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Your method is correct, but you can do it without dividing by K and then multiplying by K :

\(F\,=\,K\cdot\frac{Q_1Q_2}{r^2} \\~\\ F\,=\,\frac{K(Q_1Q_2)}{r^2}\)

                                  Multiply both sides of the equation by  r²

\(Fr^2\,=\,K(Q_1Q_2)\)

                                  Divide both sides of the equation by  F .

\(r^2\,=\,\frac{K(Q_1Q_2)}{F} \)

If we need to include both solutions for  r  ,  then we need to take the positive and negative square root of both sides.

\(r\,=\,\pm\sqrt{\frac{K(Q_1Q_2)}{F}}\)

Frst one   

Scale Factor  1/2      Dilation Center(-1, 1)

We need this formula

New Point '  = [ Old Point  - Dilation Center ] * Scale Factor  + Dilation Center

Remember that    (a , b)  -  ( c, d)  =  ( a - c, b - d)

And  C(a, b)  =  (Ca, Cb)

A'  = 

[ (-4, 4)  - ( -1, 1)]  * (1/2)  +  (-1, 1) =

 (-3, 3)* (1/2)  + (-1, 1)   =

(-3/2, 3/2) + (-1, 1)  =

(-5/2, 5/2)

B' =

[ (2, 4) - (-1, 1)]* (1/2)  + (-1, 1)  =

[ (3, 3) * (1/2) + ( -1, 1)  =

(3/2, 3/2) + (-1, 1) =

(1/2, 5/2)

C'  =  [ (-4, 1) - (-1, 1) ] * (1/2) + (-1, 1)  =

[ (-3, 0) ]* (1/2) + (-1, 1 )  =

(-3/2, 0)  + (-1, 1)  =

(-5/2, 1)

Here's a pic :

Your solution looks good, EP.

Condensed Form

\(\dfrac{1}{2} * \large (\Large \rho \normalsize \text{(two balls of same color bag A)} + \Large \rho \normalsize \text{(two balls of same color bag B)}\large)\\ \text { }\\ \large \dfrac{1}{2} \small \dfrac{\dbinom{3}{2}+\dbinom{2}{2}}{\dbinom{5}{2}} +\large \dfrac{1}{2} \small \dfrac {\dbinom{6}{2}+\dbinom{3}{2} }{\dbinom{9}{2}} =\Large \dfrac{9}{20} = \ 0.45 \)

GA

Add up all of the frequencies = 230    (the number of times she rolled the die)

46 of these were  4's    so the fraction she could expect from 690 rolls

would be   46/230  x  690 = 138 times a 4 would be rolled.

It looks like the above answer is from Mr. BB or one of his students.

For this question, it’s easier to count the negations, i.e when there are no matches.

\(\small \text {There are six ways to roll the first die. (6)}\\ \small \text {There are five ways to roll the second die for it NOT to match the first. (5)}\\ \small \text {There are four ways to roll the third die for it NOT to match the first or second. (4) }\\ \small \text {So, there are } 6*5*4 = 120 \text { ways to roll three dice with NO matches. }\\ \small\text {And there are }6^3 \text {ways to roll all combinations. (216) }\\ \large 1- \dfrac {6*5*4}{216} = \dfrac{5}{9} \approx 44.44 \% \)

GA

Is it  ?    A= 5 +.4N       A-5 = .4N       N=(A-5)/.4    hours

I think that is what is meant....then d would be

N= (10-5)/.4   hours

  N = 12.5   hours to reach  10 kgals.

Oh, my mistake, the formatting got messed up. I'll update the main post in sec

Starts with 5000 then adds 400gal/hr  (the Q asks for f(n) to give thanswer in THOUSANDS of gallons)

A= 5+ .400N    kgal     where kgal is kilo-gallon (1000 gal)

b:   Not sure if that is entered correctly...I am unclear on the notation  N=f−1(A).

c:  f(10) = 5 + .400(10) = 9.    kgal

d:   Like b above...I do not understand your notation....can you clarify for me?

Wait, why do the chooses work?? 

Thanks so much!

Clever way!

Wow! , how long did it take for you to list them out?

