Answer:
C <em>,</em><em> </em><em>negative</em><em> </em><em>two</em><em> </em><em>times</em><em> </em><em>a </em><em>number</em><em>,</em><em> </em><em>plus</em><em> </em><em>seven </em><em>,</em><em> </em><em>is </em><em>fewer</em><em> </em><em>than </em><em>twenty</em><em>-seven </em><em>and </em><em>five </em><em>tenths </em>
9514 1404 393
Answer:
(-2, 2)
Step-by-step explanation:
The orthocenter is the intersection of the altitudes. The altitude lines are not difficult to find here. Each is a line through the vertex that is perpendicular to the opposite side.
Side XZ is horizontal, so the altitude to that side is the vertical line through Y. The x-coordinate of Y is -2, so that altitude has equation ...
x = -2
__
Side YZ has a rise/run of -1/1 = -1, so the altitude to that side will be the line through X with a slope of -1/(-1) = 1. In point-slope form, the equation is ...
y -(-1) +(1)(x -(-5))
y = x +4 . . . . . . . . subtract 1 and simplify
The orthocenter is the point that satisfies both these equations. Using the first equation to substitute for x in the second, we have ...
y = (-2) +4 = 2
The orthocenter is (x, y) = (-2, 2).
Answer:
Q1) (x+7)² = 9
x = -10, -4
Q2) (x-8)² = 144
x = -4, 20
Q3) (x-1)² = 81
x = -8, 10
Step-by-step explanation:
Q1) x² + 14x + 49 = 9
x² + 2(x)(7) + 7² = 9
(x + 7)² = 9
x + 7 = +/- sqrt(9)
x + 7 = 3
x = -4
x + 7 = -3
x = -10
Q2) x² - 16x + 64 = 144
x² - 2(x)(8) + 8² = 144
(x - 8)² = 144
x - 8 = +/- sqrt(144)
x - 8 = 12
x = 20
x - 8 = -12
x = -4
Q3) x² - 2x + 1 = 81
x² - 2(x)(1) + 1² = 81
(x - 1)² = 81
x - 1 =+/- sqrt(81)
x - 1 = 9
x = 10
x - 1 = -9
x = -8
I would say 16 times the square root of 2, but I am not positive.
