$y=3cosleft(dfrac{2}{3}left(dfrac{pi}{2}+dfrac{pi}{4} right) right)+2$

$y=3cosleft(dfrac{2}{3}left( dfrac{3pi}{4}right) right)+2$

$y=3 cosleft(dfrac{pi}{2} right)+2$

$y=0+2$

$y=2$

For $x=dfrac{3pi}{4}$

$y=3cosleft(dfrac{2}{3}left(dfrac{3pi}{4}+dfrac{pi}{4} right) right)+2$

$y=3cosleft(dfrac{2}{3}(pi) right)+2$

$y=3(-0.5)+2$

$y=-1.5+2$

$y=0.5$

$y= 3 cosleft( dfrac{2}{3}left(dfrac{11pi}{6}+dfrac{pi}{4} right)right)+2$

$y=3cosleft(dfrac{2}{3}left(dfrac{25pi}{12} right) right)+2$

$y=3cosleft(dfrac{25pi}{18} right)+2$

$y=0.97394$

$k=dfrac{2pi}{24}=dfrac{pi}{12}$

$y=90sinleft(dfrac{pi}{12}x right)+30$

$k=dfrac{2pi}{3}$

$y=250cosleft(dfrac{2pi}{3}x right)+750$.

$k=dfrac{2pi}{2.5pi}=dfrac{4}{5}$

$y=-1.25sinleft( dfrac{4}{5}xright)+1.5$

$3.48$ min $< t < 4.02$ min

$5.98$ min $< t < 6.52$ min

The amplitude of the function is $dfrac{15.7-8.3}{2}=3.7$ with the equation of the axis being $y=dfrac{8.3+15.7}{2}$ or $y=12$.
period$=365$ days

$k=dfrac{2pi}{365}$

$y=3.7sinleft(dfrac{2pi}{365}x right)+12$

#### (b)

For $x=30$

$y=3.7sinleft(dfrac{2pi}{365}(30) right)+12$

$y=13.87$ hours

$k=dfrac{2pi}{365}$

$T(t)=16.2sinleft(dfrac{2pi}{365}(t-116) right)+1.4$

Graph the equation on a graphing calculator to determine when the temperature is below $0 ^{circ}C$.

$0<t<111$ and $304<t<365$

$k=dfrac{2pi}{60}=dfrac{pi}{30}$

minute hand: $D(t)=15cosleft(dfrac{pi}{30}t right)+300$;

The period of the second hand is $1$.

$k=dfrac{2pi}{1}=2pi$

second hand: $D(t)=15cos(2pi t)+300$;

The period of the hour hand is $720$.

$k=dfrac{2pi}{720}=dfrac{pi}{360}$

hour hand: $D(t)=8cosleft(dfrac{pi}{360}t right)+300$