Question

PLASE HELP Find a location that has a long fence or wall (ideally, a picket fence). Place your camera close to the fence so that it recedes into the distance.

Identify a point on the fence about 3 to 4 feet from the camera.
Focus on that point.
Set the aperture at f2.8 (or the lowest possible f-number) and set the shutter speed so you will obtain an appropriate exposure for this f-number.
Set the aperture at f5.6 and set the shutter speed to obtain an appropriate exposure time.
Set the aperture at f16 and set the shutter speed to obtain an appropriate exposure time.

Answer 1

Answer:

You are generating 5 coordinate pairs:

Tree 1 (80,0), Tree 2 (50,-100),Tree 3 (-60,-120), Tree 4 (-70,80) and Tree 5 (10,110)

You can divide each set of coordinates by 10 to plot on graph paper where each square represents 10 by 10, e.g, plot 50, -100 as 5,-10

Use distance formula to find following distances:

Tree 1 to Tree 2

Tree 2 to Tree 3

Tree 3 to Tree 4

Tree 4 to Tree 5

Tree 5 to Tree 1

example of use distance formula for Tree 1 to Tree 2: sqrt[(80-50)^2 + (0 -(-100))^2] = 104

Compute perimeter as sum of distances, say it is x.

Compute cost of company 1: 14 * x

Compute cost of company 2: 4600 + (x-500)*29

Pick whatever company has the lowest price

Explanation:

You could solve this through the pythogorous theorem...by letting the hous be at point (0,0) in the graph where the 2 x & y axis intersect. Then,

Tree(a) is at point (80,0)

Tree(b) is at point (50,-100)

Tree(c) is at point (-60,-120)

Tree(d) is at point (-70, 80)

Tree(f) is point (10,110)

Then the distance between A & B ican be calculated from the Graph as

AB = SqRT( [x coordinate distance of Tree(a) from (0,0) - x-coordinate distance of Tree(b) from 0,0)]^2 + [0-100]^2 )

= SqRT( 30^2 + (100)^2)

Similarly, from the graph ===>

BC = SqRT( (110)^2 + (20)^2)

DC = SqRT((10)^2 + (200)^2)

DF = SqRT( (30)^2 + (80)^2)

Once these distances are calculated, you can calculate the answers to the rest of the questions.