CPhill

A. AB  is the hypotenuse of a 30-60-90 right triangle....it will be twice as long as the  side opposite the 30°  angle  = CB

So....AB   = 2 CB  =  2 (271)  =  542 ft

B. We need to find  AC...it will  √3  times as long as BC   = 271√3 ft

So...using the Law of Cosines....we have

AD^2  = AC^2  + DC^2  - 2 (AC) (DC) cos ( 74°)

AD  =  √ [  (271√3)^2  + (772)^2  - 2 (271√3)(772)cos (74°) ]  =  785.2 ft

C.  We can find angle E  as follows :

DC^2  = DE^2  + CE^2  - 2(DE)(CE)cos (E)

772^2   = 543^2 + 561^2  - 2(543)(561) cos E

[ 772^2  - 543^2 - 561^2 ]  /  [ -2(543)(561) ]   = cos E

Take the inverse cosne to find E

arccos  ( [ 772^2  - 543^2 - 561^2 ]  /  [ -2(543)(561) ] )  = E  = 88.7°

You are correct, LF.....we should  use AA Similarity....not SSS (we don't know anything about the sides )

The first one can be mapped because we have Angle-Angle similarity between the two figures

The second one cannot  be mapped because the sides surrounding the 32° angle are not in proportion

AC / BC  = 4.5 / 7

But PR /PQ  = 9 / 13.5  .....and  the ratios are not the same  [ check for yourself ]

The third one can be mapped becaue we have  AA similarity, again

The last one cannot be mapped  because the sides surrounding the 35° angle are not in proportion...

Check to see that FG / FE   has a different ratio  than  BA / BC

No prob ...!!!!

The area of the first  will   be  the ratio of the square of the scale factor of the first figure to the second tmes the area of the second....so we have...

(15/21)^2 * 196  =   

(5/7)^2 * 196  =   

(25/ 49) * 196  = 

(196/4) * 25  =

4 * 25  =

100 in^2

1. The speed giving the best fuel economy wiil be the x coordinate of the vertex of this parabola

So  we have  y  = -.0075x^2 +.575x + 11.5

The x coordinate of the vertex  in the form  y  = ax^2  + bx  + cc is given by   -b/ [ 2a]

So.....plugging in what we have leads to :     -.575 / (2 * -.0075)   ≈ 38.33  ≈  38 mph 

2.  -3x^2 +2x  -3  = 0      we can us the quadratic formula to  find the solutions

x  =  (  -2 ±√ [ 2^2  - 4(-3)(-3)  ]  )  / 2 (-3)  =

=   (-2 ±√ [4 - 36 ] ) / -6  =

( -2 ±√ [ -32 ] )  /-6

(-2 ± i√  [ 16 * 2 ] )  /-6  =

( -2 ±  4i√2   ) / -6      =

( -1 ± 2i√ 2) / -3   ⇒    "C"

3.  (5x  -15)^2  = -100       take both roots

5x  - 15  = ±√(-100)

5 ( x - 3)  = ± i √ 100

5 ( x - 3)  = ±10i      divide through by 5

x  - 3   =  ±2i       add 3 to both sides

x =  3 ± 2i    ⇒   "C"    

1.   16(x - 3)  =  (y - 1)^2 

The vertex of this parabola is at  ( 3, 1)

This parabola opens to the right

In the form    4p(x - 3)  = (y - k)^2......the focus is  ( 3 + p, 1)   

We can solve for p as  4p  =16  ⇒   p  = 4

So....the focus  is  ( 3 + 4, 1)   =  (7,1)

The dirctrix lies 4 units to the left of the vertex and is given by    x  = ( 3 - p )  =   (3  - 4 )   =    -1

Here's a graph :  https://www.desmos.com/calculator/iwubncnnan

2.  y  = -1/6 x^2 + 7x  - 80     we need to complete the square on  x

y  = (-1/6)  [ x^2- 42x + 480 ]     

Take 1/2 of 42  = 21....square it  = 441....add and subtract it within the brackets

y   =    (-1/6)  [x^2 - 42x + 441 + 480  - 441 ]     factor the frist three terms....simplify the rest

y  = (-1/6) [  (x - 21)^2  + 39 ]

y = (-1/6)(x - 21)^2  - 39/6

y  = (-1/6)(x - 21)^2  - 13/2       add 13/2 to both sides

(y + 13/2) =  (-1/6)(x - 21)^2        multiply through  by -6

-6(y + 13/2)  = (x - 21)^2

Because of the presence of a "-"   in front of the "6"......this parabola turns downward

The vertex is  ( 21, - 13/2)

The focus is  given by ( 21 , -13/2  + p )

And we can find  p as    .....  4p  = -6   ⇒  p  = -6/4  = -3/2

So....the  focus  is  (21, -13/2 - 3/2)   = ( 21 , -16/2)  = (21, -8)

The directerix is given as    y  = (-13/2 - p)   = ( -13/2 - (-3/2) )   = ( -13/2 + 3/2)  = -10/2  = -5

Here's a graph :  https://www.desmos.com/calculator/g9zfw92yos

1.  y  = -1/4x^2 + 4x  - 19

The x coordinate of the vertex  is given by   -4 / (2 * -1/4)  =  -4 / (-1/2)  =   -4 * -2    =  8

And the y coordinate is given  by :

-1/4 (8)^2  + 4(8)  - 19  =

-16  + 32   - 19

-3

So....the vertex  is  (8, -3)

Here's a graph : https://www.desmos.com/calculator/z76eahtb61

2. The center of the circle  is  ( 1, 4)   and the radius  is 8

So we have

(x - 1) ^2   + ( y - 4)^2   = 8^2        expand and simplify

x^2 - 2x + 1  + y^2  - 8y + 16  = 64

x^2 + y^2  - 2x - 8y  + 17  = 64       subtract 64 from both sides

x^2  + y^2  -2x - 8y - 47   =  0 

3.  We have    y   =  1/8 x^2 + 4x + 20

We want to complete the square  on x

y = (1/8)  (x^2  + 32x + 160)

Take  1/2 of 32  = 16....square it  = 256....add and subtract it within the parentheses

y  = (1/8)  (x^2 + 32x  + 256   + 160   - 256)       factor the first three terms, simplify the last two

y = (1/8) [ (x + 16)^2  - 96 ]       apply the  1/8   over both terms in the parentheses

y  = (1/8) (x + 16)^2 - 12

(y + 12)  = (1/8)(x + 16)^2     multiply both sides by 8

8( y + 12)  = (x + 16)^2

The vertex  is at  (-16, -12)

In the form

4p ( y - k)  = (x - h)^2

4p  = 8

So....p  = 2

And the focus is given by

(-16 ,  -12 + p)   =    (-16, -12 + 2)   =   (-16, -10)

Here's the graph  :  https://www.desmos.com/calculator/cgsvcjy0ny

The original function is

f(x)  = 2x  - 1

So

f(-x)  means  that  we replace  "x"  with "-x"  in the original function

So

f(-x)  = 2(-x)  - 1  =   -2x - 1

Here are both graphs....:  

https://www.desmos.com/calculator/ob8iw0ihx3 1

Note  the  effect of replacing "x" with "-x"    .....it just "reflects" the original graph across the y axis.....!!!

I think heureka answered this here :

https://web2.0calc.com/questions/correct#r2