hectictar

\(\frac{m}{3}+\frac14>\frac{3m}{8}\)

                                               Multiply both sides of the inequality by  24  .

\(24(\frac{m}{3}+\frac14)>24(\frac{3m}{8})\)

                                               Distribute the  24  to the terms in parenthesees.

\(\frac{24m}{3}+\frac{24}4>\frac{(24)(3)m}{8}\)

                                               Simplify each fraction.

\(8m+6>9m\)

                                               Subtract  8m  from both sides of the inequality.

\(6>9m-8m\\~\\ 6>m\\~\\ m<6\)

The perimeter of the cross section is the perimeter of rectangle DEKL.

perimeter of DEKL  =  DE + EK + KL + LD

(where  DE ,  EK ,  KL , and  LD  are in meters)

KL  =  12     and     DE  =  12

To find the length of EK, look at right triangle EJK.

By the Pythagorean theorem,

EJ²  +  JK²   =   EK²

                                       Plug in  5  for  EJ  and  4  for  JK .

 5²   +   4²   =    EK²
                                       Simplify the left side of the equation.

25  +  16   =   EK²

41   =   EK²

                                       Take the positive square root of both sides.

√41  =  EK

So.....

EK  =  √41     and     LD  =  √41

perimeter of DEKL  =  DE + EK + KL + LD

                                                                       Plug in the values of  DE ,  EK ,  KL , and  LD .

perimeter of DEKL  =  12 + √41 + 12 + √41

                                                                       Combine like terms.

perimeter of DEKL  =  24 + 2√41

Thanks for the response! I didn't think anybody was going to see my question this late.

I also thought that  x ≠ -1  should be a restriction, but since it is not listed in the restrictions in the answer option I wanted to make sure... 

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.

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

Let the perimeter of the larger figure be  P .

\(\frac{25}{16}=\Big(\frac{P}{21.3}\Big)^2\\~\\ \sqrt{\frac{25}{16}}=\frac{P}{21.3}\\~\\ \frac54=\frac{P}{21.3}\\~\\ 21.3\cdot\frac54=P\\~\\ 26.625=P\\~\\ P=26.625\\~\\ P\approx26.6\)

By the Law of Sines,

\(\frac{\sin B}{8.2}=\frac{\sin 32°}{6.7}\\~\\ \sin B=\frac{8.2\sin 32°}{6.7}\\~\\ \begin{array}{lcl} B=\arcsin(\frac{8.2\sin 32°}{6.7})&\qquad\text{or}\qquad&B=180°-\arcsin(\frac{8.2\sin 32°}{6.7})\\~\\ B\approx40.433°&\text{or}&B\approx139.567° \end{array}\)

First let's use the first possible value of  B .  If  B = 40.433° ,  then...

C  =  180° - 32° - 40.433°

C  =  107.567°

And by the Law of Sines,

\(\frac{AB}{\sin107.567°}=\frac{6.7}{\sin32°}\\~\\ AB=\frac{6.7\sin107.67°}{\sin32°}\\~\\ AB\approx12.054\)

Second let's use the second possible value of  B .  If  B = 139.567° ,  then...

C  =  180° - 32° - 139.567°

C  =  8.433°

And by the Law of Sines,

\(\frac{AB}{\sin8.433°}=\frac{6.7}{\sin32°}\\~\\ AB=\frac{6.7\sin8.433°}{\sin32°}\\~\\ AB\approx1.854\)

So the two possible values of  AB  are   AB  ≈  12.054   and   AB  ≈  1.854

If   AB  =  12.054   then

the area of  △ABC  =  (1/2)(8.2)(12.054 sin 32° )

the area of  △ABC  ≈  26.189

If  AB  =  1.854   then

the area of  △ABC  =  (1/2)(8.2)(1.854 sin 32° )

the area of  △ABC  ≈  4.028

So....     AB  ≈  1.854

13.

\(\hphantom{=\,}\ \dfrac{\frac{x}{x+4}}{\frac{1}{x}+\frac{1}{x+4}}\\~\\ =\,\dfrac{\frac{x}{x+4}}{\frac{1}{x}+\frac{1}{x+4}}\cdot\dfrac{x+4}{x+4}\\~\\ =\,\dfrac{x}{\frac{x+4}{x}+1}\\~\\ =\,\dfrac{x}{\frac{x+4}{x}+1}\cdot\dfrac{x}{x}\\~\\ =\,\dfrac{x^2}{x+4+x}\\~\\ =\,\dfrac{x^2}{2x+4}\)

12.

     \( \frac{4-x}{x^2+5x-6}\div\frac{x^2-11x+28}{x^2+7x+6}\)

                                                    Factor the numerators and denominators.

=   \(\frac{-(x-4)}{(x-1)(x+6)}\div\frac{(x-7)(x-4)}{(x+1)(x+6)}\\\)

All the values of  x  that cause a zero in a denominator are:   1,  -6,  7,  4,  -1

So the restrictions are:   x ≠ 1 ,  x ≠ -6 ,  x ≠ 7 ,  x ≠ 4 ,  x ≠ -1

Now let's flip the second fraction and change the sign to multiplication.

=  \(\frac{-(x-4)}{(x-1)(x+6)}\cdot\frac{(x+1)(x+6)}{(x-7)(x-4)}\)

                                                    Multiply the fractions.

=   \(\frac{-(x-4)(x+1)(x+6)}{(x-1)(x+6)(x-7)(x-4)}\)

                                                    Cancel the common factors.

=   \(\frac{-{\color{purple}(x-4)}(x+1){\color{NavyBlue}(x+6)}}{(x-1){\color{NavyBlue}(x+6)}(x-7){\color{purple}(x-4)}}\)

=   \(\frac{-(x+1)}{(x-1)(x-7)}\)

(Hmm...  is   x ≠ -1  not actually a restriction?? Is the expression defined when  x = -1 ?)

Because an inscribed angle is half the angle measure of its intercepted arc,

m∠CAD  =  (1/2)( m arc CD )

m∠CAD  =  (1/2)( 60° )

m∠CAD  =  30°

OA and OC are both radii of circle O, so they are the same length. So △AOC is isosceles, and

m∠OCA  =  m∠CAD 

m∠OCA  =  30°

Since  ∠OBC  and  ∠OBA  form a straight line,

m∠OBC  +  m∠OBA  =  180°

m∠OBC  =  180° - m∠OBA

m∠OBC  =  180° - 60°

m∠OBC  =  120°

Since the sum of the angles in a triangle is  180° ,

m∠BOC  +  m∠OCA  +  m∠OBC  =  180°

m∠BOC  +  30°  +  120°  =  180°

m∠BOC  =  180° - 120° - 30°

m∠BOC  =  30°

Since  m∠BOC  =  m∠OCA ,  triangle OBC  is isosceles, and

the length of BC  =  the length of BO

the length of BC  =  5