CPhill

1. Opposite angles in an inscribed quadrilateral are supplementary.....thus  angles  D  and F  sum to  180°

2. Correct  !!!

3. Don't know this one

4.Since DE is parallel to AC, DE splits the sides of triangle BAC proportionally.....this means that

DA/BD  = EC/BE

5. sin C  = h/a

   The other answers are correct  !!!

6. Definition of an altitude

psin M  = h  and msinP  = h

Substitution Property of Equality

sin M / m  = sin P / p

1. Correct  !!!

2. A radius always meets a tangent at right angles.....thus HJK is a right angle

3.   We have at riangle with a base of 4 and a height  of 5

So....its area   is   (1/2) (4) (5)   = (1/2) 20  = 10 units^2

Also.....we have a semicircle  centered at ( -1, -2) and a radius of 2

So...its area  is   (1/2) (3.14) (2)^2 = 2 * 3.14  = 6.28 units ^2

So...the total area  is  [ 10 + 6.28] = 16.28 uniits^2

4. Correct !!!

5. Correct  !!!

25.

x^2    +    y^2         

___        ___      =        1

49           25

26.

The distance between  the focal points  is  given by :

2 * √ [ 10^2  - 8^2 ]  =

2 * √ [ 100  - 64 ]  =

2 * √36  =

2 * 6  = 

12 ft

21.   

This is an ellipse....in standard form, we have

x^2   +  y^2      =     1

___     ____

  1         4

The domain  is    -1  ≤ x   ≤ 1

The range  is  -2  ≤ y  ≤ 2

22.

Vertex =  (0,0)

Focus  (-4,0)

So we have

4px = -y^2         

p  = 4.......so we have

4(4)x  = -y^2

16x  = -y^2 

x  = (-1/16)y^2 

31.

Area  of the  board  =  pi * 16^2  = 256 pi

Area  of the bulls-eye  = pi *  6^2  =  36 pi

So....the probability  is     36  / 256  =   9/64  ≈  14.1%

32.

P ( prime  or > 4)   =  P(prime) + P (> 4)  - P (prime and  > 4)  =

                                   3/6           +  2/6    -  1/6   =

                                              4/6  =

                                              2/3

29.

We want to permute   all possible orders of selecting any 3 students from 10....so we have

P (10,3)   = 720 ways

30.

We have 36 possible outcomes.....the ones that sum to 6  are

1,5

5,1

4,2

2,4

3,3    =     5 ways

So...the probabilitiy  is     5 / 36

3. lim x -> 1 f(x) if f(x) {x+2 if x = 1, 2x^2 - 1 if x >1

When x approaches  1 from the left, f(x)  = (1) + 2  = 3

When x  approaches 1 from the right, f(x)  = 2(1)^2  - 1  =  1

So.....this limit  does not exist as  x approaches 1 because both limits would have to be the same approaching from the left and the right

Let's see....I'm going to use  "n"  instead of "k" to prevent confusion

Cross product

       i             j         k           i          j

   n + 1         1         -2      n + 1      1

     n             3          0         n         3

[ 0i  +  (-2n) j  + (3n + 3) k ] -  [ n k  - 6i  + 0j ] =

6i  + ( - 2n)  j + (2n + 3)k  

So.....we have  ( 6 , -2n , 3 + 2n )

So...we wnat to solve this

√ [ 6^2  + (-2n)^2  + (3 + 2n)^2  ]  = √41       square both sides

6^2  + (-2n)^2 + (3 + 2n)^2  = 41     simplify

36  + 4n^2  +  4n^2 + 12n + 9  = 41

8n^2  + 12n  + 45  = 41

8n^2 + 12n + 4  = 0

2n^2 + 3n + 1  = 0

(2n + 1) ( n + 1)  =  0

Setting  each  factor to 0   and  solving for  n we get that  n  = -1/2  or  n  = -1

So...converting back to k....then  k = -1/2  or k  = -1

2. lim x -> 0 

sin ( 1/x  )

This limit does not exist  ....see  the graph here :

https://www.desmos.com/calculator/apbhkqpybo

1. Iim x -> 4

(1)/(x-4)

This limit will not exist.....as x  approaches 4 from the left  the function becomes increasingly more negative......as  x  approaches 4 from the right, the function becomes increasingly more positive

Here's a graph to show this  :

https://www.desmos.com/calculator/sxf0xscp3a