Answer:
2/3
Step-by-step explanation:
It's difficult to make out what the force and displacement vectors are supposed to be, so I'll generalize.
Let <em>θ</em> be the angle between the force vector <em>F</em> and the displacement vector <em>r</em>. The work <em>W</em> done by <em>F</em> in the direction of <em>r</em> is
<em>W</em> = <em>F</em> • <em>r</em> cos(<em>θ</em>)
The cosine of the angle between the vectors can be obtained from the dot product identity,
<em>a</em> • <em>b</em> = ||<em>a</em>|| ||<em>b</em>|| cos(<em>θ</em>) ==> cos(<em>θ</em>) = (<em>a</em> • <em>b</em>) / (||<em>a</em>|| ||<em>b</em>||)
so that
<em>W</em> = (<em>F</em> • <em>r</em>)² / (||<em>F</em>|| ||<em>r</em>||)
For instance, if <em>F</em> = 3<em>i</em> + <em>j</em> + <em>k</em> and <em>r</em> = 7<em>i</em> - 7<em>j</em> - <em>k</em> (which is my closest guess to the given vectors' components), then the work done by <em>F</em> along <em>r</em> is
<em>W</em> = ((3<em>i</em> + <em>j</em> + <em>k</em>) • (7<em>i</em> - 7<em>j</em> - <em>k</em>))² / (√(3² + 1² + 1²) √(7² + (-7)² + (-1)²))
==> <em>W</em> ≈ 5.12 J
(assuming <em>F</em> and <em>r</em> are measured in Newtons (N) and meters (m), respectively).
<span>y^2 + 5y + 6
=</span><span>(y + 2)(y + 3)
----------------------------------</span>
A and B ............................
Answer:
You divide 5 by 47 and then see if the fourth digit after the decimal is 5 or greater or less than 5. If it's less, the digit before it remains the same and the decimal cuts off at it. If it's 5 or greater, the digit before it goes up one and the. decimal cuts off at it. Using this, you should get 0.106, I think.
