Answer:
So the parallelogram is in the first and second quadrants.
Step-by-step explanation:
From Exercise we have that the parallelogram have a vetrices:
A(negative 1,4) B(4,0) C(12,3) D(7,7). We use a site geogebra.org to plof a graph for the given parallelogram.
From the graphs we can see that the parallelogram is mostly in the first quadrant and smaller in the second quadrant. So the parallelogram is in the first and second quadrants.
Y-12=1.5(x-4)
Y-12=1.5x-6
Y=1.5x+6
Answer:
C
Step-by-step explanation:
y = (x - 2)² = x² - 4x + 4 → (1)
x + y - 2 = 0 → (2)
substitute y = x² - 4x + 4 into (2)
x + x² - 4x + 4 - 2 = 0
x² - 3x + 2 = 0
(x - 1)(x - 2) = 0
equate each factor to zero and solve for x
x - 1 = 0 ⇒ x = 1
x - 2 = 0 ⇒ x = 2
substitute these values into (2) and solve for y
x = 1 : 1 + y - 2 = 0 ⇒ y - 1 = 0 ⇒ y = 1 ⇒ (1, 1 )
x = 2 : 2 + y - 2 = 0 ⇒ y + 0 = 0 ⇒ y = 0 ⇒ (2, 0 )
the line intersects curve C at 2 points (1, 1 ) and (2, 0 )
Answer:
12 small and 7 large
Step-by-step explanation:
x = small y = large
4x + 10y = 115
