Let's start with the area of the square

now let's subtract the are of the two half circles.
two half circles are the same as one circle, and we know that the diameter of the circle is 8 (same as the side of a square) so it's radius is 8/2= 4 inches

now we just subtract and our answer is
 
        
             
        
        
        
Collinear points are two or more points that lie on the same line.
        
                    
             
        
        
        
(i) Velocity is the rate of change of position, so if
<em>r</em><em>(t)</em> = <em>b</em> cos(<em>ω t </em>) <em>i</em> + <em>b</em> sin(<em>ω t </em>) <em>j</em> + <em>v</em> <em>t</em> <em>k</em>
then
<em>v</em><em>(t)</em> = d<em>r</em>/d<em>t</em>
<em>v</em><em>(t)</em> = -<em>b</em> <em>ω </em>sin(<em>ω t</em> ) <em>i</em> + <em>b</em> <em>ω</em> cos(<em>ω</em> <em>t</em> ) <em>j</em> + <em>v</em> <em>k</em>
The speed of the particle is the magnitude of the velocity, given by
|| <em>v</em><em>(t)</em> || = √[(-<em>b</em> <em>ω </em>sin(<em>ω t</em> ))² + (<em>b</em> <em>ω</em> cos(<em>ω</em> <em>t</em> ))² + <em>v</em> ²]
… = √[<em>b </em>²<em>ω </em>² + <em>v</em> ²]
(ii) The path is a helix. Suppose you zero out the <em>k</em> component. Then the path is a circle of radius <em>b</em>, and the value of <em>ω</em> determines how quickly a particle on the path traverses the circle. Now if you reintroduce the <em>k</em> component, the value of <em>v</em> will determine how far from the plane <em>z</em> = 0 the particle moves in a helical path as <em>t</em> varies.
(iii) Acceleration is the rate of change of velocity, so
<em>a</em><em>(t)</em> = d<em>v</em>/d<em>t</em>
<em>a</em><em>(t)</em> = -<em>b</em> <em>ω </em>²<em> </em>cos(<em>ω t</em> ) <em>i</em> - <em>b</em> <em>ω</em> ² sin(<em>ω</em> <em>t</em> ) <em>j</em>
 
        
             
        
        
        
Answer:
-3
Step-by-step explanation:
I like to start by combining like terms on this one add the negatives.
3 + (-4 - 2)
3 - 6
-3
Hope this helps.
From yours truly to you,
 Que.