<span>The </span>area<span> of the </span>sector<span> of a </span>circle<span> with a </span>radius<span> of </span>8 centimeters<span> is </span>125.6 square centimeters<span>. The </span>estimated value<span> of is </span>3.14<span>. ... First find the </span>area<span> of the entire </span>circleA=3.14*82<span>=200.96 </span>cm2<span> this is the </span>area<span> for 360</span>o<span> or entire </span>circle<span>. The </span>area<span> of the </span>sector<span> represents an</span>angle<span> less then 360 and is simply the ratio of ...</span>
-1 Let's say the set of downs starts at the 20 yard line (maybe the kicking team kicked deep, the receiving team took a knee and so play starts at the 20).
Ok - we're at the 20. First down - they advance 5 yards. So we're now at the 25. We can write that mathematically as:
20
+
5
=
25
So the second play they get sacked deep and lose 6 yards. So we subtract 6:
25
−
6
=
19
So what's the change in yardage for the 2 plays? We are on the 19 and started at the 20, so we can write:
19
−
20
=
−
1
and this makes sense because we know we advanced 5 and fell back 6
Split up each force into horizontal and vertical components.
• 300 N at N30°E :
(300 N) (cos(30°) i + sin(30°) j)
• 400 N at N60°E :
(400 N) (cos(60°) i + sin(60°) j)
• 500 N at N80°E :
(500 N) (cos(80°) i + sin(80°) j)
The resultant force is the sum of these forces,
∑ F = (300 cos(30°) + 400 cos(60°) + 500 cos(80°)) i
… … … + (300 sin(30°) + 400 sin(60°) + 500 sin(80°)) j N
∑ F ≈ (546.632 i + 988.814 j) N
so ∑ F has a magnitude of approximately 1129.85 N and points in the direction of approximately N61.0655°E.
10 separated into 5 groups is 10/5
5 goes into 10 2 times.
10/5=2
X= 2.5 that is the correct answer
