CPhill

1)

The  area of an equilateral triangle   is  given by

√3/4 side^2

So

64√3  =  √3 /4  side^2

256 = side^2

16 cm  = side  of original equilateral triangle

Decreasing each side by 4 cm  produces an area of  

√3/ 4 * 12^2   =  

36√3  cm^2

So  the area decrease  is    64√3  - 36√3   =  28√3 cm^2

The pylon  cosists of a cylinder with a radius of 1 ft  and a height of 3 ft topped by a cone with a radius of 1ft  and a height of 2 ft

So.....the total volume is

pi * r^2 * h  +  pi * r^2 * h  /3  =

pi  [ 1^2 * 3   +  1^2 * 2 / 3 ]  =

pi [ 3 + 2/3 ]  =

[ 3 + 2/3] pi   ft^3

\(\frac{1}{4}<\frac{x}{7}<\frac{1}{3}\)

Multiply through by 7

7/4 < x < 7/3

1 3/4  < x <  2 1/3

The integer that satisfies this is   x  = 2

3x^2  + x  - 4

3(x^2 + (1/3)x  - 4/3)       complete the square on x

Take 1/2  of (1/3)  =   1/6

Square this.....=  1/36......add and subtract it within the parentheses

3(x^2  + (1/3)x + 1/36  - 4/3 - 1/36 )

3  [ (x + 1/6)^2   -  48/36 - 1/36]

3 [ (x + 1/6)^2  -  49/36 ]

3(x + 1/6)^2  - 49/12

So......k  =  -49/12

Notice  that we can  write  (5, 2, 4)  as the linear combination

1(5, 0, 0)  + 1(0, 2 , 0)  + 1(0, 0, 4)   =   vector (OE)  +vector( ED)   + vector (DP)

So......the length of  OE   =    5

Midpoint  of  (1,2) , (19,4)   =  ( 10, 3)

The line containing points  (0,7) and (4, - 3)  has a slope of    [ -3 - 7] / [4 - 0]  =  -10/4  = -5/2

So....a line perpendicular to this will have a slope of   2/5

So....the equation of this line is

y  = (2/5)(x - 10) + 3

y = (2/5)x - 4 + 3

y = (2/5)x - 1

So....when x  = 20

y =  (2/5)(20) - 1

y  = 8 - 1

y  = 7

EDITED TO CORRECT A SIGN ERROR

Let the height of the monument  = h

Let the distance from the base of the monument to a point directly under the apex  = x

So   we have these equations

tan(42)  =  h / x     rearrange as

x  =   h / tan (42)      (1)

We also have

tan (37.8)  =  h / (100 + x)

(100 + x) tan (37.8)   = h

100*tan(37.8)  + x *tan (37.8)  =  h     (2)

Sub  (1)  into (2)

100* tan(37.8)  + (h /tan(42)* tan (37.8)  = h

100* tan(37.8)  =  h  -  h [ tan (37.8) / tan(42) ]

100* tan (37.8)  =  h  [  1  - tan(37.8)/tan(42) ]

h  =     [ 100 * tan (37.8) ]  /  [  1 - tan(37.8)/tan(42) ] ≈  560 ft

\( 3x^2 + \left(3 + \frac 13\right)x + \frac 13\)

We can write this as

3x^2  + (10/3)x + 1/3

The discriminant  is

(10/3)^2  -  4(3)(1/3)  =

100/ 9  - 4  =

[ 100 - 36 ] / 9  =

64 / 9

x^2 - 16x + k

Take  (1/2)  of 16  = 8

Square this  =  64  =  k

\(-\frac{1}{2}-2kx = 5y \)

\( \left(\frac{1}{4},-6\right)\)

So    we can solve this  for   "k'

-1/2  - 2k(1/4)  = 5(-6)

-1/2 - (1/2)k  = - 30         multiply through by 2

-1  - k  =  -60

-k =  - 59

k   = 59