CPhill

Your answer of "D" is correct...!!!

y = (1/3) (x - 5)  + 5   multiply everything by 3

3y  = 3(1/3) (x - 5) + 3(5)  

3y  = 1 (x - 5)  + 15

3y  = x - 5 + 15

3y   =  x + 10

We have the points on the graph  (5,5)  and (20,10)

The slope between these points  is  [ 10 - 5 ] / [ 20 - 5] =   5 / 15  = 1/3

The equation of the line is

y  = (1/3) (x - 5)  + 5      multiply through by 3

3y = x - 5  + 15

3y  = x  + 10    subtract x from both sides

3y  - x   = 10   [first answer ] 

tan (30“)  =  1/sqrt(3)   ≈  .57735

The exact height  is   75 / sqrt(3)   ft

This one is a little tricky  !!

Since  surface area is in square units...we need to take the square root of the ratio of surface area of A  to the surface area of B to get the scale factor

The scale factor from A : B   =  sqrt  [ surface area of A / surface area of B ]  = 

sqrt [ 28 / 40.32 ]  =    sqrt [ 25/ 36]   =  5/6

Note that Volume of A  = (5/6)^3 * Volume of B

Note that we can set up the following  relationship :

Mass A              density * volume  of A

______  =         __________________            (cancel  the density)

Mass B             density *  volume of B 

Mass A              volume of A

______   =        ___________ 

Mass B             volume of B

Mass A           (5/6)^3 * volume of B

_______ =       _________________         (cancel volume of  B)

6912 g               volume of B

Mass A   =  (5/6)^3

______                                        (multiply both sides by 6912 g

6912 g

Mass A  =  (5/6)^3 * 6912 g    =   4000 g

Since  we are  dividing by  (x + 2)    we  need to synthetically  divide by    -2  istead of  2

The scale factor  of   B : A   = 2 : 1     = 2

Therefore...the area of the cross-section of "B'  is  Cross-Section of A * scale factor^2 =  

18 * 2^2  =  18 * 4  = 72

Therefore....the volume of  B  =  cross-section area  of B  * height of B  =   72 * 10   =  720 cm^3

Both triangles  PNQ and LPQ  have the same altitude  ( PQ)

So....triangles under the same altitude are to each other as their bases...and since  the area of LPQ is twice that of PNQ...its base (PL)  must be twice as  long as the base of PNQ  =  (NP)

Therefore....PL  = 2NP

And since PQ is parallel to LM ⇒    NL : NP  =  LM /PQ =   [ NP + PL ] / NP  = [NP + 2NP] / NP  = 3NP / NP  = 3

Therfore...LM : PQ  = 3 : 1

So...the scale factor  of triangle NLM  to triangle NPQ  is 3 : 1  = 3

And the area of triangle   NLM  = Area of triangle NPQ * scale factor^2  = 8 * 3^2  = 8 * 9  = 72 units^2

So....  Area of LQM  =    [ area of NLM  - combined areas of (PNQ + LPQ)  ] 

72  - [ 8 + 16 ]  = 

72 - 24  =

48 units^2

The y intercept  is  at  -2  = (0, - 2)

The slope of the line from this point moving to the right   is  [ change in y / change in x ]  = [2 units / 4 units] =  2/4  = 1/2

So....the equation is   y = mx + b   where m is the slope and  b  is the y intercept

So..we have

y = (1/2)x  - 2

Thanks, hectictar  !!

Another approach is  to note that a  segment drawn parallel to the base of a triangle, divides the sides in equal proportions

Therefore  for (a) 

BC /AB  = ED /AE

5 /20  = ED / 26

1/4  = ED / 26      multiply both sides by 26

26 (1/4)  = ED  = 26/4  = 13/2  = 6.5

(b)  Also...we have that...since triangle ABE is similar to triangle ACD...then

BE / AB  =  CD /AC

BE / 20  = 18/ 25     multiply both sides by 20

BE = (18/25) * 20  =  14.4