Answer:
Incident ray, the reflection ray and the normal to thereflection surface at the point of the incidence lie in the same plane
Answer:
We are given an area and three different widths and we need to determine the corresponding length and perimeter.
The first width that is provided is 4 yards and to get an area of 100 we need to multiply it by 25 yards. This would mean that our length is 25 yards and our perimeter would be 2(l + w) which is 2(25 + 4) = 58 yards.
The second width that is given is 5 yards and in order to get an area of 100 yards we need to multiply by 20 yards. This would mean that our length is 20 yards and our perimeter would be 2(l + w) which is 2(20 + 5) = 50 yards.
The final width that is given is 10 yards and in order to get an area of 100 yards we need to multiply by 10. This would mean that our length is 10 yards and our perimeter would be 2(l + w) which is 2(10 + 10) = 40 yards.
Therefore the field that would require the least amount of fencing (the smallest perimeter) is option C, field #3.
<u><em>Hope this helps!</em></u>
What question are you trying to ask?
Old scale : 1 in. to 3 ft.
length is 9 in.....9 * 3 = 27 ft (old scale length)
width is 6 in......6 * 3 = 18 ft (old scale width)
new scale : 1 in. to 4 ft
length is 9 in.....9 * 4 = 36 ft (new scale length)
width is 6 in......6 * 4 = 24 ft (new scale width)
