so i got it correct?

Answer 7

yes you did

Answer 8

yay! okay i have another one