Answer:
perimeter - 4+2+3+2+5= 16mm
Area -5×4=20mm²
 
        
             
        
        
        
<h2>Answer:</h2><h2>
</h2><h3>
No!</h3><h2>
</h2><h2>
Explanation:</h2><h2>
</h2>
The function  doesn't have inverse because it doesn't pass the horizontal line test. This test tells us that a function
 doesn't have inverse because it doesn't pass the horizontal line test. This test tells us that a function  has an inverse function<em> if and only if</em> there is no any horizontal  line that intersects the graph of
 has an inverse function<em> if and only if</em> there is no any horizontal  line that intersects the graph of  at more than one point. As you can see, from the graph of f (the red one), if you draw an horizontal line that passes through
 at more than one point. As you can see, from the graph of f (the red one), if you draw an horizontal line that passes through  then this line will touch the graph of f at three points, so the horizontal line test is not satisfied here. If you see the graph of g, this doesn't represent a function because there is at least one vertical line that touches the graph at more than one point, so this relation is not a function.
 then this line will touch the graph of f at three points, so the horizontal line test is not satisfied here. If you see the graph of g, this doesn't represent a function because there is at least one vertical line that touches the graph at more than one point, so this relation is not a function.
 
        
        
        
Hyperbola: y = 1/x
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<u>Shape:</u> open curve with two branches
<u>Domain: </u>Any non-zero real number x < 0, x > 0 or x∈ (-∞, 0) ∪ (0, +∞)
<u>Range:</u> Any non-zero real number y < 0, y > 0 or y∈ (-∞, 0) ∪ (0, +∞)
<u>Locater point:</u> Imaginary point of intersection of asymptotes (0, 0)
<u>Asymptotes:</u> x = 0 and y = 0