ElectricPavlov

4. similar triangles

so   14 is to 8   as  FM is to 9......ratio:

    14/8 = FM/9     so FM = 14/8 * 9 = 15.75 feet

Now for the other side

14/8 =HM/12     HM = 14/8 *  12 =21 feet

Total = 14 + 15.75 +21 = 50.75 ft

5. Use pythagoream theorem to find the diagonal       then take 1/2

   BD^2 = 8^2 + 14^2 = 260

BD= 16.125        1/2 BD = BE = 8.062 ft

6.  I THINK    40 is to 85    as   (40+48) is to (?+85)

      as a ratio:   40/85 = (40+48) / (?+85)     solve for ?

              (?+85) = (40+48) / (40/85)

               ?+85  = 187

                  ? =102

Volume of a square pyramid = a^2  h/3      where a=base length   h= height (I looked it up)

yields  270 cm^3   of wax needed

Each cube is  4 cm x 4 cm x 4cm = 64 cm^3

270 cm^3 / 64 cm^3/cube = 4.218 cubes......he will need 5 of the little cubes of wax.

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.

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.

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?

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....