CPhill

I think this is correct

We can place the first pawn  in  any column of the top row  = 4 ways

For the second pawn....we can place it anywhere on the second row except in the column occupied by the first pawn = 3 ways

The third pawn can then be placed in any two positions on the third row....= 2 ways

And the last pawn, by default....only has 1 way it can be placed on the fourth row

So

4 x 3 x 2  x 1   =   24 ways

Here's my best effort....although there might be an easier method !!!

Wnen  = 0   we have the point (2, 2)

When t = 1  we have the point  (6, 3)

So...the slope of the given line is   (3 - 2) / (6 - 2) =  1/4

If we let the parallel line go through the origin....we have the line  y = (1/4)x  

This means that  every point on this line has the coordinates ( x, 1/4x)

This implies that

x + (1/4)x = 10

(5/4)x = 10

x = 8   and y = (1/4)x = 2

So....the required vector  is    < 8, 2 >

The directrix lies below the focus...so...this parabola turns upward....we can immediately rule out the last two answers

The vertex lies at (0, (-3 - 2) /2) =   (0, -5/2)

We have the form

4p ( y - k) =  (x -h)^2        where (h, k) is the vertex = (0, -5/2)

p is the distance between the focus and vertex = 1/2 units

So we have

4(1/2)(y - [- 5/2] ) = (x - 0)^2

2(y + 5/2) = x^2

2y + 5 = x ^2

2y = x^2 - 5      divide both sides by 2

y = (1/2)x^2 - 5/2

6 in  = 1/2 ft

And

 1 1/2 = 3/2 ft

So...the area is base * height =   (3/2) (1/2) =   (3/4)ft^2

A = (1, 1)

B = (2, 3)

C = (4, 0)

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

CA = sqrt  [(1 - 4)^2 + (1 - 0)^2 ] = sqrt [10]

BC = sqrt [ (2 - 4)^2 + (3 - 0)^2 ] = sqrt [ 13]

Using the Law of Cosines

cos (theta) =  ( [13 - 10 - 5]/   [ - 2sqrt (50) ])

arcos =  [ -2] / [-2 sqrt (50) ] = theta

arcos  = 1 /sqrt (50)  = theta

arcos  =1 / [5sqrt(2) ] = (1/5)(1/sqrt(2) )  =  (1/5)/sqrt (2) = theta

So

cos (theta) =   (1/5) / sqrt(2)

So.... x = 1/5

If vectors are orthagonal, their dot product = 0

So

[ 4 * 3 ] + [ -7 * k] = 0

12 - 7k = 0

-7k = -12

k = 12/7

So

[2 * 3 ] + [12/7 * m] = 0

6 + (12/7)m = 0

(12/7)m = - 6

m = -6*7/12 = -7/2

THX....!!!!

[ It does avoid that nasty integral...LOL !!!   ]

If the third side = 17....we have a right triangle

So...the greatest integer length of remaining side  is 16

And because of the triangle inequality

8 + unknown side > 15

unknown side > 7

So....the possible integer lengths are  [ 8 , 16 ]

What are the statements ????

C. The points on the following shape represent the areas in which Chris will decorate his room with the ribbons. Find the area of the following shape. You must show all work to receive credit.

Figure ABCDEF is shown. A is located at 5, negative 8. B is located at 11, negative 8. C is located at 11, 0. D is located at 6, negative 3. E is located at 4, negative 3. F is located at 0, negative 6

We can use something known as "Pick's Theorem" to solve this

The area is given by

Boundary Lattice Points / 2  +  Interior Lattice Points -  1

[ A lattice point is a point that has integer coordinates ]...so we have

Boundary Lattice Points = 19

Interior Lattice Points = 43

So....the total area =    19/2 + 43 - 1   =   9.5 + 42   =  51.5 units^2