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Questions 7
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 #1
avatar+7053 
+1

The circle travels outward at a speed of 60 cm/s, which means the radius of the circle is increasing at a rate of 60 cm per second.

 

radius after  0  sec  =  0

radius after  1  sec  =  60

radius after  2  sec  =  60 + 60  =  60(2)

radius after  3  sec  =  60 + 60 + 60  =  60(3)

radius after  s  sec  =  60s

 

Or we can say...

 

radius  =  60s    , where   s  is the number of seconds after the stone hit the water.

 

Let   s  =  the number of seconds after the stone hit the water   and   a  =  the area of the circle

 

We know that the equation for the area of a circle is...

 

a  =  pi (radius)2

                            Substitute  60s  in for radius.

a  =  pi( 60s )2

 

a  =  3600 pi s2

 

This equation tells us the area, a , at any  s . But we want to know the rate of change in  a  per change in  s  at any  s . So take  d / ds  of both sides of the equation.

 

da / ds  =  d / ds [ 3600 pi s2 ]

 

da / ds  =  ( 3600 )( pi )( d/ds s2 )

 

da / ds  =  ( 3600 )( pi )( 2s )

 

da / ds  =  7200 pi s

 

To find the rate at which the area is increasing after 1 second, plug in  1 for  s  and solve for  da / ds .

 

(a)   when  s  =  1 ,          da / ds   =   7200 pi (1)   =   7200 pi     (sq cm per second)

 

(b)   when  s  =  3 ,          da / ds   =   7200 pi (2)   =   14400 pi

 

I'll let you do part  (c)  .

 

Notice that the bigger the number of seconds, the bigger the rate at which the area is increasing.

hectictar 3 hours ago
 #1
avatar+7053 
+1

To find  g(25)  , let's first find what values of  x  make  f(x) = 25

 

f(x)  =  25

                         Substitute  3x2 - 2  in for  f(x) .

3x2 - 2  =  25

                         Add  2  to both sides of the equation.

3x2  =  27

                         Divide both sides by  3 .

x2  =  9

                         Take the  ±  square root of both sides.

x  =  ±√9

 

x  =  ± 3

 

So...     f(3)  =  25     and     f(-3)  =  25

 

g( f(x) )  =  x2 + x +1

                                     Let's plug in  3  for  x .

g( f(3) )  =  32 + 3 + 1

                                     Now we can substitute  25  in for  f(3)  because we know that  f(3)  =  25 .

g( 25 )  =  32 + 3 + 1

                                     And now simplify the right side of the equation.

g( 25 )  =  13                 This is one possible value of  g( 25 ) .

 

g( f(x) )  =  x2 + x +1

                                            Now let's plug in  -3  for  x .

g( f(-3) )  =  (-3)2 + (-3) +1

                                            Substitute  25  in for  f(-3)  since  f(-3)  =  25 .

g( 25 )  =  (-3)2 + (-3) +1

                                            Simplify the right side.

g( 25 )  =  7                          This is the other possible value of  g( 25 ) .

 

The two possible values of  g( 25 )  are  13  and  7 .

 

13 + 7  =  20

hectictar 20 mai 2018
 #1
avatar+7053 
+1
hectictar 14 mai 2018
 #1
avatar+7053 
+1

\(\tan\big(\frac{7\pi}{12}\big)=\tan\big( \frac{\pi}{3}+\frac{\pi}{4} \big)\)

 

And the sum of two angles formula for tan is:

 

\(\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\)         so......

 

\(\tan\big( \frac{\pi}{3}+\frac{\pi}{4} \big)=\frac{\tan\frac{\pi}{3}+\tan\frac{\pi}{4}}{1-\tan\frac{\pi}{3}\tan\frac{\pi}{4}}\)

                                                        And we know   \(\tan\frac\pi3=\sqrt3\)   and   \(\tan\frac\pi4=1\)    so...

\(\tan\big( \frac{\pi}{3}+\frac{\pi}{4} \big)=\frac{ \sqrt3 +1}{1-(\sqrt3)(1)}\\~\\ \tan\big( \frac{\pi}{3}+\frac{\pi}{4} \big)=\frac{ \sqrt3 +1}{1-\sqrt3}\)

                                                        Multiply numerator and denominator by  \((1+\sqrt3)\) .

\( \tan\big( \frac{\pi}{3}+\frac{\pi}{4} \big)=\frac{ (\sqrt3 +1)(1+\sqrt3)}{(1-\sqrt3)(1+\sqrt3)}\\~\\ \tan\big( \frac{\pi}{3}+\frac{\pi}{4} \big)=\frac{ 2\sqrt3+4}{-2}\\~\\ \tan\big( \frac{\pi}{3}+\frac{\pi}{4} \big)=-\sqrt3-2\\~\\ \tan\frac{7\pi}{12}=-\sqrt3-2\\~\\ \text{________________________}\)

 

sin( 105° )  =  sin( 45° + 60° )

 

And the sum of two angles formula for sin is:

 

sin(α + β)  =  sin α cos β + cos α sin β         so....

 

sin(45° + 60°)  =  sin 45° cos 60° + cos 45° sin 60°    

 

sin(45° + 60°)  =  \(\big(\frac{\sqrt2}{2}\big)\big(\frac{\sqrt3}{2}\big)+\big(\frac{\sqrt2}{2}\big)\big(\frac12\big)\)

 

sin(45° + 60°)  =  \(\frac{\sqrt6}{4}+\frac{\sqrt2}{4}\)

 

sin(45° + 60°)  =  \(\frac{\sqrt6+\sqrt2}{4}\)

 

sin 105°   =   \(\frac{\sqrt6+\sqrt2}{4}\)

hectictar 14 mai 2018
 
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