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).
I believe it is 1. Function 2. Not a function 3. Function 4. Not a function. I hope this helps! :)
Given:
ΔONP and ΔMNL.
To find:
The method and additional information that will prove ΔONP and ΔMNL similar by the AA similarity postulate?
Solution:
According to AA similarity postulate, two triangles are similar if their two corresponding angles are congruent.
In ΔONP and ΔMNL,
(Vertically opposite angles)
To prove ΔONP and ΔMNL similar by the AA similarity postulate, we need one more pair of corresponding congruent angles.
Using a rigid transformation, we can prove
Since two corresponding angles are congruent in ΔONP and ΔMNL, therefore,
(AA postulate)
Therefore, the correct option is A.
Answer:
0, 10, 8, 2
Step-by-step explanation:
Simplify x3/x A.x2 B. 1/x2 - 3705076. ... + 4 f(x) = x2 – 5x + 4 f(x) = x2 + 3x – 4 f(x) = –2x2 + 10x – 8 f(x) = –4x2 – 16x – 1.
