[see picture link]Area of part 1:A1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2Area of part 2:A2=length*width=(8+2)*1=10 ft2Area of part 3:A3=length*width=8*2=16 ft2
Picture: https://us-static.z-dn.net/files/d30/9033549987fa60cf57fe05f71424f322.gif
Answer:
simplified: (16/81)
rewritten: (2/3)^4
Step-by-step explanation:
hope this helps! if wrong you can delete it :)
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).
Y^2+5y+10
To 2 Squared go together the regular y go together and then normal go together
First we will change them on the same denominator which will be 12. If we do something to the denominator we must do the same to the numerator so :
For 1/3 we get 4/12 because (1/3)*4 = 4/12
And for 2/3 we get 8/12 because (2/3)*4 = 8/12
So 1/3 is the smaller fraction, 7/12 is in the middle and 2/3 is the bigger fraction.
