Question 252614
{{{h(t)=125sin(0.157t -pi/2)+125}}} Start with the given equation.



{{{125=125sin(0.157t -pi/2)+125}}} Plug in {{{h(t)=125}}}



{{{125-125=125sin(0.157t -pi/2)}}} Subtract 125 from both sides.



{{{0=125sin(0.157t -pi/2)}}} Combine like terms.



{{{0/125=sin(0.157t -pi/2)}}} Divide both sides by 125.



{{{0=sin(0.157t -pi/2)}}} Reduce.



{{{sin(0.157t -pi/2)=0}}} Rearrange the equation



{{{0.157t-pi/2=arcsin(0)}}} Take the arcsine of both sides.



{{{0.157t-pi/2=0+2pi*n}}} or {{{0.157t-pi/2=pi+2pi*n}}} Evaluate the arcsine of 0 to get {{{0}}} or {{{pi}}}. Don't forget to add in integer multiples of {{{2pi}}}


<font color=red>---------------------------Side Note----------------------</font>


Take note how if {{{n=0}}}, then we just get 0 or {{{pi}}}. If {{{n=1}}}, then we get {{{2pi}}} or {{{3pi}}}. If we continue this, we can see that we'll hit all of the integer multiples of {{{pi}}}. Because of this we can condense the right side to just {{{pi*n}}}. 


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{{{0.157t-pi/2=pi*n}}} Condense the right side (see side note above).



{{{0.157t=pi*n+pi/2}}} Add {{{pi/2}}} to both sides.



{{{0.157t=(2pi*n+pi)/2}}} Combine like terms.



{{{t=((2pi*n+pi)/2)(1/0.157)}}} Multiply both sides by {{{1/0.157}}} to isolate 't'.



{{{t=(2pi*n+pi)/0.314}}} Multiply.



Now let's plug in some values of 'n'. If n=0, then {{{t=(2pi*0+pi)/0.314=pi/0.314=10.00507}}}


If n=1, then {{{t=(2pi*1+pi)/0.314=3pi/0.314=30.01522}}}. However, since we're only worried about the first 30 seconds, this means that {{{0<=t<=30}}} which would exclude the 't' value when {{{n=1}}}. Any other value of 'n' will generate a 't' value outside the interval {{{0<=t<=30}}}


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Answer:


So the only solution is {{{t=10.00507}}} (which is approximate) which means that at about 10.00507 seconds, the rider will be 125 feet above the ground during the first 30 seconds.



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# 2


Since the variable we want to solve for in {{{2*cos(x)+x=0}}} is buried in a trig function and is outside the trig function, there is no way to solve for it exactly. 



So we have to use a graphing calculator to approximate the roots of {{{2*cos(x)+x}}}. Graphing the given expression, we get


{{{ drawing(500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,2*cos(x)+x)

)}}}


Graph of {{{y=2*cos(x)+x}}}



Now use the graphing calculator's root/zero function to find the approximate root of {{{x=-1.029866529}}} which rounds to {{{x=-1.03}}} to the nearest hundredth.



So the answer is {{{x=-1.03}}}