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Answer:

Set your compass to a radius of 6, put the center point at #(6, 0)#, and draw a circle.

Explanation:

Multiply both sides of the equation by r:

#r^2 = 12rcos(theta)#

Substitute #x^2 + y^2# for #r^2# and #x# for #rcos(theta)#

#x^2 + y^2 = 12x#

The standard form of this type of equation (a circle) is:

#(x - h)^2 + (y - k)^2 = r^2#

To put the equation in this form, we need to complete the square for both the x and y terms. The y term is easy, we merely subtract #0# in the square:

#x^2 + (y - 0)^2 = 12x#

To complete the square for the x terms, we add #-12x + h^2# both sides of the equation:

#x^2 - 12x + h^2 + (y - 0)^2 = h^2#

We can use the pattern, #(x - h)^2 = x^2 - 2hx + h^2# to find the value of h:

#x^2 - 2hx + h^2 = x^2 - 12x + h^2#

#-2hx = -12x#

#h = 6#

Substitute 6 for h into the equation of the circle:

#x^2 - 12x + 6^2 + (y - 0)^2 = 6^2#

We know that the first three terms are a perfect square with h = 6:

#(x - 6)^2 + (y - 0)^2 = 6^2#

This is a circle with a radius of 6 and a center point #(6, 0)#

Answer:

#2sin((29pi)/24)e^(((25pi)/24)i)#

Explanation:

we know euler's theorem states that #e^(itheta)=costheta+isintheta#
so,#e^(7pi/6i)-e^(19pi/8 i)#
=#cos(7pi/6)+isin(7pi/6)-(cos(19pi/8)+isin(19pi/8))#
=#cos(7pi/6)-cos(19pi/8)+i(sin(7pi/6)-sin(19pi/8))#
now #cosc-cosd=2sin((c+d)/2) sin((d-c)/2)#
#sinc-sind=2cos((c+d)/2) sin((c-d)/2)#
so we get #cos((7pi)/6)-cos((19pi)/8)#
=#2sin((85pi)/24)sin((29pi)/24)#
again similarly we get #i(sin((7pi)/6)-sin((19pi)/8))#
=#2icos((85pi)/24)sin((29pi)/24)#
so,#e^((7pi)/6)-e^((19pi)/8)#=#2sin((29pi)/24)(sin((85pi)/24)-icos((85pi)/24))#
=#2sin((29pi)/24)[sin(2pi+((37pi)/24))-icos(2pi+((37pi)/24))]#
=#2sin((29pi)/24)[sin((37pi)/24)-icos((37pi)/24)]#
=#2sin((29pi)/24)[cos((25pi)/24)+isin((25pi)/24)]#
=#2sin((29pi)/24)e^(((25pi)/24)i)#.

drawn

Let CB be the cliff . From point A the angle of elevation of the peak C of the cliff is #/_CAB=theta#'. After going up a distance , #AD=k# #m# towards the top of the cliff at an angle,#/_DAE =phi#,it ls found the angle of elevation #/_CDF=alpha#.

DF and DE are perpendiculars drawn from D on CB and AB.

Now #DE =ksinphi and AE =kcosphi#

Let
#h="CB the height of the cliff" and b=BA#

For #DeltaCAB," "(CB)/(BA)=tantheta#

#=>b/h=cottheta=>b=hcottheta#

Now #DF=BE=BA-AE=b-kcosphi#

#CF=CB-FB=CB-DE=h-ksinphi#

For #DeltaCDF," "(CF)/(DF)=tanalpha#

#=>(h-ksinphi)/(b-kcosphi)=tanalpha#

#=>(h-ksinphi)/(hcottheta-kcosphi)=tanalpha#

#=>h-htanalphacottheta=ksinphi-kcosphitanalpha#

#=>h=(k(sinphi-cosphitanalpha))/(1-tanalphacottheta)#

#=>h=(ksintheta(sinphicosalpha-cosphisinalpha))/(sinthetacosalpha-costhetasinalpha)#

#=>h=(ksinthetasin(phi-alpha))/(sin(theta-alpha))#

Answer:

Start from a "basic cycle" for the #tan# function to obtain the results below.

Explanation:

Starting with the "basic cycle" for #tan(theta)# i.e. for #theta in [-pi/2,+pi/2]#
enter image source here

Then consider what values of #x# would place #(x+pi)# in this same range, i.e. #(x+pi) in [-pi/2,+pi/2]#
(Yes; I know: because #tan# has a cycle length of #pi# we could recognize that the cycle will repeat at exactly the same place, but let's do this for the more general case.)
#(x+pi)in[-pi/2,pi/2]color(white)("X"rarrcolor(white)("X")x in [-(3pi)/2,-pi/2]#
Giving us the "basic cycle" for #tan(x+pi)#:
enter image source here

Adding #5# to this to get #y=5+tan(x+pi)# simply shifts the points upward #5# units:

In the image below, I have added the "non-basic cycles" as well as the "basic cycle" used for analysis:
enter image source here

Answer:

#((3pi)/2, 2)#

Explanation:

The maximum and minimum of any #cos# graph on its own will be #+1# or #-1#. It doesn't matter so much what's inside the #cos# function, only what's outside.

Multiply these by #-2#, the outside of the equation, to get maximum/minimum points of #-2# and #+2#.

The maximum is therefore #+2#.

To find at which point this occurs, work backwards.

#+2 = -2cos(x-pi/2)#

Divide both sides by #-2#

#-1 = cos(x-pi/2)#

Do a backwards #cos# (#cos^-1#) to get rid of the forwards #cos#

#cos^-1(-1)=x-pi/2#
#pi = x-pi/2#
#x = (3pi)/2#

Now we have our #x# and #y# values:

#((3pi)/2, 2)#

drawn The given function representing the height #h# in cm of the person above the ground with time #t# is given by

#h=15cos((2pi)/3t)+65#

(a) h is a cosine function of t, So it will have maximum value when #cos((2pi)/3t)=1=cos0=>t=0# s
and the maximum height of the swing becomes

#h_(max)=15cos((2pi)/3xx0)+65=80cm#

(b) it takes #t=0#sec to reach maximum height after the person starts.

c) The minimum height of the swing will be achieved when #cos((2pi)/3t)=-1#

Minimum height

#h_(min)=15xx(-1)+65=50cm#

(d). Again #cos((2pi)/3t)=-1=cos(pi)=>t=3/2=1.5#s

Here #t=1.5#s represents the minimum time required to achieve the minimum height after start.

e) For #h=60# the equation becomes

#60=15cos((2pi)/3t)+65#

#=>60-65=15cos((2pi)/3t)#

#=>cos((2pi)/3t)=-1/3#

#=>t=[1/120(2nxx180pmcos^-1(-1/3))] sec" where " n in ZZ#

#=>t=(3npm0.912)#sec

when #n=0, t=0.912s# this is the minimum time to reach at height #h=60 cm # after start. just after that height #h=60 cm # will be achieved again for #n=1 #

#=>t=(3xx1-0.912)=2.088#sec

So in #(2.088-0.912)=1.176s# sec (the minimum time difference) within one cycle the swing will be less than 60 cm above the ground.

f) The height of the swing at 10 s can be had by inserting #t=10#s in the given equation

#h_(10)=15cos((2pi)/3xx10)+65=57.5#cm

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