Answer:
Step-by-step explanation:
Answer:So first, I found the length of the sides and the diagonal of the square, which are 18−−√ and 6 respectively. By graphing, I know the solution is (0,−1). Then, I assume that since the length between (3,2) and (−3,2) is the diagonal, then the distance between (0,5) and the remaining vertex must be the diagonal too. And since the length of the side is 6, then the distance between the vertex and either (3,2) or (−3,2) must be 6. So:
(x−3)2+(y−2)2−−−−−−−−−−−−−−−√=18−−√
(x−0)2+(y−5)2−−−−−−−−−−−−−−−√=6
Which gives (after a bit of cleaning up):
x2+y2−10y=11
x2−6x+y2−4y=5
Then, replacing the second expression into the first one:
x2−6x+y2−4y=5⇒x2=5+6x−y2+4y
5+6x−y2+4y+y2−10y=11
5+6x+4y−10y=11
6x−6y+6
x−y=1
x=1+y
Up to this point, I know I'm not entirely wrong because the expression is true for the actual coordinates of the vertex, because 0=1+(−1) is true. But I wouldn't know how to proceed if I hadn't known the answer beforehand. I need to find both x and y, is there a linear equation I'm missing to find the exact coordinates of the last vertex? Is my process okay or is there a simpler way to do it?
Step-by-step explanation:
Answer:
See the attached for a graph.
Step-by-step explanation:
The edge of the solution space is marked by the line ...
... y = 4 - 2x
This is in slope-intercept form. It has a y-intercept of 4 and a slope of -2. Because the relation symbol is "less than" (with no "or equal to"), the line itself is not part of the solution space. It is graphed as a dashed line for that reason.
The values of y in the solution space are those <em>less than</em> the ones that make up the line, so the shading is of the half-plane <em>below</em> the line.
72 = 1/2 (16x) because the area of a triangle is 1/2 (bh) and the height is x and the base is 16 and the area = 72 so solve for x to get base.
