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deleted.

 Jun 11, 2019
edited by sinclairdragon428  Nov 20, 2019
 #1
avatar+8966 
+3

1.

 

distance between  (1, 1)  and  (4, 7)  =  \(\sqrt{(4-1)^2+(7-1)^2}\ =\ \sqrt{9+36}\ =\ \sqrt{45}\ =\ 3\sqrt5\)

 

2.

 

Let  x  be the thickness of the walls in feet.  
(12 - 2x)(17 - 2x)  =  176

 

 

204 - 24x  - 34x + 4x2  =  176  
4x2 - 58x + 204  =  176

 

 

4x2 - 58x + 28  =  0  
4x2 - 2x - 56x + 28  =  0

 

 

2x(2x - 1) - 28(2x - 1)  =  0  
(2x - 1)(2x - 28)  =  0

 

 

x = 1/2     or     x = 14  
x  can't be  14  because that would make the walls thicker than the dimensions of the exterior of the room.

 

 

x  =  1/2  is the solution.  
 Jun 11, 2019
 #2
avatar+8966 
+3

3.

 

midpoint of  (6, 12)  and  (0, -6)  =  \(\Big( \frac{6+0}{2},\frac{12+-6}{2}\Big)\ =\ \Big(\frac62,\frac62\Big)\ =\ (3,3)\)

 

sum of coordinates of  (3, 3)  =  3 + 3  =  6

 Jun 11, 2019

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