Work = force x distance

            6 N x 8 M = 48 N-M

    Kinteic enrgy = 1/2 mv^2 = work = 48 N-M       (the m in 1/2 mv^2 is the mass)

                          = 1/2 (.5 kg) v^2 = 48 N-M

                             so v= 13.86 M/s

2:    KE =  1/2 1080 kg *( 96 km/hr * 1000m/km * 1hr/3600sec)^2 = 384 kj 

    (the stuff in red is converting to m/s from km/hr     kj=kilojoule)

     The work required to stop the auto is equal to the K.E    384 kj

y=  a (x-h)^2  +k        and

y= -2 (x-3)^2 +10

Now can you see that a =-2    h=3   k = 10  ???

Here is a graph  

Updated your question....see it there.....    

The roots are given   as   3+-sqrt5

so one part would be

x+(- 3+sqrt5) =0  

x-3+sqrt5  = 0     solve for x   (add  (3 and subtract sqrt5) from both sides )

x= 3 -sqrt5

   and the other would be  x+( -3-sqrt5)  =0

            x-3-sqrt5    solve for x  (add 3 and sqrt 5 to both sides  

                                          x = 3+sqrt5

These are the roots given in the question....

1. 9x^2-18x      +9y^2 + 36y      = -44    divide thru by 9

      x^2 - 2x        + y^2 +4          = -44/9    'complete the square' for x and y

    ( x-1)^2          + (y+2)^2         = -44/9  + 1 + 4

     (x-1)^2 + (y+2)^2   =   1/9      Now this is standard form for a circle (x-h)^2 + (y-k)^2 = r^2 

where center is h,k  and r = radius   

center =  1,-2    radius = 1/3

2. Let's set this up as vetex form for a parabola     h, k = vertex = 1,3  (from graph)

   so      a(x-1)^2  + 3 =y   is the equation for this parabola...now let's find 'a'

           a convineint point which must satisfy this equation (on the graph) is  0,1 ...substitute this

          a (0-1)^2 + 3 = 1

              a+3 = 1

               a= -2      so the equation becomes  y=   -2 (x-1)^2 +3

Numbers beginning with 1 = 8C0 =1

Numbers beginning with 2 = 8C1 = 8

Numbers beginning with 3 = 8C2 = 28

Numbers beginning with 4 = 8C3 = 56

Numbers beginning with 5 = 8C4 = 70

Numbers beginning with 6 = 8C5 = 56

Numbers beginning with 7 = 8C6 = 28

Numbers beginning with 8 = 8C7 = 8

Numbers beginning with 9 = 8C8 = 1

Total of all numbers listed by "heureka"(count them!) =1+8+28+56+70+56+28+8+1 = 256 numbers. 2^8 =256 is purely coincidental !!!!!!!!!!!. This is the solution in book!!!!.

6.

There is discontinuity at any x value that causes a zero in the denominator.

There is discontinuity when...

x + 9  =  0       and       x + 7  =  0

x  =  -9            and       x  =  -7

7.

Since  (x + 1)  is in the denominator but not in the numerator, there is an asymptote at   x  =  -1

Since  (x + 4)  is in the denominator and in the numerator, there is a hole at   x  =  -4

Shouldnt it be (-3,10)?

OK...here is MY guesstimate.....I think people are just waiting for ME to humiliate myself with a wild answer !

Bag One    how many ways can you withdraw 2 of the balls out of  5 c 2 = 10 ways

                     how many ways to pair two whites    3 c 2 =  3

                       How many ways to pair two reds     2 c 2 = 1

      Summary bag one      (3+1)/10     and we have a 1/2 chance of grabbing bag 1  

             1/2  (4/10)  = 4/20

Bag 2           how many ways to withdraw two out of 9 balls    9 c 2  = 36 ways

                        how many ways to pair two whites   6 c 2 = 15

                           how many ways to pair 2 reds       3 c 2 = 3

      Summary bag 2    (15 +3)/36 = 18/36   and we have a 1/2 chance of grabbing bag 2

            1/2 (18/36) = 18/72   

Now let's add them together:  4/20 + 18/72 = 648/1440 =  9/20

(Well...right or wrong...that is what I derived !)     ~EP

Should it be (x+(3-sqrt 5))(x+(3+sqrt5))?

A. Height of tree = 130 ft  x tan theta

B.  130 * tan (25) = 60.62 ft

C  Tan theta = (tree height) / (shadow length)

D   tan theta =   100/130

         theta = arctan (100/130) = 37.57 dgrees