Help please!

Question

0

875

2

+379

deleted.

edited by sinclairdragon428 Nov 20, 2019

hectictar · Answer 1 · 2019-06-11T01:36:34

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.

hectictar · Answer 2 · 2019-06-11T01:38:31

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