CPhill

x^2          y^2

___  +    ____   =    1

50              9

The  eccentricity, e,   is given by  c  / a

c  = sqrt  [ a^2  - b^2 ]    =   sqrt  [ 50  - 9 ]  =  sqrt (41)

a  =  sqrt (50)

So

e  = sqrt (41) / sqrt (50)   ≈  .906  = .91

As e approaches 0,  we have  a more circular than eongated ellipse

As e approaches 1, we have  a more eongated than circular ellipse

So....this one is more elongated than circular

See the graph here   :  https://www.desmos.com/calculator/u6nvuljikf

If x + y = 3, y + z = 5, and z + x = 4, then find x + y + z.

x +  y  =   3        (1)

y +  z  =   5        (2)

x + z  =    4        (3)

Subtract   (3)  from (2)   and we get that

y - x  =  1    ⇒    -x  + y  =   1     (4)

Add (1)  and (4)

2y  =  4

y  = 2

And   x  + y  = 3    implies that x  = 1

And  x + z =  4     implies that z  = 3

So

x + y +  z  =   1   + 2  +   3    =   6

Area  of a trapezoid  =  (1/2) (height) ( sum of parallel  base lengths)

So

145.5  =  (1/2) (10) (19.1  +  other base length)     

145.5 =  5 (19.1  +  other base length)      divide both sides by 5

29.1  =  19.1  +   other base length      subtract   19.1  from both sides

10 miles   = other base length

They will  be   R distance apart   whenever   we have  an equilateral triangle  with one vertex at the center of the circle and the other two vertexes  on the circle.......the sides will = R

And the  central angle  formed  by the two sides of R each will be 60°

So.....the probability  that the distance between them will  be  >  R  is given by

1 -  60/360   =

1  -  1/6  =

5/6

Second one

The vertex  of the parabola  is  (0, 1)

So   A  = (-2, 5)     B  = (1, 2)     C  = ( 0,1)

We can let AB  be the base of the triangle

 AB  =  sqrt [ ( -2 - 1)^2 + (5 - 2)^2 ]  =  sqrt [ 3^2  + 3^2]  = sqrt  [18]  =  3sqrt (2)

And   we can use the formula for the distance that  (0, 1)  is from the line  x + y - 3 =  0.....this will be the height of ABC

So we have

l Ax + By - C  l

_____________    =

sqrt ( A^2 + B^2)

l   1(0)  + 1(1)   -  3   l                  l  -2  l                2

_________________  =           _______  =       ______  =     sqrt  (2)

   sqrt  [ 1^2 + 1^2]                      sqrt(2)             sqrt (2)

So.....the area of  ABC  =    (1/2) AB * height of ABC  =  (1/2)(3 sqrt (2) ) ( sqrt (2) )  =  (1/2) (3) (2)  =  

3 units^2  =  [ ABC  ]

x +  y  =  3   ⇒  y =  3 - x      (1)

y  = x^2 + 1    (2)

Equate (1)  and (2)  and we have that

3 - x  = x^2 + 1   rearrange as

x^2 + x - 2   =  0          factor

(x + 2) ( x - 1)  =  0   Set each factor to 0 and solve for  x  and we have that

x = -2     and x  =  1

So  when x  = -2,  x + y  = 3  implies that y = 5

And when x = 1,  x + y  = 3  implies that y  = 2

So   A  = ( -2, 5)   and B  = ( 1, 2)

7x^2  +  9x  +  21  = 0

And

( m + 7)(n + 7)   =    (mn)^2  + 7 (m + n)   +  49     (1)

By Vieta,

m + n  = - 9/7         (2)

And

mn  = 21/7  = 3   square both sides

(mn)^2  = 9        (3)

Sub (2)  and (3)  into  (1)    and we have that

(m + 7) ( n +7)   =  (9)  + 7(-9/7)  + 49   =      49

One of the roots of x^3 - 5x^2 + 5x - 1 = 0 is 2 - sqrt(3).  What are the other two roots?

One of the other roots  is    2  + sqrt (3)

Notice that if  we  can add   the coefficients and the constant and we get 0 as a result, then 1 is also a root

So

1  - 5 + 5   - 1    =   0

So  the remaining root is      1

11y -3x = 8

-3x = -16y + 33

Rearrange  these  as

-3x + 11y   =  8

-3x + 16y  =  33           subtract te second equation from the first

-5y   =  -25         divide   both sides by  -5

y  = 5

And   we can find   x  by  using   

 -3x +  11(5) =  8

-3x + 55    = 8          subtract 55 from both sides

-3x  = -47                divide both sides by -3

x = 47/3

Question 2

The scale on the  blueprint is  1/10   that of the true dimensions

So....the radius  of the circle on the blueprint  = (1/10) (60) =  6 ft

And the center  ofthe crcle is   ( 15,30)

So.....the equation of the  flower garden  is

(x - 15)^2  + ( y - 30)^2   =  6^2

(x - 15)^2  + ( y - 30)^2  =  36